# We Knew Warmers Were Wacko...But Really?



## SSDD (Jul 12, 2015)

I have never had any doubt that warmers were a bit off...strike that...way off.  But it seems that they are wackier than even I imagined.  The weak minded simpletons have apparently bought into their own farce and are now suffering from....are you ready for this?....really....I mean, are you really ready for this?

Pre-Traumatic Stress Syndrome.  

Pre-traumatic stress syndrome for Pete's sake.  What sort of weak mind suffers from pre traumatic stress syndrome?  How much of a whining wimp and vapor prone pu$$y does one have to be to suffer from pre traumatic stress?

Seems that climate scientists...being the screaming ninnies that they are have their own psychologists to help them with their weak minded hysteria.  Here...read all about it.

Climate Scientists Face Psychological Problems -- Science of Us

What else is there really to say?  Pre traumatic stress syndrome....a syndrome associated with a trauma that hasn't happened...a syndrome as a result of one's imagination.  And these are supposed to be scientists.  We have people who call themselves scientists who are suffering syndromes over a panic response to warming that is WELL within natural variation.

Just the thought of someone suffering from a thing as patently ridiculous as pre traumatic stress syndrome makes me ask...in all seriousness, what sort of psychological state must someone be in to suffer from such a thing?  What sort of psychology would predispose someone to fall victim to such a ludicrous ailment?  

I don't know what else to say other than I am not in the least surprised...   Us skeptics have often discussed what might happen to the warmer wackaloons when the wheels finally came off and the crazy train went over the cliff...Could this be the result of model results finally getting so far away from reality that the practitioners of the idiot religion of AGW can't help but see that they have failed miserably and are going to have to start explaining their failure....could this be an avoidance mechanism?

Pre traumatic stress syndrome....can climate science possibly get any more warped?


----------



## Toddsterpatriot (Jul 12, 2015)

Are they talking about smart photons now?


----------



## tinydancer (Jul 12, 2015)

Holy toledo! Cue twilight zone music. That's freaking out there.


----------



## Missourian (Jul 12, 2015)




----------



## Stephanie (Jul 12, 2015)

those poor dear climate Scientist needs to RETIRE and go off to the loony farm

We tried to tell people they were NUTS and so where those who belonged in their CULT


----------



## skookerasbil (Jul 12, 2015)

[URL=http://s42.photobucket.com/user/baldaltima/media/Cartman20Kicks20a20Field20Goal.gif.html]
	
[/URL]


----------



## mamooth (Jul 12, 2015)

"Confirmed wacko declares everyone else is a wacko. Film at 11."


----------



## CrusaderFrank (Jul 12, 2015)

The AGWCult is wack.


----------



## Billy_Bob (Jul 12, 2015)

SSDD said:


> I have never had any doubt that warmers were a bit off...strike that...way off.  But it seems that they are wackier than even I imagined.  The weak minded simpletons have apparently bought into their own farce and are now suffering from....are you ready for this?....really....I mean, are you really ready for this?
> 
> Pre-Traumatic Stress Syndrome.
> 
> ...



I am SPEECHLESS!

PRE...   which means they haven't suffered it yet,  but WILL....??? This is kin to a 2 year old throwing a tantrum for not getting their way.. IF they dont get their way they is gonna have traumatic stress..

I'll give them traumatic stress... a butt paddling!

SO this is the new name for alarmists throwing a hussy fit for not getting their way?

ETA: I had to go back and check that this was a serious paper, not a hoax, MY GOD ITS REAL, done by a real psychologist. (this psych doc must really be hard up for clients)


----------



## CrusaderFrank (Jul 12, 2015)

mamooth said:


> "Confirmed wacko declares everyone else is a wacko. Film at 11."



^ Serious psychological problems: Exhibit A


----------



## HereWeGoAgain (Jul 12, 2015)

Billy_Bob said:


> SSDD said:
> 
> 
> > I have never had any doubt that warmers were a bit off...strike that...way off.  But it seems that they are wackier than even I imagined.  The weak minded simpletons have apparently bought into their own farce and are now suffering from....are you ready for this?....really....I mean, are you really ready for this?
> ...



Another inventive way for liberals to go on disability.


----------



## SSDD (Jul 13, 2015)

Toddsterpatriot said:


> Are they talking about smart photons now?



Interesting....trying to derail a thread with your theory that things must be "smart" in order to obey the laws of physics.  Why not start a thread if you feel you must get it off your chest?


----------



## SSDD (Jul 13, 2015)

mamooth said:


> "Confirmed wacko declares everyone else is a wacko. Film at 11."



As usual, you couldn't be more wrong if you tried....I didn't declare them wacko..they psychologist they turn to with their whiney hysterical fears declared them wacko....I suppose you fear that you will begin to or already suffer from pre traumatic stress syndrome as well....or at least feel for the poor weak minded losers who do.


----------



## SSDD (Jul 13, 2015)

Billy_Bob said:


> SSDD said:
> 
> 
> > I have never had any doubt that warmers were a bit off...strike that...way off.  But it seems that they are wackier than even I imagined.  The weak minded simpletons have apparently bought into their own farce and are now suffering from....are you ready for this?....really....I mean, are you really ready for this?
> ...



Right now, they are suffering over the fact that they will suffer...I guess it is like when you were a kid and someone ran up to you and told you that your mom knew what you were doing and you were in big trouble and she wanted you home right now....they are suffering that long walk back home to get their punishment for lying, faking data, manufacturing results, calling computer model output real data, and promoting alarmism which cost hundreds of billions.

When we were kids, we just called it being in trouble...wacko climate scientists have to call it pre traumatic stress syndrome.  What a bunch of losers with a great big capital flashing red neon "L"


----------



## SSDD (Jul 13, 2015)

CrusaderFrank said:


> mamooth said:
> 
> 
> > "Confirmed wacko declares everyone else is a wacko. Film at 11."
> ...



Does the hairball ever have anything to say any more that actually reflects the topic being discussed or has she been relegated by her peers to nothing more than inane drive by sniping?


----------



## Toddsterpatriot (Jul 13, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Are they talking about smart photons now?
> ...


 
*Interesting....trying to derail a thread with your theory*

Mentioning your wacky theory on a thread about wackos isn't really derailing.

*Why not start a thread*

I prefer to mock your idiocy where I find it.


----------



## IanC (Jul 13, 2015)

I know lots of people who drive themselves crazy by worrying about imaginary or unlikely things. 

Giving a name to it is pretty funny though. Pre traumatic stress syndrome. Hahahahaha.


----------



## Old Rocks (Jul 13, 2015)

Toddsterpatriot said:


> Are they talking about smart photons now?


LOL


----------



## SSDD (Jul 13, 2015)

Toddsterpatriot said:


> SSDD said:
> 
> 
> > Toddsterpatriot said:
> ...


*
*
Smart particles is your hypothesis..not mine...and if they aren't the topic of the OP, then sure it is to derail a thread


----------



## Crick (Jul 13, 2015)

Given that the topic of this thread is nothing but ad hominem attacks and prejudice, it could use some derailment.


----------



## Toddsterpatriot (Jul 13, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > SSDD said:
> ...


 
*Smart particles is your hypothesis..not mine...*

It's your theory that says a photon can never travel from a cooler to a warmer object.
That requires either smart photons or electrons smart enough to know whether or not to emit a photon or smart enough to know in which direction to emit a photon.

So which is it?


----------



## SSDD (Jul 13, 2015)

Crick said:


> Given that the topic of this thread is nothing but ad hominem attacks and prejudice, it could use some derailment.



Talking about the mental state of your priests bothers you?


----------



## SSDD (Jul 13, 2015)

Toddsterpatriot said:


> SSDD said:
> 
> 
> > Toddsterpatriot said:
> ...



Niether....just requires that they obey the laws of physics.  You remain confused regarding the SB law...Of course a radiator radiates equally in every direction if it is in a vacuum at 0 degrees kelvin...but when other objects are present, a different equation is used whereby the amount a radiator radiates is governed by the difference in the radiator and its surroundings.  Couldn't be simpler and that is why a different equation is used from that of a radiator in a vacuum at 0 degrees kelvin....set the temperature of the radiator and its surroundings to the same temperature and tell me what P equals.


----------



## Toddsterpatriot (Jul 13, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > SSDD said:
> ...


* 
Of course a radiator radiates equally in every direction if it is in a vacuum at 0 degrees kelvin*

Excellent! I agree!
But wait, what if a warmer object is 1 light year away?
Your theory is this radiator "knows" the temperature of that distant object and won't radiate in that direction.

*but when other objects are present, a different equation is used*

Actually, the same equation is used, for each object.

* Couldn't be simpler*

That's why your confusion is so amusing.

*set the temperature of the radiator and its surroundings to the same temperature and tell me what P equals*

The relationship governing the net radiation from hot objects is called the Stefan-Boltzmann law:







When both temperatures are the same, the object absorbs and radiates the exact same amounts of energy. The net radiation, P, is zero.
Constantly, happily radiating away, and never getting hotter or colder.
As opposed to your claim, that both radiator and surroundings cease radiating altogether. Something only objects at absolute zero would do.


----------



## IanC (Jul 13, 2015)

Hahahahaha. Officially derailed.


----------



## SSDD (Jul 15, 2015)

Toddsterpatriot said:


> *
> Of course a radiator radiates equally in every direction if it is in a vacuum at 0 degrees kelvin*
> 
> Excellent! I agree!
> ...




Already been through what reality is like for a photon...if they exist.  I didn't make up the rules...I just understand them.  To a photon, that object a light year away is the same as direct contact...same rules as for conduction.  No back conduction...sorry you can't wrap your head around it.  




Toddsterpatriot said:


> Actually, the same equation is used, for each object.



Actually, it isn't.  Perhaps that is part of your problem...not knowing when to apply which equation.



Toddsterpatriot said:


> When both temperatures are the same, the object absorbs and radiates the exact same amounts of energy. The net radiation, P, is zero.



Show me any mention of net in the second law of thermodynamics.  Failure to do that is.......well.......failure.


----------



## SSDD (Jul 15, 2015)

IanC said:


> Hahahahaha. Officially derailed.



Yeah.....sometimes he is as bad as mammoth.  Parents didn't teach him the difference between appropriate conversation....staying on topic....and inappropriate conversation.....changing the topic out of the blue.  Very rude...and immature but what can you do with kids these days.  Basic failure of manners.  You have to blame the parents.


----------



## SSDD (Jul 15, 2015)

Crick said:


> Given that the topic of this thread is nothing but ad hominem attacks and prejudice, it could use some derailment.



Where is the ad hominem or the prejudice?  Ad hom is what you are so good at...a topic is presented and you attack the source rather than address the topic....and prejudice?  Don't see that one here at all.  The topic is that climate scientists are suffering from *snicker**  pre traumatic stress syndrome...sorry I just can't help but laugh over that particular string of words and what they imply...nevertheless...that is the topic.  I didn't make it up....I just reported it.


----------



## Toddsterpatriot (Jul 15, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *
> ...


 
*To a photon, that object a light year away is the same as direct contact..*

And, apparently, an object that isn't there now, but will be in a year. Or if the object is cooler now, but in 363 days suddenly becomes warmer..
It's almost as if the photon can predict the future movement of every object in the universe, no matter how far away, as well as the temperature today and far into the future, before it decides whether and which direction it will travel.

*Actually, it isn't.*

Actually, it is. Use the formula to calculate energy loss for a 100K object with 0K  surroundings. Do the same for a 50K object with 0K surroundings.
Now put them next to each other.
Is it a coincidence that the 100K object now loses less energy, the exact reduction perfectly matching the energy the cooler object emitted in the earlier calculation?

*Show me any mention of net in the second law of thermodynamics.*

As soon as you show me proof that an object above 0K stops radiating when a warmer object is near, or even light years away.
Or even if an object of the same temperature is near.


----------



## mamooth (Jul 15, 2015)

SSDD said:


> Yeah.....sometimes he is as bad as mammoth.  Parents didn't teach him the difference between appropriate conversation....staying on topic....and inappropriate conversation.....changing the topic out of the blue.  Very rude...and immature but what can you do with kids these days.  Basic failure of manners.  You have to blame the parents.



I don't blame your parents. I blame you for being a squealing coward. That's entirely on you. You deliberately chose to guzzle cult piss by the gallon and self-lobotomize yourself.

Now, what do you plan to do about me calling you out on your pathetic behavior? Maybe cry at me even harder? Because that's worked so well for you in the past.

Why does the dickless brigade think that crying at me will suddenly convince me to go easy on them? They should know by now that it just encourages me. If they're crying, I know I'm hitting a sore spot, and therefore should keep hitting it.


----------



## Kosh (Jul 15, 2015)

You have to remember that "Climate Change" is blamed for everything, like more pimples and more prostitution. That is why the AGW is a cult and not real and actual science.


----------



## SSDD (Jul 15, 2015)

Toddsterpatriot said:


> *To a photon, that object a light year away is the same as direct contact..*
> 
> And, apparently, an object that isn't there now, but will be in a year. Or if the object is cooler now, but in 363 days suddenly becomes warmer..
> It's almost as if the photon can predict the future movement of every object in the universe, no matter how far away, as well as the temperature today and far into the future, before it decides whether and which direction it will travel.




Again...I don't make the rules for how a photon experiences the universe....according to physics...they exist at all points between their origin and their destination at the same time....If you are going to believe in photons, then you must accept the rules made up by the people who invented them.  I didn't....I am just reporting what physicists claim.  Again..sorry you can't wrap your head around it.  It must suck....but such is life.





Toddsterpatriot said:


> Actually, it is. Use the formula to calculate energy loss for a 100K object with 0K  surroundings. Do the same for a 50K object with 0K surroundings.



Actually, it isn't....your surroundings in both instances are 0 degrees K.  Are you really this dense?  

This   
	

	
	
		
		

		
		
	


	




  is the formula for a radiator....any radiator in a vacuum at 0 degrees K.  

This, on the other hand,   
	

	
	
		
		

		
		
	


	




   is the formula for a radiator not in a vacuum radiating to surroundings that are above 0 degrees K.  Now again, set T and Tc to the same temperature and tell me what P equals.




Toddsterpatriot said:


> As soon as you show me proof that an object above 0K stops radiating when a warmer object is near, or even light years away.
> Or even if an object of the same temperature is near.



If the second law isn't enough for you...again,  look at the equation above...set T and Tc to the same temperature and tell me what P equals...in case you can't manage it, P - zero.  Do you know what zero means?

Now we are back to you providing a statement of the second law of thermodynamics which mentions net energy transfer and you know as well as I do that you can't provide one.  Like it or not, the laws of physics, and the equations that describe them are on my side.


----------



## SSDD (Jul 15, 2015)

mamooth said:


> I don't blame your parents. I blame you for being a squealing coward. That's entirely on you. You deliberately chose to guzzle cult piss by the gallon and self-lobotomize yourself.



I do blame your parents......look how you turned out.  Look at your mode of conversation...look how vulgar you are...does that make you proud?...think it would make your parents proud?



mamooth said:


> Now, what do you plan to do about me calling you out on your pathetic behavior? Maybe cry at me even harder? Because that's worked so well for you in the past.



My behavior is fine....yours, on the other hand is deplorable...butting in to a discussion.... vulgar language...etc.  You are pitiable......what else is there to say about you?   You are the sort to be held up as an example of bad behavior....go about the threads and look at the content of your posts... Your parents failed and raised a miserable human being.[/QUOTE]


----------



## Toddsterpatriot (Jul 15, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *To a photon, that object a light year away is the same as direct contact..*
> ...


 
*according to physics...they exist at all points between their origin and their destination at the same time.*

So they can predict what will be in their path, 1 million light years away, in 1 million years, and depending on the temperature of that object, decide today if they will travel that way.
I'm not sure if that makes them smart or magic.

*I am just reporting what physicists claim.* 

I've never seen a physicist make your photon (or wave) claims before.


----------



## SSDD (Jul 15, 2015)

Toddsterpatriot said:


> *according to physics...they exist at all points between their origin and their destination at the same time.*
> 
> So they can predict what will be in their path, 1 million light years away, in 1 million years, and depending on the temperature of that object, decide today if they will travel that way.
> I'm not sure if that makes them smart or magic.



If you were already at your destination you wouldn't need to "predict" anything.  Again...sorry you can't wrap your head around this but like it or not, that is how photons experience the universe if they actually exist as physics claims.  



Toddsterpatriot said:


> I've never seen a physicist make your photon (or wave) claims before.



Guess you never looked.  There are plenty who postulate what it would be like to travel at the speed of light.  According to physics, two observers anywhere along a photon's path could see the photon simultaneously if it were possible to see a photon.  Again...sorry this is to big an idea to wrap your mind around...I don't believe in photons personally, but if they exist as science claims and you believe in them, then you are stuck believing in them as science claims.  I am just pointing out what the universe is like for a photon according to science..

Your buddy Ian already commented on this and agreed that photons exist at every point along their path...that would mean that they are at their destination before they ever leave.


----------



## HenryBHough (Jul 15, 2015)

A true believer would have not the slightest hesitation in licking a lamp post in winter to prove it isn't cold.  Regardless of the alleged temperature - 'cause ever'body knows it's too damn hot.


----------



## Toddsterpatriot (Jul 15, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *according to physics...they exist at all points between their origin and their destination at the same time.*
> ...


 
*If you were already at your destination you wouldn't need to "predict" anything.* 

But we're not talking about being at your destination, immediately, we're talking about knowing what will be at your destination, millions, hell billions, of years from now.

*that is how photons experience the universe*

Right, photons can experience not only the universe, but the future, as if they know every movement of every atom, as well as its temperature, now and at every moment in the future. LOL!

*Guess you never looked.*

I have looked. I've never seen a physicist claim a photon will not travel from a cooler object to a warmer one. 

Let's try another experiment. I travel into Earth orbit with a laser. I calculate where the Sun will be in the time it takes the light from my laser to travel 1 AU. I fire my laser in that direction. Will the laser's photons (or waves, I don't care) leave the laser, or will they simply refuse to be emitted?

*photons exist at every point along their path...that would mean that they are at their destination before they ever leave.*

That is awesome! So how do they know if an alien, 1 million years from now, will move a hotter object into their path?


----------



## Toddsterpatriot (Jul 15, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *To a photon, that object a light year away is the same as direct contact..*
> ...


 
*Actually, it isn't....your surroundings in both instances are 0 degrees K.*

Actually, it is. Simply put 0 K into the second formula, what answer do you get?

*Now again, set T and Tc to the same temperature and tell me what P equals.*

P, net energy loss, equals zero. Even though both objects keep radiating.

*If the second law isn't enough for you..*

It is enough for me, because I understand it doesn't cause objects above 0K to stop radiating.

*Do you know what zero means?*

No net energy loss (or gain).

*Now we are back to you providing a statement of the second law of thermodynamics which mentions net energy transfer*

Wasn't the SB formula I provided enough for you? Are you really this dense?


----------



## SSDD (Jul 16, 2015)

Toddsterpatriot said:


> *If you were already at your destination you wouldn't need to "predict" anything.*
> 
> But we're not talking about being at your destination, immediately, we're talking about knowing what will be at your destination, millions, hell billions, of years from now.



No we aren't...we are talking about already being at your destination....Again, according to physics, a photon exists at EVERY point along its path at the same time...that includes the origin and the final destination.  Again, sorry you are having so much trouble wrapping your head around this.




Toddsterpatriot said:


> Right, photons can experience not only the universe, but the future, as if they know every movement of every atom, as well as its temperature, now and at every moment in the future. LOL!



Future is an irrelevant term to an entity that doesn't experience time.  Truly sorry you can't comprehend this.




Toddsterpatriot said:


> I have looked. I've never seen a physicist claim a photon will not travel from a cooler object to a warmer one.



No need....the second law does that.  




Toddsterpatriot said:


> That is awesome! So how do they know if an alien, 1 million years from now, will move a hotter object into their path?



No need to know...they are already there.  They exist at every point along their path from origin to destination at the same time.  Sorry this is so hard for you but you are really beginning to sound like a 5 year old asking why repeatedly...you are still trying to impose time on an entity that doesn't experience time...or distance....or space.


----------



## SSDD (Jul 16, 2015)

Toddsterpatriot said:


> Actually, it is. Simply put 0 K into the second formula, what answer do you get?



This is all to hard for you, isn't it....put 0 in place of T and Tc and P=0...like I said.  Now, what do you think 0 means?




Toddsterpatriot said:


> P, net energy loss, equals zero. Even though both objects keep radiating.



Nothing about net in either the SB equation or the Second Law of Thermodynamics....Making up stuff to insert into them doesn't make it true.  Let me know when either is amended to say anything at all about net energy flow.




Toddsterpatriot said:


> It is enough for me, because I understand it doesn't cause objects above 0K to stop radiating.



So you say, but that isn't what the equations say.  




Toddsterpatriot said:


> No net energy loss (or gain).



Again...nothing about net anything in either law.  They are descriptions of gross energy flows.




Toddsterpatriot said:


> Wasn't the SB formula I provided enough for you? Are you really this dense?



Sorry you don't understand that any more than the idea that photons don't experience time or distance or space...you see the equation and feel like you need to add something that isn't there...ie net.  There is nothing about net in the SB law or the second law...they describe gross energy flows.  Saying net when neither even mentions net is meaningless. If you can't work within what the laws actually say, then you are clearly wrong.[/QUOTE]


----------



## Crick (Jul 16, 2015)

SSDD said:


> No we aren't...we are talking about already being at your destination....Again, according to physics, a photon exists at EVERY point along its path at the same time...that includes the origin and the final destination.  Again, sorry you are having so much trouble wrapping your head around this.



I'm throwing a bullshit flag on that one.

Let's see your source for that idea.  You say "according to physics...".  What physics would that be?


----------



## SSDD (Jul 16, 2015)

Crick said:


> SSDD said:
> 
> 
> > No we aren't...we are talking about already being at your destination....Again, according to physics, a photon exists at EVERY point along its path at the same time...that includes the origin and the final destination.  Again, sorry you are having so much trouble wrapping your head around this.
> ...



That would be the physics of relativity....sorry you can't wrap your mind around it any more than toddster.   Do a bit of reading on relativistic time dilation....that is the reason that a photon exists simultaneously at every point along its path.


----------



## Toddsterpatriot (Jul 16, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *If you were already at your destination you wouldn't need to "predict" anything.*
> ...


 
I notice you clipped the experiment that proves how silly your claims are.

Let's try another experiment. I travel into Earth orbit with a laser. I calculate where the Sun will be in the time it takes the light from my laser to travel 1 AU. I fire my laser in that direction. Will the laser's photons (or waves, I don't care) leave the laser, or will they simply refuse to be emitted?


----------



## Toddsterpatriot (Jul 16, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Actually, it is. Simply put 0 K into the second formula, what answer do you get?
> ...


 
*This is all to hard for you, isn't it....put 0 in place of T and Tc and P=0...like I said. Now, what do you think 0 means?*

No, just in the place of Tc.


----------



## IanC (Jul 17, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *according to physics...they exist at all points between their origin and their destination at the same time.*
> ...




I agree that from the photon's frame of reference there is no time or distance.

the electric force and magnetism could not work without the virtual photon being able to 'test the waters' and find out whether a suitable particle was available to transfer energy to. otherwise it simply ceases to exist, but its lifetime is not long enough to actually travel from one particle to the other in our frame of reference, only in the timeless distanceless world of photons.

one of Feynman's more wacky ideas was that inertia stems from breaking the intricate web of photon paths that have been put in action but not completed. jus' sayin'


----------



## Crick (Jul 17, 2015)

SSDD said:


> Crick said:
> 
> 
> > SSDD said:
> ...



Let's see a quote from a reputable source supporting your claim.


----------



## IanC (Jul 17, 2015)

oh, I forgot to add that I totally disagree with  SSDD's smart photon theory. thermodynamics is statistical in nature. it has nothing to do with individual atomic scale interactions.


----------



## SSDD (Jul 17, 2015)

Toddsterpatriot said:


> I notice you clipped the experiment that proves how silly your claims are.



If your claims had any merit, you could provide actual measured observation instead of a mind experiment.  Clearly, you can't .



Toddsterpatriot said:


> Let's try another experiment. I travel into Earth orbit with a laser. I calculate where the Sun will be in the time it takes the light from my laser to travel 1 AU. I fire my laser in that direction. Will the laser's photons (or waves, I don't care) leave the laser, or will they simply refuse to be emitted?



Lets try providing some actual observed evidence rather than fantasy....making up mind experiments for a situation that you obviously can't get your head wrapped around is pointless.  By the way, are you under the impression that the laser itself doesn't constitute work?  The second law says that energy won't move from cool to warm spontaneously...without some work having been done to make the move happen....a laser is the addition of work.

Show an observed, measured example of energy moving spontaneously from a cool object to a warm object.


----------



## SSDD (Jul 17, 2015)

Toddsterpatriot said:


> *This is all to hard for you, isn't it....put 0 in place of T and Tc and P=0...like I said. Now, what do you think 0 means?*
> 
> No, just in the place of Tc.



Then you have a radiator radiating into cooler surroundings....nothing strange going on there.  What's your point homer? As Tc increases P decreases till finally P=0.....that's the whole point that you don't seem to be able to grasp.


----------



## SSDD (Jul 17, 2015)

IanC said:


> I agree that from the photon's frame of reference there is no time or distance.
> 
> the electric force and magnetism could not work without the virtual photon being able to 'test the waters' and find out whether a suitable particle was available to transfer energy to. otherwise it simply ceases to exist, but its lifetime is not long enough to actually travel from one particle to the other in our frame of reference, only in the timeless distanceless world of photons.
> 
> one of Feynman's more wacky ideas was that inertia stems from breaking the intricate web of photon paths that have been put in action but not completed. jus' sayin'



A virtual particle to test the water for a theoretical particle....and you don't question QM?


----------



## SSDD (Jul 17, 2015)

Crick said:


> Let's see a quote from a reputable source supporting your claim.



Look it up yourself....I gave you all the information you need.  I understand that relativistic time dilation is something that you can't grasp but there is plenty of information out there....you have already proven that if you get information that proves you wrong, you ignore it and continue to spout your absurdities anyway....find your own...I'm not wasting my time on you.

Ignorance on the topic of relativity and the effects of moving at the speed of light is unusual for someone who claims to be an engineer...care to explain that?  Never had a physics course?


----------



## SSDD (Jul 17, 2015)

IanC said:


> oh, I forgot to add that I totally disagree with  SSDD's smart photon theory. thermodynamics is statistical in nature. it has nothing to do with individual atomic scale interactions.



Nothing smart about it....as you acknowledge...the photon exists at every point between its origin and its destination simultaneously...no need to know anything, or be in the least bit smart...radiative transfer is the same as conductive transfer...there is no such thing as  back conduction....do you claim that energy must be smart to know that it can't back conduct?

These claims of smart particles are a logical fallacy on your part...it is an appeal to ridicule.  You claim smart particles and snicker which you believe relieves you from addressing the original point....You really don't believe that particles must be smart in order to obey the laws of physics, but since there is nothing in the second law...or the SB law about net energy movement...or statistics....you offer up a logical fallacy and bypass the fact that the laws in question do not support your position.

I have asked before for one of you guys to let me know when the second law is rewritten to state that it is describing a statistical energy movement.....or net energy movements....or that they do not apply to atomic scale energy movements.....none of you have informed me of any such change so they still say what they say.  Clearly it is a belief that you are repeating and not fact since the laws haven't been rewritten to support your claims.  They still support mine.


----------



## IanC (Jul 17, 2015)

SSDD said:


> IanC said:
> 
> 
> > I agree that from the photon's frame of reference there is no time or distance.
> ...




I dont mind you quoting scientists from 150 years ago, and I dont mind you accepting their explanations of what they observed, but I do object to your belief that they magically had knowledge of atomic scale interactions even though they were unknown at the time. they measured the net results of large numbers of individual interactions and made generalized statements. we now know more.

QM is obviously odd and even counter intuitive at times. that's why so many experiments have been performed. unlike climate science QM has come through with flying colours.


----------



## SSDD (Jul 17, 2015)

IanC said:


> I dont mind you quoting scientists from 150 years ago, and I dont mind you accepting their explanations of what they observed, but I do object to your belief that they magically had knowledge of atomic scale interactions even though they were unknown at the time. they measured the net results of large numbers of individual interactions and made generalized statements. we now know more.



Since the statements of the laws have not been changed to state that they don't apply at the sub atomic level....or that they are only describing statistical estimations it is obvious that "new" science hasn't superseded old science.  Till the laws are reworded they are as they are....and they are laws of nature...not laws of larger than atomic level interactions.

What you are claiming is hypothetical...unobserved.....unmeasured......untestable.......not real.  You are telling me what you and others believe....not what has been tested, observed, and proven.  It is fine to have a belief.....it isn't fine to state your belief as if it were fact.



IanC said:


> QM is obviously odd and even counter intuitive at times. that's why so many experiments have been performed. unlike climate science QM has come through with flying colours.



Which explains why QM is in a constant state of crisis.  Get over it Ian....QM is chock full of contradiction....it is an ad hoc explanation for things we are just now starting to wonder about...it isn't a proven branch of science....and no new physical laws have emerged from it and no old physical laws have been altered because of it....it is, at present groping in the dark, through a hole, with a blindfold covering the eyes and 3 layers of oven mitts covering the hands.  It's great that we are trying to learn, but your belief that QM represents some sort of absolute, concrete, observed, tested, proven knowledge is analogous to the climate scientists belief in models....  Obviously, we don't know enough about the movement of energy in the atmosphere to put together any sort of reliable model and we don't know nearly enough about what happens at the sub atomic level to even begin to paint an actual picture of what is going on there....we are performing experiments and making up, what we think are plausible explanations for what we are seeing in the experiments...NOTHING MORE....so get over yourself.


----------



## IanC (Jul 17, 2015)

SSDD said:


> IanC said:
> 
> 
> > oh, I forgot to add that I totally disagree with  SSDD's smart photon theory. thermodynamics is statistical in nature. it has nothing to do with individual atomic scale interactions.
> ...




radiative transfer is not the same as conductive transfer. how on earth did you come up with THAT one?

my example of reactive (therefore virtual) photons in electric and magnetic force transfer does not carry over to radiative photons that are created simply to shed energy. there is no attractive or repulsive coloured light, there is no coloured attractive or repulsive force photons.


----------



## IanC (Jul 17, 2015)

SSDD said:


> IanC said:
> 
> 
> > I dont mind you quoting scientists from 150 years ago, and I dont mind you accepting their explanations of what they observed, but I do object to your belief that they magically had knowledge of atomic scale interactions even though they were unknown at the time. they measured the net results of large numbers of individual interactions and made generalized statements. we now know more.
> ...




just about every technology that incorporates electricity leads back to QM.

Newton's Laws were proven wrong by Relativity, but they are still useful for most general tasks. the Laws of Thermodynamics are proven wrong at the atomic scale by QM, but are still useful for most general conditions.


----------



## SSDD (Jul 17, 2015)

IanC said:


> radiative transfer is not the same as conductive transfer. how on earth did you come up with THAT one?



Since the distance between the origin of a photon and its destination is zero, the same rules for energy transfer would apply....energy doesn't know whether it is in the form of a photon or a wave, or a vibration...it is simply energy and it moves downhill constantly to a state of greater entropy.......law of nature ian.....not law of statistics....or law of systems.


By the way, regarding QM....from the Encyclopedia of Mathematical Physics (2006) states in Introductory Article: Quantum Mechanics:

_QM in its present formulation is a refined and and successful instrument for the description of the non relativistic phenomena at the Planck scale, *but its internal inconsistency is still standing on shaky ground.*_

It is a description of observations...and even has some predictive value, but as I said, it is not much more than stories made up to explain the results of experiments.....we still have virtually no idea of what is really happening...best plausible explanations is the current state of QM like it or not.


----------



## IanC (Jul 17, 2015)

SSDD said:


> IanC said:
> 
> 
> > radiative transfer is not the same as conductive transfer. how on earth did you come up with THAT one?
> ...




Conduction is not propagated by photons, and certainly not at the speed of light.

QM describes basic things exceedingly well.

If there is confusion at the outer edges that is to be expected.


----------



## SSDD (Jul 17, 2015)

IanC said:


> just about every technology that incorporates electricity leads back to QM.



And about every conversation on catastrophic anthropogenic climate change leads back to CO2....it is entirely possible to see a thing happen and make up an entirely incorrect story about why it happens....  As I said, QM is ok as a predictor...but the stories we make up about what is happening at the sub atomic level are just that...best guesses about things we are just beginning to comprehend...to believe that we have any real understanding of what is actually happening at the sub atomic level is pure faith.



IanC said:


> the Laws of Thermodynamics are proven wrong at the atomic scale by QM, but are still useful for most general conditions.



Nothing at the sub atomic level is proven...you throw that word around as if using it makes it so...it isn't.  If any such thing had been proven, then the laws of physics would be rewritten to reflect this new "proof".  They haven't.  Suspicion is not a basis for proof..


----------



## Toddsterpatriot (Jul 17, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > I notice you clipped the experiment that proves how silly your claims are.
> ...


 
*The second law says that energy won't move from cool to warm spontaneously...without some work having been done to make the move happen....a laser is the addition of work.*
So it is possible for a photon to move from cooler to hotter.
Thanks for playing.


----------



## Toddsterpatriot (Jul 17, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *This is all to hard for you, isn't it....put 0 in place of T and Tc and P=0...like I said. Now, what do you think 0 means?*
> ...


 
*Then you have a radiator radiating into cooler surroundings....nothing strange going on there.* 

Who said anything about strange? I simply said it was the same formula.
Glad you finally realize that.

*As Tc increases P decreases till finally P=0.....that's the whole point that you don't seem to be able to grasp.*

And according to your silly theory, when P=0, both objects stop emitting.


----------



## SSDD (Jul 17, 2015)

Toddsterpatriot said:


> *The second law says that energy won't move from cool to warm spontaneously...without some work having been done to make the move happen....a laser is the addition of work.*
> So it is possible for a photon to move from cooler to hotter.
> Thanks for playing.



I never said that it didn't if one inputs work into the system to make it happen...that however does not include back radiation from the cooler atmosphere to the warmer earth which is what this entire topic is about....Energy can't move spontaneously from the cool atmosphere to the warmer surface of the earth.

I had assumed (apparently incorrectly) that you actually knew what the second law said regarding the fact that energy can not move from cool to warm spontaneously....I am sure that I posted a statement of the second law that said explicitly that energy could move from cool to warm if one input work to the system.

No back radiation from the atmosphere to the surface however...that one is still a loser.


----------



## SSDD (Jul 17, 2015)

Toddsterpatriot said:


> Who said anything about strange? I simply said it was the same formula.
> Glad you finally realize that.



It is the same formula if the surroundings are at zero K....as soon as you rise above zero K, however, P starts to decrease.  Sorry this is all so difficult for you.




Toddsterpatriot said:


> And according to your silly theory, when P=0, both objects stop emitting.



Not my theory...the SB law.  That number that P represents has a meaning.....if P=0 that means something and there is no mention in the SB law about net energy transfer....P represents gross energy flow and when P=0 that means that P isn't radiating....

Again, all objects radiate when in a vacuum at 0 degrees K....things start changing when they are no longer in a vacuum at 0 degrees K as evidenced by the equation.

Is this really so far past your ability to understand?


----------



## mamooth (Jul 17, 2015)

And to think that every university in the world with a physics department still teaches Statistical Mechanics, even though SSDD says it's all totally wrong. SSDD really needs to set them all right.


----------



## jc456 (Jul 17, 2015)

mamooth said:


> And to think that every university in the world with a physics department still teaches Statistical Mechanics, even though SSDD says it's all totally wrong. SSDD really needs to set them all right.


just curious, you can prove he's wrong?


----------



## Toddsterpatriot (Jul 17, 2015)

jc456 said:


> mamooth said:
> 
> 
> > And to think that every university in the world with a physics department still teaches Statistical Mechanics, even though SSDD says it's all totally wrong. SSDD really needs to set them all right.
> ...


 
If his claims about photons aren't enough for you........


----------



## Toddsterpatriot (Jul 17, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *The second law says that energy won't move from cool to warm spontaneously...without some work having been done to make the move happen....a laser is the addition of work.*
> ...


 
*I never said that it didn't if one inputs work into the system to make it happen...*

So a photon, or wave, from a candle flame can travel to the much hotter Sun, because work was done?

*No back radiation from the atmosphere to the surface however...that one is still a loser.*

No work was done to heat the atmosphere? Are you sure?


----------



## Toddsterpatriot (Jul 17, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Who said anything about strange? I simply said it was the same formula.
> ...


 
*Not my theory...the SB law.* 

The SB law says they stop emitting? Show me.

*there is no mention in the SB law about net energy transfer....*

You're claiming that old Science article I referenced was wrong?
You'd think they'd know at least as much as you, especially back in the 60s, when AGW wasn't confusing the issue.


----------



## IanC (Jul 18, 2015)

thermodynamic theory says every object of sufficient size gives off radiation proportional to its temperature. there is no arbitrary choice of whether to radiate or not.

the power, energy loss (gain) per second per meter squared, is a summation of object radiation loss minus object radiation gain from the surroundings. both the object and the surroundings radiate according to their temperatures. 

in the special case where both the object and the surroundings are the same temperature, both continue to fully radiate according to their temperature but the power is zero because they are both gaining energy as fast as they are radiating it away.

SSDD purposely confuses the basic principle of 'every object radiates according to its temperature' with the equation for Power/energy transfer that is dependent on a temperature differential. he declares that the power describes the actual radiation, eg the warmer object emits less than the basic principle demands and the cooler object doesnt radiate at all.

SSDD cannot let go of this folly because if he did then his worldview of thermodynamics would fall apart.

my first sentence contained 'every object of sufficient size'. temperature is the measurement of average kinetic energy. obviously a single molecule can have any speed but it is not a 'temperature'. only when large amounts of molecules are present do you get the collisions necessary to produce the 'radiation proportional to its temperature'.


----------



## SSDD (Jul 18, 2015)

mamooth said:


> And to think that every university in the world with a physics department still teaches Statistical Mechanics, even though SSDD says it's all totally wrong. SSDD really needs to set them all right.



Actually, they don't hairball...a survey of physics texts reveals that physics for the soft sciences teaches back radiation..physics for the hard sciences does not.


----------



## SSDD (Jul 18, 2015)

Toddsterpatriot said:


> No work was done to heat the atmosphere? Are you sure?



Yep...sorry you aren't.  Absorption and emission do not constitute work.  I really thought that you had some clue about this and were just being obtuse.  The more we talk, the more apparent it becomes that you really don't have a clue.

work has a meaning.....force times distance??.....ever hear of it?  Absorption and emission are not work....I really do apologize for some of the things I have been thinking about you and the apparent dishonest nature of your posts....I see now that what I was mistaking for dishonesty was, in fact, abject ignorance.


----------



## SSDD (Jul 18, 2015)

Toddsterpatriot said:


> The SB law says they stop emitting? Show me.



Already did...multiple times.  Sorry you failed to understand.  One more time....as if showing you again after all these times will make a difference.  This is real simple math so you shouldn't have a problem with it, although you have repeatedly shown that you do.  Make T any positive number you wish....make Tc the same number....the resulting number within the parentheses will be zero.....the fact that the zero is in parenthesis in the equation means that you multiply the other parts of the equation by the number in the parentheses....zero times anything equals zero.  So when you have T = to Tc P=0.  And again, the SB law doesn't mention net energy flows anywhere  The SB law is a statement on the rate of * outward radiative energy *(per unit area)...









Toddsterpatriot said:


> You're claiming that old Science article I referenced was wrong?



Actually, you and the science article are claiming that the SB law is wrong...I didn't see any evidence there that it was wrong...and surely something as big as learning that one of the laws of physics was wrong would have made headlines......can you show me a reputable source stating that the SB law is wrong?



Toddsterpatriot said:


> You'd think they'd know at least as much as you, especially back in the 60s, when AGW wasn't confusing the issue.



You would think so...wouldn't you.[/QUOTE]


----------



## SSDD (Jul 18, 2015)

IanC said:


> thermodynamic theory says every object of sufficient size gives off radiation proportional to its temperature. there is no arbitrary choice of whether to radiate or not.



When the object is a black body  in a vacuum at 0 degrees K.  You guys always fail to mention that fact.  Once you are no longer talking about a black body and it is out of the vacuum and the surroundings are above 0 degrees K...the equation clearly shows that things begin to change,



IanC said:


> in the special case where both the object and the surroundings are the same temperature, both continue to fully radiate according to their temperature but the power is zero because they are both gaining energy as fast as they are radiating it away.



Yeah....heard it before.  The thing is though, that according to the SB law, P does not represent "net" energy flow.  

SSDD purposely confuses the basic principle of 'every object radiates according to its temperature' with the equation for Power/energy transfer that is dependent on a temperature differential. he declares that the power describes the actual radiation, eg the warmer object emits less than the basic principle demands and the cooler object doesnt radiate at all.

The Stefan Boltzmann Law

The Stefan-Boltzman Law for Black Body Radiation

http://www.public.asu.edu/~hhuang38/mae578_lecture_03.pdf

Here is an informative clip from that last one



> This is known as Stefan-Boltzmann law, which states that the *rate of outward radiative energy *(per unit area) emitted by an object with temperature T is proportional to the 4th power of T



Note that the SB law is, in fact, talking about rates of outward radiative energy....not net energy flows or any other addendum you would care to add to the law in an effort to make it support your beliefs.  Simply rate of outward energy flow....no more...no less.



IanC said:


> SSDD cannot let go of this folly because if he did then his worldview of thermodynamics would fall apart.



It isn't me, Ian, who is adding words like net, and statistical to the actual statements of the law in an effort to support my argument...that is you and yours.  I am satisfied with the laws as they are stated and until you and yours actually prove that they are wrong, they will remain as they are and state what they state and you will still be making statements that don't agree with the laws.  When QM actually proves that energy movement is a net proposition, the laws of thermodynamics will be changed and rewritten to reflect that proof.  As of now, they don't which makes me right and you wrong.


----------



## IanC (Jul 18, 2015)

did you actually read that third link? the lecture? hahahahaha. OMG that's so funny. it is polar opposite to what SSDD says. hahahahahahaha.


----------



## SSDD (Jul 18, 2015)

IanC said:


> did you actually read that third link? the lecture? hahahahaha. OMG that's so funny. it is polar opposite to what SSDD says. hahahahahahaha.



Have to wonder if you read it ian....or read into it what you wish it said.


----------



## Crick (Jul 18, 2015)

Does anyone here think SSDD's interpretation of basic thermodynamics is correct?


----------



## Toddsterpatriot (Jul 18, 2015)

SSDD said:


> IanC said:
> 
> 
> > thermodynamic theory says every object of sufficient size gives off radiation proportional to its temperature. there is no arbitrary choice of whether to radiate or not.
> ...


 
*This is known as Stefan-Boltzmann law, which states that the rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T*

Did you notice, it doesn't say, "Unless a warmer object is nearby, in that case it stops radiating" It also failed to say, "Unless an object above 0K is nearby, in which case it emits more slowly"


----------



## SSDD (Jul 18, 2015)

Crick said:


> Does anyone here think SSDD's interpretation of basic thermodynamics is correct?



That's the thing...I seem to be the only one who is not interpreting....  I take the statement of the law at face value...without adding to it or rewriting it in any way....it is you guys who are interpreting....not me.


----------



## SSDD (Jul 18, 2015)

Toddsterpatriot said:


> *This is known as Stefan-Boltzmann law, which states that the rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T*
> 
> Did you notice, it doesn't say, "Unless a warmer object is nearby, in that case it stops radiating" It also failed to say, "Unless an object above 0K is nearby, in which case it emits more slowly"



They assume you can grasp the equation one uses when the radiator is not a perfect black body in a vacuum at 0 degrees K....in which P decreases as the difference between T and Tc decreases till P = 0.


Again...sorry this is all so difficult for you to understand.


----------



## Toddsterpatriot (Jul 18, 2015)

IanC said:


> did you actually read that third link? the lecture? hahahahaha. OMG that's so funny. it is polar opposite to what SSDD says. hahahahahahaha.


 
Think of a small area element at the surface of the sun: It emits radiation with the intensity of
_B_(_T_)
in all direction, thereby producing a total energy flux of 4
_B_(_T_), where 4 is the total solid angle of a sphere enclosing that area element. Note, however, that a half of this energy flux goes back into the interior of the sun.

Holy crap! Energy moving from cooler to hotter.
And from his own link.


----------



## Toddsterpatriot (Jul 18, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *This is known as Stefan-Boltzmann law, which states that the rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T*
> ...


 
*rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T*

Again, no caveats...sorry this is all so difficult for you to understand


----------



## Toddsterpatriot (Jul 18, 2015)

SSDD said:


> IanC said:
> 
> 
> > thermodynamic theory says every object of sufficient size gives off radiation proportional to its temperature. there is no arbitrary choice of whether to radiate or not.
> ...


 
Page 16 of your lecture link shows energy emitted by cloudy atmosphere toward Earth (+67) and  energy emitted by cloudless atmosphere toward Earth (+34).


----------



## SSDD (Jul 18, 2015)

Toddsterpatriot said:


> Think of a small area element at the surface of the sun: It emits radiation with the intensity of
> _B_(_T_)
> in all direction, thereby producing a total energy flux of 4
> _B_(_T_), where 4 is the total solid angle of a sphere enclosing that area element. Note, however, that a half of this energy flux goes back into the interior of the sun.
> ...



Again...you are introducing work into the equation....the second law says that energy won't move spontaneously from cool to warm....that precludes back radiation...and instead of endless mind games...why don't you think of providing some actual observed, measured instance of energy moving spontaneously from cool to warm....oh...that's right...there are none so you are left with endless mind experiments that remain unobservable...unmeasurable...and untestable...


----------



## SSDD (Jul 18, 2015)

Toddsterpatriot said:


> Page 16 of your lecture link shows energy emitted by cloudy atmosphere toward Earth (+67) and  energy emitted by cloudless atmosphere toward Earth (+34).



And if the figure for the radiation emitted from the surface of the earth is higher than either of those numbers, no energy will move to the surface of the earth....P is greater than 0 so long as T is greater than Tc.

Again...so very sorry that this all is so difficult for you.


----------



## Toddsterpatriot (Jul 18, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Think of a small area element at the surface of the sun: It emits radiation with the intensity of
> ...


 
*Again...you are introducing work into the equation....the second law says that energy won't move spontaneously from cool to warm*

Work? Are you claiming fusion at the surface of the sun causes energy to move from the cooler surface to the hotter core? 

*why don't you think of providing some actual observed, measured instance of energy moving*

Why do I need to provide anything?  I simply point out where your own source contradicts your silly claim.


----------



## Toddsterpatriot (Jul 18, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Page 16 of your lecture link shows energy emitted by cloudy atmosphere toward Earth (+67) and  energy emitted by cloudless atmosphere toward Earth (+34).
> ...


 
*And if the figure for the radiation emitted from the surface of the earth is higher than either of those numbers, no energy will move to the surface of the earth*

But of course the surface emits more. So you're contradicting your own source, again.
Thanks for the laugh.

*Again...so very sorry that this all is so difficult for you.*

Pointing out your ignorance is not difficult at all.


----------



## IanC (Jul 19, 2015)

Toddsterpatriot said:


> SSDD said:
> 
> 
> > Toddsterpatriot said:
> ...




hey Todd....was that an epic face plant putting up that lecture link or what? I was worried I'd piss myself I was laughing so hard.


----------



## SSDD (Jul 19, 2015)

Toddsterpatriot said:


> Work? Are you claiming fusion at the surface of the sun causes energy to move from the cooler surface to the hotter core?




Is your small area element powered?...if it is, then we are not talking about spontaneous energy movement..




Toddsterpatriot said:


> Why do I need to provide anything?  I simply point out where your own source contradicts your silly claim.



Your example didn't stipulate the temperature of the surface of the earth....so you didn't point out anything...if the energy radiated from the surface of the earth was greater than that radiated from the atmosphere...none of it was absorbed by the earth...you must have all the numbers to plug into the equation before you can claim anything more than an incomplete example.


----------



## SSDD (Jul 19, 2015)

Toddsterpatriot said:


> But of course the surface emits more. So you're contradicting your own source, again.
> Thanks for the laugh.




Since the source never stipulated the amount being radiated from the surface...the equation was never completed...not sure what you are laughing at unless you find your own ignorance funny....I don't....it's a bit sad really.


----------



## Crick (Jul 19, 2015)

What's it like to have dug yourself into a hole as deep as this one?  Do you stay awake nights trying to dream up a way out?


----------



## SSDD (Jul 19, 2015)

IanC said:


> hey Todd....was that an epic face plant putting up that lecture link or what? I was worried I'd piss myself I was laughing so hard.



I see that you think an incomplete energy balance equation is funny also...that is very interesting...just for the sake of discussion, lets assume that the temperature of the temperature of the surface is greater than that of the atmosphere....within the SB equation for radiators other than black bodies not in a vacuum at 0 degrees K...how much radiation from the cooler atmosphere do you think reaches the surface of the earth?


----------



## Crick (Jul 19, 2015)

Every bit that 's headed that way.


----------



## SSDD (Jul 19, 2015)

Crick said:


> God are you stupid.  What's it like to have dug yourself into a hole as deep as this one?  Do you stay awake nights trying to dream up a way out?



Sorry crick...the adults are talking...I can't help but note that you have had exactly zero to say regarding the actual math and physics that are being discussed here....we all know why.  Are you now doing mammoths job for her...drive by sniping with no mention at all on the topic that is being discussed?

Feel free to actually join in and speak in your own words on the topic....everyone here knows that the chance of that is about zero....


----------



## SSDD (Jul 19, 2015)

Crick said:


> Every bit that 's headed that way.



Based on which part of the SB equations?


----------



## Crick (Jul 19, 2015)

The topic of this particular thread, in my humble opinion, is itself a violation of the rules and should have been chopped at birth.  

I tired of explaining to you what sort of idiocy it takes to hold the views on thermodynamics that you hold.  Ian and Todd have been doing a bang up job in my stead.  I've had thermodynamics and heat transfer.  You, VERY obviously, have not.  When you have, come back and apologize to us for making us suffer through your desperate cling to ignorance.


----------



## SSDD (Jul 19, 2015)

Crick said:


> The topic of this particular thread, in my humble opinion, is itself a violation of the rules and should have been chopped at birth.



Like I said...you really don't know what the hell we are talking about and just decided to do a bit of drive by sniping....We all know that you aren't going to engage on the topic....asking which part of the SB equation supported your claim was about as easy a question as I could have possibly asked and you....predictably....dodged.



Crick said:


> I tired of explaining to you what sort of idiocy it takes to hold the views on thermodynamics that you hold.



More dishonesty on your part....we all know that you never "explain" anything.  You often go grab a bit of cut and paste which rarely supports your claim or a graph which you can't read...but you never engage in any discussion of actual science or math in your own words and crick....we all know why.



Crick said:


> Ian and Todd have been doing a bang up job in my stead.



You really wouldn't know would you...the fact that you can't point out which part of a very simple equation supports your claim regarding how much energy from the atmosphere would actually reach the earth literally screams that you don't know who is right and who is wrong on this....you believe that they are supporting your faith so you go with them...and that is about the extent of your contribution here....you fail to note that neither toddster nor Ian will step up and state which part of the equation supports their claim that the energy reaches the surface either....why you might ask?.....because the equations don't support that claim.



Crick said:


> I've had thermodynamics and heat transfer.



You haven't had jack...your inability to actually discuss any subject regarding physics or mathematics makes that perfectly clear...the fact that you can't read and comprehend the simplest of graphs slams a great big exclamation point on the fact that you haven't had jack.



Crick said:


> You, VERY obviously, have not.  When you have, come back and apologize to us for making us suffer through your desperate cling to ignorance.



Unlike you, I am prepared to discuss the topic and use the laws concerned to support my claims...to date all that either toddster or Ian has rebutted me with is their own faith,....feel free to look back and bring forward any actual observed, measured examples of their claims....you will find none while every observation and measurement ever made supports my claim since I am merely stating the second law of thermodynamics and the SB law.

Now run along if you can't even answer the most simple mathematical question....you have no place here.  mamooth is the resident inane sniper...we don't need another.  Maybe you could wear a duck beak and big red shoes and do a happy dance or some such thing to provide comic relief.


----------



## IanC (Jul 19, 2015)

SSDD said:


> IanC said:
> 
> 
> > hey Todd....was that an epic face plant putting up that lecture link or what? I was worried I'd piss myself I was laughing so hard.
> ...




your question is kinda convoluted but....all of the radiation emitted towards the surface reaches the surface, unless some other particle gets in the way. the other half of the radiation goes up.


----------



## SSDD (Jul 19, 2015)

IanC said:


> SSDD said:
> 
> 
> > IanC said:
> ...



Sorry Ian, but it doesn't..  Here's the equation.....
	

	
	
		
		

		
		
	


	




   First how much radiation is being emitted from the surface of the earth if there were no atmosphere...you will want to know that so you can see the amount of change.....then put the temperature of the surface of the earth in place of T....then put the temperature of the sky with or with out clouds in place of Tc...P is the amount of radiation being emitted from the surface of the earth....and again, the SB equation is all about outward bound radiation from the radiator...there is no net energy transfer....its all gross energy flows.  You have no observation of net flows...you have no measurements of net flows...you have no proof of net flows...net flow is a theory that is unobservable.....unmeasurable...and untestable.  The laws in question don't even mention net energy flows...

Either you can prove your point with the laws and equations as they are written and stated or you can't...if you can't then you fail.  Sorry.


----------



## Toddsterpatriot (Jul 19, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Work? Are you claiming fusion at the surface of the sun causes energy to move from the cooler surface to the hotter core?
> ...


 
*Is your small area element powered?...*

Only if you think fusion takes place at the surface. Do you?

*Your example didn't stipulate the temperature of the surface of the earth*

My example? No, it was your example.

*so you didn't point out anything...*

I pointed out that your example showed back radiation. 

*if the energy radiated from the surface of the earth was greater than that radiated from the atmosphere...none of it was absorbed by the earth*

Radiation from the atmosphere is magically different than other radiation and somehow is not absorbed? You make less sense every time you build upon your special theory.

*you must have all the numbers to plug into the equation before you can claim anything more than an incomplete example*

Your link didn't include a complete example?


----------



## Toddsterpatriot (Jul 19, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > But of course the surface emits more. So you're contradicting your own source, again.
> ...


 
*not sure what you are laughing*

I'm laughing at your silly theory getting refuted by your own link.


----------



## SSDD (Jul 19, 2015)

Toddsterpatriot said:


> Only if you think fusion takes place at the surface. Do you?




Your mind experiment...either your small area element is powered or it isn't...interesting that you can't even give a straight answer to whether or not it is powered.



Toddsterpatriot said:


> My example? No, it was your example.



I would never have mentioned it since all of the required information wasn't there....you picked it out and brought it forward as if it proved anything more than that you didn't understand what was missing.




Toddsterpatriot said:


> I pointed out that your example showed back radiation.



My example pointed out radiation from the atmosphere...it didn't say that any reached the surface of the earth...and if you plugged the numbers into the SB equation, you would see for yourself that none reached the surface of the earth...of course now we know that plugging in numbers is over your head.




Toddsterpatriot said:


> Radiation from the atmosphere is magically different than other radiation and somehow is not absorbed? You make less sense every time you build upon your special theory.



Of course not...your logical fallacy fails every time...it is no different from any radiation which doesn't move from cool to warm.  Interesting that you believe intelligence is required to obey the laws of physics....according to you water must be smart to run downhill....and a rock must be smart to fall down instead of up...and electricity must be smart to know to run in only one direction down a wire...



Toddsterpatriot said:


> Your link didn't include a complete example?



Sorry you could't tell.  It must suck to not be able to speak about the topic in anything more than what you believe are smart assed quips....we all know why people respond with smart assed quips rather than engage in a substantive discussion...your answers are much like those crick and mammoth give....skirting any actual discussion of the topic...no defense of your claims...no actual engaging on the topic...at least Ian discusses the topic....he won't go into the nuts and bolts of what the laws say...or the mathematical formulae that describe them.. but he doesn't do it because he at least knows enough to know that to do so would do nothing but support my position....he spends his time telling me what he wished the laws said rather than what they actually say.

You are becoming increasingly tedious and boring...now that I know that this is all past your ability to understand, there isn't much point in continuing...you are never going to be any more interesting than you are now...which isn't very.


----------



## Toddsterpatriot (Jul 19, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Only if you think fusion takes place at the surface. Do you?
> ...


 
*Your mind experiment...*

My experiment? It was from your link. Are you stoned?

*either your small area element is powered or it isn't*

Do you think the surface of the Sun is powered?

*I would never have mentioned it since all of the required information wasn't there....*

Your own link showed back radiation. 

*My example pointed out radiation from the atmosphere...it didn't say that any reached the surface of the earth..*

So, your smart waves radiated from the cooler atmosphere to the warmer surface, because they knew they wouldn't actually reach the warmer surface? What prevented them from reaching the warmer surface?

*it is no different from any radiation which doesn't move from cool to warm.* 

Except, apparently, the radiation shown in _your_ source. LOL!

*You are becoming increasingly tedious and boring...*

Having your idiocy rubbed in your face must be very exhausting for you. 
Especially when you do it to yourself. Thanks again for those links.


----------



## SSDD (Jul 20, 2015)

Since it is clear that you really can't talk about the topic...and aren't going to show how any of the physical laws in question support your claims....I suppose the conversation has gone as far as it will ever go....just a side note on the topic however...

A paper was recently published in the Journal of the Royal Meterological Society finding that GHG's at present earth temperatures are providing a negative feedback and causing cooling rather than the warming you warmers and luke warmers believe.  This paper supports my position that the climate sensitivity to GHG's is in fact zero or LESS than ZERO and would explain why the tropospheric hot spot which would necessarily exist if back radiation existed has never appeared.  There is no back radiation and the absence of the hot spot proves it.


----------



## Toddsterpatriot (Jul 20, 2015)

SSDD said:


> Since it is clear that you really can't talk about the topic...and aren't going to show how any of the physical laws in question support your claims....I suppose the conversation has gone as far as it will ever go....just a side note on the topic however...
> 
> A paper was recently published in the Journal of the Royal Meterological Society finding that GHG's at present earth temperatures are providing a negative feedback and causing cooling rather than the warming you warmers and luke warmers believe.  This paper supports my position that the climate sensitivity to GHG's is in fact zero or LESS than ZERO and would explain why the tropospheric hot spot which would necessarily exist if back radiation existed has never appeared.  There is no back radiation and the absence of the hot spot proves it.


 
*A paper was recently published in the Journal of the Royal Meterological Society finding that GHG's at present earth temperatures are providing a negative feedback and causing cooling rather than the warming you warmers and luke warmers believe.*

That's interesting and certainly could be the case.

It also has nothing to do with your claim there is no back radiation or your idiocy concerning SB.


----------



## SSDD (Jul 20, 2015)

Toddsterpatriot said:


> It also has nothing to do with your claim there is no back radiation or your idiocy concerning SB.



There is no spontaneous back radiation...never has been.  Not one observation ever of energy moving spontaneously from a cool area to a warm area.


----------



## Toddsterpatriot (Jul 20, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > It also has nothing to do with your claim there is no back radiation or your idiocy concerning SB.
> ...


 
*There is no spontaneous back radiation*

That's so weird, your own source showed there was.

*Not one observation ever of energy moving spontaneously from a cool area to a warm area.*

I've always heard that heat won't do that, never saw the claim, except from you, that energy won't.


----------



## SSDD (Jul 20, 2015)

Toddsterpatriot said:


> That's so weird, your own source showed there was.



Actually toddster...without completing the equation..it showed nothing at all....the numbers are there...go ahead and complete the equation and tell me how much back radiation they show.




Toddsterpatriot said:


> I've always heard that heat won't do that, never saw the claim, except from you, that energy won't.



Is heat a form of energy itself or is heat merely the physical evidence of energy moving from one place to another?  Important question...got an answer?


----------



## Toddsterpatriot (Jul 20, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > That's so weird, your own source showed there was.
> ...


 
Actually, your link did show that. In the diagram on the last page.

So you don't actually have a source that says that about energy.


----------



## SSDD (Jul 20, 2015)

Toddsterpatriot said:


> Actually, your link did show that. In the diagram on the last page.
> 
> So you don't actually have a source that says that about energy.



Is heat a form of energy or is heat the evidence that energy is moving from one place to another?


----------



## SSDD (Jul 20, 2015)

Toddsterpatriot said:


> Actually, your link did show that. In the diagram on the last page.
> 
> So you don't actually have a source that says that about energy.



Didn't see any numbers plugged into the SB equations....All I saw is the same sort of magical assumptions you and ian have been making all along...plug the numbers into the SB equation and tell me how much radiation you see racing the surface of the earth.


----------



## Crick (Jul 20, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Actually, your link did show that. In the diagram on the last page.
> ...



What would you like it to be Sid?


----------



## Toddsterpatriot (Jul 20, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Actually, your link did show that. In the diagram on the last page.
> ...


 
*Didn't see any numbers plugged into the SB equations....*

No numbers, and yet they still know more about energy than you.


----------



## Toddsterpatriot (Jul 20, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Actually, your link did show that. In the diagram on the last page.
> ...


 
Tell me again about your link that said energy from the Sun's surface is emitted toward the core.


----------



## SSDD (Jul 20, 2015)

Crick said:


> SSDD said:
> 
> 
> > Toddsterpatriot said:
> ...



As usual, a mamootesque  comment meaning nothing and not moving the conversation in any direction whatsoever....typical for someone who really isn't prepared to discuss the topic...no surprise there.


----------



## SSDD (Jul 20, 2015)

Toddsterpatriot said:


> Tell me again about your link that said energy from the Sun's surface is emitted toward the core.



Is heat a form of energy or evidence that energy is moving from one place to another....what's the matter toddster.....cat got your tongue?  Which is it?


----------



## Crick (Jul 20, 2015)

Which one are you talking about?


----------



## Toddsterpatriot (Jul 20, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Tell me again about your link that said energy from the Sun's surface is emitted toward the core.
> ...


 
*Which is it?*

You're the one with the unique interpretation of the 2nd Law.
And Stefan-Boltzmann. And relativity. You tell me.
And then explain why your links all contradict your claims.
Is the surface of the Sun powered? LOL!


----------



## SSDD (Jul 21, 2015)

Toddsterpatriot said:


> *Which is it?*
> 
> You're the one with the unique interpretation of the 2nd Law.
> And Stefan-Boltzmann. And relativity. You tell me.
> ...



So you are quite sure that energy can move from cool to warm but unsure as to whether heat is energy or an indicator that energy is moving from one place to another....how can you be so sure of one when you are clueless as to the other....answer = faith.


----------



## Crick (Jul 21, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *Which is it?*
> ...



Since neither will behave as you claim they will, your attempted semantic detour here is simply another failure.


----------



## IanC (Jul 21, 2015)

I have to agree with crick here. SSDD has lost the war on simple radiation so he wants to switch the discussion to a vastly more complex subject where no one can define the terms let alone describe the process.


----------



## Toddsterpatriot (Jul 21, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *Which is it?*
> ...


 
*So you are quite sure that energy can move from cool to warm*

For hot objects other than ideal radiators, the law is expressed in the form:







where e is the emissivity of the object (e = 1 for ideal radiator). If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form





Stefan-Boltzmann Law

You'll notice they use the phrase "net radiation loss".
And no mention of the radiation slowing or stopping when another object is nearby.

You may have noticed that none of your sources mention slowing or stopping.
In fact, your sources mention the cooler surface of the Sun radiating toward the hotter core.
They also mentioned our cooler atmosphere radiating toward the warmer surface of Earth.
In direct contradiction to your claim that back radiation does not exist.

*....how can you be so sure of one*

Because the Stefan-Boltzmann Law and all your sources said so.


----------



## SSDD (Jul 21, 2015)

Toddsterpatriot said:


> SSDD said:
> 
> 
> > Toddsterpatriot said:
> ...



I am sure that you are unaware of this but when you calculate net loss of anything....your equation must show the amount of the thing coming in and the amount of the thing going out....Can you point out in the above equation where anything at all is gained?  That equation shows P changing based on the difference between T and Tc....where do you see anything whatsoever about energy being gained in that equation.....there is only one term in that equation representing energy and it is for outgoing energy....where is the term for incoming energy?


----------



## Toddsterpatriot (Jul 21, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > SSDD said:
> ...


 
The net energy out from the warmer object is the net energy into the cooler surroundings.
Why do you feel I would be unaware of that?

From the same source as above......
_While the typical situation envisioned here is the radiation from a hot object to its cooler surroundings, the Stefan-Boltzmann law is not limited to that case. If the surroundings are at a higher temperature (TC > T) then you will obtain a negative answer, implying net radiative transfer to the object._


----------



## IanC (Jul 21, 2015)

SSDD purposely ignores that two processes are going on at the same time. Both objects are radiating, the warmer one more than the cool one. For any period of time measured there is a net excess of radiation from warm to cool. You cannot separate the outcome of the two processes but you can calculate the individual strengths.

Pnet = Pwarm - Pcool

This is now where SSDD says mathematics is wrong and the associative rule produces a tainted result.

You simply can't argue with SSDD on this. He has a blind spot that he will protect at all costs.


----------



## Crick (Jul 21, 2015)

Several.


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> The net energy out from the warmer object is the net energy into the cooler surroundings.
> Why do you feel I would be unaware of that?



You don't seem to understand that the term net has a meaning....it is the total of something coming in minus the amount of something going out.  The equation describes something going out...but doesn't describe anything coming in.  I can have an output of anything from gas, to liquid, to solid, to electricity and do any number of things to alter the amount going out without ever altering the amount of that same thing coming in.  The changes I make to the output are just that...changes to gross flow...in order to have a net flow...I need to alter the amount of whatever I have outgoing with an incoming amount.



Toddsterpatriot said:


> From the same source as above......
> _While the typical situation envisioned here is the radiation from a hot object to its cooler surroundings, the Stefan-Boltzmann law is not limited to that case. If the surroundings are at a higher temperature (TC > T) then you will obtain a negative answer, implying net radiative transfer to the object._



Again...net means something...it isn't just a meaningless term thrown in....net is incoming minus outgoing...the SB equation in question only shows outgoing...where is the incoming term that would alter P from a simple statement of gross flow to a more complex term of gross flow?


----------



## SSDD (Jul 22, 2015)

IanC said:


> SSDD purposely ignores that two processes are going on at the same time. Both objects are radiating, the warmer one more than the cool one. For any period of time measured there is a net excess of radiation from warm to cool. You cannot separate the outcome of the two processes but you can calculate the individual strengths.



I am not ignoring anything...I am looking at the equations for what they are.... I believe you purposely ignore the fact that net has a specific meaning....net is incoming minus outgoing...or outgoing minus incoming...a statement of change in outgoing or incoming is a gross change.  It is only a net change if there is a change in the amount of whatever I have incoming or outgoing as a result of some additional incoming or outgoing that would not be shown in a simple equation showing gross movement.  P(+-)P1 =Twarm=Tcool is a statement of net change while P=Twarm=Tcool is a statement of gross change.  The math is what it is Ian and as I have said all along...mathematical statements have meaning.  Simply saying that something means net when the equations don't support your statement doesn't make them mean net.



IanC said:


> Pnet = Pwarm - Pcool



Saying net doesn't make it so till you show where you are altering P by some incoming energy...P = Twarm - Tcool is a statement of gross change....you need an addition to or a subtraction of something from P besides Twarm and Tcool in order to have a statement of net change rather than gross change.  Just saying net doesn't make it net because net means something...unless you are saying that in climate science net doesn't mean incoming minus outgoing as it does in every other branch of science.



IanC said:


> This is now where SSDD says mathematics is wrong and the associative rule produces a tainted result.



I am not wrong and you remain unable to prove that I am wrong...you simply say you are right and think that is enough to make you right.



IanC said:


> You simply can't argue with SSDD on this. He has a blind spot that he will protect at all costs.



I am afraid that it is you who has the blind spot ian.....you can't see that there is a difference between P+P1=Twarm - Tcool which is a statement of net change in P.... and P=Twarm-Tcool which is a statement of gross change of P.


----------



## SSDD (Jul 22, 2015)

Crick said:


> Several.


As if you had a clue.

Maybe you would like to take a shot at showing where the incoming energy is shown in the equation which would alter the equation from a statement of gross energy flow to one of net energy flow.  P=A-B is a statement of gross change where I have simply done something with A and B that results in less P......P(=-)P1=A-B is a statement of net change where P has been changed by gaining or losing something aside from the simple subtraction of A-B.  In the SB equation...P is changed by the subtraction of T from Tc...that is a gross change in energy flow....in order to have a net change the equation would have to show P(=-)P1=T-Tc.  There is a difference and one represents gross change while the other represents net change...words mean things.


----------



## Crick (Jul 22, 2015)

I think it has been AMPLY demonstrated here that you are a complete idiot and there's simply no hope for you.


----------



## SSDD (Jul 22, 2015)

Crick said:


> I think it has been AMPLY demonstrated here that you are a complete idiot and there's simply no hope for you.



So you believe that given the equation P=Twarm-Tcool   you get to just decide whether or not it means net or that you can just write it as Pnet=Twarm-Tcool and it magically becomes an equation of net even if you have no expression demonstrating a change of P other than the simple Twarm-Tcool? 

Is that what you are saying crick?  Let me guess...it is and your position is based entirely on faith rather than even basic mathematical knowledge.

Funny...on another thread Frank pointed out that if one asks you a mathermatical question all one can expect from you is an insult....I asked you a mathematical question and rather than you pointing out some error that I have made...what do you do...throw out an insult...You are no engineer....you are no educated anything....my bet is that you flip burgers or push a broom....you certainly don't do anything that would require even the most basic education in math.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > The net energy out from the warmer object is the net energy into the cooler surroundings.
> ...


 
*You don't seem to understand that the term net has a meaning....it is the total of something coming in minus the amount of something going out.* 

Excellent. You got one right.

*The equation describes something going out...but doesn't describe anything coming in.*

The equation describes the net.

*Again...net means something...it isn't just a meaningless term thrown in....net is incoming minus outgoing*

Exactly.

*the SB equation in question only shows outgoing*

The equation shows the net.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> IanC said:
> 
> 
> > SSDD purposely ignores that two processes are going on at the same time. Both objects are radiating, the warmer one more than the cool one. For any period of time measured there is a net excess of radiation from warm to cool. You cannot separate the outcome of the two processes but you can calculate the individual strengths.
> ...


 
Did you ever come up with an explanation why one of your links shows energy moving from our cooler atmosphere to the warmer surface of the Earth?

Or why another of your links said energy goes from the cooler surface of the Sun toward its warmer core?


----------



## IanC (Jul 22, 2015)

SSDD is not looking at the equations for what they are. Generalizations from 100 years ago to get the basic ideas described. For example, the process obviously needs a calculus term as power from both sides changes with respect to time.

The idea that a hundred year old first approximation can be used as proof that radiation from an individual particle is prohibited is obvious nonsense.


----------



## rdean (Jul 22, 2015)

Stephanie said:


> those poor dear climate Scientist needs to RETIRE and go off to the loony farm
> 
> We tried to tell people they were NUTS and so where those who belonged in their CULT


When did the Republicans suddenly start believing the earth is billions of years old?


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> The equation describes the net.




So where is the expression of incoming energy altering P?


*the SB equation in question only shows outgoing*



Toddsterpatriot said:


> The equation shows the net.



Point out the expression which alters P aside from T-Tc which is not an expression of incoming energy changing P...T-Tc is akin to reducing the size of an arpeture in which gas is escaping....it changes the gross flow of gas, but does not alter the original quantity of P which would be required in order for the equation to express a net change.


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> [
> 
> Did you ever come up with an explanation why one of your links shows energy moving from our cooler atmosphere to the warmer surface of the Earth?



They didn't "show" anything...stating that it reaches the surface is no more accurate than claiming that the SB equation expresses net...simply saying a thing doesn't make it so.


----------



## SSDD (Jul 22, 2015)

IanC said:


> SSDD is not looking at the equations for what they are. Generalizations from 100 years ago to get the basic ideas described. For example, the process obviously needs a calculus term as power from both sides changes with respect to time.



Imagine that....looking at an equation for what it is and stating what it says....as I have pointed out repeatedly....it is you guys who are interpreting the equations to be saying something that they aren't....When net energy exchange is observed, measured and proven, I am sure that the second law of thermodynamics and all laws that derive from it will be rewritten to state that they are talking about net energy exchanges....today, however, they aren't...and aren't likely to be altered anytime in the near future.



IanC said:


> The idea that a hundred year old first approximation can be used as proof that radiation from an individual particle is prohibited is obvious nonsense.



You are kidding...right?  The whole AGW farce is based on 100 year old science...

And since there never has been any proof that energy moves from cool to warm, the laws, and accompanying equations say what they have always said....there is no back radiation and it is good to see that at last you can finally admit that the equations say as much...maybe you can explain it to toddster and crick since clearly neither of them have the slightest idea of what the equations say or don't say.  Think you can bring yourself to do that?


----------



## SSDD (Jul 22, 2015)

rdean said:


> Stephanie said:
> 
> 
> > those poor dear climate Scientist needs to RETIRE and go off to the loony farm
> ...



All along....those few people who claim a young earth are mostly southern backwoods cross burning, sheet wearing democrats whose family have been democrats since slave days....the same ones who pitched over teaching evolution.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > The equation describes the net.
> ...


 
*So where is the expression of incoming energy altering P?*

rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T

*P = εAσT^4*

The above equation shows the outgoing radiation of the cooler object.
It also shows the outgoing radiation of the warmer object.
Pretty weird, right?
To get the net energy lost (by the warmer object) and gained (by the cooler object) you have to add the 2 numbers.

The below formula does it in a single step.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > [
> ...


 
*They didn't "show" anything...*

They show that even your own sources disagree with your claim.

*stating that it reaches the surface is no more accurate*

They were your sources. Why don't you post a couple that actually agree with you?

*...simply saying a thing doesn't make it so.*

I know, and we're trying to educate you, but it's not working yet, despite the links you posted.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> IanC said:
> 
> 
> > SSDD is not looking at the equations for what they are. Generalizations from 100 years ago to get the basic ideas described. For example, the process obviously needs a calculus term as power from both sides changes with respect to time.
> ...


 
*there is no back radiation*

Despite your own links that say otherwise. Priceless!

I decided to try to help you with your heat confusion. Took a trip to the library yesterday.
Schaum's Outlines
College Physics 11th Edition.
Chapter 18, page 213

*Heat* (Q) is thermal energy in transit from a system (or aggregate of electrons, ions, and atoms) at one temperature to a system that is in contact with it but is at a lower temperature. 

Chapter 18, page 222

*Radiation* is the mode of transport of radiant electromagnetic energy through vacuum (e.g., the space between atoms). Radiant energy is distinct from heat, though both correspond to energy in transit. Heat is heat; electromagnetic radiation is electromagnetic radiation-don't confuse the two.

All objects whose temperature is above absolute zero radiate energy. When an object at absolute temperature T is in an environment where the temperature is Tₑ, the* net* energy radiated per second by the 
object is P=εAσ(T⁴-Tₑ⁴)

The author is Eugene Hecht, Ph.D.
Says he's from Adelphi University

EUGENE HECHT Faculty Profiles Adelphi University

That's his profile page.
Maybe you could email him and point out his errors?


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> *So where is the expression of incoming energy altering P?*
> 
> rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T
> 
> *P = εAσT^4*



The above equation shows the outgoing radiation of the cooler object.[/quote]

You really don't have a clue...do you.  P=the power of the radiation emitting from the radiator....which equals e which is the emissivity of the radiator * A which is the area of the radiator * sigma which is the SB constant * T^4 which is the temperature of the radiator...  Tc is the expression for the cooler temperature of the surroundings....there is nothing in that expression that says anything whatsoever about the temperature of the cooler surroundings.



Toddsterpatriot said:


> It also shows the outgoing radiation of the warmer object.
> 
> Pretty weird, right?



It only shows the outgoing radiation of the radiator....and yes....its pretty weird that you have pretended to have a clue this long only to find out that you don't have the first clue....anyone with even a rudimentary knowledge of the SB law knows that the above expression says nothing about the temperature of the cooler surroundings.



Toddsterpatriot said:


> To get the net energy lost (by the warmer object) and gained (by the cooler object) you have to add the 2 numbers.



You were better off playing cricks game....snide remarks but never actually touching the math....once you do that, you let everyone know what you know...and it is painfully obvious now that you don't know this topic...  There is no expression of the temperature or the radiation emitted by the cooler surroundings....an expression of temperature would be meaningless.  In order to have an expression of net energy...you would have to have a number that actually alters P before you subtract Tc from T.  The equation is not an expression of net anything...it is describing gross energy flow.



Toddsterpatriot said:


> The below formula does it in a single step.



Sorry, but it doesn't...that expression says that the radiating power of P is equal to the emissivity of the radiating object * the SB constant * the area of the radiator * the temperature to the 4th power of the radiator minus the temperature to the 4th power of the temperature of the surroundings....

That's all it says and all it ever will say....there is nothing there that would suggest net anything....T-Tc is analogous to partially closing the gates on a dam...they change the gross flow of the water, but have nothing to do with a net change in the water behind the dam....in order to calculate a net change in the water behind the dam...you need to know how much water you have....and how much water is coming in from rivers and the amount of water escaping through the dam.

Busted toddster...you don't know jack about the math involved and now everyone who does knows that you don't.


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> Despite your own links that say otherwise. Priceless!




Saying a thing doesn't make it so...the equations don't describe net energy exchanges...even Ian has admitted as much.

I decided to try to help you with your heat confusion. Took a trip to the library yesterday.
Schaum's Outlines
College Physics 11th Edition.
Chapter 18, page 213

*


Toddsterpatriot said:



			Heat
		
Click to expand...

*


Toddsterpatriot said:


> (Q) is thermal energy in transit from a system (or aggregate of electrons, ions, and atoms) at one temperature to a system that is in contact with it but is at a lower temperature.



So are you saying that heat is a form of energy or not?  

Theoretical Physics Second Edition on page 514 says that "heat is not “a form of energy” but a mode of energy transmission}

and then

Environmental Physics, on page 77 says that "Heat is a form of energy that is present in all matter."

So which is it?


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> I know, and we're trying to educate you, but it's not working yet, despite the links you posted.



At this point...it is clear that you aren't educating anyone....unless you consider that demonstrating how much you don't know about basic math is educational...you were busted when you made the claim that *P = εAσT^4* says anything at all about the temperature of the cooler surroundings in the SB law.  You may as well have stood at the podium shouted at the top of your lungs through a microphone that you don't understand this topic.


----------



## Toddsterpatriot (Jul 22, 2015)

*It only shows the outgoing radiation of the radiator....*

No, it shows the outgoing radiation of any object above absolute zero.
Remember......_rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T_

*anyone with even a rudimentary knowledge of the SB law knows that the above expression says nothing about the temperature of the cooler surroundings.*

Of course it says nothing about the surroundings, we're discussing outgoing radiation at the moment. Try to focus. If you want to talk about net energy lost or gained, then you need to know about the surroundings.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Despite your own links that say otherwise. Priceless!
> ...


 
*So are you saying that heat is a form of energy or not?*

*Heat* (Q) is thermal energy in transit from a system (or aggregate of electrons, ions, and atoms) at one temperature to a system that is in contact with it but is at a lower temperature

The author is Eugene Hecht, Ph.D.

I don't find anything in his description to disagree with.
If you understood it, you might sound less stupid.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > I know, and we're trying to educate you, but it's not working yet, despite the links you posted.
> ...


 
*you were busted when you made the claim that P = εAσT^4 says anything at all about the temperature of the cooler surroundings in the SB law.*

Since I never claimed that formula said anything about the temperature of the surroundings whether cooler, hotter or the same, you're arguing with yourself.
Is that the only time you win?

Maybe you should email the professor I mentioned, or your 2 sources that contradicted your claims, and explain to them how they are wrong?
Please be sure to post their responses.
Thanks.


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> *It only shows the outgoing radiation of the radiator....*
> 
> No, it shows the outgoing radiation of any object above absolute zero.
> Remember......_rate of outward radiative energy (per unit area) emitted by an object with temperature T is proportional to the 4th power of T_




You just keep digging deeper and deeper.  The expression you provided was only for A radiator.  If you think there is something in there about a cooler radiator feel free to point it out.



Toddsterpatriot said:


> Of course it says nothing about the surroundings, we're discussing outgoing radiation at the moment. Try to focus. If you want to talk about net energy lost or gained, then you need to know about the surroundings.



You may as well quit pretending that you grasp any of this...you are busted toddster....There is nothing whatsoever in either expression of the SB law that says anything at all about incoming energy from another radiator.  The fact that you think there is is ample proof that you don't have a clue.


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> *So are you saying that heat is a form of energy or not?*
> 
> *Heat* (Q) is thermal energy in transit from a system (or aggregate of electrons, ions, and atoms) at one temperature to a system that is in contact with it but is at a lower temperature
> 
> ...



Back to weaseling I see....I don't blame you....trying to say that the SB equation had an expression concerning incoming energy from another radiator pretty much lost you any credibility you had with regard to the math...clearly it is past your understanding...

Being afraid to say whether you think heat is a form of energy or a means of transferring energy speaks volumes.


----------



## SSDD (Jul 22, 2015)

Toddsterpatriot said:


> *you were busted when you made the claim that P = εAσT^4 says anything at all about the temperature of the cooler surroundings in the SB law.*
> 
> Since I never claimed that formula said anything about the temperature of the surroundings whether cooler, hotter or the same, you're arguing with yourself.
> Is that the only time you win?




Digging deeper and deeper and deeper....the SB equation only speaks of a radiator radiating into its surroundings....it doesn't matter whether the surroundings are full of other radiators or a completely empty vacuum...the equation describes a gross energy flow from a radiator which varies with the difference between the temperature of the radiator and its surroundings.   If you thought that it spoke to anything other than that then chalk it up to just one more thing you don't know about the SB law.




Toddsterpatriot said:


> Maybe you should email the professor I mentioned, or your 2 sources that contradicted your claims, and explain to them how they are wrong?



No need...the equations speak for themselves....if you choose to believe something that the equations don't say feel free.  Appeals to authority are not valid arguments.....if you think the equation speaks to incoming radiation from another radiator step right on up and prove that you and the professor are right and point out where within the SB equation the expression for that incoming radiation is.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *It only shows the outgoing radiation of the radiator....*
> ...


 
*The expression you provided was only for A radiator.*

And as you know, everything above 0K is a radiator.
I'm kidding, it seems you don't know that.

*you are busted toddster*

Says the guy who keeps posting sources that directly contradict him.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *So are you saying that heat is a form of energy or not?*
> ...


 
*trying to say that the SB equation had an expression concerning incoming energy from another radiator pretty much lost you any credibility you had*

Says the guy who thinks objects stop radiating, when a warmer object is near. LOL!


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *you were busted when you made the claim that P = εAσT^4 says anything at all about the temperature of the cooler surroundings in the SB law.*
> ...


* 
No need...the equations speak for themselves....*

And they, and your own sources, prove you're wrong.


----------



## Toddsterpatriot (Jul 22, 2015)

SSDD said:


> IanC said:
> 
> 
> > thermodynamic theory says every object of sufficient size gives off radiation proportional to its temperature. there is no arbitrary choice of whether to radiate or not.
> ...


 

Look at your second source.

Radiation emitted by objects

_All objects that have a temperature greater than 0 K emit radiation_

No exceptions listed.

Do you have a source that says, "All objects that have a temperature greater than 0 K emit radiation, unless a warmer object is nearby"?

How about a source that says, "All objects that have a temperature greater than 0 K emit radiation, unless their surroundings are warmer"?

I wonder why you've failed to produce such a source?


----------



## SSDD (Jul 23, 2015)

Toddsterpatriot said:


> *The expression you provided was only for A radiator.*
> 
> And as you know, everything above 0K is a radiator.
> I'm kidding, it seems you don't know that.



So now you are back to weasel words pretending to know what the hell you are talking about when you, I and anyone who bothered to read your posts knows that you don't....You claimed that the expression you provided gave information about the cooler radiator which was supposed to be radiating towards the warmer radiator...your words...You said:



			
				toddster said:
			
		

> *P = εAσT^4*
> 
> The above equation shows the outgoing radiation of the cooler object.
> It also shows the outgoing radiation of the warmer object.p



Regarding the cooler radiator...it shows no such thing....you don't have a clue.....and at this point, you are not much more than comic relief....pretending to know that you understand the math when you don't even know what the above expression is referencing.[/quote]


----------



## SSDD (Jul 23, 2015)

Toddsterpatriot said:


> *trying to say that the SB equation had an expression concerning incoming energy from another radiator pretty much lost you any credibility you had*
> 
> Says the guy who thinks objects stop radiating, when a warmer object is near. LOL!



All I think is what the equations relating to the SB law say is happening....as you have seen and I have proven, the SB equations are not showing any net energy exchange but are describing gross energy flows...you don't believe what the SB law says and are actually calling it magic and claim that something other than what the equations describe is happening...so which one of us is actually out there.


----------



## SSDD (Jul 23, 2015)

Toddsterpatriot said:


> *
> No need...the equations speak for themselves....*
> 
> And they, and your own sources, prove you're wrong.



Still waiting for you to show where, in any of the expressions of the SB law there is information about incoming radiation from a cooler radiator...till you can provide that information, I remain correct and you or any source remain wrong.  The equations say what they say and no source has the power to override physical laws....you show where the expressions in the SB law show incoming radiation from a cooler source and you win...fail to do that and you just fail.


----------



## Crick (Jul 23, 2015)

Idiot.

Here you go.  

Stefan-Boltzmann Law


----------



## SSDD (Jul 23, 2015)

Toddsterpatriot said:


> Look at your second source.
> 
> Radiation emitted by objects
> 
> _All objects that have a temperature greater than 0 K emit radiation_


_

The equations of the SB law describe a one way gross flow of energy...which is correct...a physical law or a source on the internet._



Toddsterpatriot said:


> Do you have a source that says, "All objects that have a temperature greater than 0 K emit radiation, unless a warmer object is nearby"?



All I have is the SB law which says that the radiating power of P is equal to the emissivity of the radiating object * the SB constant * the area of the radiator * the temperature to the 4th power of the radiator minus the temperature to the 4th power of the temperature of the surroundings....

That statement explicitly states that the Radiating power of P becomes smaller as the difference between the temperature of the radiator and the temperature of its surroundings decrease...Are you saying that the SB law is wrong?



Toddsterpatriot said:


> How about a source that says, "All objects that have a temperature greater than 0 K emit radiation, unless their surroundings are warmer"?



The equations of the SB law itself aren't a good enough source for you?  They say explicitly that the radiating power of P is equal to the emissivity of the radiating object * the SB constant * the area of the radiator * the temperature to the 4th power of the radiator minus the temperature to the 4th power of the temperature of the surroundings....

If you could do math you could keep raising the temperature of Tc (the radiator's surroundings) till the temperature of the surroundings were the same temperature as the object and the radiating power of the object would be zero.  What other source could I provide that is more authoritative than the physical law that states exactly that?



Toddsterpatriot said:


> I wonder why you've failed to produce such a source?



I wonder why you fail to recognize that the physical law itself is the supreme source...any source that doesn't agree with what the physical law says is clearly wrong...now you prove that there is a mathematical statement in the SB law regarding incoming radiation from the background which the radiator is absorbing and you win....fail to do that and again...you just fail.  I may not have an advanced degree in math but I can recognize what simple equations like those associated with the SB law say...and I understand that equations associated with physics are describing things that are happening in the real world....and the equations associated with the SB law are describing gross energy flows as there is no expression describing incoming radiation from any other source altering P...  

I also understand that if I say anything...or make any claim not supported by the physical law and the equations associated with it that I am wrong till such time as the law is rewritten...you are making claims that the equations of the law don't support...you are claiming that the law is incorrect....in short...you are wrong.  The law itself is the ultimate source and it supports my position regardless of what any other source says.


----------



## SSDD (Jul 23, 2015)

Crick said:


> Idiot.



Says the timid little titmouse who is still afraid to engage the actual topic and show where in the SB law there is any expression at all describing incoming radiation from any other source....

We both know you won't because you don't have the education you claim and can't even speak to very simple math like that involved in the expressions associated with the SB law.

You say I'm an idiot but can't even begin to say where I have made any error in my description of the equations describing the SB law...you are, in fact, the idiot and your fear of actually speaking to the math proves it every time you do it.


----------



## Crick (Jul 23, 2015)

Here you go.

Stefan-Boltzmann Law

Idiot


----------



## jc456 (Jul 23, 2015)

well, in one of the links in your link I found this snippet from a sublink:
Heat Transfer from Cold to Warmer Region

abstract:
"Although internal energy will not spontaneously flow from a cold region to a hot region, it can be forced to do so by doing work on the system. Refrigerators and heat pumps are examples of heat engines which cause"

I thought this is what SSDD stated in about 20 plus posts in here.  Just saying here's a link Crickster provided in an argument against SSDD that agrees with SSDD.  So which is it? Crick?

I know you avoid big words, but the one of interest for me is the word "spontaneously"


----------



## Toddsterpatriot (Jul 23, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *The expression you provided was only for A radiator.*
> ...


 
*You claimed that the expression you provided gave information about the cooler radiator*

I claimed no such thing. How can the formula, with no specific variables in it give those variables? You're becoming less rational with every post.
Every object above 0 K radiates.

*P = εAσT^4*

The above formula shows the rate, when radiating in 0 K surroundings.
If you are only interested in outgoing radiation, that's what you want to look at.
If the surroundings are above 0 K, the object cools more slowly, or even warms, because the surroundings, or nearby object, is radiating toward the object.
There is no dimmer switch, magically slowing the outgoing radiation, except in your confused imaginings.

*Regarding the cooler radiator...it shows no such thing*

Since all objects above 0 K radiate, regardless of their surrounding,s you are incorrect.
For proof, look at your link that shows the cooler atmosphere radiating to the warmer surface of the Earth.
For more proof,  look at your link that explained the Sun's surface radiating toward the hotter core of the Sun.
You could even find the old link of mine to the Science article that explained how a human body both radiates energy to the surrounding room,  and receives energy radiating from the room.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> well, in one of the links in your link I found this snippet from a sublink:
> Heat Transfer from Cold to Warmer Region
> 
> abstract:
> ...


 
*I thought this is what SSDD stated in about 20 plus posts in here.*

SSDD is confusing movement of heat with radiation. Among other things.

*Just saying here's a link Crickster provided in an argument against SSDD that agrees with SSDD.* 

Who is Crickster and what is the specific post you feel is somehow refuted by that snippet?


----------



## Toddsterpatriot (Jul 23, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Look at your second source.
> ...


 
*That statement explicitly states that the Radiating power of P becomes smaller as the difference between the temperature of the radiator and the temperature of its surroundings decrease...*

And they say it without claiming that any object above 0 K stops radiating. Ever.

*Are you saying that the SB law is wrong?*

No, just you.

*If you could do math you could keep raising the temperature of Tc (the radiator's surroundings) till the temperature of the surroundings were the same temperature as the object and the radiating power of the object would be zero.*

Radiating power? LOL!
That's net radiated power.

Thermal radiation is energy transfer by the emission of electromagnetic waves which carry energy away from the emitting object. For ordinary temperatures (less than red hot"), the radiation is in the infrared region of the electromagnetic spectrum. The relationship governing the net radiation from hot objects is called the Stefan-Boltzmann law:










While the typical situation envisioned here is the radiation from a hot object to its cooler surroundings, the Stefan-Boltzmann law is not limited to that case. If the surroundings are at a higher temperature (TC > T) then you will obtain a negative answer, implying net radiative transfer to the object.

Stefan-Boltzmann Law


----------



## SSDD (Jul 23, 2015)

Crick said:


> Here you go.
> 
> Stefan-Boltzmann Law
> 
> Idiot



And still the timid little titmouse is afraid to actually engage in the conversation...if you believe that there is anything whatsoever in the SB law that speaks to radiation incoming to the radiator from another source...point it out.


----------



## SSDD (Jul 23, 2015)

jc456 said:


> well, in one of the links in your link I found this snippet from a sublink:
> Heat Transfer from Cold to Warmer Region
> 
> abstract:
> ...



I have been telling them for pages...actually since I got here..spontaneously is the key word...and spontaneously is speaking to the so called back radiation that they claim is coming back to the surface of the earth.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > well, in one of the links in your link I found this snippet from a sublink:
> ...




so all of these pages of mathematical examples and things and you can't figure out who crickster is.  hmmmmm


----------



## jc456 (Jul 23, 2015)

SSDD said:


> jc456 said:
> 
> 
> > well, in one of the links in your link I found this snippet from a sublink:
> ...


And I agree with no back radiation.  No one can prove it.  so it's just magic. and has magic powers.  I just found it interesting that crick's link backed your statement and felt obligated to let him know it.


----------



## SSDD (Jul 23, 2015)

Toddsterpatriot said:


> *You claimed that the expression you provided gave information about the cooler radiator*
> 
> I claimed no such thing.



Now you are just a liar...here again is your quote from post 139



			
				toddster from post 139 said:
			
		

> *P = εAσT^4*
> 
> The above equation shows the outgoing radiation of the cooler object.



So yes, you claimed that the expression above gave information about the outgoing radiation of a cooler object than the radiator P.  Lies aren't helping you out here.



Toddsterpatriot said:


> How can the formula, with no specific variables in it give those variables? You're becoming less rational with every post.



Just quoting you as I have again...you were the one who claimed that the mathematical expression gave information about a cooler radiator...




Toddsterpatriot said:


> The above formula shows the rate, when radiating in 0 K surroundings.
> If you are only interested in outgoing radiation, that's what you want to look at.



Of course it is an expression of gross energy flow.



Toddsterpatriot said:


> If the surroundings are above 0 K, the object cools more slowly, or even warms, because the surroundings, or nearby object, is radiating toward the object.



Again...an expression of gross energy flow....the same sort of equation could be used to calculate gross water flows over a dam....subtract the smaller arpeture of the open gate from the original arpeture and you will see that the gross flow over the dam decreases....if you want any information about net flow leaving the lake behind the dam you need to know how much water is coming in from tributary streams...



Toddsterpatriot said:


> There is no dimmer switch, magically slowing the outgoing radiation, except in your confused imaginings.



You are the only one claiming magic here...I am just pointing out that the SB law is describing gross energy movement...you are the one that claims that the radiator must be doing some magic in order to obey the laws of physics.




Toddsterpatriot said:


> Since all objects above 0 K radiate, regardless of their surrounding,s you are incorrect.




Sorry, but again you are wrong....the expression is for a radiator..not for the radiator and some nearby cooler object...fabricating won't help you out here...but if you think the expression is taking about more than radiator P, then by all means point out where in the expression information about some secondary radiator is to be found....mathematical expressions mean things and if there were information about a secondary radiator.....it would be evident in the expression...so where is it?




Toddsterpatriot said:


> For proof, look at your link that shows the cooler atmosphere radiating to the warmer surface of the Earth.p.quote]
> 
> So again..you are saying that source is more credible than the SB law itself?
> 
> ...


----------



## SSDD (Jul 23, 2015)

Toddsterpatriot said:


> *I thought this is what SSDD stated in about 20 plus posts in here.*
> 
> SSDD is confusing movement of heat with radiation. Among other things.




The only mention of temperature in the equations we are discussing is the temperature of the radiator and the temperature of its surroundings...P is not an expression of temperature....P is an expression of radiating power...  I am not confused....You on the other hand are quickly tying yourself in knots resulting from your weasel speak.

The only confusion here is radiating from you...I know precisely what the SB law says and what each term in the equations represents...and none of them speak to incoming radiation from another source which would be required to formulate an equation of net energy movement.


----------



## Wuwei (Jul 23, 2015)

SSDD said:


> I have been telling them for pages...actually since I got here..spontaneously is the key word...and spontaneously is speaking to the so called back radiation that they claim is coming back to the surface of the earth.


We already covered this a while ago and you must have forgotten: 

The cosmic ray background (CRB) was discovered with a radio telescope. The CRB is a *cold 2.725 deg K*. Radio telescopes are at ambient outdoor temperatures, averaging *15 deg C*. The *very cold* radiation from the CRB must strike the *much warmer* parabola dish in order for it to strike a detector at the focal point.

This illustrates that photons from a cold substance can strike a much warmer substance and be detected to have done so.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
If you mean Crick, just say Crick.
If you think a certain post had the link that helps poor SSDD, reply to that post.


----------



## Toddsterpatriot (Jul 23, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *You claimed that the expression you provided gave information about the cooler radiator*
> ...


 
*So yes, you claimed that the expression above gave information about the outgoing radiation of a cooler object than the radiator P. Lies aren't helping you out here.*

The expression is about the outgoing radiation, not the object itself. Not lying, trying to translate your confused ramblings.

*Just quoting you as I have again...you were the one who claimed that the mathematical expression gave information about a cooler radiator...*

Since every object above 0 K radiates, that formula can be used to calculate the outgoing radiation of any object. Whether it's all alone or near warmer objects. Or cooler objects. Or objects at the same temperature.

*So again..you are saying that source is more credible than the SB law itself?*

The source and the SB law are credible, your misinterpretation is not.

*Again..you are claiming that the source is more credible than the physical law?*

The law, all your sources and all my sources are credible.
Your belief that objects above 0 K stop radiating is not credible.
Why don't you find a link that actually, explicitly states what you claim is true?
Be careful, every source you've posted here has proven your claim was wrong.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> SSDD said:
> 
> 
> > jc456 said:
> ...


 
*And I agree with no back radiation.*

Because you're ignorant.

* No one can prove it.* 

Funny that SSDD's sources said there is.

* I just found it interesting that crick's link backed your statement*

Which link in what post do you mistakenly feel did that?


----------



## Toddsterpatriot (Jul 23, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *I thought this is what SSDD stated in about 20 plus posts in here.*
> ...


 
* I know precisely what the SB law says*

Where does it say an object above 0 K stops radiating....ever?


----------



## IanC (Jul 23, 2015)

Toddsterpatriot said:


> SSDD said:
> 
> 
> > Toddsterpatriot said:
> ...




It doesn't. The obvious case is two objects of the same temp. Do red hot iron bars go dark if they are placed side by side? No. Still radiating kT^4.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


I did and I did!


----------



## jc456 (Jul 23, 2015)

Wuwei said:


> SSDD said:
> 
> 
> > I have been telling them for pages...actually since I got here..spontaneously is the key word...and spontaneously is speaking to the so called back radiation that they claim is coming back to the surface of the earth.
> ...


I'll disagree, the warm is radiating and therefore finds the cold substance.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > SSDD said:
> ...


Sir, I will politely tell you that there is no evidence of back radiation.  You can post on this forum as many times as you like, but the fact is there is no evidence of it. None.

It's Crickster's link.  It was embedded in his link. I thought I included it in my post with the abstract.

Edit: yep I did. post #162.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Wuwei said:
> 
> 
> > SSDD said:
> ...


 
*I'll disagree, the warm is radiating and therefore finds the cold substance.*

How does the "warm radiating" measure the temperature of the Universe?


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
No you didn't.
Post #162 was not a reply to another post.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
*Sir, I will politely tell you that there is no evidence of back radiation.* 

Since all objects above 0K radiate, always, you are mistaken.
Even SSDD's own link showed radiation moving from the cooler atmosphere to the warmer surface of the Earth.
In post #73.
This link.... http://www.public.asu.edu/~hhuang38/mae578_lecture_03.pdf
Page 16 of the pdf.

*You can post on this forum as many times as you like, but the fact is there is no evidence of it. None.*

Except for SSDD's link. And the SB Law, of course.

*It's Crickster's link. It was embedded in his link.*

Since you didn't respond to Crick's post, your claim is difficult to prove.

*Edit: yep I did. post #162*

If you don't respond to a post, it's hard to know what you're talking about.
Try again?


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Wuwei said:
> ...


it goes to radio shack and gets a meter I supposed.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


you didn't ask me that.  You asked for the link and post.  I gave you the post number I put the link information i referenced in. And since reading is a skill and actually using a brain function to understand, my response was to Crickster and I mentioned min and his link, that it wouldn't take too many cells to understand the post.  I guess not.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


*"Except for SSDD's link. And the SB Law, of course."*
oh how funny, and of course you have evidence of it?  How do you know the equation is even accurate if you don't have evidence to support the equation.  I'm sure you have it.  So please post that and I will gladly tell you I was in error.
*
"This link.... http://www.public.asu.edu/~hhuang38/mae578_lecture_03.pdf
Page 16 of the pdf."*

So what is it in this link on page 16 that you wish me to look at?  Proof of back radiation from CO2?  Where? I see Clouds radiating, but I do not see any source stating CO2. and the radiation is not absorbed in the oceans.  So I still don't know if that is showing back radiation adding heat and I still don't know how you prove infrared out of a cloud back to the surface.

*"Since you didn't respond to Crick's post, your claim is difficult to prove."*

Yeah right, that's so difficult. The post right after his mentioning his and using a link from his.  So that tells me A. you didn't look at Crick's link.  so yeah you probably would be lost. but see, I added the link and the abstract and stated it backed SSDD.  So whether or not you agree or not with what I wrote, Crickster posted it, look at post #161, and it was a sublink in the link he provided.  Whew, that sure took some brain work, LOL.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


* 
you didn't ask me that.*

If you want your ramblings to be understood, in context, you'll respond to the actual post that you feel you're responding to, that way, everyone can see what Crick said, in response to whatever SSDD idiocy he was refuting, and then we can see if your rambling nonsense actually helped SSDD's point or not.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
*So what is it in this link on page 16 that you wish me to look at?*

The diagram that shows energy moving from our cooler atmosphere to the warmer surface of the Earth. You know, backradiation. And it was from SSDD's own link. LOL!

* So that tells me A. you didn't look at Crick's link.*

Since you didn't indicate which of Crick's posts you were responding to, how am I supposed to know what you're talking about?
If you meant post #161, that was originally a link I posted.

*I added the link and the abstract and stated it backed SSDD.* 

Since you didn't explain what post or comment of SSDD's you feel it supported, how am I supposed to know what you're talking about?

* Whew, that sure took some brain work,*

You'd better rest then. I know you're not used to working it very hard.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
*  How do you know the equation is even accurate if you don't have evidence to support the equation.*






You want me to prove this equation?


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


hey ramboo, what is it you didn't understand.  It had a link and an abstract paragraph.  I also added the main word from the abstract.  What was difficult for ya? holy crap, now I'm not sure you're smart enough to be spewing all this math in here.  Got that experimental back radiation evidence yet?


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


 I want you to prove back radiation, and you put up the equation.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


see post #162.


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
*hey ramboo, what is it you didn't understand.*

What post were you responding to? What did you refute? What did you support?


----------



## Toddsterpatriot (Jul 23, 2015)

jc456 said:


> Toddsterpatriot said:
> 
> 
> > jc456 said:
> ...


 
Here is the proof of back radiation.


----------



## jc456 (Jul 23, 2015)

Toddsterpatriot said:


> jc456 said:
> 
> 
> > Toddsterpatriot said:
> ...


what I said in the post #162.  Go read it.


----------



## Wuwei (Jul 23, 2015)

jc456 said:


> Wuwei said:
> 
> 
> > SSDD said:
> ...


I don't think you understand what you are typing.  It is the cold background radiation that is striking the much warmer radio telescope.


----------



## SSDD (Jul 24, 2015)

Toddsterpatriot said:


> Since all objects above 0K radiate, always, you are mistaken.



You are always saying that....lets test your claim,.  We will kill two birds with one stone...we will test both your junior high level math skills and see what sort of character you possess simply by looking at your answer which, if you possess good character need only be one word.

Yo claim all objects above 0K radiate all the time.  Here is the equation that describes the radiating power of an object above 0K






   lets tale the discussion out of the hypothetical and make it real....lets plug some numbers into it representing actual temperatures






Solve for P which is the radiating power of our radiator,  What is the radiating power of P.  Hint:  It is a one word answer.


----------



## Toddsterpatriot (Jul 24, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Since all objects above 0K radiate, always, you are mistaken.
> ...


 
Since all objects above 0K radiate, always,

*You are always saying that....lets test your claim,.*

It's not my claim, it's common knowledge.

*




Solve for P which is the radiating power of our radiator,*

You aren't solving for radiating power, because that would be
P = εAσ(255^4)
Your equation is solving for net power radiated.

You need to get over your "dimmer switch" view of radiation.


----------



## SSDD (Jul 24, 2015)

Toddsterpatriot said:


> *Solve for P which is the radiating power of our radiator,*
> 
> You aren't solving for radiating power, because that would be
> P = εAσ(255^4)




So you are no good at math.....you don't have a clue as to what the equations are saying....you don't know what an equation describing net looks like and you your character is wanting....disappointing, but not really surprising.

The above equation describes a radiator radiating into a vacuum at 0k



Toddsterpatriot said:


> Your equation is solving for net power radiated.



Since we have already been through that and know that there is no expression there that references incoming radiation to the radiator from any other source which would be required for an equation describing a net change....we have an equation which describes nothing more than a change in outgoing radiation....much like partially closing the gates on a dam...which is a gross change. 

The equation I provided calculates the radiating power of a radiator radiating into something other than a radiator radiating into a vacuum at 0K as evidenced by the expression describing the surrounding temperature.  I believe you know this but lack the character required to admit it.  How does it feel to be as dishonest as the true warmer wackos?

The equations say what they say.....and there is nothing there that would indicate net change.  I didn't make them up...but I, unlike you do know enough math to know what they are describing.


----------



## IanC (Jul 24, 2015)

Everything radiates at P=kT4, all the time. That is the universal law. Everything is trying to shed energy as fast as it can . a game of hot potato. Entropy.

If another object is nearby, it is also radiating at P=kT4. The net energy exchange is Pnet= Pwarm - Pcool. Both objects are still radiating at P=kT4.

SSDD has a fundamental disconnect because he gives properties of matter to photons. Water in a pipe moves in the direction of highest pressure because matter cannot coexist in the same place at the same time. Photons fly through other photons as if there was nothing there, eg light fibers that sent information both directions at the same time.

One of the fatal flaws in SSDD's version is that variable radiation would decrease the amount of momentum exchanged between the objects. This would decrease entropy. A big no-no in the real world.


----------



## Wuwei (Jul 24, 2015)

SSDD said:


> lets tale the discussion out of the hypothetical and make it real....lets plug some numbers into it representing actual temperatures
> 
> 
> 
> ...


You still don't understand what everyone is telling you. This is what the hyperphysics site - A Good Absorber is a Good Emitter - says about your formula. The site says that T is the temperature of an object. However they give the only valid definition of Tc as:
"_*In this relationship the term with Tc represents the energy absorbed from the environment. This expression explicitly assumes that the same coefficient e applies to both the emission into the environment and the absorption from the environment*_."​
SSDD, Every object simultaneously emits and absorbs radiation The object does *not *just exhibit a reduced emission.


----------



## SSDD (Jul 27, 2015)

IanC said:


> If another object is nearby, it is also radiating at P=kT4. The net energy exchange is Pnet= Pwarm - Pcool. Both objects are still radiating at P=kT4.



So you say...simply tacking net onto an expression, however, does not mean that you are calculating net.  Where is the expression for incoming radiation from that other object that changes P?  I don't see it...where is it?  If we are looking at an equation describing a net energy exchange, there must be an expression for energy coming in as well as going out...where is it?

Either you are saying that the SB law is wrong and you are right...or that there is an expression of incoming radiation in that equation that I am not seeing....which is it?


----------



## SSDD (Jul 27, 2015)

Wuwei said:


> You still don't understand what everyone is telling you. This is what the hyperphysics site - A Good Absorber is a Good Emitter - says about your formula. The site says that T is the temperature of an object. However they give the only valid definition of Tc as:
> "_*In this relationship the term with Tc represents the energy absorbed from the environment. This expression explicitly assumes that the same coefficient e applies to both the emission into the environment and the absorption from the environment*_."​
> SSDD, Every object simultaneously emits and absorbs radiation The object does *not *just exhibit a reduced emission.



I understand precisely what everyone is telling me....I also understand that the equation 
	

	
	
		
		

		
		
	


	




 is a description of a gross energy flow...the power of which is determined by the difference in the temperature of the radiator and the temperature of its surroundings.  Now, either you can state where within that equation one would find the expression describing the amount of radiation coming in from another radiator which would indeed make it an expression describing net energy flow...or you can't, in which case, my claim that it is an expression of gross energy flow is correct.  In that case you and Ian and others are making the claim that the Stefan Boltzman law is incorrect and that the equations associated with it are wrong.  Surely there is a Nobel in it for someone who can prove that a physical law is incorrect.  

So take the above equation apart and tell me where the information describing the incoming energy from another radiator is hiding...because that information is necessary if you are going to have an equation describing net energy flow.


----------



## SSDD (Jul 27, 2015)

By the way....Tc = the temperature of the radiator's surroundings...there is no information there whatsoever about energy being absorbed from the environment...and certainly nothing there regarding any amount being absorbed which would be required for an equation revealing a net energy flow.  Simply saying a thing does not make it true.


----------



## Toddsterpatriot (Jul 27, 2015)

SSDD said:


> By the way....Tc = the temperature of the radiator's surroundings...there is no information there whatsoever about energy being absorbed from the environment...and certainly nothing there regarding any amount being absorbed which would be required for an equation revealing a net energy flow.  Simply saying a thing does not make it true.


 
_Science 24 May 1963: 
Vol. 140 no. 3569 pp. 870-877 
DOI: 10.1126/science.140.3569.870 

In a practical situation and room-temperature setting, humans lose considerable energy due to thermal radiation. However, the energy lost by emitting infrared light is partially regained by absorbing the heat flow due to conduction from surrounding objects, and the remainder resulting from generated heat through metabolism. Human skin has an emissivity of very close to 1.0 . Using the formulas below shows a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts. _


----------



## SSDD (Jul 28, 2015)

Toddsterpatriot said:


> SSDD said:
> 
> 
> > By the way....Tc = the temperature of the radiator's surroundings...there is no information there whatsoever about energy being absorbed from the environment...and certainly nothing there regarding any amount being absorbed which would be required for an equation revealing a net energy flow.  Simply saying a thing does not make it true.
> ...



All you need do is show me where in the SB equations there is an expression that states how much radiation is coming in from another source besides the radiator referenced by P....such an expression would be necessary for the equation to be one describing net energy flow...absent an expression describing incoming radiation the equation only describes gross energy flow.

You can provide as many thought experiments as you like but it is not going to alter the fact that at present, you are claiming that the equations describing the SB law and, in fact, the SB law itself is incorrect.  You claim it is describing net energy movements but the equations associated with it only describe gross energy movement.  Either you or the SB law is wrong...I am going to guess that it is you since we both know that you can't point to any reference within the law to incoming radiation from another source.


----------



## IanC (Jul 28, 2015)

perhaps it is time to just play SSDD's game of semantics. so what *IS *the Stefan-Boltzmann Law? it describes how all objects warmer than zero degrees Kelvin radiate. all the other knock-on equations are just mathematical variations, and not the S-B Law.



Radiation Heat Transfer

The radiation energy per unit time from a *blackbody* is proportional to the fourth power of the absolute temperatureand can be expressed with *Stefan-Boltzmann Law *as

_q = σ T4 A_ _  (1)_

_where_

_q_ _= heat transfer per unit time (W)_

_σ_ _= 5.6703 10-8 (W/m2K4) - *The* *Stefan-Boltzmann Constant*_

_T_ _= absolute temperature Kelvin (K)_

_A_ _= area of the emitting body (m2)

_
SSDD's favourite version, 






is not the S-B Law.

it is a mathematically rearranged equation to give the net radiation transfer

*Net Radiation Loss Rate*
If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

_q = ε σ (Th4 - Tc4) Ac_ _  (3)_

_where_

_Th_ _= hot body absolute temperature (K)_

_Tc_ _= cold surroundings absolute temperature (K)_

_Ac = area of the object  (m2)_



I looked at quite a few references for the S-B Law, and they all described how an object radiated according to its temperature. there was no discussion at all about the temperature of the surroundings, just the object's temp and emissivity.


----------



## Toddsterpatriot (Jul 28, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > SSDD said:
> ...


 
Wuwei already showed you.

"_*In this relationship the term with Tc represents the energy absorbed from the environment."*_


----------



## SSDD (Jul 28, 2015)

IanC said:


> perhaps it is time to just play SSDD's game of semantics. so what *IS *the Stefan-Boltzmann Law? it describes how all objects warmer than zero degrees Kelvin radiate. all the other knock-on equations are just mathematical variations, and not the S-B Law.



The law is described by the associated equations...no more...no less.  If the law meant something else, then the equations would reflect something else.



Radiation Heat Transfer

The radiation energy per unit time from a *blackbody* is proportional to the fourth power of the absolute temperatureand can be expressed with *Stefan-Boltzmann Law *as

_


IanC said:



			q = σ T4 A
		
Click to expand...

_


IanC said:


> _  (1)_


_
_
That equation describes a perfect black body radiating into a vacuum at 0 degrees K.  You always fail to mention that.



IanC said:


> SSDD's favourite version,
> 
> 
> 
> ...



But it is derived from the law and represents a radiator other than a perfect black body radiating into surroundings that are above 0 degrees K...in essence...the real world as opposed to the fantasy of a black body radiating into perfectly empty space.



IanC said:


> it is a mathematically rearranged equation to give the net radiation transfer



No it isn't and you know perfectly well that it isn't.  You are well versed enough in mathematics to know that that equation is not describing a net energy transfer...if it were, there would be an expression representing the amount of incoming energy from an outside source....it isn't there and therefore it is not, and never was intended to give net anything.



IanC said:


> I looked at quite a few references for the S-B Law, and they all described how an object radiated according to its temperature. there was no discussion at all about the temperature of the surroundings, just the object's temp and emissivity.



Then you didn't look very hard...In the above equation P is the radiation emitted from the radiator and  P varies with the difference in the temperature between itself T and its surroundings Tc....like it or not, that is what the equation says and the fact that it doesn't jibe with your claims makes your claims suspect while the equation supports my position 100%


----------



## SSDD (Jul 28, 2015)

Toddsterpatriot said:


> Wuwei already showed you.
> 
> "_*In this relationship the term with Tc represents the energy absorbed from the environment."*_



wuwei is wrong...and his source is wrong.  Describe how the difference between the temperature of a radiator and the temperature of its surroundings describes incoming radiation.  Like it or not, the information is not there...  If the equation were about incoming radiation, then the expression would be for incoming radiation rather than simply the temperature of the surroundings...the equation is describing gross energy movement...not net energy transfers.


----------



## Toddsterpatriot (Jul 28, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > Wuwei already showed you.
> ...


 
You have yet to provide a single source that explicitly states an object above 0K ever stops radiating.
I've provided multiple sources that state objects receive energy radiated from sources cooler than themselves.
In fact, several of your recent sources have disagreed with your claim and agreed with me.

Go find a source that says an object stops radiating near a warmer object or admit your error.


----------



## SSDD (Jul 28, 2015)

Toddsterpatriot said:


> You have yet to provide a single source that explicitly states an object above 0K ever stops radiating.
> I've provided multiple sources that state objects receive energy radiated from sources cooler than themselves.
> In fact, several of your recent sources have disagreed with your claim and agreed with me.



Of course I have...you failed to solve for P in the equation above...solve for P and P =0 whether you like it or not.  If the equations associated with the law itself aren't good enough for you then what is?  The fact that you can't even do such basic math...or that you refuse to accept the answer you get when you do the math is irrelevant....the answer is zero and the equation is describing gross energy flow...since none of you seem to be able to show any real expression of incoming radiation which would be required to formulate an equation showing net energy flow.


----------



## Toddsterpatriot (Jul 28, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > You have yet to provide a single source that explicitly states an object above 0K ever stops radiating.
> ...


 
* Of course I have*

I must have missed it, provide it again.

*you failed to solve for P in the equation above...solve for P and P =0 whether you like it or not.* 

When net equals zero, neither object stops radiating.

*If the equations associated with the law itself aren't good enough for you then what is?*

The equations are fine, your confusion is not.

*the answer is zero and the equation is describing gross energy flow..*

You just won't provide a source that says so explicitly.

*...since none of you seem to be able to show any real expression of incoming radiation*

But of course I did. It's the same equation for both objects, because both objects above 0k radiate continuously. 
You have to add them together to get the net. Addition, apparently, is too difficult for you.

Still waiting for your source that says objects above 0K stop radiating. Ever.
It will take me only moments to find sources that disagree with your claim, why can't you find any that agree?


----------



## SSDD (Jul 28, 2015)

Toddsterpatriot said:


> I must have missed it, provide it again.



In your eyes an equation, derived from the SB law where P=0 is not proof to you?  So you are saying that the SB law is incorrect.  OK.



Toddsterpatriot said:


> When net equals zero, neither object stops radiating.



Since you could provide no evidence of an expression in the equation which gave any actual amount of incoming radiation from another source which would be required if one were going to write an equation showing a net energy exchange...it is clear that the equation is a description of gross energy flow.  Your failing at basic math doesn't change the fact.....the equation is not showing net anything.




Toddsterpatriot said:


> The equations are fine, your confusion is not.



Of course the equations are fine...and my position on them is fine...your insistence that they are describing net energy flow, when you are unable to show any expression that could be construed as providing precise information about incoming radiation from another source clearly demonstrates that it is you who is confused....or perhaps just not grown up enough to admit that you are wrong.  Continuing to claim that the equations are describing net energy flow when it is obvious to anyone who has even a small grasp of math only makes you look like a whining baby demanding that up is down and right is left.  Grow up and describe how one might derive any information whatsoever about the output of an unknown radiator or radiators simply by having the temperature of the surroundings.

Here...the surrounding temperature is 58 degrees F.   What is the amount of radiation that one, or any number of radiators are putting out.  Clearly, you can't have any idea and yet, you believe that you can provide just that from nothing more than a temperature within an equation....that assumption and it is an assumption based on an unsupportable belief that the equation is providing a net energy flow even though it is not written in any form that would provide information about a net flow...it is no better than writing Pnet=Tnet-Tcnet....simply saying net does not make it net.  If you can prove net with the law itself and not your unfounded claims, then by all means do it by telling me how much radiation the unknown radiator or radiators are putting out in order to make the air temperature 58 degrees.  You might start by telling me the area of the radiators...that would be a good start...how will you get that from nothing other than the temperature?




Toddsterpatriot said:


> You just won't provide a source that says so explicitly.



The equation is the ultimate source...the only source that is unimpeachable...sorry that you aren't able to recognize that fact.



Toddsterpatriot said:


> But of course I did. It's the same equation for both objects, because both objects above 0k radiate continuously. You have to add them together to get the net. Addition, apparently, is too difficult for you.
> 
> 
> 
> ...


----------



## Toddsterpatriot (Jul 28, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > I must have missed it, provide it again.
> ...


 
*In your eyes an equation, derived from the SB law where P=0 is not proof to you?*

That is proof. Proof that net energy transfer is zero.
Still waiting for you to prove that means both stopped radiating.

*Since you could provide no evidence of an expression in the equation which gave any actual amount of incoming radiation from another source which would be required if one were going to write an equation showing a net energy exchange*

P = εAσ(255^4)

This is the equation showing the energy radiated by an object at 255K into a vacuum with surroundings at 0K.

P = εAσ(250^4)

This is the equation showing the energy radiated by an object at 250K into a vacuum with surroundings at 0K.

When you put them next to each other,  P = εAσ(255^4 - 250^4) shows the net energy lost by the warmer object and gained by the cooler object.
Because they are both radiating, all the time.
And you have yet to provide a source that agrees with your silly claim that any object over 0K stops radiating, ever.
Why such a long term failure to find a source that agrees with you?
I mean, even your own sources disagreed with you.
That's hilarious!

*Of course the equations are fine...and my position on them is fine...*

You just can't find a single source that agrees with you.

*The equation is the ultimate source*

If your confusion was correct, you'd have thousands of sources describing objects above 0K that suddenly stopped radiating, instead of zero sources.

*any source that disagrees with the fact that P=0 in the example I gave you*

I don't have any sources that disagree that net energy exchange is zero in the example you gave. Do you?

If only you had a source that agrees all radiating ceased in your example.......


----------



## IanC (Jul 29, 2015)

the original data for development of the S-B Law came from heating (and cooling) a crucible with a small aperture. the inside was coated in carbon soot to make it as close as possible to a blackbody, the radiation leaving the aperture was analyzed for amount and wavelength.

trial and error led to the discovery that the radiation was proportional to the 4th power of temperature measured in Kelvin which starts at absolute zero (-273.15 C). a lot of scientists were involved, including Max Planck who introduced a 'granularity' constant to get rid of the ultraviolet catastrophe,  which set the whole quantum theory ball rolling.

the point is.....all objects radiate according to their temperature, at all times. THAT is the S-B law.

all the other variations in the equations for S-B are just that, variations.

SSDD's favourite is 






which is the same as P= kAT^4 - kATcool^4 , no difference just rearranged terms

the visual representation of that is -






everything under the pink curve is a one-to-one exchange between the warm object and the cool one. the area between the blue and pink curves represents the amount of radiation that is flowing from warm to cool. the overlapped area on the curves doesnt disappear, its effects just cancel out, leaving zero net change.


----------



## IanC (Jul 29, 2015)

the other thing to think about is what causes objects to shed radiation.

kinetic movement forces particles to collide, or at least graze each other. kinetic energy is turned into potential energy by distorting the electron clouds and such. when the particles separate the distortions return to normal releasing the potential energy as photons. that's why cool objects have an upper limit to the wavelength they can emit. slower moving particles produce lower energy photons, on average.

SSDD says the radiation stops. but that means the collisions stop. and that makes no sense unless the temperature drops to absolute zero.

every facet of SSDD's explanation leads to logical exclusion. all objects radiate all the time, just like the S-B Law states.


----------



## SSDD (Jul 29, 2015)

Toddsterpatriot said:


> *In your eyes an equation, derived from the SB law where P=0 is not proof to you?*
> 
> That is proof. Proof that net energy transfer is zero.
> Still waiting for you to prove that means both stopped radiating.



I see you couldn't tell me how much radiation an unknown radiator or radiators were emitting when the temperature of the air is 58 degrees.  I guess you can't derive any such information from a mere temperature which is all that the SB equation gives about its surroundings....clearly it is not an equation that gives information about net anything...simply claiming that it is net when the equation clearly is not net just keeps on giving up information about you.

*


Toddsterpatriot said:



			Since you could provide no evidence of an expression in the equation which gave any actual amount of incoming radiation from another source which would be required if one were going to write an equation showing a net energy exchange
		
Click to expand...

*


Toddsterpatriot said:


> P = εAσ(255^4)
> 
> This is the equation showing the energy radiated by an object at 255K into a vacuum with surroundings at 0K.
> 
> ...



No...those both only give you information about how much energy they are radiating into a vacuum at 0 degrees K...neither gives any information at all about any sort of incoming radiation...they are radiating into an empty vacuum so there is no need for information about any other radiator.....if there were two radiators in the vacuum, then the temperature wouldn't be zero....would it?...and another equation would be required.



Toddsterpatriot said:


> they are both radiating, all the time.p.quote]
> 
> If each is a perfect black body alone in a vacuum at zero degrees K....you really don't know what the equations are saying do you?
> 
> ...


----------



## SSDD (Jul 29, 2015)

IanC said:


> the original data for development of the S-B Law came from heating (and cooling) a crucible with a small aperture. the inside was coated in carbon soot to make it as close as possible to a blackbody, the radiation leaving the aperture was analyzed for amount and wavelength.
> 
> trial and error led to the discovery that the radiation was proportional to the 4th power of temperature measured in Kelvin which starts at absolute zero (-273.15 C). a lot of scientists were involved, including Max Planck who introduced a 'granularity' constant to get rid of the ultraviolet catastrophe,  which set the whole quantum theory ball rolling.



So you are acknowledging that the equation in question is deriving information about gross energy flow.  Thanks.



IanC said:


> the point is.....all objects radiate according to their temperature, at all times. THAT is the S-B law.



When they are perfect black bodies radiating along in a vacuum at 0 degrees K...and even then not only according to their temperature but their size as well. Take them out of the vacuum and raise the temperature of their surroundings and then P varies with the temperature of its surroundings...not simply according to its own temperature.

all the other variations in the equations for S-B are just that, variations.



IanC said:


> SSDD's favourite is
> 
> 
> 
> ...



And yet, the amount of radiation emitted from P varies with that temperature...and the equation is still only describing gross energy movement from P.  There is not enough information there to derive net anything.

t


----------



## IanC (Jul 29, 2015)

now you are telling us their experiments were done in a vacuum at zero Kelvin? hahahahaha

the gross flow of radiation from any object is kT^4. it is always there.

net flow is kT^4 minus the amount of radiation the object receives back from the environment.


----------



## SSDD (Jul 29, 2015)

IanC said:


> now you are telling us their experiments were done in a vacuum at zero Kelvin? hahahahaha
> 
> the gross flow of radiation from any object is kT^4. it is always there.
> 
> net flow is kT^4 minus the amount of radiation the object receives back from the environment.



Just stating what the equation represents....if you think that is funny...OK...but it is what the equation represents none the less and you know it.

As to net...prove mathematically that the equation is one that provides information about net radiation flow rather than gross radiation flow...you and I both know that if it were an equation providing information regarding net flows it would be constructed differently but go ahead and prove mathematically that it is calculating net flow....'

Or grow up enough to admit that you know it is an equation that calculates gross flow but that you disagree with it....which is closer to the truth, isn't it.  You would rather lie about what the equations represent than allow that you have been making statements that you can't support...are you prepared to lie indefinitely?....because we both know that there is no mathematical proof that can make an equation for gross energy flow into one expressing net energy flow.   Net is an assumption....an unobserved, unmeasured, untestable assumption....nothing more.


----------



## Toddsterpatriot (Jul 29, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > *In your eyes an equation, derived from the SB law where P=0 is not proof to you?*
> ...


 
*No...those both only give you information about how much energy they are radiating into a vacuum at 0 degrees K*

The equation tells you how much they are radiating. If there is no object radiating back, it also tells you their energy loss.
If there is an object radiating back, you need to add the energy gained to find the net energy loss.

*if there were two radiators in the vacuum, then the temperature wouldn't be zero....would it?...and another equation would be required.*

Or you would have to use the complicated mathematical operation known as addition.

*The law itself agrees with me...what else do I need*

You need a source that says objects above 0K stop radiating. Ever.
Your failure to provide one is obvious.

*Since there is no information about incoming radiation from another radiator*

Plug in the radiator info into the SB formula. That will give you the incoming radiation.

*lets see the proof that the equation is deriving net. Not just saying net but mathematically proving net*

Let's see the proof any object above 0K stops radiating. 
Not just your misinterpretation of the SB.


----------



## Toddsterpatriot (Jul 29, 2015)

IanC said:


> now you are telling us their experiments were done in a vacuum at zero Kelvin? hahahahaha
> 
> the gross flow of radiation from any object is kT^4. it is always there.
> 
> net flow is kT^4 minus the amount of radiation the object receives back from the environment.


 
The only thing he's shown in this thread is that his brain operates in a vacuum at 0K.


----------



## SSDD (Jul 29, 2015)

Toddsterpatriot said:


> The only thing he's shown in this thread is that his brain operates in a vacuum at 0K.



Laugh it up buwheeet....what you haven't done is shown any sort of mathematical proof that the equation in question is calculating net anything.....a point upon which your whole argument rests....fail to do that and you just fail...You can say whatever you care to say....proving it correct on the other hand seems to be problematic for you....

Lets see the mathematical proof that the equation is calculating net.  Laugh it up while you fail...


----------



## Toddsterpatriot (Jul 29, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > The only thing he's shown in this thread is that his brain operates in a vacuum at 0K.
> ...


 
Still no proof objects above 0K stop radiating.
What was your explanation for our measurement of the temperature of the CBR?
I'm sure it's just as funny as the rest of your confusion.


----------



## Toddsterpatriot (Jul 29, 2015)

SSDD said:


> Toddsterpatriot said:
> 
> 
> > The only thing he's shown in this thread is that his brain operates in a vacuum at 0K.
> ...


* 
Lets see the mathematical proof that the equation is calculating net.*

*Human body emission[edit]*






Much of a person's energy is radiated away in the form of infrared light. Some materials are transparent in the infrared, but opaque to visible light, as is the plastic bag in this infrared image (bottom). Other materials are transparent to visible light, but opaque or reflective in the infrared, noticeable by darkness of the man's glasses.
As all matter, the human body radiates some of a person's energy away as infrared light.
The net power radiated is the difference between the power emitted and the power absorbed:





Applying the Stefan–Boltzmann law,




The total surface area of an adult is about 2 m2, and the mid- and far-infrared emissivity of skin and most clothing is near unity, as it is for most nonmetallic surfaces.[34][35] Skin temperature is about 33 °C,[36] but clothing reduces the surface temperature to about 28 °C when the ambient temperature is 20 °C.[37] Hence, the net radiative heat loss is about




Black-body radiation - Wikipedia the free encyclopedia

Net? That's weird. Any luck finding a source that agrees with your claim that matter above 0K stops emitting? Ever? LOL!


----------

