Tropospheric Hot Spot- Why it does not exist...

[SSDD]

All bodes radiate according to their area, their emissivity. Surely it must irk you to add complications to this scenario that accomplish nothing save to make you and your ideas look as stupid as stupid can be. Matter cannot control, throttle or aim its emissions. You cannot violate special relativity. You are a fool to think otherwise.

I am not irked in the least...I am amused by your inability to read a mathematical formula and have any grasp at all of what it says...it only highlights the fact that your claims of being an engineer of any sort are bullshit...Below is the SB law as it applies to a theoretical perfect black body in the presence of other matter...the expression Tc to the 4th power refers to the temperature of the background...which would be anywhere other than in the perfect vacuum of space...

Now demonstrate those claimed mathematical skills you have and state in your own words what that equation says... You have claimed that all bodies radiate according to their area and emissivity....what does the equation above state that P is dependent upon?

Or don't and demonstrate, as I have stated that you are no more an engineer than my dog and are, in fact, unable to read even the most basic sort of equation.

 

[Wuwei]

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

We covered this many times and gave a dozen references on what science has discovered. What you are inventing here was wrong the first dozen times you brought it up, it is still wrong, and will always be wrong.

Here it is again: There is nothing in the entropy concept of the 2nd law that prevents two objects from radiating energy toward each other, as long as the hotter object always radiates more energy to the colder object than the colder object radiates to the hotter object. In radiation energy exchange, entropy will increase and satisfy the 2nd law.

 

[Wuwei]

Now demonstrate those claimed mathematical skills you have and state in your own words what that equation says... You have claimed that all bodies radiate according to their area and emissivity....what does the equation above state that P is dependent upon?

It really doesn't matter what Crick or I say. It's what the science over the last 100 years says:
The formula is the difference between radiation emitted and radiation received.
Also note this:
There is nothing in the entropy concept of the 2nd law that prevents two objects from radiating energy toward each other, as long as the hotter object always radiates more energy to the colder object than the colder object radiates to the hotter object. In radiation energy exchange, entropy will increase and satisfy the 2nd law.

 

[Toddsterpatriot]

The thermal energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by

etc...............

Finally you are talking language that JC will be able to understand.

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

 

[jc456]

All bodes radiate according to their area, their emissivity. Surely it must irk you to add complications to this scenario that accomplish nothing save to make you and your ideas look as stupid as stupid can be. Matter cannot control, throttle or aim its emissions. You cannot violate special relativity. You are a fool to think otherwise.

dude just post up that experiment and we'll all bow to you.

 

[jc456]

The thermal energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by

etc...............

Finally you are talking language that JC will be able to understand.

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

so what is the answer to that if P=0?

 

[Toddsterpatriot]

The thermal energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by

etc...............

Finally you are talking language that JC will be able to understand.

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

so what is the answer to that if P=0?

The answer would be, net radiation loss is 0.

 

[jc456]

The thermal energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by

etc...............

Finally you are talking language that JC will be able to understand.

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

so what is the answer to that if P=0?

The answer would be, net radiation loss is 0.

yep zero, why?

 

[Toddsterpatriot]

Finally you are talking language that JC will be able to understand.

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

so what is the answer to that if P=0?

The answer would be, net radiation loss is 0.

yep zero, why?

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

 

[jc456]

Guess you are both unaware that that formula applies to a theoretical black body in a vacuum....take it out of the vacuum and put it in the presence of other matter and the formula changes to reflect the fact that in the presence of matter, the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings...Make its own temperature and the temperature of its surroundings the same and P=0

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

so what is the answer to that if P=0?

The answer would be, net radiation loss is 0.

yep zero, why?

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

so no radiation.

how did they know that?

 

[Toddsterpatriot]

the black body radiates according to its area, its emissivity, and the difference between its own temperature and the temperature of its surroundings

That's funny!

Make its own temperature and the temperature of its surroundings the same and P=0


Yup, the net radiation loss rate takes the form

Nothing there about matter above 0K stopping all emissions.

so what is the answer to that if P=0?

The answer would be, net radiation loss is 0.

yep zero, why?

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

so no radiation.

how did they know that?

so no radiation.

Incorrect.

 

[jc456]

so what is the answer to that if P=0?

The answer would be, net radiation loss is 0.

yep zero, why?

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

so no radiation.

how did they know that?

so no radiation.

Incorrect.

so zero isn't zero? ok talk about your smart photons.

 

[Wuwei]

so zero isn't zero? ok talk about your smart photons.

X - X = 0
no matter what x is.

 

[jc456]

so zero isn't zero? ok talk about your smart photons.

X - X = 0
no matter what x is.

yep, you got the picture now.

 

[SSDD]

so zero isn't zero? ok talk about your smart photons.

X - X = 0
no matter what x is.

yep, you got the picture now.

Just goes on and on...even when the equation clearly states that the amount of radiation emitted by the radiator equals the emissivity times the SB constant times the area of the emitter times the difference between the temperature of the emitter and its surroundings....

Guys can't even read a simple algebraic equation and grasp what it says...if there is any thing at all that can alter the value of P, then clearly the radiator doesn't radiate according to its own area and temperature if it is in the presence of other matter....and if all of science denies the SB law as they claim...then why hasn't the law been changed?....or why hasn't the second law of thermodynamics been changed?....

 

[SSDD]

We covered this many times and gave a dozen references on what science has discovered. What you are inventing here was wrong the first dozen times you brought it up, it is still wrong, and will always be wrong.

What does this equation say?

 

[Wuwei]

We covered this many times and gave a dozen references on what science has discovered. What you are inventing here was wrong the first dozen times you brought it up, it is still wrong, and will always be wrong.

What does this equation say?

 

[Toddsterpatriot]

The answer would be, net radiation loss is 0.

yep zero, why?

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

so no radiation.

how did they know that?

so no radiation.

Incorrect.

so zero isn't zero? ok talk about your smart photons.

so zero isn't zero?

Net radiation loss of zero means zero net radiation loss.
You'd have to be stupid to believe it means no energy radiation above 0K.

 

[jc456]

yep zero, why?

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

so no radiation.

how did they know that?

so no radiation.

Incorrect.

so zero isn't zero? ok talk about your smart photons.

so zero isn't zero?

Net radiation loss of zero means zero net radiation loss.
You'd have to be stupid to believe it means no energy radiation above 0K.

or you'd be stupid if you thought zero meant something more than zero.

 

[Toddsterpatriot]

Because the radiation out is matched by the radiation in.
No need for smart photons and no need for matter to use a dimmer switch to raise and lower emissions.

so no radiation.

how did they know that?

so no radiation.

Incorrect.

so zero isn't zero? ok talk about your smart photons.

so zero isn't zero?

Net radiation loss of zero means zero net radiation loss.
You'd have to be stupid to believe it means no energy radiation above 0K.

or you'd be stupid if you thought zero meant something more than zero.

Where did you get the idea that zero net energy loss means anything other than zero net energy loss?