## lördag 17 juli 2010

### Non-Backradiation: Measuring Temperature at Distance

Suppose a body B absorbs light from a source A through a medium and as a response emits light into the medium. In a discussion of backradiation, B can be the top of the atmosphere, A the Earth surface and the medium a transparent atmosphere between B and A.

We ask if B is able to heat A by backradiation, assuming that B has lower temperature than A.

We recall Planck's Law stating that the intensity R_B(f) of the radiation of a light of frequency f from a blackbody B of temperature T_B is given by (with suitable normalization)
• R_B(f) = T_B f^2 C(f,T_B)
where C(f,T_B) is an exponential cut-off factor given by
• C(f,T_B) = f / T_B(exp(f/T_B) - 1) ~ 1 for f/T_B small, and ~ 0 for f/T_B large,
assuming that the cut-off of the incoming light being absorbed by B occurs at higher
frequency, that is, originates from a body of higher temperature.

The cut-off frequency proportional to temperature is referred to as Wien's displacement law, thus tells the temperature of the the emitting body. This is why by looking at the spectrum of the light emitted by (the surface of) the Sun, we can tell that its temperature is about 5778 K.

Emitted light thus carries a signature about its emission temperature from a source A, which a blackbody B records when absorbing and then reemitting the light, as if B can measure the temperature of A.

Now to the question: Can B at lower temperature than A, heat A? Can a cold top of the atmosphere by backradiation heat a warm Earth surface? Can a fool teach a wise something?

We seek an answer in the mathematical model of blackbody radiation studied in Computational Black Body Radiation, which consists of a wave equation for B combined with an equation for the temperature T_B of B as a sum over frequencies:
• dT_B/dt = sum (R_A^2 - R_B^2)
where R_A=R_A(f) is the intensity of incoming light from A and R_B=R_B(f) the intensity of emitted light from B. This equation tells that as long a T_A is bigger than T_B, the body B will be heated by A. The corresponding model for the body A includes the temperature equation
• dT_A/dt = sum (R_B^2 - R_A^2)
In the model, B and A will reach temperature equilibrium with T_B = T_A with the direction of the energy transfer given by the sign of sum (R_A^2 - R_B^2).

We thus see that in the model there is no backradiation: The temperature of B will continue to rise as long as T_B is bigger than T_A as if the body B can compare its own temperature with
that of the distant object A.

It is like two rivals measuring strength by sight at distance.

In a more realistic model the body B will loose heat to a sink S and A will be heated by a source H
• dT_B/dt = sum (R_A^2 - R_B^2) - S
• dT_A/dt = sum (R_B^2 - R_A^2) + H
and equilibrium can be reached with T_B smaller than T_A. But the sign of the energy transfer sum (R_A^2 - R_B^2) keeps its significance and does not allow backradiation from cold to hot.

#### 21 kommentarer:

1. In your example, if A is colder than B the term "sum R_A^2" represents backradiation flowing from a colder body to a warmer one. (The *net* flow will always be from warmer to colder, of course).

2. Am I the only person to have discovered that there is no net heat flow from the surface to the top of the atmosphere averaged over a day-night cycle?

3. I have changed to sum (R_A^2 - R_B^2) from sum R_A^2 - sum R_B^2 and in this more correct setting the backradiation term sum R_A^2 does not appear.

4. Cleas, of course it's still there even if you try to hide it inside those parenthesis. Look at your last two equations with A the surface, B the atmosphere and H energy from the sun being absorbed at the surface. Then R_B is the backradiation from the atmosphere, and it's easy to see that this will increase the surface temperature, i.e. cause a greenhouse effect.

5. Of course the presence of B has an influence on A but not by transferring energy to A in some form of nonphysical backradiation.

6. Claes, it doesn't matter if you think there is radiation from B to A or if B through some unspecified mechanism makes A radiate less heat. The net effect in both cases is that A heats up, i.e., you get a greenhouse effect.

7. Yes, it makes a difference, because the nature of the mechanism comes in.

For example, if A and B have the same temperature, I say that they do not exchange heat energy, while back radiation says that the there is a two-way flow of energy back and forth. My idea is that such two-way flow is unphysical because it is unstable, and to insist that it is physical is confusing.

8. Cleas, where does it make a difference? Can you explain how you can measure the difference between the situations? And, in particular, can you explain why this difference matters when it comes to calculations of the greenhouse effect.

With your belief it seems as if an atmosphere with more CO2 should make the surface radiate less heat, i.e. have to become warmer to match the incoming energy.

Not that your idea on backradiation make much sense, but you should at least try to present a consistent view of what measurable effect it has.

9. It makes a difference e.g. if A and B have the same temp. If you insist that
there is a two-way heat transfer between A and B, then you have to explain how. If you simply say that there is no heat transfer, then there is nothing to explain. Ockhams razor.

10. There is a two way transfer of energy because all bodies emit radiation according to Stefan-Boltzmann law. It is you who have to explain how body A knows about B and is able to adjust how much radiation it emits accordingly.

In any case, this *still* isn't an argument against the greenhouse effect.

11. I just wrote a couple of posts explaining how an absorbing body knows the temperature of the emitting body, by the Planck blackbody spectrum. Read.

12. Ockham's razor is certainly useful. It tells me that a system of physics in which a body must somehow know where it absorbed radiation from, and not to re-emit any in that direction, is unrealistic.

You have never explained how bodies can know not to emit radiation in certain directions.

13. The spectrum of sunlight tells that the emission temp is 5778 K.

14. It is interesting to note that the blackbody temperature of the sun is only around 6000 K. Yet, the kernel of the sun maintains a much higher temperature, up to several millions degrees K if I remember right. How is this possible? Because of the Greenhouse Effect maybe?

15. Claes, those posts explaining how an absorbing body knows the temperature of the emitting body seems to have gone missing.

16. As I said, the spectrum of the incoming light tells at which temp it was emitted.

17. And what tells bodies absorbing sunlight that they can't emit radiation towards the Sun?

18. Claes, but there *is* no incoming light according to you! If B faces a cooler object A, there will be no radiation from A to B according to you, yet B is supposed to know the temperature of A to adjust the amount of energy radiated. You again assume the existence of backradiation while claiming you don't believe in it.

And you *still* refuse to explain what quantitative as opposed to philosophical difference your hypothesis would make to the greenhouse effect on Earth.

Anders, the sun is hotter at the center because that's where the heat is produced and because the outer layers aren't perfect heat conductors. No mystery there. Some basic physics training would solve a lot of problems for both you and Claes.

19. The emission necessarily has lower temp than incoming light and thus the flow of energy is from the Sun to the Earth: no backradiation.

20. It's like talking to a brick wall with you, Claes. You seem unable to grasp concepts that young children have no trouble with. The net flow of energy is from Sun to Earth, yes. This means only that the flow from Sun to Earth is greater than that from Earth to Sun, not that there is no energy being emitted by the Earth.

It is hard to believe that you can't understand this. I think there are two possible explanations. First, you are genuinely stupid. Second, you are simply being deliberately contrary. If the former, then I feel quite sorry for you. If the latter, I only hope you are having fun.

21. Thomas, aren't you exchanging cause and effect regarding the sun. The energy is produced because of the immense temperature, fusion requires a lot of heat doesn't it. Now something prevents the sun from simply exploding doesn't it?