RWturner wrote: “No, Gord is not confused. The filament has a constant power supplied to it, PLUS, the erroneously assumed ABSORBED back radiation. If the filament has the initial power PLUS the back radiation, it has a higher temperature and emits more radiation, if it emits more radiation, then the back radiation increases, so on and so forth. I think the major paradox in back radiation absorption hypothesis is that there would never be a point of equilibrium.”

If you surround a filament with a perfect insulator and you provide a constant input of work (electricity) then the temperature will increase without limit (actually until something melts, vaporizes, or otherwise fails). There will never be a point of equilibrium. There is nothing paradoxical in that. There is no violation of the Second Law. The nature of the insulator does not matter. There is, of course, no such thing as a perfect insulator (or a perfect reflector) but that is immaterial to the hypothetical argument.

Gord seems to claim that the runaway will occur even without an input of energy. That is wrong, as pointed out by Jenn.

RWturner wrote: “I mean, this is SO easy to demonstrate experimentally by taking two very hot objects and moving them closer to each other, if this blog is correct then the two objects should heat up. No assumptions are needed, just a simple experiment with a very sensitive thermal camera. Anyone know of an actual experiment demonstrating this?”

Is there a constant input of energy to the hot objects, with a matching constant loss? In that case they will heat up. Nothing strange there. Pack a whole bunch of people into a poorly ventilated room. It will get quite warm. I have participated in the experiment. đź™‚

RWturner wrote: “Where are you all wrong?”

We aren’t. You are.

]]>RWturner,

You seem to have a problem differentiating between power and energy. Power is Watts, or joules/second. Energy is joules. Energy is conserved, not power.

If an object is radiating a constant 1000W/m², then something must be supplying that power or it will cool because it’s losing energy. If you put that object along with its power supply into a perfectly reflecting cavity, the temperature will indeed increase. Note ‘with its power supply’. It’s the power supply that is causing the temperature to increase, not just reflection and absorption.

If you put an object at a specific temperature into a perfectly reflecting cavity, the temperature won’t go up or down. The energy radiated from the object will be exactly matched by the energy reflected from the walls and absorbed again.

]]>I ran the numbers and found that the above math resulted in the creation of 151,992,878 W after the stars became closer. Though, this could have simply been rounding errors, using scientific figures would have resulted in 0.00 W increase of EM 0.00 K increase in temperature.

]]>No, Gord is not confused. The filament has a constant power supplied to it, PLUS, the erroneously assumed ABSORBED back radiation. If the filament has the initial power PLUS the back radiation, it has a higher temperature and emits more radiation, if it emits more radiation, then the back radiation increases, so on and so forth. I think the major paradox in back radiation absorption hypothesis is that there would never be a point of equilibrium.

I mean, this is SO easy to demonstrate experimentally by taking two very hot objects and moving them closer to each other, if this blog is correct then the two objects should heat up. No assumptions are needed, just a simple experiment with a very sensitive thermal camera. Anyone know of an actual experiment demonstrating this?

Where are you all wrong? With the assumption that the back radiation is simply absorbed by the molecules in the hotter object. Thermal radiation can be absorbed by a molecule ONLY when said absorption results in the molecule vibrating to a higher quantized state, otherwise the molecule is transparent to that radiation. Therefore, your error in the math above is the assumption that the two stars are absorbing all of the incident radiation from each other. Admittedly, QM is difficult to understand, so if there is something missing then please enlighten.

]]>DeWitt,

It’s also my understanding that only a fraction of the lines has been measured, while the most by count have been only calculated. All the strongest lines have, however, been both measured and calculated as far as I have understood.

There are 7,400,447 spectral lines in the present database. Measuring most of those would be obvious waste of effort.

The HITRAN2012 database contains about 300.000 lines for CO2 (128,170 for the main isotopes). About 140,000 measured line positions and 44,000 measured intensities of various isotopologues were used. Thus only 15% of the intensities are based on measurements, while the share is close to 50% for the positions.

]]>Pekka,

Using that approach or something like it, it’s possible to calculate the frequencies and Einstein coefficients for the vibrational and rotational transitions of simple molecules like CO2 *ab initio*. In fact, I believe that’s been done for the majority of lines in the HITRAN database.

DeWitt and Frank,

The derivation of your link is based on accepting some formulas and using consistency requirements to figure out, what the relationships between the coefficients must be. That’s the original approach of Einstein in 1917, when Quantum Mechanics was not available for understanding from the basics, what determines the constants and gives then values that satisfy those relationships.

The situation is not fully satisfactory without the QM derivation of the coefficients. This is not the place to go deep enough to QM, but some idea of that can be obtained from a href=”http://electron6.phys.utk.edu/qm2/modules/m10/einstein.htm”>this presentation.

There may be better explanations, but this the first one that I found.

Basically the result is based on the time reversal symmetry of QM and on the properties of electromagnetic radiation and its interaction with matter.

The formulas contain transition amplitudes that are unchanged, when the initial and final states are interchanged. The bosonic nature of photos plays also an important role in more advanced derivations.

]]>Frank,

On the relationship of the three coefficients, see here. As Pekka stated, the coefficients are not independent. If you know one, you know all three.

Therefore, for any two energy levels of a particular ionization state of a particular element,

only one of the three Einstein coefficients is independent. Once its value is known, the values of all three coefficients are determined. The values of the two unspecified coefficients are dependent on the value of the specified coefficient, the statistical weights of the two energy levels, and the energy difference between the two levels.

The simple form of Schwarzschild’s equation assumes thermal *and* radiative equilibrium. Radiative equilibrium in the troposphere is only a good assumption for the most strongly absorbing frequencies because then the temperature difference between emitting and absorbing molecules is small. All this is taken into account in a line-by-line radiative transfer program like LBLRTM.

Frank,

The coefficients of induced and spontaneous emission and absorption are closely related. Thus determining one of them or a combination of them determines all.

DeWitt: I did learn more from re-reading section 2.3.2. To see if I understand the situation correctly and if you may have oversimplified, let’s consider several cases:

The difference between rotational energy levels is small enough that they are almost equally populated at atmospheric temperature. Absorption of a microwave photon will change from one rotational state to another. Since these states are almost equally populated (and assuming equal degeneracy), the rate of induced emission and absorption should be similar (See equation 14). If collisional relaxation didn’t exist, we could deduce that the rate of spontaneous emission was small compared with induced emission.

The difference between vibrational energy levels is much larger, so they are not equally populated at atmospheric temperatures. Therefore, the rate of induced emission is much less than the rate of absorption. If emission and absorption were in equilibrium, most emission would be due to spontaneous emission (not induced emission). However, in the atmosphere, absorption and emission are NOT in equilibrium at many wavelengths – that is why we observe TOA radiation intensities appropriate for blackbody radiation with temperatures from 200 to 300 degK. At these wavelengths, we rely upon collisions (not radiation) to ensure a Boltzmann distribution of vibrational energy levels (LTE). If we go above the stratosphere, collisions aren’t frequent enough to maintain LTE.

The difference between electronic energy levels is even larger. Collisions rarely produce electronically excited states in the earth’s atmosphere. So the atmosphere is not in LTE with the visible and UV radiation passing through it.

(There is a gradual change from one regime to another that doesn’t depend on the nature of the excited state. By chance, the temperature and density of the earth’s atmosphere allows me to ascribe these differences in behavior to rotational, vibrational and electronic excited states. On a very cold planet, collisions won’t be energetic enough to create a Boltzmann distribution of vibrationally excited states.)

You said: “At LTE, absorption must equal emission whether there is stimulated emission or not.” As best I can tell, the troposphere is in LTE because of collisions, but emission and absorption are not in equilibrium. Furthermore, it is the relaxation rate, not simply the collision rate, that determines whether LTE exists.

What I still don’t fully understand is how we can use the absorption cross-section we measure in a laboratory in the emission term – n*o*B(lamba,T) – of the Schwarzschild eqn. When infrared emission is mostly spontaneous, it depends on A_21; whereas absorption in a laboratory spectrophotometer depends mostly on B_12. The reference I provided is pretty indirect on this subject, partially due to line width. And one might hope for a more direct path from the absorption coefficient to the lifetime of an excited state/emission rate.

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