You’re abusing the math here. You’ve written down the expanded form of the Stefan-Boltzmann equation, which assumes a very specific relationship between temperature and emitted spectrum (which you say yourself). Then you write the temperature in terms of everything else in the equation, and assume a completely different emitted spectrum that invalidates the original equation that you derived T from in the first place.
What you’re doing isn’t math, it’s just meaningless symbolic manipulation, and it has no relationship to the actual physics.
If you insist that this photon flux has a temperature,
Of course it does—this is a very very common and useful concept in physics—and when you say it doesn’t this betrays lack of familiarity with physics. Photon gas most certainly has temperature, in exactly the same way as a gas of anything else has temperature. Not only that, but it also has pressure and entropy.
In fact, in some situations, like an exploding hydrogen bomb, a photon gas has considerable temperature and pressure, far in excess of say the temperature in the center of the sun or the pressure in the center of the Earth.
You say yourself that you assume ‘local equilibrium with it’s incoming irradiance’. Firstly, you’re using the word ‘local’ incorrectly. I assume you mean ‘equilibrium at each wavelength’. If so, this is the very definition of being at the same temperature as the incoming irradiance (assuming everything here is thermal in nature) and, again, there is nothing more to say. http://physics.info/temperature/
You’re abusing the math here. You’ve written down the expanded form of the Stefan-Boltzmann equation, ..
EDIT: I fixed the equations above, replaced with the correct emission function for a grey body.
Yes I see—the oT^4 term on the left needs to be replaced with the black body emission function of wavelength and temperature—which I gather is just Planck’s Law. Still, I don’t (yet) see how that could change the general conclusion.
Where E_{lambda}. is the incoming irradiance distribution as a function of wavelength, and the rest should be self-explanatory. For any incoming irradiance spectrum, the steady state temperature will depend on the grey body emissivity function and in general will differ from that of a black body.
Photon gas most certainly has temperature, in exactly the same way as a gas of anything else has temperature.
I’m aware that the concept of temperature is applied to photons—but given that they do not interact this is something very different than temperature for colliding particles. The temperature and pressure in the examples such as the hydrogen bomb require interactions through intermediaries.
The definition of temperature for photon gas in the very page you linked involves a black body model due to the lack of photon-photon interactions—supporting my point about photon temps such as the CMB being defined in relation to the black body approx.
You say yourself that you assume ‘local equilibrium with it’s incoming irradiance’. Firstly, you’re using the word ‘local’ incorrectly
I meant local in the physical geometric sense—as in we are modelling only a local object and the space around it over small smallish timescale. The meaning of local equilibrium should thus be clear—the situation where net energy emitted equals net energy absorbed. This is the same setup as the examples from wikipedia.
I’m sorry that you’re so insistent on your incorrect viewpoint that you’re not even willing to listen to the obvious facts, which are really very simple facts.
Actually I’ve updated numerous times during this conversation—mostly from reading the relevant physics. I’ve also updated slightly on answers from physicists which reach my same conclusion.
I’ve provided the radiosity equations for a grey body in outer space where the temperature is driven only by the balance between thermal emission and incoming irradiance. There are no feedback effects between the emission and the irradiance, as the latter is fixed—and thus there is no thermodynamic equilibrium in the Kirchoff sense. If you still believe that you are correct, you should be able to show how that math is wrong and what the correct math is.
This grey body radiosity function should be able to model the temp of say an icy earth, and can show how that temp changes as the object is moved away from the sun such that the irradiance shifts from the sun’s BB spectrum to that of the CMB.
We know for a fact that real grey body objects can have local temps lower than the black body temp for the irradiance of an object near earth. The burden of proof is now on you to show how just changing the irradiance spectrum can somehow lead to a situation where all possible grey body materials have the same steady state temperature.
You presumably believe such math exists and that it will show the temp has a floor near 2.7K for any possible material emissivity function, but I don’t see how that could possibly work.
I assume you mean ‘equilibrium at each wavelength’
No. In retrospect I should not have used the word ‘equilibrium’.
As you yourself said, the earth is not in thermodynamic equilibrium with the sun, and this is your explanation as to why shielding works for the earth.
Replace the sun with a distant but focused light source, such as a large ongoing explosion. The situation is the same. The earth is never in equilibrium with the explosion that generated the photons.
The CMB is just the remnant of a long gone explosion. The conditions of thermodynamic equilibrium do not apply.
If you are correct then you should be able to show the math. Provide an equation which predicts the temp of an object in space only as a function of the incoming (spectral) irradiance and that object’s (spectral) emissivity.
You’re abusing the math here. You’ve written down the expanded form of the Stefan-Boltzmann equation, which assumes a very specific relationship between temperature and emitted spectrum (which you say yourself). Then you write the temperature in terms of everything else in the equation, and assume a completely different emitted spectrum that invalidates the original equation that you derived T from in the first place.
What you’re doing isn’t math, it’s just meaningless symbolic manipulation, and it has no relationship to the actual physics.
Of course it does—this is a very very common and useful concept in physics—and when you say it doesn’t this betrays lack of familiarity with physics. Photon gas most certainly has temperature, in exactly the same way as a gas of anything else has temperature. Not only that, but it also has pressure and entropy.
In fact, in some situations, like an exploding hydrogen bomb, a photon gas has considerable temperature and pressure, far in excess of say the temperature in the center of the sun or the pressure in the center of the Earth.
https://en.wikipedia.org/wiki/Photon_gas
You say yourself that you assume ‘local equilibrium with it’s incoming irradiance’. Firstly, you’re using the word ‘local’ incorrectly. I assume you mean ‘equilibrium at each wavelength’. If so, this is the very definition of being at the same temperature as the incoming irradiance (assuming everything here is thermal in nature) and, again, there is nothing more to say. http://physics.info/temperature/
I’m sorry that you’re so insistent on your incorrect viewpoint that you’re not even willing to listen to the obvious facts, which are really very simple facts.
EDIT: I fixed the equations above, replaced with the correct emission function for a grey body.
Yes I see—the oT^4 term on the left needs to be replaced with the black body emission function of wavelength and temperature—which I gather is just Planck’s Law. Still, I don’t (yet) see how that could change the general conclusion.
Here is the corrected equation:
Where E_{lambda}. is the incoming irradiance distribution as a function of wavelength, and the rest should be self-explanatory. For any incoming irradiance spectrum, the steady state temperature will depend on the grey body emissivity function and in general will differ from that of a black body.
I’m aware that the concept of temperature is applied to photons—but given that they do not interact this is something very different than temperature for colliding particles. The temperature and pressure in the examples such as the hydrogen bomb require interactions through intermediaries.
The definition of temperature for photon gas in the very page you linked involves a black body model due to the lack of photon-photon interactions—supporting my point about photon temps such as the CMB being defined in relation to the black body approx.
I meant local in the physical geometric sense—as in we are modelling only a local object and the space around it over small smallish timescale. The meaning of local equilibrium should thus be clear—the situation where net energy emitted equals net energy absorbed. This is the same setup as the examples from wikipedia.
Actually I’ve updated numerous times during this conversation—mostly from reading the relevant physics. I’ve also updated slightly on answers from physicists which reach my same conclusion.
I’ve provided the radiosity equations for a grey body in outer space where the temperature is driven only by the balance between thermal emission and incoming irradiance. There are no feedback effects between the emission and the irradiance, as the latter is fixed—and thus there is no thermodynamic equilibrium in the Kirchoff sense. If you still believe that you are correct, you should be able to show how that math is wrong and what the correct math is.
This grey body radiosity function should be able to model the temp of say an icy earth, and can show how that temp changes as the object is moved away from the sun such that the irradiance shifts from the sun’s BB spectrum to that of the CMB.
We know for a fact that real grey body objects can have local temps lower than the black body temp for the irradiance of an object near earth. The burden of proof is now on you to show how just changing the irradiance spectrum can somehow lead to a situation where all possible grey body materials have the same steady state temperature.
You presumably believe such math exists and that it will show the temp has a floor near 2.7K for any possible material emissivity function, but I don’t see how that could possibly work.
No. In retrospect I should not have used the word ‘equilibrium’.
As you yourself said, the earth is not in thermodynamic equilibrium with the sun, and this is your explanation as to why shielding works for the earth.
Replace the sun with a distant but focused light source, such as a large ongoing explosion. The situation is the same. The earth is never in equilibrium with the explosion that generated the photons.
The CMB is just the remnant of a long gone explosion. The conditions of thermodynamic equilibrium do not apply.
If you are correct then you should be able to show the math. Provide an equation which predicts the temp of an object in space only as a function of the incoming (spectral) irradiance and that object’s (spectral) emissivity.