5 votes

What is the meaning of a thermal equilibrium between matter and radiation?

There are many questions here. I will answer the one in the title. We imagine a system consisting of a chunk of matter surrounded with a bath of photons. The atoms forming the skin of the chunk of ...
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4 votes

Why Doesn't the Stefan-Bolzmann Law Deter Controlled Fusion Engineers From Seeking Extreme Temperatures To Maximize the $nτT$ Triple Product?

The Stefan-Bolzmann law only applies to opaque objects. The plasma in a tokamak is almost perfectly transparent. If the Stefan-Bolzmann law applied to the plasma in ITER, it would have the luminosity ...
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  • 5,508
3 votes
Accepted

Can plasmas be black bodies?

Plasma in many concrete cases often is not a black body, e.g. plasma in Earth's ionosphere, or in a discharge lamp, or in tokamak. This is because plasma in these cases is very thin (rarified gas), ...
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3 votes

What is the meaning of a thermal equilibrium between matter and radiation?

A "Stationary radiation field" means that the intensity and spectrum of the radiation is not changing with time. For the matter and radiation to also be in thermal equilibrium, requires that ...
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  • 113k
2 votes

Can plasmas be black bodies?

The argument is silly if the claim is that plasmas cannot appear anything like blackbodies, since there are observable examples like the Sun. To be a blackbody, a volume of plasma needs to come into ...
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  • 113k
2 votes

Why Doesn't the Stefan-Bolzmann Law Deter Controlled Fusion Engineers From Seeking Extreme Temperatures To Maximize the $nτT$ Triple Product?

The energy losses don't scale as $T^4$ - that would be appropriate for a blackbody, but the fusion plasma is optically thin, does not absorb incident radiation and therefore cannot be a blackbody. At ...
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1 vote

What would a piece of high emissivity polished metal inside a blackbody cavity look like? Darker than the walls of the cavity but the same color?

Though I've never seen Planck's law written with the emissivity (neither ϵ(λ) or just a constant ϵ) in front, it seems to me that it needs to be there Yes, multiply by emissivity. and the ...
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  • 669
1 vote

What is the definition of the Rayleigh-Jeans tail?

The Rayleigh-Jeans tail (of a blackbody distribution) is simply where you can assume $h\nu \ll k_BT$, where $\nu$ is the frequency. In this regime, you can simplify the Planck function, by allowing $\...
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  • 113k
1 vote

What is the definition of the Rayleigh-Jeans tail?

As far as I know, the two ends of the black-body radiation curve is historically described using the Rayleigh-Jeans law and the Wien law, as seen on Wikipedia for example The Rayleigh-Jeans law is ...
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  • 106
1 vote

Why Doesn't the Stefan-Bolzmann Law Deter Controlled Fusion Engineers From Seeking Extreme Temperatures To Maximize the $nτT$ Triple Product?

Yet we see many if not most of the major efforts to achieve controlled nuclear fusion as relatively uninterested in increasing density, inertial confinement being the exception I'm not sure you could ...
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1 vote
Accepted

Doubt regarding rate of loss of heat due to radiation

The absorptive power is dependant on $a$. However, from Kirchoff's Law we know that a good absorber is a good emitter i.e. $a=e$ which simplifies your equation into the one given. Also see Why do dark ...
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  • 188

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