黑体辐射热力学，教材打架时相信谁? 精选

p=u/3                        (1)

第二步：可以证明空腔中的热辐射能量密度仅仅和温度相关，

u=u(T)                       (2)

第三步：空腔中，由于没有其它物质，光压(电磁辐射压)必定就是热辐射压。故，压强仅仅依赖于温度，

p(T)=u(T) /3               (3)

u(T)=a T4                 (4)

(第一步)Maxwell has demonstrated within his theory of electromagnetism that a ray of light or heat radiation exerts a pressure on a surface element which equals the energy of a volume element of Ether supporting that radiation, if the ray is orthogonal to the surface.

(第二步)Let an absolutely empty space enclosed by walls impenetrable for heat radiation.The walls should have absolute temperature t. Let us denote the energy density of the heat radiation [inside the walls] with \psi(t). We have to consider that not all heat rays are perpendicular to each wall. The simplest way is to apply an argument used by Kroenig for gas theory and to consider a cube whose surfaces are parallel to the three mutually perpendicular axes of a coordinateframe. If one assumes that a third of the total radiation propagates parallelto each of the three axes, one obtains a result best suited to the average situation.

(第三步)Then one third of all rays will press on each wall and the pressure per surface element of one of the walls will be according to Maxwell's law

f(t)=(1/3)\psi(t)

http://onlinelibrary.wiley.com/doi/10.1002/andp.18842580616/pdf

“Despite an earlier heuristic classical derivation by Boltzmann, this law is, in fact, incompatible with classical thermodynamics of the electromagnetic spectrum. It is therefore only historical justice that this work earned Planck the sobriquet, “father of the quantum”. ”

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Dear Professor D. Mattis,
……

I paid special attention to your comment on the Boltzmann's derivation from Maxwell’s electromagnetic theory to the pressure- energy density relationship for the blackbody radiation: p=u/3. It is INDEPENDENT of temperature, but Boltzmann substituted it into thermodynamic relation where it requires p=u(T)/3 to reach the Stefan-Boltzmann law.  Your comment is "Despite an earlier heuristic classical derivation by Boltzmann,..." p.93, first edition.

Can I understand your comment in the following? The relation p=u(T)/3 can only be taken as an assumption, and it can be derived from the first principles: statistical physics.

Please let me know whether I correctly understand you.

Thank you.
Sincerely yours,
Q. H. Liu

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Mattis教授的回信铿锵有力：“你完全正确地理解了这一评述”。
Sorry for delay in answering you.
You did understand the remark correctly.
Best,
Daniel Mattis

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Daniel Mattis是统计物理学界的一位“老司机”。http://nanoinstitute.utah.edu/profiles/mattis.php
https://de.wikipedia.org/wiki/Daniel_Mattis

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Q.H.

Since the Stefan-Boltzmann constant depends on hbar, a full understanding of the Stefen-Boltmann "law" cannot come from"classical" physics along.

However, it is commonly argued that the result for the power radiated,

P ~ T^4,

can come from a "classical" argument.

https://en.wikipedia.org/wiki/Stefan-Boltzmann_law

In this argument, if one accepts that the radiation pressure p is related to the energy density u of radiation according to

p = u / 3,

then one arrives at

P ~ T^4

via a "thermodynamic" argument.

So, the key issue is whether or not a"classical" argument leads to

p = u / 3

for "blackbody radiation"

I gave a typical "classical" argument for thison pp. 129-130 of my handwritten E&M notes,

http://physics.princeton.edu/~mcdonald/examples/ph501/ph501lecture11.pdf

Looking over my notes, I see that it is possible to beskeptical about this argument, since it seems to assume that a blackbody absorbs radiation, but does not emit it (which is not the case).

The wiki page cites a different argument, based on the Maxwell stress tensor,

T_ij = (E_iE_j + B_i B_j) / 4 pi - delta_ij (E^2 + B^2) /8 pi,

in Gaussian units.

https://en.wikipedia.org/wiki/Maxwell_stress_tensor

For a surface element on the wall of a cavity that is perpendicular to z, the pressure on it is

P = - T_zz = - (E_z^2 + B_z^2) / 4 pi + u

noting that u = E^2 + B^2 / 8 pi

For isotropic radiation, on average

E_z^2 = E^2 / 3

B_z^2 = B^2 / 3,

so

P = - (E^2 + B^2) / 12 pi + u

= - 2 u / 3 + u

= u / 3.

Thus, it seems that the wiki version of the argument is perhaps more sound than that I gave in my notes.

--Kirk

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http://blog.sciencenet.cn/blog-3377-1089832.html