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国内《热力学》教材对有些问题的处理,给人的印象是热力学是1900年及以前经典物理学的一部分。热力学就是热力学,能把所有宏观热现象一网打尽。而理解这些热现象,没有量子力学(量子统计),常常难以获得到位、深刻的结论。
一、国内热力学教材处理黑体辐射只有一个声音
这个声音是属于巨人王竹溪先生。
把王先生的分析进行分解,可发现由四步组成。
第一步:电磁学可以给出光压公式,
即,空腔中电磁辐射的压强p等于电磁能量密度u的三分之一。
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)
注意,在这个公式中,压强没有独立的定义。
第四步:认为黑体辐射是一个PVT系统。利用热力学基本方程(考虑到空腔中的电磁波需要在器壁中不断吸收发射),得一个微分方程。把 u=u(T) 和 p(T)=u(T) /3 代入热力学基本方程,得到一个一阶线性常微分方程,参考下面的截图中王竹溪先生的证明。得能量密度正比于温度的四次方,
u(T)=a T4 (4)
其中a为积分常数。
王先生(《热力学》(1955)p.112)在这个地方有两句话(红笔和篮笔)值得注意。
王先生之后,上面用红笔和篮笔标记的两句话,林宗涵、汪志诚先生省略了前一句,有些作者(例如苏汝铿《统计物理学》(第二版)P.157)把这两句话都省掉了。直接把(1)当成(2)代入热力学基本方程,是一件匪夷所思的事!
二、王先生的声音是玻尔兹曼1884年声音的回声
把玻尔兹曼1884年推导Stefan-Boltzmann定律的德文原文找出来,请德国物理学教授帮忙,把关键的部分翻译成英文。可以找到四句话,分别和前面的一~四步对应。原文的开头几句,正好和前面的一~三步对应。
(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)
这就是今天的热辐射压公式。玻尔兹曼1884年的文章可以自由下载,链接是:
http://onlinelibrary.wiley.com/doi/10.1002/andp.18842580616/pdf
三、可是,玻尔兹曼的推导仅仅具有“启发性”!
所谓“启发性的观点”或“启发性的证明”,即一种实用主义的解法,即非最优也非完美。但是,这一方法能够用来加速寻找满意解的过程。请看Daniel C. Mattis 《statistical mechanics made simple》(1st ed. p.93)。在利用统计物理推导出光压公式之后,Mattis写道:
“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”. ”
这句话的含义是,电磁谱的经典热力学结果具有紫外灾难。如果非要从麦克斯韦的电磁波理论出发,压强就是一个无穷大!在1884年年代,即使玻尔兹曼意识到了无限大困难,也会认为在经典电磁理论范围内可以解决!伟大如普朗克,在他在1900年获得了正确的结果、成为“量子之父”之后,一直到死,他念念不忘的还是经典电磁理论!
问题还可以更为深入!
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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
----
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
四、玻尔兹曼—普朗克—王竹溪的作法出现在1900年以前才算合理
玻尔兹曼的处理被普朗克细致化,写入了《The Theory of Heat Radiation》(1914)。王竹溪的《热力学》也引用了普朗克的著作。可惜,尽管普朗克是量子力学之父,却不是彻底的革命家。况且,《The Theory of Heat Radiation》(1914)之后10年,也就是1924年,才出现玻色-爱因斯坦统计!
如何理解黑体辐射的热力学? Mattis说得很明白: stefan-boltzmann定律“is, in fact, incompatible with classical thermodynamics of the electromagnetic spectrum.” 在玻尔兹曼—普朗克—王竹溪一派的做法中,第三步是“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。”这一步逻辑上不严谨!除了不能排除能量密度为无穷,还不能排除能量为常数甚至为零的情况!此时,压强必须单独定义!
按照今天的理解,从经典的电磁理论出发,即使可以得到辐射压强p(T)=u(T)/3,也是一个无限大!理解黑体辐射的热力学,非量子统计不可! 当然,在热力学层面,完全可以不追求深入的理解,仅仅把p(T)=u(T)/3理解为一个实验结果、或者干脆假设如此!
总而言之,玻尔兹曼—普朗克—王竹溪的作法出现在1900年以前,马马虎虎!出现在今天,历史的灰尘太厚了!
请看另一位大牛Princeton大学Kirk T. Mcdonald的看法!和Mattis角度不同,但是结论一样!
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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
——————
另外,当你看到一位大学教授的备课本,全都是用英文大写字母、工工整整书写时,能不能献上你的膝盖?
看看人家的讲义:http://physics.princeton.edu/~mcdonald/examples/ph501/ph501lecture11.pdf
反正我是五体投地了!
五、王竹溪先生留下锦囊否?
再读王先生的两句话,顺序很重要!第一句是p(T)=u(T)/3,第二句是这个结果被实验验证了!
事实上,Lebedew实验上验证的是,不含温度的光压公式p=u/3,和热平衡无关,和黑体辐射无关。这一点,普朗克的书《The Theory of Heat Radiation》(1914)说的很清楚,
那么,为什么王竹溪先生说Lebedew验证的不是p=u/3,而是p(T)=u(T)/3? 唯有这样,王先生接下来行文才是可以接受的! 难道说,王先生预见到了科学的发展,一定会证明p(T)=u(T)/3?
今天,由于辐射压公式p(T)=u(T)/3获得了充分的检验,可以把当成一个实验定律来用了! 就像利用理想气体的Boyle定律去理解理想气体一样!
也就是,今天,我们不必再为p(T)=u(T)/3的经典电磁波理论推导而烦恼,当成实验定律直接取过来就是。 这样,stefan-boltzmann定律就能顺理成章了。
玻尔兹曼、普朗克、王竹溪诸位巨人基于经典电磁波理论推导出的p(T)=u(T)/3过程,只能从历史的角度去欣赏其“启发性”。
六、文献相互打架时相信谁?
读几本新书,读几篇当代文献,和国际上的一些还在拉风的专家交换想法,就能知道文献之间、名家之间的区别,就能有所收获!
虽不至,亦不远也!
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