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derivation of wiens displacement law and wiens distribution law from thermodynamic principle.but not from plancks law
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Wien's law
- Thread starter koustav
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- #2
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Your post is ambigous. Are you asking for something, or do you want to present something to us?
- #3
atyy
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As Wikipedia says, a derivation of Wien's law without assuming quantum mechanics is found in Wannier's book. I don't know this important derivation off the top of my head, so I'd have to look it up.
- #4
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The most simple derivation I know is using dimensional analysis.
Let's assume we can work with classical physics alone and ask for the mathematical form of the black-body spectrum. It can be obtained as the em. radiation in a cavity in thermal equilibrium at a given temperature determined by heating the walls and keeping it at a certain temperature ##T## in a thermostat. Now it is clear that this problem has only the intensive parameter ##T## (temperature of the walls), because the radiation is isotropic and homogeneous (otherwise it couldn't be in thermal equilibrium). The energy density per unit volume and per unit frequency ##\omega## of the radiation is what we look for. There cannot be a chemical potential for radiation, because it doesn't carry any intrinsic ("charge like") conserved quantity.
Further some fundamental constants can be involved: The speed of light (as the fundamental constant inherent in electromagnetism; here I use CGS units, because this is the most simple way for theoretical classical electromagnetism), the Boltzmann constant ##k_B##. Thus in our "pool" of variables to express ##u## we have:
##T##: Dimension K (Kelvin)
##k_B##: Dimension erg/K=g cm^2/(s^2 K)
##\omega##: Dimension s (seconds)
##c##: Dimension cm/s
##u##: Dimension ##\text{erg} \text{s}/\text{cm}^3=\text{g}/(\text{s} \text{cm})## (energy per volume and per ##\mathrm{d} \omega##)
Since there is no dimension K anywhere except in the temperature, only the quantity ##k_B T## can enter the law. To build a dimensionless quantitiy (which must be a constant), we have to build this quantity, I'll call ##C##, from ##k_B T##, ##V##, ##\omega##, ##c##, and ##u## (and ##u## must be involved in order to get a relation for this quantity). So we can put the relation into the form
##C=u (k_B T)^{\alpha} \omega^{\gamma} c^{\delta}.##
Since g only appears in ##u## and ##T## we must have ##\alpha^{-1}## and the rest of the dimensional analysis gives uniquely ##\delta=3, \quad \gamma=-2.##
Thus we have
$$u=C \frac{k_B T \omega^2}{c^3}.$$
This gives the obviously wrong result because of the well-known Rayleigh-Jeans UV catastrophy.
The answer is that we have to consider quantum theory, and this adds the additional universal quantity ##\hbar## of dimension erg s to the game. Thus we have
$$C=u (k_B T)^{\beta} \omega^{\gamma} c^{\delta} \hbar^{\epsilon}.$$
leading to
$$[C]=\text{erg}^{\beta+\epsilon} \text{s}^{1-\gamma+\epsilon} \text{cm}^{\delta-3}.$$
This leads to
$$\delta=3, \quad \beta+\epsilon=-1, \quad \gamma-\epsilon=1.$$
This is an underdetermined set of equations. So we need to find another dimensionless constant that doesn't depend on ##u## to determine the general law
$$C'=(k_B T)^{\beta'} \omega^{\gamma'} c^{\delta'} \hbar^{\epsilon'}.$$
Then
$$u=\frac{k_B T \omega^2}{c^3} f(C'),$$
where ##f## is some arbitrary function. In order that this solves the UV problem, ##C'## must depend on ##\omega##, so that we can assume ##\gamma'=1##. The dimensional analysis then leads to
$$\epsilon=-\beta=\gamma=1.$$
Thus we have
$$u=\frac{k_B T \omega^2}{c^3} f \left (\frac{\hbar \omega}{k_B T} \right ).$$
The displacement law can now derived as follows. For the frequency, where one has maximal energy density per frequency we have
$$\partial_{\omega} u(\omega)=\frac{2 \omega k_B T}{c^3} f(\xi) + \frac{\hbar \omega^2}{c^3} f'(\xi), \quad \xi=\frac{\hbar \omega}{k_B T}.$$
Setting this to 0 leads to a necessary condition for a maximum. Thus the maximum frequency is determined by a solution of the equation
$$2f(\xi)+\xi f'(\xi)=0.$$
For any solution ##\xi_0## of this equation, one finds
$$\xi_0=\frac{\hbar \omega_m}{k_B T}.$$
So we have ##\omega_m = \xi_0 k_B T/\hbar##, which is Wien's displacement law (in the frequency domain).
The constant ##\xi_0## can only be determined when one has the unknown function ##f##. This one only gets with additional assumptions. Planck first guessed it in a clever way using the high-precision data by Rubens und Kurlbaum on the spectrum of the black-body radiation for both high and low frequencies. In an effort to find a theoretical explanation of this law, which fitted the data very accurately, Planck came to the discovery of quantum theory, because he needed the assumption that electromagnetic radiation of frequency ##\omega## is absorbed and emitted by the walls of the cavity only in portions ##n \hbar \omega##, where ##n \in \mathbb{N}##. This introduced the Planck constant as a new fundamental constant into physics, and this we used above in our dimensional analysis to solve the UV problem.
Nowadays, with the full knowledge about QED we can derive Planck's law very easily just using quantum statistics (with the em. field necessarily to be quantized as bosons due to the spin-statistics theorem).
Let's assume we can work with classical physics alone and ask for the mathematical form of the black-body spectrum. It can be obtained as the em. radiation in a cavity in thermal equilibrium at a given temperature determined by heating the walls and keeping it at a certain temperature ##T## in a thermostat. Now it is clear that this problem has only the intensive parameter ##T## (temperature of the walls), because the radiation is isotropic and homogeneous (otherwise it couldn't be in thermal equilibrium). The energy density per unit volume and per unit frequency ##\omega## of the radiation is what we look for. There cannot be a chemical potential for radiation, because it doesn't carry any intrinsic ("charge like") conserved quantity.
Further some fundamental constants can be involved: The speed of light (as the fundamental constant inherent in electromagnetism; here I use CGS units, because this is the most simple way for theoretical classical electromagnetism), the Boltzmann constant ##k_B##. Thus in our "pool" of variables to express ##u## we have:
##T##: Dimension K (Kelvin)
##k_B##: Dimension erg/K=g cm^2/(s^2 K)
##\omega##: Dimension s (seconds)
##c##: Dimension cm/s
##u##: Dimension ##\text{erg} \text{s}/\text{cm}^3=\text{g}/(\text{s} \text{cm})## (energy per volume and per ##\mathrm{d} \omega##)
Since there is no dimension K anywhere except in the temperature, only the quantity ##k_B T## can enter the law. To build a dimensionless quantitiy (which must be a constant), we have to build this quantity, I'll call ##C##, from ##k_B T##, ##V##, ##\omega##, ##c##, and ##u## (and ##u## must be involved in order to get a relation for this quantity). So we can put the relation into the form
##C=u (k_B T)^{\alpha} \omega^{\gamma} c^{\delta}.##
Since g only appears in ##u## and ##T## we must have ##\alpha^{-1}## and the rest of the dimensional analysis gives uniquely ##\delta=3, \quad \gamma=-2.##
Thus we have
$$u=C \frac{k_B T \omega^2}{c^3}.$$
This gives the obviously wrong result because of the well-known Rayleigh-Jeans UV catastrophy.
The answer is that we have to consider quantum theory, and this adds the additional universal quantity ##\hbar## of dimension erg s to the game. Thus we have
$$C=u (k_B T)^{\beta} \omega^{\gamma} c^{\delta} \hbar^{\epsilon}.$$
leading to
$$[C]=\text{erg}^{\beta+\epsilon} \text{s}^{1-\gamma+\epsilon} \text{cm}^{\delta-3}.$$
This leads to
$$\delta=3, \quad \beta+\epsilon=-1, \quad \gamma-\epsilon=1.$$
This is an underdetermined set of equations. So we need to find another dimensionless constant that doesn't depend on ##u## to determine the general law
$$C'=(k_B T)^{\beta'} \omega^{\gamma'} c^{\delta'} \hbar^{\epsilon'}.$$
Then
$$u=\frac{k_B T \omega^2}{c^3} f(C'),$$
where ##f## is some arbitrary function. In order that this solves the UV problem, ##C'## must depend on ##\omega##, so that we can assume ##\gamma'=1##. The dimensional analysis then leads to
$$\epsilon=-\beta=\gamma=1.$$
Thus we have
$$u=\frac{k_B T \omega^2}{c^3} f \left (\frac{\hbar \omega}{k_B T} \right ).$$
The displacement law can now derived as follows. For the frequency, where one has maximal energy density per frequency we have
$$\partial_{\omega} u(\omega)=\frac{2 \omega k_B T}{c^3} f(\xi) + \frac{\hbar \omega^2}{c^3} f'(\xi), \quad \xi=\frac{\hbar \omega}{k_B T}.$$
Setting this to 0 leads to a necessary condition for a maximum. Thus the maximum frequency is determined by a solution of the equation
$$2f(\xi)+\xi f'(\xi)=0.$$
For any solution ##\xi_0## of this equation, one finds
$$\xi_0=\frac{\hbar \omega_m}{k_B T}.$$
So we have ##\omega_m = \xi_0 k_B T/\hbar##, which is Wien's displacement law (in the frequency domain).
The constant ##\xi_0## can only be determined when one has the unknown function ##f##. This one only gets with additional assumptions. Planck first guessed it in a clever way using the high-precision data by Rubens und Kurlbaum on the spectrum of the black-body radiation for both high and low frequencies. In an effort to find a theoretical explanation of this law, which fitted the data very accurately, Planck came to the discovery of quantum theory, because he needed the assumption that electromagnetic radiation of frequency ##\omega## is absorbed and emitted by the walls of the cavity only in portions ##n \hbar \omega##, where ##n \in \mathbb{N}##. This introduced the Planck constant as a new fundamental constant into physics, and this we used above in our dimensional analysis to solve the UV problem.
Nowadays, with the full knowledge about QED we can derive Planck's law very easily just using quantum statistics (with the em. field necessarily to be quantized as bosons due to the spin-statistics theorem).
Last edited:
- #5
atyy
Science Advisor
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This gives the obviously wrong result because of the well-known Rayleigh-Jeans UV catastrophy.
The answer is that we have to consider quantum theory, and this adds the additional universal quantity ##\hbar## of dimension erg s to the game.
What you write is correct. However, quantum theory isn't necessary for Wien's law. Wien's law is a result of classical thermodynamics. IIRC, it's one of the amazing things that all (or most) of the results of classical thermodynamics applied to black bodies are correct. Only classical statistical mechanics is wrong, and has to be replaced by quantum statistical mechanics. It's rare to find the derivation from classical thermodynamics nowadays, but the "ancients" knew it, and I assume that Planck knew it. Wannier's book has this marvellous derivation, which I've read, but cannot reproduce off the top of my head.
- #6
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I've no access to this book, but from this dimensional analysis (I forgot where I read this one, and it's so easy that one can reproduce it as you see ;-)) the clue is that there must be another fundamental constant of a dimension in the game to cure the UV catastrophy. Nowadays we know, it's Planck's constant ##\hbar##, which leads to the correct functional form of the law but not to the Bose-Einstein distribution. So there must be something assumed in addition.
As I said, nowdays it's clear what it is, namely that the absorption and emission of "photons" with frequency ##\omega## to and from the wall is only possible in integer multiples of ##\hbar \omega##. Doing naive kinetic theory for this, you'd conclude from the "collision term" that in equilibrium you have a Maxwell-Boltzmann distribution. To see this take as the simplest process the photo effect, i.e.,
$$\text{Atom+gamma} \leftrightarrow \text{excited Atom}$$
The naive equilibrium condition thus reads
$$f_{A^*}=f_{\gamma} f_A.$$
From the theorem on summational invariants, taking the log of this expression immediately yields
$$f_j(E_j)=\exp[-E_j/(k_B T)].$$
So the energy-spectral function is the one of the general Wien form with
$$f=\frac{2\hbar \omega}{2 \pi^2 k_B T} \exp \left (-\frac{\hbar \omega}{k_B T} \right ).$$
The factor 2 in the numerator comes from the two polarization states of each photon then the number density must be multiplied by the energy ##\hbar \omega## to get the energy density and finally there is a factor ##\mathrm{d}^3 k/(2 \pi)^3=\mathrm{d} \omega \omega ^2/(2 \pi^3 c^3)## from the conversion from wave numbers to frequency (the integration over the solid angle is trivial leading to a factor ##4 \pi##, included in this formula). So we get
$$u(\omega)=\frac{\hbar \omega^3}{2 \pi^2 c^3} \exp \left (-\frac{\hbar \omega}{k_B T} \right ).$$
Of course, this solves the UV problem, but Rubens and Kurlbaum found that this law is wrong at low frequencies (long wave lengths).
This puzzle is solved by Einstein's finding (1917) of spontaneous emission (or Bose enhancement), which is the weirdest of all the quantum effects included in Planck's Law. Nowadays we know that it comes from the demand that many-body states must be either symmetric (bosons) or antisymmetric (fermions) under exchange of particles. It can also be derived from the kinetic-theory argument but now one has to use the BUU collision term. The atoms can be either bosons (upper sign) or fermions (lower sign) depending on whether their total spin is integer or half-integer. The photons have integer spin, so they are bosons. Thus the equilibrium condition now reads
$$f_{A^*}(1+f_{\gamma})(1 \pm f_A) - f_{\gamma} f_A (1 \pm f_{A^*})=0.$$
Now from the theorem on summational invariants it follows that
$$\ln \left (\frac{1+f_{\gamma}}{f_{\gamma}} \right )=\frac{\hbar \omega}{k_B T} \; \Rightarrow \; f_{\gamma}=\frac{1}{\exp[\hbar \omega/(k_B T)]-1}.$$
The rest is the same as above. So we finally get the correct energy spectrum
$$u=\frac{\hbar \omega^2}{\pi^2 c^3} \frac{1}{\exp[\hbar \omega/(k_B T)]-1}.$$
So there seems to be no way to get the correct Planck spectrum without quantum theory.
Wien's displacement law, of course, holds for both Wien's and Planck's spectrum.
As I said, nowdays it's clear what it is, namely that the absorption and emission of "photons" with frequency ##\omega## to and from the wall is only possible in integer multiples of ##\hbar \omega##. Doing naive kinetic theory for this, you'd conclude from the "collision term" that in equilibrium you have a Maxwell-Boltzmann distribution. To see this take as the simplest process the photo effect, i.e.,
$$\text{Atom+gamma} \leftrightarrow \text{excited Atom}$$
The naive equilibrium condition thus reads
$$f_{A^*}=f_{\gamma} f_A.$$
From the theorem on summational invariants, taking the log of this expression immediately yields
$$f_j(E_j)=\exp[-E_j/(k_B T)].$$
So the energy-spectral function is the one of the general Wien form with
$$f=\frac{2\hbar \omega}{2 \pi^2 k_B T} \exp \left (-\frac{\hbar \omega}{k_B T} \right ).$$
The factor 2 in the numerator comes from the two polarization states of each photon then the number density must be multiplied by the energy ##\hbar \omega## to get the energy density and finally there is a factor ##\mathrm{d}^3 k/(2 \pi)^3=\mathrm{d} \omega \omega ^2/(2 \pi^3 c^3)## from the conversion from wave numbers to frequency (the integration over the solid angle is trivial leading to a factor ##4 \pi##, included in this formula). So we get
$$u(\omega)=\frac{\hbar \omega^3}{2 \pi^2 c^3} \exp \left (-\frac{\hbar \omega}{k_B T} \right ).$$
Of course, this solves the UV problem, but Rubens and Kurlbaum found that this law is wrong at low frequencies (long wave lengths).
This puzzle is solved by Einstein's finding (1917) of spontaneous emission (or Bose enhancement), which is the weirdest of all the quantum effects included in Planck's Law. Nowadays we know that it comes from the demand that many-body states must be either symmetric (bosons) or antisymmetric (fermions) under exchange of particles. It can also be derived from the kinetic-theory argument but now one has to use the BUU collision term. The atoms can be either bosons (upper sign) or fermions (lower sign) depending on whether their total spin is integer or half-integer. The photons have integer spin, so they are bosons. Thus the equilibrium condition now reads
$$f_{A^*}(1+f_{\gamma})(1 \pm f_A) - f_{\gamma} f_A (1 \pm f_{A^*})=0.$$
Now from the theorem on summational invariants it follows that
$$\ln \left (\frac{1+f_{\gamma}}{f_{\gamma}} \right )=\frac{\hbar \omega}{k_B T} \; \Rightarrow \; f_{\gamma}=\frac{1}{\exp[\hbar \omega/(k_B T)]-1}.$$
The rest is the same as above. So we finally get the correct energy spectrum
$$u=\frac{\hbar \omega^2}{\pi^2 c^3} \frac{1}{\exp[\hbar \omega/(k_B T)]-1}.$$
So there seems to be no way to get the correct Planck spectrum without quantum theory.
Wien's displacement law, of course, holds for both Wien's and Planck's spectrum.
- #7
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And BUU comes from?
Thanks,
Thanks,
- #8
atyy
Science Advisor
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So there seems to be no way to get the correct Planck spectrum without quantum theory.
Wien's displacement law, of course, holds for both Wien's and Planck's spectrum.
Yes, I'm referring to Wien's displacement law that can be derived from classical thermodynamics. The correct spectrum requires quantum mechanics, and was first derived by Planck.
- #9
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Boltzmann-Uehling-Uhlenbeck equation. Uehling and Uhlenbeck introduced the Bose-enhancement and Pauli-blocking factors for bosons and fermions in the collision terms.
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