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The Rayleigh-Jeans law does a good job of describing the spectral radiance of a black body at low frequencies: $$B_{\nu}(T)=\frac{2kT\nu^2}{c^2}$$ with $T$ the temperature and $\nu$ the frequency...
#3: Post edited
by
HDE 226868
·
2022-01-21T04:16:22Z (almost 3 years ago)
- The [Rayleigh-Jeans law](https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law) does a good job of describing the spectral radiance of a black body at low frequencies:
- $$B_{\nu}(T)=\frac{2kT\nu^2}{c^2}$$
with $T$ the temperature and $u$ the frequency. There are a couple of ways to derive it. One (presented rather nicely in Section 2.4 of [_Elements of Radio Astronomy_](https://www.cv.nrao.edu/~sransom/web/Ch2.html)) involves analyzing standing wave modes in a cubical cavity of side length $a$, subject to appropriate boundary conditions. You can show easily that- $$B_{\nu}(T)=\frac{c}{4\pi}\frac{N_{\nu}(\nu)\langle E\rangle}{a^3}$$
- with $N_{\nu}(\nu)$ the number of modes per unit frequency at a frequency $\nu$ and $\langle E\rangle$ the average density of a mode. Assuming that the mode energies follow a continuous Boltzmann distribution (the classical approach), you find that $\langle E\rangle=kT$; substituting in the proper expression for $N_{\nu}(\nu)$ derived by studying the modes leads you to the Rayleigh-Jeans law. Of course, if you assumed that the mode energies are quantized (the quantum approach), you would end up with the full Planck function:
- $$B_{\nu}(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$
- The second approach is to start from the Planck function (derived via your method of choice) and take the low-frequency limit, where $h\nu\ll kT$, so
- $$e^{h\nu/kT}\approx1+\frac{h\nu}{kT}$$
- which, when substituted into the Planck function, yields the Rayleigh-Jeans law.
- In short, two different assumptions lead us to the same result: one is that photon energies aren't quantized; the other is that we're working at low frequencies. They seem a bit different on the surface, but I assume they're related. Are they, and if so, how? My guess is that it boils down to the fact that at low frequencies, adjacent modes of energies $E=nh\nu$ and $E=(n+1)h\nu$ are only separated by a small amount of energy (as $h\nu$ is smaller than at high frequencies), so differences between energy levels are small and the spectrum of energies can be approximated as being continuous - but I'm not sure. I might be overthinking this a bit.
- The [Rayleigh-Jeans law](https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law) does a good job of describing the spectral radiance of a black body at low frequencies:
- $$B_{\nu}(T)=\frac{2kT\nu^2}{c^2}$$
- with $T$ the temperature and $
- u$ the frequency. There are a couple of ways to derive it. One, requiring no explicit assumptions about the energy range of the photons (and presented rather nicely in Section 2.4 of [_Elements of Radio Astronomy_](https://www.cv.nrao.edu/~sransom/web/Ch2.html)), involves analyzing standing wave modes in a cubical cavity of side length $a$, subject to appropriate boundary conditions. You can show easily that
- $$B_{\nu}(T)=\frac{c}{4\pi}\frac{N_{\nu}(\nu)\langle E\rangle}{a^3}$$
- with $N_{\nu}(\nu)$ the number of modes per unit frequency at a frequency $\nu$ and $\langle E\rangle$ the average density of a mode. Assuming that the mode energies follow a continuous Boltzmann distribution (the classical approach), you find that $\langle E\rangle=kT$; substituting in the proper expression for $N_{\nu}(\nu)$ derived by studying the modes leads you to the Rayleigh-Jeans law. Of course, if you assumed that the mode energies are quantized (the quantum approach), you would end up with the full Planck function:
- $$B_{\nu}(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$
- The second approach is to start from the Planck function (derived via your method of choice) and take the low-frequency limit, where $h\nu\ll kT$, so
- $$e^{h\nu/kT}\approx1+\frac{h\nu}{kT}$$
- which, when substituted into the Planck function, yields the Rayleigh-Jeans law.
- In short, two different assumptions lead us to the same result: one is that photon energies aren't quantized; the other is that we're working at low frequencies. They seem a bit different on the surface, but I assume they're related. Are they, and if so, how? My guess is that it boils down to the fact that at low frequencies, adjacent modes of energies $E=nh\nu$ and $E=(n+1)h\nu$ are only separated by a small amount of energy (as $h\nu$ is smaller than at high frequencies), so differences between energy levels are small and the spectrum of energies can be approximated as being continuous - but I'm not sure. I might be overthinking this a bit.
#2: Post edited
by
HDE 226868
·
2022-01-21T04:13:56Z (almost 3 years ago)
- The [Rayleigh-Jeans law](https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law) does a good job of describing the spectral radiance of a black body at low frequencies:
- $$B_{\nu}(T)=\frac{2kT\nu^2}{c^2}$$
- with $T$ the temperature and $\nu$ the frequency. There are a couple of ways to derive it. One (presented rather nicely in Section 2.4 of [_Elements of Radio Astronomy_](https://www.cv.nrao.edu/~sransom/web/Ch2.html)) involves analyzing standing wave modes in a cubical cavity of side length $a$, subject to appropriate boundary conditions. You can show easily that
- $$B_{\nu}(T)=\frac{c}{4\pi}\frac{N_{\nu}(\nu)\langle E\rangle}{a^3}$$
- with $N_{\nu}(\nu)$ the number of modes per unit frequency at a frequency $\nu$ and $\langle E\rangle$ the average density of a mode. Assuming that the mode energies follow a continuous Boltzmann distribution (the classical approach), you find that $\langle E\rangle=kT$; substituting in the proper expression for $N_{\nu}(\nu)$ derived by studying the modes leads you to the Rayleigh-Jeans law. Of course, if you assumed that the mode energies are quantized (the quantum approach), you would end up with the full Planck function:
- $$B_{\nu}(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$
- The second approach is to start from the Planck function (derived via your method of choice) and take the low-frequency limit, where $h\nu\ll kT$, so
- $$e^{h\nu/kT}\approx1+\frac{h\nu}{kT}$$
- which, when substituted into the Planck function, yields the Rayleigh-Jeans law.
In short, two different assumptions lead us to the same result: one is that photon energies aren't quantized; the other is that we're working at low frequencies. They seem a bit different on the surface, but I assume they'd related. Are they, and if so, how? My guess is that it boils down to the fact that at low frequencies, adjacent modes of energies $E=nhu$ and $E=(n+1)hu$ are only separated by a small amount of energy (as $hu$ is smaller than at high frequencies), so differences between energy levels are small and the spectrum of energies can be approximated as being continuous - but I'm not sure. I might be overthinking this a bit.
- The [Rayleigh-Jeans law](https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law) does a good job of describing the spectral radiance of a black body at low frequencies:
- $$B_{\nu}(T)=\frac{2kT\nu^2}{c^2}$$
- with $T$ the temperature and $\nu$ the frequency. There are a couple of ways to derive it. One (presented rather nicely in Section 2.4 of [_Elements of Radio Astronomy_](https://www.cv.nrao.edu/~sransom/web/Ch2.html)) involves analyzing standing wave modes in a cubical cavity of side length $a$, subject to appropriate boundary conditions. You can show easily that
- $$B_{\nu}(T)=\frac{c}{4\pi}\frac{N_{\nu}(\nu)\langle E\rangle}{a^3}$$
- with $N_{\nu}(\nu)$ the number of modes per unit frequency at a frequency $\nu$ and $\langle E\rangle$ the average density of a mode. Assuming that the mode energies follow a continuous Boltzmann distribution (the classical approach), you find that $\langle E\rangle=kT$; substituting in the proper expression for $N_{\nu}(\nu)$ derived by studying the modes leads you to the Rayleigh-Jeans law. Of course, if you assumed that the mode energies are quantized (the quantum approach), you would end up with the full Planck function:
- $$B_{\nu}(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$
- The second approach is to start from the Planck function (derived via your method of choice) and take the low-frequency limit, where $h\nu\ll kT$, so
- $$e^{h\nu/kT}\approx1+\frac{h\nu}{kT}$$
- which, when substituted into the Planck function, yields the Rayleigh-Jeans law.
- In short, two different assumptions lead us to the same result: one is that photon energies aren't quantized; the other is that we're working at low frequencies. They seem a bit different on the surface, but I assume they're related. Are they, and if so, how? My guess is that it boils down to the fact that at low frequencies, adjacent modes of energies $E=nh
- u$ and $E=(n+1)h
- u$ are only separated by a small amount of energy (as $h
- u$ is smaller than at high frequencies), so differences between energy levels are small and the spectrum of energies can be approximated as being continuous - but I'm not sure. I might be overthinking this a bit.
#1: Initial revision
by
HDE 226868
·
2022-01-21T04:10:02Z (almost 3 years ago)
How are the assumptions behind two ways of deriving the Rayleigh-Jeans law related?
The [Rayleigh-Jeans law](https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law) does a good job of describing the spectral radiance of a black body at low frequencies:
$$B_{\nu}(T)=\frac{2kT\nu^2}{c^2}$$
with $T$ the temperature and $\nu$ the frequency. There are a couple of ways to derive it. One (presented rather nicely in Section 2.4 of [_Elements of Radio Astronomy_](https://www.cv.nrao.edu/~sransom/web/Ch2.html)) involves analyzing standing wave modes in a cubical cavity of side length $a$, subject to appropriate boundary conditions. You can show easily that
$$B_{\nu}(T)=\frac{c}{4\pi}\frac{N_{\nu}(\nu)\langle E\rangle}{a^3}$$
with $N_{\nu}(\nu)$ the number of modes per unit frequency at a frequency $\nu$ and $\langle E\rangle$ the average density of a mode. Assuming that the mode energies follow a continuous Boltzmann distribution (the classical approach), you find that $\langle E\rangle=kT$; substituting in the proper expression for $N_{\nu}(\nu)$ derived by studying the modes leads you to the Rayleigh-Jeans law. Of course, if you assumed that the mode energies are quantized (the quantum approach), you would end up with the full Planck function:
$$B_{\nu}(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$
The second approach is to start from the Planck function (derived via your method of choice) and take the low-frequency limit, where $h\nu\ll kT$, so
$$e^{h\nu/kT}\approx1+\frac{h\nu}{kT}$$
which, when substituted into the Planck function, yields the Rayleigh-Jeans law.
In short, two different assumptions lead us to the same result: one is that photon energies aren't quantized; the other is that we're working at low frequencies. They seem a bit different on the surface, but I assume they'd related. Are they, and if so, how? My guess is that it boils down to the fact that at low frequencies, adjacent modes of energies $E=nh\nu$ and $E=(n+1)h\nu$ are only separated by a small amount of energy (as $h\nu$ is smaller than at high frequencies), so differences between energy levels are small and the spectrum of energies can be approximated as being continuous - but I'm not sure. I might be overthinking this a bit.