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I saw this question: Trouble in deriving the Rayleigh-Jeans formula from the steps shown on HyperPhysics site, but I don't consider it answered, so I'd like to dwell a little more into it.

After dealing with the condition defining standing waves for the solution of the wave equation and accepting, as an approximation, that the number of modes can be expressed in terms of volume, one obtains the following expression for that ``volume'' $$ V = \frac{8 \pi}{3} \frac{L^3}{\lambda^3}, \tag 1$$where $L$ is the length of cube's edge, while $\lambda$ is the wavelength. So far, so good. However, when you try to find what that expression referred to unit wavelength is what's obtained is the following $$ \frac{dV}{d \lambda} = \frac{d}{d \lambda} \left( \frac{8 \pi}{3} \frac{L^3}{\lambda^3} \right) = - \frac{8 \pi L^3}{\lambda^4}. $$As seen, an annoying minus sign pops up. I looked around to see if it were noticed at all but it was glossed over by all authors I checked, including Rayleigh in his classical 1900 paper -- in it, he simply ignores the minus sign when presenting the formula containing $\lambda$ in terms of $\nu$. That ignoring of the minus sign is unacceptable. Other authors obviously follow suit and also ignored that minus sign, proceeding with the absolute value of the magnitude. There was one place where the explanation for this minus was that it indicates that the number of modes decreases with the increasing wavelength. That's an unacceptable explanation -- the decrease of the expression's value with the increase of $\lambda$ will take place anyway, $\lambda$ is in the denominator. Let alone that similar reasoning should apply, in a reverse sense, also to $ \nu $ and it does because when $\frac{c}{\nu}$ is replaced for $\lambda$ in eq.(1) no negative sign is needed (and indeed there's no such negative sign to begin with) $$ -\frac{8 \pi L^3}{\left( \frac{c}{\lambda} \right)^4} d \left( \frac{c}{\lambda} \right) = \frac{8 \pi L^3 \nu^2}{c^3} d \nu \tag 2$$to know that the decrease of $\nu$ corresponds to the decrease of number of modes.

Of course, if that minus sign problem shouldn't deter one from accepting such derivation there's an even bigger question as to why was this referral to a unit frequency (unit wavelength) at all needed? Well, apparently, the idea was to apply the equipartition theorem (which in the case of a linear harmonic oscillator, having two degrees of freedom, amounts to $kT$) and in such a way calculate the energy of an oscillator having a given frequency. However, the energy of a given oscillator depends not only on its frequency but also on its amplitude and therefore the energy of an individual oscillator, having frequency $\nu$, is by no means $kT$. Well, it's true that for statistically significant number of particles, say one mole, one derives the statistical fact that the energy amounts to $\frac{1}{2}RT$ per degree of freedom and that may be divided by the number $N$ of the participating particles in one mole, to get what the average energy of an individual particle per degree of freedom is; namely, $\frac{1}{2}kT$. However, again, that is an average value per particle, and, as said, doesn't constitute the energy of an oscillator having a given $\nu$. Will be glad if someone finds these questions interesting and share his or her opinion. Thanks.

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  • $\begingroup$ Don't understand what you're trying to do with Eqn. #1. The link that you provided says that the expression on the right side is the number of modes for a given $\lambda$. It's not a volume, and doesn't even have the dimensional units of a volume. As for the negative sign for the derivative, the website page you linked says why the sign is negative. $\endgroup$ – Samuel Weir Jan 1 '17 at 9:01
  • $\begingroup$ The website I cited says why it's negative but that explanation doesn't stand scrutiny, as I explained. As for the volume, that's part of the assumptions during the derivation. Not shown here. Don't think the whole derivation is necessary because it's not the issue in this question. $\endgroup$ – ganzewoort Jan 1 '17 at 9:27
  • $\begingroup$ I don't understand why this question was downvoted (-1 as I write this). Clearly, there has been a real effort explaining the problem. This is not a unique case and so I'd like to ask the PhysicsSE community to be a little bit more tolerant and helpful. $\endgroup$ – cinico Jan 2 '17 at 8:21
  • $\begingroup$ "...the decrease of the expression's value with the increase of λλ will take place anyway, λλ is in the denominator..." Think about that statement again for a minute. How can a (differentiable) function decrease while its derivative is positive...? $\endgroup$ – Nephente Jan 2 '17 at 8:52
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Suppose I drive down a highway with trees planted at regular intervals, at one mile per minute. I'm interested in knowing the density of the trees, the number per mile.

At the beginning of the highway, there are 100 trees in front of me. After 50 minutes of driving, there are 0 trees in front of me. Therefore the density of trees is $$\frac{0 - 100}{50} = -2 \text{ trees per mile}.$$ The minus sign here is clearly meaningless. It's impossible to have a negative density of trees, just like it's impossible to have a negative number of trees. What it really means is that I was counting the number of trees backwards -- I really should have been counting the number of trees behind me, not the number in front of me. The true density is $2$, not $-2$.

The minus sign you're worrying about is the exact same thing. Here, $N(\lambda)$ is "the number of modes with larger wavelength then $\lambda$". Its derivative is negative, for the exact same reason that "the number of trees ahead of me" decreases as I move forward. But the minus sign is meaningless, because modes, like trees, are a positive quantity.

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