Differential form of Planck's Distribution Law interpretation

Question

So I didn't encounter differentials that often until now, I was taught that the seperate parts of $dy/dx$ for example are not supposed to have any sort of independent existence - ok.

(Calculus, 4th edition, p. 155, Spivak)

Then I've found the following mathematical representation of Plank's distribution law (in Physical Chemistry: A Molecular Approach, Chapter 1, McQuarrie):

$$d\rho(\nu, T) = \rho_\nu(T) d\nu = \frac{8\pi h \nu^3}{c^3} \frac{d\nu}{e^{\frac{h\nu}{k_B T}} - 1}$$

With $d\rho$ and $d\nu$.

So I've read up about differentials in Stewarts Calculus Book which gives following definition:

$$dy = f'(x) \cdot dx$$

If I think about planck's law in those terms, my "$f'(x)$" would equal:

$$\rho_\nu(T) = \frac{8\pi h \nu^3}{c^3} \frac{1}{e^{\frac{h\nu}{k_B T}} - 1}$$

right? So to get a better understanding about what is changing and not only how it changes I thought about what $f(x)$ might be? So integrating $\rho_\nu(T)$ (computer calculated):

$$\rho(T) = \frac{8 \pi^5 k_B^4 T^4}{15 h^3 c^3}$$

Is this a more fundamental formula on the way to planks law? One which Planck used to further deriver his formula, which was there further at the beginning?

I have generally a problem to inteprete all of this in terms of physics (and maybe also mathematically), I would really appreciate if someone could clear things up and tell me if my thoughts are even correct.

I first thought $dx$ and $dy$ must be especially small but they can actually be any real number if we treat them like that (separated):

(Calculus, 8ed, James Stewart, p.190)

And sorry if this belongs more in math stackexchange.

Edit: I would also appreciate some book recommendations about the problem of dy/dx and dy, dx notation and how to interprete it correctly. Maybe then I could interprete the above formulation of the Distribution law better.

Answer 1

Your question is not fully well-formed. Because of that, I cannot know if you actually had an answerable question underneath it, but I can make some assumptions and try to answer it.

So integrating $\rho_\nu(T)$ (computer calculated): $$\rho(T)=\frac{8\pi^5k_B^4T^4}{15h^3c^3}$$ Is this a more fundamental formula on the way to [Planck's] law?

You have somewhat of a mal-formed statement here, but this cannot be more fundamental. The integrated thing is no longer a function of $\nu$ at all, and thus necessarily has less information in it than the original. Planck's law not only gives you the Stefan-Boltzmann law (which is related to the integrated form), but even derives the constant in front of it, but also tells you how the dependence upon $\nu$ goes, an information that is totally missing inside Stefan-Boltzmann law. So, if you were asking about this, you are wrong.

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There is a bit of extremely unfortunate history, and incredible physics inertia, that we are still stuck with bad notation from the pioneers. They did not know what they were writing down, so that we had to contend with silly stuff like $\rho_\nu(T)\,\mathrm d\nu$. They reason they had to consider $\rho_\nu(T)\,\mathrm d\nu$ and not just $\rho_\nu(T)$ alone, is because they discovered that when converting to wavelength $\lambda$, the correct conversion formula is $\rho_\lambda(T)=\rho_\nu(T)\left|\frac{\mathrm d\nu}{\mathrm d\lambda}\right|$ with appropriate substitutions on the RHS to get rid of $\nu$ dependency in favour of $\lambda$

This is absolutely silly. The mathematically correct thing is just that $\rho_\nu(T)=\frac{\mathrm dI}{\mathrm d\nu}$ for some function $I$ that is the integral you expressed as $\rho(T)$. Once you see this, it is obvious that $$\frac{\mathrm dI}{\mathrm d\lambda}=\frac{\mathrm dI}{\mathrm d\nu}\left|\frac{\mathrm d\nu}{\mathrm d\lambda}\right|$$ by the standard Jacobian conversion formula. Physics is just littered with bad notation.

Answer 2

In case of a function of one variable, $dx$ and $dy$ are full differentials, and the derivative is indeed their ratio. THis, allows mathematical manipulations such as, e.g., deriving the derivative for the inverse function: $$ y'=\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{x'},$$ or transforming to new variables in a probability distribution: $$ w_x(x)dx=w_y(y)dy\Rightarrow w_y(y)=w_x(x(y))\frac{dy}{dx} $$ (where the dependence $y(x)$ is assumed to be monotonuous.) This is however not true for the functions of many variables, where differential and partial derivative are not equivalent, so that $$\frac{partial f}{\partial x}$ is not a ratio!

See for more details this answer: Physical interpretation of total derivative

Remark
This is a kind of mistakes/errors that do occur in general textbooks with titles like calculus or physics, aiming at giving basic background. It is much less likely in more specialized texts like differential calculus or something like that.