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I saw a book on my brother's bookshelf and I decided to look through it and to me it looked pretty complicated (it's physics). On the first page I encountered the Rayleigh-Jeans formula:

$$u(v) \space dv = \dfrac{8\pi v^2 kT}{c^3} dv$$

I have a couple of (mathematical) questions regarding this:

  • What kind of formula is this? I understand the fraction, the $u(v)$ part resembles the regular $f(x) = ..$ you encounter in basic math often, but the $dv$ in the beginning and end confuse me. I am familair with integrals, but this seems to be something else. A differential equation?

  • How do you work with these kinds of formulas? At my level of physics/mathematics you just have to plug in data (like $k, T, v, c, \pi$) and you get an answer. What do you have to do here, with the $dv$ part, in order to get the 'answer'?

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physics.SE is the place to ask this ;) – Jorge May 18 '13 at 18:44
@Nivalth If you read the question, you'd know that everything I asked here is about the math, not the physics. It just happens that this equation is actually from physics. – fssalk May 18 '13 at 18:46
There is a derivation here. – oldrinb May 18 '13 at 18:54
$dv$ is the size of the interval of the quantity $v$ in consideration. In physics it represents a small quantity and in math it is called a differential, a topic you see towards the end of calculus I. – Maesumi May 18 '13 at 19:23

The Rayleigh Jeans formula is the classical (non-quantum mechanical in this context) result for the black body radiation.

The quantity $u(\nu)d\nu$ is the spectral radiance $u(\nu)$ per unit frequency $d\nu$. You can see that $u(\nu)=K\nu^2$ for some $K>0$ so

$$ \lim_{\nu\to\infty} u(\nu)=+\infty$$

These infinite radiance result is not experimentally allowed, and it's called the ultraviolet catastrophe. Its solution is due to Max Planck, via Planck's formula, one of the first (if not the first) quantum mechanical results ever.

I still propose the migration to physics.SE because it is a physics question (hence the physics tag) and people there will provide you much more insightul answers than mine.

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How can we see that $u(v) = kv^2$ for some $K>0$? Wouldn't this mean that 8$\pi$ and $c^3$ and $dv$ are equal to 0 (which is impossible?)? – fssalk May 18 '13 at 19:10
@fssalk No, it just means that $$K=\frac{8\pi k T}{c^3}$$ Note the capital letter $K$, which should not to be confused with $k$ Boltzmann's constant – Jorge May 18 '13 at 19:42

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