# Why don't the derivatives of the blackbody spectrum over frequency and wavelength match?

The question is: The radiation emitted by a black body can be represented either by the energy distribution over the wavelength or by that over the frequency. In the first case the maximum energy corresponds to $\lambda(m)$ , while in the second case there is a maximum corresponding to frequency $\nu(m)$. Is $\nu(m)$= $c$/$\lambda(m)$?

Now I have basically solved the problem by showing that $dE(\nu)/d(\nu)$= $0$ does not correspond to $dE(\lambda)/d(\lambda)$=$0$.

My question is: how do I account for this result physically?

• Related: physics.stackexchange.com/q/13611/2451 , physics.stackexchange.com/q/91192/2451 and links therein. – Qmechanic Mar 19 '14 at 11:48
• Also there's a good answer here. – Ruslan Mar 19 '14 at 17:43
• This link gives a good explanation. To quote "Estimated spectral shifts are caused by nonlinear frequency or wavelength 'gauge' relations to the experimentally accessible parameter (an intensity within an interval of such parameter)." – noir1993 Mar 23 '14 at 7:47

Spectral power density expressed as a function of wavelength $E(\lambda)$ must satisfy

$$P=\int_0^\infty E(\lambda)\; d\lambda,$$

where $P$ is total power. On the other hand, if we express it as a function of frequency $E'(\nu)$, we have:

$$P=\int_0^\infty E'(\nu)\; d\nu.$$

And the thing is that $E(\lambda)\ne E'(\nu)$ for $\lambda=\frac{c}{\nu}$. Let's try to see this in terms of differential of power $dP$:

$$dP(\lambda)=E(\lambda)\; d\lambda=E\left(\frac{c}{\nu}\right)\; \frac{d\left(\frac{c}{\nu}\right)}{d\nu}d\nu=-E\left(\frac{c}{\nu}\right)\frac{c}{\nu^2}\;d\nu.$$

Thus we can see that

$$E'(\nu)=-E\left(\frac{c}{\nu}\right)\frac{c}{\nu^2}=-E(\lambda)\lambda^2/c.$$

This is because $E(\lambda)$ is not a usual function, it's power density function. It will be different for different scales of its argument. It's related to the fact that $$\lambda_1-\lambda_2\ne \frac{c}{\nu_2-\nu_1}.$$

• Can you please give a physical interpretation for your result? at first when i saw this question, i thought that the derivatives of the wavelength and the frequency should match, if I think of a single blackbody. But the math showed otherwise. But I can't account for this result physically. – Pallavi Roy Mar 19 '14 at 13:06

Ruslan already gave a rigorous derivation of the inequality, so here is an intuitive physical explanation:

What does $E(\lambda)$ mean? For concreteness, let's use units of nanometers. $E(\lambda)$ means the radiated intensity per nanometer measured. For example, if we have a glass filter that removes all light except that between 600nm and 601nm, then the power density surviving through the filter will be $\approx E(600.5\text{nm}).$

What does $E(\nu)$ mean? For concreteness, let's use units of gigahertz. $E(\nu)$ means the radiated intensity per gigahertz measured. For example, if we have a glass filter that removes all light except that between 500,000GHz (which is the frequency of 600nm light) and 500,001GHz, then the power density surviving through the filter will be $\approx E(500,000.5\text{GHz}).$

Do you expect these numbers to be the same? No. Why? Because the filter width in the two cases is different. You can check for yourself that the interval between 600nm and 601nm is actually 831GHz wide. So $E(600.5\text{nm})$ should be roughly 831 times as large as $E(500,000.5\text{GHz})$.

So they're really describing physically different ideas, which is why the formulas on the Wikipedia page for Planck's law are different for wavelength and frequency.

But that's not all. Since the wavelength interval width converted to GHz also depends on what wavelength you're at, the multiplication factor varies depending on what part of the spectrum you're looking at. So there is no reason to expect that the maximum location of $E(\lambda)$ should be the same as the maximum of $E(\nu)$.