# Why does Planck's Law have different peaks?

1. Nov 12, 2013

### lugita15

Suppose we have a blackbody at temperature T. Then if we write Planck's law for wavelength, and find the wavelength corresponding to the peak, we get a certain value lambda_max. If, on the other hand, you wrote Planck's law for frequency, and we found the frequency corresponding to peak of that, we get a certain value nu_max.

But if you multiply lambda_max and nu_max, you do not get the speed of light. In other words, the peak of Planck's law for wavelength does not correspond to the peak of Planck's law for frequency. How is this possible?

Any help would be greatly appreciated.

2. Nov 12, 2013

### Rajini

Hello,
Have a look at the following article:
Title: A better presentation of Planck’s radiation law
Author: Jonathan M. Marr and Francis P. Wilkin.
Cheers, Rajini.

3. Nov 13, 2013

### phyzguy

It comes about from the inverse relation between wavelength and frequency. Try thinking about the following situation, which might be easier to visualize. Suppose you are running a race which is 100 m long, and you run it many times with a distribution of velocities which peaks at 10 m/s. Try choosing some distribution of velocities (say a Gaussian) then calculate the time it takes to run the race, and plot the distribution of times. You will find that the time it takes to run the race does not peak at 10 s, as you might think, but peaks at a slightly longer time. This is because the slower races take a longer time, and shift the distribution to longer times. The inverse relation between velocity and time causes the same shift as the inverse relation between wavelength and frequency.

4. Nov 13, 2013

### kith

There's also a paper which discusses the spectral optimization of the human eye wrt to this and argues that it isn't adapted to the solar spectrum as well as one might naively think.

Some paradoxes, errors, and resolutions concerning the spectral optimization of human vision
B. H. Soffer, D.K. Lynch,
Am. J. Phys. 67 (11), November 1999

5. Nov 13, 2013

### D H

Staff Emeritus
Not a Gaussian! Given some random variable X with a Gaussian distribution X~N(μ,σ2), a random variable Y=1/X will have a rather ill-behaved, Cauchy-like distribution. The problem is that because X is normally distributed, values of X can get very close to zero with non-zero probability.

Pick a better distribution, one for which the pdf is zero for all x less than or equal to 0.

Suppose you do have a random variable X whose reciprocal results in a well-behaved distribution. The mean of 1/X will not be 1/E[X]. It will depend on E[X] and on all higher moments of X about the mean:
$$E\left[\frac 1 X\right] = \frac 1 {\bar X} \left(1 + \sum_{k=2}^{\infty} (-1)^k \frac {E\left[(X-\bar X)^k\right]} {\bar X ^k}\right)$$