# A paradox in Planck's Law?

1. Jul 2, 2008

### cmos

Hi all,

Physical law:
I understand the derivation of the Planck law for the blackbody spectrum and why it takes slightly different forms whether you are doing the analysis in the frequency domain or the wavelength domain. That is to say, you cannot simply invoke the Planck relation ($$E=h\nu=hc/\lambda$$) if you want to convert the final result between frequency and wavelength domains.

My problem:
What is the physical implication of this? Maybe another way of saying this is, do we detect wavelength or frequency?

An example:
Consider a blackbody at 6000 K. From the Planck law (derived in the frequency domain), the spectral radiance of emitted light will peak at 353 THz. From the Plack law (derived in the wavelength domain), the spectral radiance of emitted light will peak at 483 nm. Clearly the Planck relation does not prevail.

As a check, my results correspond with those obtained from the displacement law of Wien.

The two numbers above correspond to 1.46 eV and 2.57 eV, respectively. Suppose I had a friendly and very much alive cat put into some ungodly contraption, a box. There are two photon counters, A and B. A will accept only photons of 1.46 eV energy and B will only accept photons of 2.57 eV (maybe put in a plus or minus several meV for all us Heisenberg buffs).

The contraption is designed so that once activated, once detector A receives 1000 photons, it will crack a vial of cyanide thus ending the life of our friendly companion. However, if detector B should receive 1000 photons first, then detector A will be deactivated and our friendly companion will happily survive and join us for some future thought experiment.

So, does Planck's cat live or die?

2. Jul 2, 2008

### George Jones

Staff Emeritus
Last edited by a moderator: Apr 23, 2017
3. Jul 2, 2008

### cmos

Dr. Jones,

Your comments are always welcomed; however, I think you misunderstood my dilemma. I understand the mathematics to derive either form of the Planck law. One such way is simply by the link you posted; you show how you transform between the two domains.

The problem I presented is in the physical interpretation. Please look at the numbers in my first post, they speak for themselves. In my "paradox," I outline how the peaks obtained from the two domains lead to two vastly different results. Should our kitten live or die?

I feel that after 107 years, that this paradox must have been resolved; I have just been unable to ascertain the means by which to do so.

4. Jul 2, 2008

### malawi_glenn

I have never noticed that, are you sure you have done correct?

5. Jul 2, 2008

### Zizy

Using ln, this problem doesnt occur and peak also matches eye maximum sensitivity. Loose translation of a statement of my professor - this could be used as a proof that the god invented logarithm before humans.

EDIT: Also, I do not think math thing behind has any physical significance.

6. Jul 2, 2008

### gel

If I understand you correctly, if the photon counters only accept photons of exactly the frequencies you state, then the experiment will never end because they will never detect any photons.

If they accept a range of photons, you just multiply the Planck density by this width, being consistent between using the correct units (energy/wavelength/frequency) for the density and the width of the range. Then there's no problem, and it is all consistent because

$$I(\nu,T)\,d\nu=-I(\lambda,T)\,d\lambda.$$

See http://en.wikipedia.org/wiki/Planck%27s_law" [Broken].
A probability distribution, and its peak, is entirely dependent on the variable you express it in terms of. So, no paradox.

edit: If you use logs as Zizy suggests then it just converts the non-linear relation between the frequency and wavelenth to a linear one, so the issue desn't arrive. However you still have to multiply the density by the range of log(frequency) or log(wavelength) that the sensor detects to get the probability. That's just what probability density means.

Ah, I see you said maybe put in plus or minus a few meV. That's the whole issue, you have to (nothing to do with Heisenberg). And if the range is expressed in meV, then you should use the Planck distribution in meV (proportional to frequency).

Last edited by a moderator: May 3, 2017
7. Jul 2, 2008

### cmos

gel,

I had a feeling that this problem could be resolved by the fact that we are dealing with distributions; however, I still do not accept your answer. I will pose the problem slightly different, but first some results I worked on today....

I based my original post on counting photons from the peaks of the Planck energy density distributions in the two domains. I realized soon thereafter that directly relating the peaks of these distributions to photon number was not valid. I have since come up with expressions for photon flux from a blackbody radiator. I list these in case I have made a mistake in my transformations:

$$\dot n(\nu) = \frac{2\pi\nu^2}{c^2} \frac{1}{e^{h\nu/kT}-1}$$
$$\dot n'(\lambda) = \frac{2\pi c}{\lambda^4} \frac{1}{e^{hc/\lambda kT}-1}$$

Please note that I use the 'prime' symbol to denote that we are working in a different domain; it does not represent differentiation. Edit: The above expressions are for the flux through 2 pi steradians of space.

Sadly, the results still give two different peaks. In the frequency domain: 199 THz. In the wavelength domain: 612 nm.

Suppose I pose the "paradox" as so:
Let detector A be a photon counter that works in the frequency domain while detector B is a photon counter that works in the wavelength domain. These detectors accept ALL photons and records the count in a histogram. After an hour a computer looks at the photon count at 199 THz from detector A and the count at 612 nm from detector B. If A is greater than B, then kitty lives, otherwise kitty dies.

Thoughts?

Last edited: Jul 2, 2008
8. Jul 2, 2008

### cesiumfrog

Isn't this just something really simple?

Due to the inverse (non-linear) relationship of frequency and wavelength, consider two "equal chunks of frequency space" namely 1-2Hz and 3-4Hz. Now, if we convert to wavelength space, the first box is twice as large as the second box. Agreed?

So why should it surprise you if the peak in frequency space doesn't quite match the peak in wavelength space? The OP's paradox is that saying
is the same as saying "plus or minus a nm or so" for one detector but "plus or minus half a nm" for the other. No wonder the first detector seems to get too much signal in the wavelength picture: it's less selective.

9. Jul 2, 2008

Staff Emeritus
I think that all the OP is saying is that the average of the reciprocal is not the reciprocal of the average. It has nothing really to do with QM, or Stat Mech, or really, anything except arithmetic.

10. Jul 2, 2008

### cmos

Which is why I amended the "paradox" at the end of post 7; so that we may look at the final count at the specified peak position.

11. Jul 2, 2008

### cmos

12. Jul 2, 2008

Staff Emeritus
There is no physical interpretation. It's just arithmetic. For example, equal sized bins in wavelength are not equal sized bins in frequency, so you don't expect the peaks to be in the same place in the two domains.

13. Jul 2, 2008

### cesiumfrog

14. Jul 2, 2008

### cmos

If there is no physical interpretation, then you are clearly lacking in the physics. Also, your explanation does not resolve the apparent "paradox" I have posted.

15. Jul 2, 2008

### cmos

cesiumfrog,

I suggest carefully re-reading the end of post #7 where I amend my question to properly make use of photon counting. None on this thread, thus far, have been able to explain this.

16. Jul 3, 2008

### malawi_glenn

I think this is the answer:

$$I(\nu,T)\,d\nu=-I(\lambda,T)\,d\lambda$$

This gives you the true relation between the two distributions=

$$I(\nu,T)=I(\lambda,T)\frac{\lambda ^2}{c}$$

you think it is: $$I(\nu,T)=I(\lambda,T)$$, which is wrong, since we are dealing with probabilty densities.

Last edited by a moderator: Jul 3, 2008
17. Jul 3, 2008

Staff Emeritus
Hmmm...one of us has a PhD in physics. The other is making an elementary arithmetic mistake, calling it a paradox, and refusing to believe the correct explanation. Which one of us is lacking in physics?

18. Jul 3, 2008

### Redbelly98

Staff Emeritus
This is an "apples vs. oranges" comparison. One number is the peak power emitted per unit frequency, and the other is the peak power emitted per unit wavelength. In other words, the two spectra are really of different quantities, Watts-per-Hz and Watts-per-nm, and have different peaks.

19. Jul 3, 2008

### cmos

Well then, I strongly question the rigor your institution put you through.

20. Jul 3, 2008

### cmos

In post # 7, I accounted for this (somewhat?) by expressing the Planck law in terms of (spectral) photon flux as opposed to spectral radiance. But this still does make sense, b/c now we are looking at flux-per-Hz and flux-per-nm. A peculiarity I can see falling out of this is that the true peak flux (as opposed to spectral flux) may not coincide with either of the peaks I previously mentioned.

I also thought last night how one might make and invoke a spectral radiometer. Similar to a scattering experiment where you must measure over an interval $$d\theta$$, by using a prisim you would effectively measure of an interval $$d\lambda$$. This leads me to wonder if it even makes sense to talk about a "true peak flux." More on all of this when I have had time to think about it.