What is the paradox in Planck's Law?

[malawi_glenn]

cmos, the "spectral radiance", i.e I(lamda,T) and I(nu,T), do they have the units "number of photons per unit wavelenght/frequency"?

Ans: No they don't, the two distributions you have are not answering the number of photons/wavelenght(freq.) - so I you must go back some steps in the derivation of the Planck distribution in order to perform the photon counting experiment of yours.

 

[lightarrow]

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.

A paradox?:
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?

I have the copyright of this paradox! https://www.physicsforums.com/showthread.php?t=132258

 

[Cthugha]

Not true, the peak (actually the distribution itself) is inherent to the physical system, regardless of your measurement. In terms of photon counting, photon number is a correlation to intensity, i.e. irradiance magnitude. So in collecting photons from a blackbody source, there must be a point frequency/wavelength at which you collect the most.

You are lacking some basic understanding of coordinate transformations and phase space (no offense intended).

Choosing the scaling of a detector equals choosing your set of coordinates to describe the problem. The problem is now, that the unit volumina of both choices of coordinates (linear in wavelength and linear in frequency) do not have equal size in various regions of the phase space. This is like going from cartesian coordinates to spherical coordinates, where you get the area element $$r^2 sin \theta$$ telling you, that the unit volume in this choice of coordinates scales with $$r^2$$ and $$\theta$$.

So if you want frequency and wavelength of the peak to "match", you need another set of coordinates. I suppose one, which is chosen in a way that equal volumina in phase space contain equal numbers of states should work. However, this will be a nonlinear scale in both wavelength and frequency, but you will get a peak position, from which you can extract a "matching" frequency and wavelength.

 

[Redbelly98]

Just a side-note on this whole discussion:

Yet another way to plot a spectrum is to graph the flux within a narrow range of, say, 1% of the wavelength or frequency. This will have a maximum somewhere between the maxima for the per-wavelength and per-frequency curves.

At any rate, for this plot it doesn't matter if you base the 1% range on the wavelength or the frequency, since the range is a small fraction of the wavelength. The maximum occurs at the same wavelength in either case.

I could elaborate or explain this in more detail, if anyone doesn't understand what I am getting at.

 

[cmos]

cmos, the "spectral radiance", i.e I(lamda,T) and I(nu,T), do they have the units "number of photons per unit wavelenght/frequency"?

Ans: No they don't, the two distributions you have are not answering the number of photons/wavelenght(freq.) - so I you must go back some steps in the derivation of the Planck distribution in order to perform the photon counting experiment of yours.

I mean no offense to this, but are you even bothering to read my posts? In post #7 (the post I have referred to numerous times), I address that mistake and give the proper terms for spectral flux, i.e. photon number per unit area per unit time per unit {wavelength or frequency}.

 

[cmos]

You are lacking some basic understanding of coordinate transformations and phase space (no offense intended).

Choosing the scaling of a detector equals choosing your set of coordinates to describe the problem. The problem is now, that the unit volumina of both choices of coordinates (linear in wavelength and linear in frequency) do not have equal size in various regions of the phase space. This is like going from cartesian coordinates to spherical coordinates, where you get the area element $$r^2 sin \theta$$ telling you, that the unit volume in this choice of coordinates scales with $$r^2$$ and $$\theta$$.

So if you want frequency and wavelength of the peak to "match", you need another set of coordinates. I suppose one, which is chosen in a way that equal volumina in phase space contain equal numbers of states should work. However, this will be a nonlinear scale in both wavelength and frequency, but you will get a peak position, from which you can extract a "matching" frequency and wavelength.

What I was trying to get at when you quoted me is that the peak of your distribution should not be affected by the dimensions of your detector. I did not mean for "scaling" to refer to the domain that we are working in.

 

[Redbelly98]

cmos,

The histograms you refer to in Post #7 would be for a certain "binning scheme". That is, each count total in the histogram addresses the number of photons detected over some range of wavelengths (or frequencies). The counts do not represent the number of photons at any particular wavelength.

I may be wrong, but you are probably envisioning constant-wavelength increments for the wavelength histogram, and constant-frequency bins for the frequency histogram. In which case the two histograms represent two different quantities: photons per unit wavelength in one case, and photons per unit frequency in the other. As you've already come to realize, these are different.

Hope that helps clear things up somewhat.

 

[cmos]

From post #7:



You will detect ZERO photons at exactly 199 THz, and ZERO photons at exactly 612 nm.

To detect any photons at all, a range of wavelengths or frequencies must be specified.

Edit:
You may want to think about how those histograms are to be "binned". If you were to make the bins of the two distributions to directly correspond with each other, the bin with the highest photon count would be the same for the two detectors.

I don't completely agree with this, but your edit does help to vindicate it. The accuracy of your detector will determine what it interprets. So if the sensitivity of your detector is in the GHz range, then 199.0002 THz will probably be interpreted as exactly 199 THz.

I see no problem, for this type of thought experiment, with saying that the sensitivities are negligible. If the spread is small enough, then there should be negligible error in counting the photons at the two point previously specified.

 

[cmos]

Just a side-note on this whole discussion:

Yet another way to plot a spectrum is to graph the flux within a narrow range of, say, 1% of the wavelength or frequency. This will have a maximum somewhere between the maxima for the per-wavelength and per-frequency curves.

At any rate, for this plot it doesn't matter if you base the 1% range on the wavelength or the frequency, since the range is a small fraction of the wavelength. The maximum occurs at the same wavelength in either case.

I could elaborate or explain this in more detail, if anyone doesn't understand what I am getting at.

Redbelly,

I see you have already re-posted while I try to address each of the previous nights arguments one-by-one. I will ignore that post since this is the one I think that really hits the spot.

What you suggest in the above quote does resolve the dilemma. I want to commend you on being one of the few to actually contribute to this discussion. Well done!

 

[cesiumfrog]

I see no problem[..] with saying that the sensitivities are negligible. If the spread is small enough, then there should be negligible error

This is where you are mistaken. (Hint: the ratio of two infinitesimal quantities is not always unity.)

 

[gallusquidam]

If one only replaces λ with c/ν in Planck´s equation, then on maximizing the frequency equation by putting the derivative equal to zero, frequency and wavelength correspond to the simple equation: (c = λν). However, if one wishes to integrate the frequency equation and obtain the same integrated result, then dλ must be transformed to dν [ie. dλ = -c dν/νν. The range of the integral must then also be transformed in accordance with (c = λν). This procedure is what is taught in basic calculus. There is then no discrepancy in the calculations.

 

[reilly]

This discussion is amazing. the whole issue is merely an exercise in the chain rule of differentiation; perfect for a first year calculus course exam. Of course , the peaks for frequency and wave length occur at different places. Just check it out in a beginning calculus text.

gallusquidam is correct.

The mystery is why this discussion has gone on so long over a basic idea that goes back a few hundred years or so, and which we all learned in beginning calculus.

Regards,
Reilly Atkinson

 

[lightarrow]

This discussion is amazing. the whole issue is merely an exercise in the chain rule of differentiation; perfect for a first year calculus course exam. Of course , the peaks for frequency and wave length occur at different places. Just check it out in a beginning calculus text.

gallusquidam is correct.

The mystery is why this discussion has gone on so long over a basic idea that goes back a few hundred years or so, and which we all learned in beginning calculus.

You are right, however, from a physical point of view, the answer doesn't seem so obvious to me. Let's make this example: considering the Sun's surface at 5780K, in one case you have its spectrum's peak at 502 nm = green (from the radiance as function of wavelenght), in another case at 885 nm = infrared (from the radiance as function of frequency). Why do we see the Sun's light as yellow and not as red?

 

[Redbelly98]

I see the sun's light as white, because the human visual system adjusts to whatever light source is present in determining the color of objects.

 

[gallusquidam]

It was nice of you to say I am right but I am not sure I was understood. Changing the variable with which a function is defined is a separate operation from integrating the function with respect to the new variable. When you change the variable from λ to ν, forget about the dλ = -c dν/νν part. In other words, leave out the factor (-c/νν) in the new frequency function. When this new frequency function is maximized, the frequency maximum will correspond with the wavelength maximum ie. ν(max) = c/λ(max). Planck's function is clouded in mystery for some but it is just another function. It may be necessary to illustrate this operation with a simpler function. Let me know if this is needed. My point is that this operation must make the frequency maximum the same as the wavelength maximum.

 

[lightarrow]

I see the sun's light as white, because the human visual system adjusts to whatever light source is present in determining the color of objects.

Ok, forget the Sun. You have two light sources (lamps) which you can approximate as blackbodies; one peaks at 502 nm, the other at 885 nm. What colour do you see them? Do you see them of the same colour?

 

[Redbelly98]

Are you referring to the peak of the radiant flux per unit wavelength, or per unit frequency?

 

[lightarrow]

Are you referring to the peak of the radiant flux per unit wavelength, or per unit frequency?

Exactly. But is it always specified in the books of physics, when they talk about light sources?

 

[Almanzo]

With quantum mysteries, it is sometimes a good idea to see if one can state a similar problem in the classical domain.

Suppose we have a bag full of ball bearings in many different but macroscopic sizes. (A few millimeters to a centimeter.) We proceed to measure their diameters, using a screw-micrometer, and their volumes, using the displacement of some liquid. Two histograms may be created, each of which we will assume to have just one peak.

Now, for every individual ball, the diameter has a certain relationship to the volume. This is the same relationship (the same formula) for each of the balls, provided that they are exactly spherical. The question is now whether that same relaionship must exist between the peak-diameter and the peak-volume of our two histograms.

The cat paradox does now translate into a device which accepts the ball bearings one by one, and counts those who fall within two specified categories, one corresponding to the peak in the diameter-histogram, the other to the peak in the volume-histogram. The cat dies, or is set free according to the first category into which a thousand balls have been counted.

 

[cesiumfrog]

Nice analogy, Almanzo, I like how you incorporated quanta.

 

[Redbelly98]

Exactly. But is it always specified in the books of physics, when they talk about light sources?

If it's not, I assume it's per whatever variable is used for the x-axis.

 

[lightarrow]

If it's not, I assume it's per whatever variable is used for the x-axis.

Of course, when there is a diagram of the spectrum.

 

[Redbelly98]

Please let me ask, what is your point? (I am losing track of where our conversation is going.)

 

[lightarrow]

Please let me ask, what is your point? (I am losing track of where our conversation is going.)

In wikipedia, for example, it's written:
http://en.wikipedia.org/wiki/Wien's_displacement_law

Light from the Sun and Moon. The surface temperature (or more correctly, the effective temperature) of the Sun is 5778 K. Using Wien's law, this temperature corresponds to a peak emission at a wavelength of 2.89777 million nm K/ 5778 K = 502 nm = about 5000 Å. This wavelength is (not by accident) fairly in the middle of the most sensitive part of land animal visual spectrum acuity.

Now, while it's quite clear from the context that they are talking about the radiance as a function of wavelenght, sometimes in the books of physics or in scientific magazines, what I quoted from wiki, is written without specifying the spectrum domain and this has no physical meaning.
I just wanted to point it.

 

[Redbelly98]

Valid point. And even more so when you realize the human visual response curve does not have the "per-wavelength or per-frequency?" issue like an emission spectrum does. Change the sun' spectrum to the "per frequency" variety, and it's peak will be somewhere else relative to the visual response curve.

 

[Almanzo]

The problem with a histogram is the arbitraryness of the division of parameter space into different slots. This arbitraryness can be removed -- for the classical case, at any rate -- by using a accumulation curve.

In the ball-bearing experiment this curve would have height zero for diameter or volume zero, and increase by one step at every diameter or volume which is found. (If two or more balls have exactly the same diameter or volume, the curve would increase by two or more steps.) At infinite diameter or volume the height of the curve would be equal to the number of balls tested.

Now, instead of finding a peak in the distribution, one has to find the region of greatest steepness in the accumulation curve. Finding this region for the volume-curve and the diameter curve, we find a volume and a diameter. The question is now whether the relation between this diameter and volume are the same relation which we would expect in any individual ball.

In the quantum case, one would have to test two equivalent samples, measuring wavelength for one sample and frequency for the other one. In principle one of them should be measurable exactly if one forgoes measuring the other one. But if exact measurements are not possible, the accumulation curve will change from a sharp curve consisting of little steps into a more fuzzy, "S-shaped" band. (Not really S-shaped, of course; more like a hillside.) A region of greatest steepness should still be there to be found, though not to be pinpointed exactly.

 

[apolski]

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

How you found that expression?

 

[apolski]

$$\dot n(\nu,T) = \frac{J(\nu,T)}{h\nu}$$

But why?