What is the paradox in Planck's Law?

[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.

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?

 

[George Jones]

It's just integration by substitution.

I don't know if https://www.physicsforums.com/showthread.php?p=934184#post934184" will help.

 

[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.

 

[malawi_glenn]

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

 

[Zizy]

Using ln, this problem doesn't 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.

 

[gel]

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?

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" .
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).

 

[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?

 

[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

plus or minus several meV for all us Heisenberg buffs

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.

 

[Vanadium 50]

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.

 

[cmos]

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?

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.

 

[cmos]

Vanadium,

The please enlighten us with your physical interpretation, I am curious.

 

[Vanadium 50]

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.

 

[cesiumfrog]

cmos, I suggest carefully re-reading the thread. Four different people now have answered correctly.

 

[cmos]

Vanadium,

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.

 

[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.

 

[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.

You can also read about your paradox here: http://en.wikipedia.org/wiki/Wien's_displacement_law (see Frequency form)

 

[Vanadium 50]

If there is no physical interpretation, then you are clearly lacking in the physics.

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?

 

[Redbelly98]

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.

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.

 

[cmos]

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?

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

 

[cmos]

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.

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.

All in all, thank you Redbelly for your comments. Well done.

 

[cmos]

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.

You can also read about your paradox here: http://en.wikipedia.org/wiki/Wien's_displacement_law (see Frequency form)

We took this into account from the very beginning.

 

[malawi_glenn]

We took this into account from the very beginning.

No you did not, you compared the max of I(lamda) vs max of I(nu), which is, as Redbelly formulated it -> comparing apples against organges.

 

[cmos]

No you did not, you compared the max of I(lamda) vs max of I(nu), which is, as Redbelly formulated it -> comparing apples against organges.

Sure I did, otherwise I wouldn't have found the two different peaks. What you show is the transformation from one distribution to the other, which leads to the two different peaks. I've already commented on Redbelly's post. My comments may lead to a resolution; I have not had a chance to try it yet.

 

[cmos]

malawi,

I noticed in post #4 that you stated you never noticed this before. Forgive me for asking, but are you aware that it is commonly known that the peak positions of the Planck distribution do not correspond by the simple relation \nu=c/\lambda? The reason for this is because of the way the distributions transfer, as you and George Jones have shown.

In post #7 I show that the same holds true when expressing the Planck distributions in terms of photon flux as opposed to energy density. My comments in post #20 is what I hope will resolve this issue.

 

[malawi_glenn]

If you think this is a true paradox, write an article and send it to a referee.

However, this is not a new discovery, since it is mentioned in the wiki-article.

 

[cmos]

It seems that you have been on this forum a while so I don't really know what your problem is. I have not claimed to make a new discovery nor am I refuting physical law. I have simply posed a question that does not have a plug-and-chug answer. It require actual thinking. In addition to that, my "paraodx" was not posted in any wiki-article. It is something I came up as a way to analyze common results which are also in a wiki-article (I myself prefer textbooks and my own re-derivations).

It seems to me, that in your inability to help solve the problem, you are defaulting to trying to trivialize my statements. If you are here to help, then please do so; that is why I brought this problem to these forums. If you only like to help with plug-and-chug solutions, then there is nothing wrong with that as people also need that type of help; this however is not one of those cases.

 

[malawi_glenn]

No your paradox was not the in the wiki-article, but the 'physical problem' that the paradox is not new.

My point that you should write an article is that you did not seem to get any help here, it was just a tip how you can proceed.

The energy in the two distributions in a infinitesimal interval should equal each other, no matter what quantity you are measuring: wavelenght vs. frequency.

 

[Redbelly98]

Perhaps "paradox" was not the best choice of words here, but what cmos has brought up here is a common conceptual error made by many people with a technical education.

At a previous job, I took an in-house "physics of lighting" course with weekly homework problems. One problem was to give the relationship between the peak frequency of the flux-per-frequency spectrum, and the peak wavelength of the flux-per-wavelength spectrum of a blackbody. Many in the course simply answered with

<br /> \nu_{peak} \ \lambda_{peak} = c<br />

Some of these folks got quite angry when told that this was incorrect. It was very tough to try to explain this to them, as you not only had to overcome this very common misconception, but also (at that point) some angry tension at having gotten the problem wrong.

Anyway, I am glad to see that cmos is more open to our explanations than were those former coworkers of mine!

 

[cmos]

I use the word paradox because the problem I have posed seems to lead to two very different outcomes depending on the domain you decide to work with. I want to emphasize here: the problem I have presented deals with photon counting. I refer you to my post #7 again.

You can think of the problem in this way: Given a blackbody at some temperature, at what wavelength do you expect to receive the largest number of photons? How about frequency?

Knowing that the peaks in the blackbody spectrum do not simply correspond by the Plack relation, I then asked myself: What if we setup two detectors in parallel, one detecting wavelength and one detecting frequency? Should these not correspond?

By calling it a paradox, I do not mean to say there is no solution. I mean to say that there is an apparent contradiction that, by realizing some subtlety, will be delightfully resolved by finding a solution.

What perplexes me is that out of all the people who have replied to this thread more than once, none have accepted the question. All the replies give answers to a different question: What is the peak wavelength and frequency of the Planck distribution?
Note that I was the first person in this thread to answer this very question.

 

[cmos]

Note:

I have since solved the problem I have presented. The brief discussion with Redbelly proved the most useful in this regard (this is a hint). I hope that some of you, or others to happen to cross this thread, will try the problem for yourselves. For this reason I will not post my solution. Please private message me to verify your answers (or if you get stuck).

So, is kitty alive or dead?

 

[Cthugha]

Note:

I have since solved the problem I have presented. The brief discussion with Redbelly proved the most useful in this regard (this is a hint). I hope that some of you, or others to happen to cross this thread, will try the problem for yourselves. For this reason I will not post my solution. Please private message me to verify your answers (or if you get stuck).

So, is kitty alive or dead?

I still don't see your point. You have been told about 5 times or so in several different forms, that a linear scaling in frequency automatically implies a nonlinear scaling in wavelength and vice versa and that's it. In terms of photon counting it is the same case: the scaling of your detector determines, where the maximum will be.

As a visualization, you can look here:

http://de.wikipedia.org/wiki/Bild:PlanckDist_ny_lambda_de.png

unfortunately, it is in German, but if you know that "Wellenlänge" means wavelength and "Frequenz" means frequency. It should be easy to see, why different maxima arise.

 

[Redbelly98]

You can think of the problem in this way: Given a blackbody at some temperature, at what wavelength do you expect to receive the largest number of photons? How about frequency?

Hi,

There is a problem in how the above question is asked (and perhaps you have resolved it already when you solved the problem later). Photons cannot be not detected "at" any specific wavelength, or frequency. Rather they are detected within some range of wavelengths or frequencies.

Two better ways to ask the question are:

1. At what wavelength would you detect the most photons within a narrow, fixed wavelength range, say +/- 0.01 nm? Answer: at the peak of the photons-per-wavelength spectrum.

2. At what frequency would you detect the most photons within a narrow, fixed frequency range, say +/- 1 Hz? Answer: at the peak of the photons-per-frequency spectrum.

At any rate, the answer to #1 is at a different frequency than the answer to #2, since the "fixed wavelength range" corresponds to a varying frequency range. (And vice versa.)

 

[cmos]

I still don't see your point. You have been told about 5 times or so in several different forms, that a linear scaling in frequency automatically implies a nonlinear scaling in wavelength and vice versa and that's it.

Funny, because I think I said that in my first post.

In terms of photon counting it is the same case: the scaling of your detector determines, where the maximum will be.

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.

 

[cmos]

There is a problem in how the above question is asked (and perhaps you have resolved it already when you solved the problem later). Photons cannot be not detected "at" any specific wavelength, or frequency. Rather they are detected within some range of wavelengths or frequencies.

Yes, but in post #7 (sorry for not reiterating it), I allowed for all photons to be collected and the results plotted to a histogram. Then a computer decides the fate of the cat based on which ever detector has collected the most photons at the two different peaks. The two peaks corresponding the Planck distributions in the form of photon flux.

 

[Redbelly98]

From post #7:

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

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.

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.

 

[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.