I Different Blackbody Intensity Peaks, how do we measure it?

[greswd]

Your $B(\lambda, T)$ above is my $L(\lambda, T)$. (precisely).

So does this $B(\lambda, T)$ refer to $B_\lambda$ or $B_\nu$?

Why is it one and not the other?

 

[Charles Link]

So does this $B(\lambda, T)$ refer to $B_\lambda$ or $B_\nu$?

Why is it one and not the other?

In a much earlier post above, I used $L(\lambda,T)$ and I have it memorized from many years of spectroscopic measurements and calculations. The $B_{\lambda}(\lambda,T)$ that the OP listed matched it precisely. For a $B_{v}(\nu, T)$, you always need to check the defintion: Is $\nu=1/\lambda$ or is $\nu=c/\lambda$ ? For the wavelength function there is no ambiguity. The one item that is also important here though, is the blackbody function given as $M(\lambda, T)$ or $L(\lambda, T)$ ? The $B$ notation is not standard, and I needed to determine if it was the $L$ form or the $M$ form ($M=L*\pi$). It was indeed the $L$ form.

 

[tade]

In a much earlier post above, I used $L(\lambda,T)$ The $B_{\lambda}(\lambda,T)$ that the OP listed matched it precisely.

Is there a scientific reason as to why the one that matches it precisely is $B_{\lambda}$ and not $B_{\nu}$ ?

I believe that this must be due to the type of apparatus used.

 

[Charles Link]

Is there a scientific reason why the one that matches it precisely is $B_{\lambda}$ and not $B_{\nu}$ ?

I believe that this must be due to the type of apparatus used.

I never memorized a $B_{\nu}(\nu)$ blackbody form. With a diffraction grating based spectrometer, the spectroscopist usually presents spectra in wavelength. There is a FTIR (Fourier Transform spectrometer) with a moving Michelson mirror, and researchers often present those spectra in wavenumber. Both forms are appropriate, although it can be somewhat confusing when first learning the subject. I did check you $B_{\nu}(\nu)$from the other (wavelength) form. You correctly had the $1/\lambda^2$ factor (actually $c/\lambda^2$for your $\nu=c/\lambda$). Your definition uses $\nu=c/\lambda$. It makes a difference whether you graph a spectrum in $\nu=1/\lambda$ or $\nu=c/\lambda$ for your units on the y-axis.

 

[tade]

I never memorized a $B_{\nu}(\nu)$ blackbody form. With a diffraction grating based spectrometer, the spectroscopist usually presents spectra in wavelength.

$B_{\lambda}$ peaks at 524 nm and $B_{\nu}$ peaks at 923 nm.

You say that with a diffraction grating based spectrometer, we can measure $L(\lambda,T)$, which then gives a peak at 524 nm.

Is there any instrument that gives the raw data of $L(\lambda,T)$ which peaks at 923 nm?

 

[Charles Link]

$B_{\lambda}$ peaks at 524 nm and $B_{\nu}$ peaks at 923 nm.

You say that with a diffraction grating based spectrometer, we can measure $L(\lambda,T)$, which then gives a peak at 524 nm.

Is there any instrument that gives the raw data of $L(\lambda,T)$ which peaks at 923 nm?

Once you have $R(\lambda)$ measured with a calibration run, you can measure any blackbody of any temperature that gives off enough energy to measure. When you report your results, for example for a T= 5525 blackbody, if you have units of wavelength on the x-axis it will peak at 524 nm. If you have units of wavenumber on the x-axis, it will peak at a wave number that has corresponding wavelength of 923 nm. The raw data peak is of little significance. There are too many other spectral factors, (detector spectral response, optics, and spectrometer diffraction grating throughput) that affect this.

 

[greswd]

The peak occurs from a result of the function $I_{\lambda}(\lambda)$ or $I_{\nu}(\nu)$. The spectrometer voltage $V(\lambda)$ is likely to also peak somewhere in the vicinity of both of these

So for the Sun which is at 5525 K, at what wavelength does $V(\lambda)$ peak?

 

[Charles Link]

So for the Sun which is at 5525 K, at what wavelength does $V(\lambda)$ peak?

A major factor here is the detector response. If you use a silicon photodiode or any other photodide, they respond to photons so the detector response is typically proportional to $\lambda$ in the region of responsivity. A thermopile type detector (e.g. pyroelectric) will have a flat spectral response. These two detector types will have peaks in the raw voltage spectrum that are quite different with the peak from the silicon photodiode occurring at a longer wavelength. The precise position of this raw voltage peak is really of little significance.

 

[greswd]

A major factor here is the detector response. If you use a silicon photodiode or any other photodide, they respond to photons so the detector response is typically proportional to $\lambda$ in the region of responsivity. A thermopile type detector (e.g. pyroelectric) will have a flat spectral response. These two detector types will have peaks in the raw voltage spectrum that are quite different with the peak from the silicon photodiode occurring at a longer wavelength. The precise position of this raw voltage peak is really of little significance.

With different equipment giving different peaks, how do we end up with the one peak of 524 nm?

 

[Charles Link]

With different equipment giving different peaks, how do we end up with the one peak of 524 nm?

Please read the previous postings far above. That's what the calibration spectral run is used for. Each detector employed needs a separate calibration run.

 

[greswd]

Please read the previous postings far above. That's what the calibration run is used for. Each detector employed needs a separate calibration run.

I see.

Ultimately, what function describes the number of photons of a particular wavelength emitted by the blackbody per unit time? It can't be both $B_{\lambda}$ and $B_{\nu}$

 

[The Bill]

With different equipment giving different peaks, how do we end up with the one peak of 524 nm?

By dividing the voltage by the response function for that particular detector.

 

[Charles Link]

By dividing the voltage by the response function for that particular detector.

@greswd See in particular post #9 above. And yes, with a different detector, it is essentially a different spectrometer.

 

[greswd]

@greswd See in particular post #9 above. And yes, with a different detector, it is essentially a different spectrometer.

Is the answer to my question in #61?

 

[Charles Link]

Is the answer to my question in #61?

For number of photons from a blackbody, F.Reif Statistical Physics does a good job. The derivation is done in "k" space. A photon energy $e_p=hc/\lambda$ gets multiplied to the photon number density. The spectrum can be converted from k-space to the appropriate x-axis you want to use.

 

[greswd]

For number of photons from a blackbody, F.Reif Statistical Physics does a good job. The derivation is done in "k" space. A photon energy $e_p=hc/\lambda$ gets multiplied to the photon number density. The spectrum can be converted from k-space to the appropriate x-axis you want to use.

By appropriate x-axis you mean wavelength or frequency? Thank you. I think we've finally reached it.

 

[greswd]

For number of photons from a blackbody, F.Reif Statistical Physics does a good job. The derivation is done in "k" space. A photon energy $e_p=hc/\lambda$ gets multiplied to the photon number density. The spectrum can be converted from k-space to the appropriate x-axis you want to use.

I have found a copy of that book
<<link deleted>>

which page is it at? Is it Page 387? (as printed on the page itself)

 

[Charles Link]

I have found a copy of that book
<<link deleted>>

which page is it at? Is it Page 387? (as printed on the page itself)

I was unable to read the pdf, but from the picture of the cover, yes, that's the book. And looking back over the complete postings, I hope most of the questions were answered. It really can be a subject of some confusion when first encountered, but once it is studied in detail, it becomes a little more routine.

 

[Charles Link]

Looking back over the complete postings, hopefully most of the questions were answered. It really can be a subject of some confusion when first encountered, but once it is studied in detail, it becomes a little more routine.

 

[greswd]

Looking back over the complete postings, hopefully most of the questions were answered. It really can be a subject of some confusion when first encountered, but once it is studied in detail, it becomes a little more routine.

After you click on the link, it says "2 Items". Then click on "Reif Fundamentals" which is on the left. It should open the book right away.

The reason I have to ask you this is because I want make sure that what you said in #65 is not just a derivation of Planck's Law

 

[Charles Link]

After you click on the link, it says "2 Items". Then click on "Reif Fundamentals" which is on the left. It should open the book right away.

The reason I have to ask you this is because I want make sure that what you said in #65 is not just a derivation of Planck's Law

I remember/memorized the derivation quite a few years ago. Reif's book is where I learned it. The derivation has a number of steps in it. First it counts photon modes in k-space. Then it converts to spherical k-space coordinates. It picks up a factor of 2 for polarization. This is all inside an enclosed cavity. Then it puts in the Bose factor for the mean occupation number at energy E. Next it computes by the effusion formula $R=nv_{mean}/4$ for the number of particles (photons) per unit time per unit area that will emerge from a small aperture (The effusion formula works for any gas where collisions are minimal. Treating the enclosed radiation as a collection of photons (particles) gets the correct answer. In this case $v_{mean}=c$). Finally, this is converted to an energy spectrum using $e_p=hc/\lambda$. The result is the Planck spectral blackbody function. (One sort of unanswered question in Reif's presentation is why the chemical potential $\mu$ in the Bose factor is necessarily zero for a collection of photons. In any case, it gets the correct result.)

 

[greswd]

effusion formula $R=nv_{mean}/4$ $v_{mean}=c$
$e_p=hc/\lambda$.

these are the notations Reif used right? I'm trying to locate it in the book.

 

[greswd]

I was unable to read the pdf, but from the picture of the cover, yes, that's the book.

How come you can't open the book?

I can't seem to find the part you described. The Blackbody section is from pg 373 (as printed on the page itself)

 

[tade]

For number of photons from a blackbody, F.Reif Statistical Physics does a good job. The derivation is done in "k" space. A photon energy $e_p=hc/\lambda$ gets multiplied to the photon number density. The spectrum can be converted from k-space to the appropriate x-axis you want to use.

What is the name of this formula? Can it be found on Wikipedia?

 

[tade]

I remember/memorized the derivation quite a few years ago. Reif's book is where I learned it. The derivation has a number of steps in it. First it counts photon modes in k-space. Then it converts to spherical k-space coordinates. It picks up a factor of 2 for polarization. This is all inside an enclosed cavity. Then it puts in the Bose factor for the mean occupation number at energy E. Next it computes by the effusion formula $R=nv_{mean}/4$ for the number of particles (photons) per unit time per unit area that will emerge from a small aperture (The effusion formula works for any gas where collisions are minimal. Treating the enclosed radiation as a collection of photons (particles) gets the correct answer. In this case $v_{mean}=c$). Finally, this is converted to an energy spectrum using $e_p=hc/\lambda$. The result is the Planck spectral blackbody function. (One sort of unanswered question in Reif's presentation is why the chemical potential $\mu$ in the Bose factor is necessarily zero for a collection of photons. In any case, it gets the correct result.)

Please help me Charles, I can't find it in the book.

 

[Charles Link]

I located my copy of Reif. pp.386-388 covers some of it, but p.376 and thereabouts does the energy density part inside the enclosure. The derivation as I know it I uses much of Reif's calculations, but instead of his more difficult detailed balance arguments, the effusion rate gets you there quicker. The effusion rate is a formula I first learned in a chemistry course, and I later derived the effusion formula later in a physics class.

 

[tade]

I located my copy of Reif. pp.386-388 covers some of it, but p.376 and thereabouts does the energy density part inside the enclosure. The derivation as I know it I uses much of Reif's calculations, but instead of his more difficult detailed balance arguments, the effusion rate gets you there quicker. The effusion rate is a formula I first learned in a chemistry course, and I later derived the effusion formula later in a physics class.

Idk why you can't open the pdf. So I've turned it into images. Are these the pages you are referring to? In your textbook what is the title or heading?

After the third image, Reif just goes on to derive Planck's Law for frequency. I don't see anything new unfortunately.

 

[tade]

Charles, can you take a photo of the relevant page? That would be very helpful thank you.

 

[Charles Link]

Idk why you can't open the pdf. So I've turned it into images. Are these the pages you are referring to? In your textbook what is the title or heading?

After the third image, Reif just goes on to derive Planck's Law for frequency. I don't see anything new unfortunately.

It looks like the very same book. Go back to p.373 etc. That is the start of the blackbody discussion. The wavelength formula, along with other frequency units is an addition you can do with a little calculus. Much of the derivation is found on pp.373-376. With a little modification, you can include the effusion rate formula and bypass the lengthy "principle of detailed balance" arguments to get the emitted radiation. .. editing...from p.387 it appears your pdf is in fact the same edition of the textbook. (The text was a perfect match.) It's much easier to turn the pages of a book then to sift through pdf images. Anyway, try reading pp.373-376 Black*Body Radiation (9.13) Electromagnetic Radiation in Thermal equilibrium inside an enclosure. Reif counts the states in k-space at the bottom of p.374 and at the top of p.375. It's a little bit of work, but it could be very good for you to study it in detail.

 

[tade]

For number of photons from a blackbody, F.Reif Statistical Physics does a good job. The derivation is done in "k" space. A photon energy $e_p=hc/\lambda$ gets multiplied to the photon number density. The spectrum can be converted from k-space to the appropriate x-axis you want to use.

On which page is the information you mentioned in the quote above? What is the page number and heading name?

It looks like the very same book. Go back to p.376, etc. That is the start of the blackbody discussion. The wavelength formula, along with other frequency units is an addition you can do with a little calculus.

It would be great if you could take a photo of the relevant page. Thanks.

 

[tade]

you can use http://postimage.org/

its very convenient

 

[Charles Link]

you can use http://postimage.org/

its very convenient

See my edited previous post.

 

[tade]

It's much easier to turn the pages of a book then to sift through pdf images.

Oh Charles you fogey

It looks like the very same book. Go back to p.373 etc. That is the start of the blackbody discussion. The wavelength formula, along with other frequency units is an addition you can do with a little calculus. Much of the derivation is found on pp.373-376. With a little modification, you can include the effusion rate formula and bypass the lengthy "principle of detailed balance" arguments to get the emitted radiation. .. editing...from p.387 it appears your pdf is in fact the same edition of the textbook. (The text was a perfect match.) It's much easier to turn the pages of a book then to sift through pdf images. Anyway, try reading pp.373-376 Black*Body Radiation (9.13) Electromagnetic Radiation in Thermal equilibrium inside an enclosure. Reif counts the states in k-space at the bottom of p.374 and at the top of p.375. It's a little bit of work, but it could be very good for you to study it in detail.

All I see is Planck's Law for frequency (9.13.10)

Then I think it wouldn't be the intensity, just the spectral radiance for frequency.

 

[Charles Link]

Oh Charles you fogey
All I see is Planck's Law for frequency (9.13.10)

Then I think it wouldn't be the intensity, just the spectral radiance for frequency.

That is the energy density per unit frequency interval inside the cavity. The radiated part doesn't get computed until around p.387. It takes a fair amount of effort to really get proficient at computing some of these things. I did spend dozens upon dozens of hours studying Reif's book which we used in a senior undergraduate course. Some of the pages, such as the blackbody derivation I think I spent as much as 2-3 hours per page, and perhaps more. To learn it in detail, it does take work.

 

[Charles Link]

I would like to take a couple minutes to show where my 5:3 ratio came from in the earlier discussion in the peak of the wavelength found in the frequency vs. wavelength spectrum that you may find of interest. A little calculus in taking derivatives of the wavelength Planck function and setting it equal to zero to get the peak gives $exp(-\alpha)=1-\alpha/5$ where $\alpha=hc/(\lambda kT)$ The solution for this is $\alpha=0$ (extraneous), and $\alpha=5$ (approximately). (More precisely it is something like 4.96). (Incidentally, this is where Wien's law, $\lambda_{max}T=2898$ microns deg K, comes from.) With the frequency Planck function derivative, you get $exp(-\alpha)=1-\alpha/3$. Thereby $\alpha=3$ (approximately). A slight arithmetic error resulted because the $\alpha$ solution for this case may be more like 2.9 or thereabouts.

 

[tade]

That is the energy density per unit frequency interval inside the cavity. The radiated part doesn't get computed until around p.387. It takes a fair amount of effort to really get proficient at computing some of these things. I did spend dozens upon dozens of hours studying Reif's book which we used in a senior undergraduate course. Some of the pages, such as the blackbody derivation I think I spent as much as 2-3 hours per page, and perhaps more. To learn it in detail, it does take work.

I've uploaded Pg 387 in a previous post.

After that they just derive Planck's Law for frequencies again

 

[Charles Link]

I've uploaded Pg 387 in a previous post.

After that they just derive Planck's Law for frequencies again

The concept of energy spectral density, which is really what your OP was about, is more of a calculus problem than something they are going to work out in an advanced physics text. Hopefully you can recognize why the following equation must hold for these spectral densities:
$|L_{\lambda}(\lambda,T)d\lambda|=|L_{\nu}(\nu,T)d\nu|$. Reif basically derives $L_{\nu}(\nu,T)$. With a little calculus you can make any necessary conversions.

 

[tade]

The concept of energy spectral density, which is really what your OP was about, is more of a calculus problem than something they are going to work out in an advanced physics text. Hopefully you can recognize why the following equation must hold for these spectral densities:
$|L_{\lambda}(\lambda,T)d\lambda|=|L_{\nu}(\nu,T)d\nu|$. Reif basically derives $L_{\nu}(\nu,T)$. With a little calculus you can make any necessary conversions.

This I understand, but I don't think that my questions have been conclusively answered yet.I have to ask you this, if we have an IDEAL detector, which has no flaws whatsoever, and is perfect in every way for its job, and we measure the Sun's spectrum, at which wavelength will the readings peak?

 

[Charles Link]

With a diffraction grating spectrometer, to a good approximation the spread of wavelengths $\Delta \lambda$ at each point in the spectrum is given by $\Delta \lambda=(d) (\Delta x)/f$ where $d$ is the distance between lines on the grating, $\Delta x$ is the slit width and $f$ is the focal length of the focusing optics in the spectrometer. This $\Delta \lambda$ is nearly constant in a spectral run, so that the spectrum would be found to peak at or near the peak in the wavelength blackbody function. For T=6000 K, this would be near $\lambda=500$ nm. ((2898/6000)*1000)nm to be more precise.

 

[tade]

With a diffraction grating spectrometer, to a good approximation the spread of wavelengths $\Delta \lambda$ at each point in the spectrum is given by $\Delta \lambda=(d) (\Delta x)/f$ where $d$ is the distance between lines on the grating, $\Delta x$ is the slit width and $f$ is the focal length of the focusing optics in the spectrometer. This $\Delta \lambda$ is nearly constant in a spectral run, so that the spectrum would be found to peak at or near the peak in the wavelength blackbody function. For T=6000 K, this would be near $\lambda=500$ nm. ((2898/6000)*1000)nm to be more precise.

Just to make things very clear, I am talking about an IDEAL spectrometer.

Where does 2898 come from?

 

[Charles Link]

Just to make things very clear, I am talking about an IDEAL spectrometer.

Where does 2898 come from?

Wien's law for a blackbody is $\lambda_{max}T=2898$ microns deg Kelvin (see also post #85-if you take the derivative of the Planck wavelength function and set it equal to zero, you can derive Wien's law). Meanwhile one other item is the points in a diffraction grating spectral run are normally equally spaced in wavelength, so that with a flat detector response, your raw data would look very much like the processed wavelength spectrum. The optical layout (insides) of a diffraction grating spectrometer is fairly standard. There are a couple of minor variations, but basically the standard version is an Ebert spectrometer. Another similar variety if I remember correctly is called a Czerny-Turner. The optics basically send a collimated beam (parallel rays=plane wave) incident on the grating by having the entrance slit of the spectrometer at the focal point of the first parabolic or spherical mirror and then the "far field pattern" off of the grating is created in the focal plane of a second focusing mirror=parallel rays at the same angle $\theta$ onto the mirror focus at the same position $x=f \cdot \theta$ in the focal plane etc. The exit slit of the spectrometer is placed in the focal plane of the second focussing optics. Normally both slits are chosen to have the same slit width. There is often a trade-off between resolution by going to a very narrow slit versus the amount of energy that reaches the detector. Wide slit=plenty of energy, but lower resolution. ... editing...Additional item worth mentioning is the grating is rotated during a spectral run to get the spectrum. The textbook equation applies $m\lambda=d (\sin(\theta_i)+\sin(\theta_r) )$ for the location of a given wavelength. (The spectrum is most often observed with m=1, but the higher orders can be used.) Most often the grating is of the reflective variety (the lines on the grating can be thought of as "Huygens mirrors" (as opposed to Huygens sources from the slits of a transmission grating.) The appropriate gear mechanism is usually designed into the rotation of the grating so that the grating steps through equal wavelength increments as it steps through the spectrum.

 

[tade]

Remember Charles, we're sticking to ideal spectrometers.

Wien's law for a blackbody is $\lambda_{max}T=2898$ microns deg Kelvin

For $B_\lambda$ it is 2898 microns, approximately 3000. For $B_\nu$, it is 5099 microns, approximately 5000. This is where you get 3/5, as mentioned earlier.

For T=6000 K, this would be near $\lambda=500$ nm. ((2898/6000)*1000)nm to be more precise.

I asked you about an ideal spectrometer. Why would the measurement be consistent with 2898 microns and not 5099 microns?

 

[Charles Link]

Remember Charles, we're sticking to ideal spectrometers.
For $B_\lambda$ it is 2898 microns, approximately 3000. For $B_\nu$, it is 5099 microns, approximately 5000. This is where you get 3/5, as mentioned earlier.I asked you about an ideal spectrometer. Why would the measurement be consistent with 2898 microns and not 5099 microns?

The energy that comes though the slit is proportional to $L_{\lambda}(\lambda,T)\Delta \lambda =L_{\nu}(\nu,T)\Delta \nu$ where the $\Delta$ 's are the wavelengths or wavenumbers covered by the slit. If $\Delta \lambda$ remains constant throughout the spectral run, the (raw) spectrum will look like $L_{\lambda}(\lambda,T)$. If $\Delta \lambda$ is constant, you can compute the variation of $\Delta \nu=\Delta \lambda/\lambda^2$ . Notice we can really use either side of the first equation above to do our computation of the energy that comes through. Alternatively, if $\Delta \nu$ remained constant throughout the run, then $\Delta \lambda$ would not. The observed spectrum would then be $L_{\lambda}(\lambda,T)\lambda^2$. Note though, even in an ideal spectrometer, $\Delta \lambda$ will not be precisely constant throughout the spectral run. The equation $\Delta \lambda =(d)(\Delta x)/f$ is an approximation and not exact. That's why a calibration spectrum is usually necessary for accurate results.

 

[vanhees71]

drvrm, vanhees71, The Bill, any suggestions?

I've still no clue, what this discussion is about. The Planck spectrum as an energy distribution per frequency looks different from the same distribution expressed in terms of wave length with the relation $\lambda=c/\omega$. This is so because a distribution transforms as a distribution under coordinate transformations. In this case of only one independent variable ($\nu$ or $\lambda$) it's simply the chain rule
$$\frac{\mathrm{d} f}{\mathrm{d} \nu}=\frac{\mathrm{d} f}{\mathrm{d} \lambda} \frac{\mathrm{d} \lambda}{\mathrm{d} \nu}.$$
Now when you calculate the value $\nu_{\text{max}}$, where the frequency distribution peaks and compare it to the $\nu_{\text{max}}'=c/\lambd_{\text{max}}$ where $\lambda_{\text{max}}$ is the wavelength where the wave-length distribution peaks, are slightly different. This is not very surprising since the distribution functions look different depending on which variable, $\nu$ or $\lambda$, you relate it to.

To measure the spectrum you simply measure the intensity of the black-body radiation and either plot it against $\nu$ or $\lambda$ respectively. There is no principle problem in determining either of the two "maxima" of the corresponding functions. There is not too much physics in this, except that it characterizes roughly the relevant order of magnitude of frequencies/wave lengths which are dominating the spectrum of a black body. There's also Wien's displacement law about the location of $\lambda_{\text{max}}$ as function of $T$:

https://en.wikipedia.org/wiki/Wien's_displacement_law

 

[greswd]

The Planck spectrum as an energy distribution per frequency looks different from the same distribution expressed in terms of wave length with the relation $\lambda=c/\omega$. This is so because a distribution transforms as a distribution under coordinate transformations.
Now when you calculate the value $\nu_{\text{max}}$, where the frequency distribution peaks and compare it to the $\nu_{\text{max}}'=c/\lambd_{\text{max}}$ where $\lambda_{\text{max}}$ is the wavelength where the wave-length distribution peaks, are slightly different. This is not very surprising since the distribution functions look different depending on which variable, $\nu$ or $\lambda$, you relate it to.

The maths is pretty clear cut, I think I understand the OP's concern clearly, and its not the math.

I think he is confused about the physical aspects of the spectrum, which is why he's asking about experimental results.

Physically, at which wavelength is there the greatest number of photons? In reality, there can only be one wavelength at which this occurs, not two.

And let's assume that the detector is ideal, so that we don't have to deal with all the optical defects and corrections etc.

 

[greswd]

If $\Delta \lambda$ remains constant throughout the spectral run, the (raw) spectrum will look like $L_{\lambda}(\lambda,T)$.
Alternatively, if $\Delta \nu$ remained constant throughout the run, then $\Delta \lambda$ would not. The observed spectrum would then be $L_{\lambda}(\lambda,T)\lambda^2$.

Why would the value of the peak wavelength change depending on whether $\Delta \lambda$ or $\Delta \nu$ was constant?

Doesn't the spectrum only peak at one wavelength, therefore making it physically impossible to measure two different peak wavelengths?

 

[Charles Link]

Why would the value of the peak wavelength change depending on whether $\Delta \lambda$ or $\Delta \nu$ was constant?

Doesn't the spectrum only peak at one wavelength, therefore making it physically impossible to measure two different peak wavelengths?

You need to study the posts above in more detail. I think if you do, you will conclude they are correct, including #93. The OP has asked some very good questions-similar ones that I have had from other students who were first learning the details of spectroscopy and diffraction grating spectrometers. The OP appears to have figured out a good portion of it. It does take work to figure out the details, and the subject does contain quite a number of details.

 

[tade]

You need to study the posts above in more detail. I think if you do, you will conclude they are correct, including #93. The OP has asked some very good questions-similar ones that I have had from other students who were first learning the details of spectroscopy and diffraction grating spectrometers. The OP appears to have figured out a good portion of it. It does take work to figure out the details, and the subject does contain quite a number of details.

But this quote above is not an attempt to answer greswd's question in #96. This quote does not follow.

He understands what I'm trying to get at.

 

[greswd]

You need to study the posts above in more detail. I think if you do, you will conclude they are correct, including #93. The OP has asked some very good questions-similar ones that I have had from other students who were first learning the details of spectroscopy and diffraction grating spectrometers. The OP appears to have figured out a good portion of it. It does take work to figure out the details, and the subject does contain quite a number of details.

He's right, I feel that you're not moving in the direction of trying to answer the question.

 

[Charles Link]

I answered the questions the best I can. In asking about the raw spectrum from a spectrometer, it was necessary to include some finer details about a typical spectrometer. One additional item with these spectrometers is there is an overlap of the orders (different m's and different wavelengths) so that a couple of long pass order sorting filters typically need to be employed in a spectral run. In any case, if you managed to pick up on a couple of the concepts, I think it was a very worthwhile discussion. Even the blackbody Planck function contains plenty of detail that can take a lot of time to digest.