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

[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?