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grav
2009-Nov-18, 01:30 PM
This Wiki link (a) shows the Wien displacement for a black-body spectrum that is used to find the peak wavelength and peak frequency of the CMB pressure. The problem is that when one finds the peak wavelength in terms of dP/dw and then finds the peak frequency in terms of dP/df and then finds the peak wavelength using that with w = c/f after changing all of the wavelengths to frequencies in the formula in this manner, then the graph for the same CMB pressure will look slightly different when plotted out each of these two ways and dividing the peak wavelength by the peak frequency will give a constant speed of light that is much less than c. Dividing c by the peak frequency, then, will give a different peak wavelength depending upon which way we do it. For instance, finding the peak wavelength directly for T = 2.72839 K gives around .0019 meters, while finding it in terms of frequency first and then using w = c/f gives .001062 meters.

However, at the very end of the link, it also provides a "compromise" number of 4, that between 5 for the wavelength and 3 for the frequency, which when used, provides the exact same graph for both and we find w * f = c. That gives a single peak wavelength of .001345 meters either way. I had found that using another method as well, and I think it makes the most sense, so wouldn't that be the most likely choice to use as the solution for the peak wavelength and frequency, then, since there is only one graph for one pressure, the same one for both? I know it would depend upon how the peaks are defined, but wouldn't this be the most obvious choice of definition, then, since it provides a single consistent peak wavelength instead of two possibilities depending upon whether we find it directly or through the peak frequency found in this manner?

tusenfem
2009-Nov-18, 02:23 PM

grav
2009-Nov-18, 03:10 PM
there is no wiki linkWhoops, sorry about that. :shifty: Here (http://en.wikipedia.org/wiki/Wien's_displacement_law) it is.

ngc3314
2009-Nov-18, 03:11 PM
This sounds like it may be a case of F-nu to F-lambda conversion . In transforming between representations of a spectrum in flux per unit frequency (F-nu) and per unit wavelength (F-lambda), not only does frequency need to be transformed into wavelength, but the transformation has to include d-nu/d-lambda since the bin size varies as (c/lambda^2). As a result, the peak in one representation falls in a different place than in the other.

Amber Robot
2009-Nov-18, 04:07 PM
This sounds like it may be a case of F-nu to F-lambda conversion . In transforming between representations of a spectrum in flux per unit frequency (F-nu) and per unit wavelength (F-lambda), not only does frequency need to be transformed into wavelength, but the transformation has to include d-nu/d-lambda since the bin size varies as (c/lambda^2). As a result, the peak in one representation falls in a different place than in the other.

Yeah, that never fails to trip up the students.

grav
2009-Nov-18, 04:19 PM
This sounds like it may be a case of F-nu to F-lambda conversion . In transforming between representations of a spectrum in flux per unit frequency (F-nu) and per unit wavelength (F-lambda), not only does frequency need to be transformed into wavelength, but the transformation has to include d-nu/d-lambda since the bin size varies as (c/lambda^2). As a result, the peak in one representation falls in a different place than in the other.I'm not sure what that means exactly. Since it is the same graph for the same pressure in both cases, wouldn't it be better to find the peak wavelength and peak frequency independently of dw and df altogether, gaining identical graphs for both where the peak wavelength correlates with the peak frequency with just w * f = c? Wouldn't that be the best way to make any real sense out of it and gain only a single associated peak wavelength / frequency?

Jeff Root
2009-Nov-18, 11:12 PM
I haven't made any attempt to understand this in depth, and my background
knowledge is sparse... but ...

Is this a consequence of the fact that Wein's law is really only a crude
approximation of what the peak should be? The whole Planck curve needs
to be considered, and Wein's law doesn't give that. It should give a value
close to the true peak, but there is no guarantee on how close it will be.

As I very crudely understand it.

-- Jeff, in Minneapolis

Spaceman Spiff
2009-Nov-19, 12:42 AM
I haven't made any attempt to understand this in depth, and my background
knowledge is sparse... but ...

Is this a consequence of the fact that Wein's law is really only a crude
approximation of what the peak should be? The whole Planck curve needs
to be considered, and Wein's law doesn't give that. It should give a value
close to the true peak, but there is no guarantee on how close it will be.

As I very crudely understand it.

-- Jeff, in Minneapolis

No, it's precise, for a blackbody of a single T.

The position of the Wien peak depends entirely on what units are being considered, and there are many.

F_nu vs. wavenumber, k = 2pi/lambda
F_nu vs. nu (frequency)
F_lambda vs. lambda
F_lambda vs. nu
nuF_nu vs. nu
nuF_nu vs. wavenumber
photon flux per unit whatever interval vs. whatever....

This is nothing other than the property of a mathematical function. To provide an example: the two most commonly quoted forms of the Wien peak are for F_nu vs. nu (frequency) and F_lambda vs. lambda. If you convert the peak frequency computed from the first, and then convert this frequency into wavelength (lambda = c/nu), it is not the same as the peak computed from the function of F_lambda vs. lambda.
:)

Amber Robot
2009-Nov-19, 12:46 AM
No, it's precise, for a blackbody of a single T.

Technically it does involve an approximation and is valid when h*nu >> kT.

grav
2009-Nov-19, 01:17 AM
I haven't made any attempt to understand this in depth, and my background
knowledge is sparse... but ...

Is this a consequence of the fact that Wein's law is really only a crude
approximation of what the peak should be? The whole Planck curve needs
to be considered, and Wein's law doesn't give that. It should give a value
close to the true peak, but there is no guarantee on how close it will be.

As I very crudely understand it.

-- Jeff, in MinneapolisYes, perhaps a few more specifics are in order. It is precise, but it is based upon a particular way the peak wavelengths and peak frequencies are determined, which give two different values for the same graph. The pressure is found with

P = Int 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1]

That is the pressure found using the wavelengths of light from zero to infinity over small intervals of dw. In terms of frequency, we have f = c / w and dw becomes

dw = w' - w = c/f' - c/f = c (1/f' - 1/f) = c (f - f') / (f f')

figuring for very small intervals of df, so dw = c (f' - f) / (f' f) = c df / f^2. The pressure in terms of frequency, then, is found with

P = Int 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1]
= Int 8 pi h c [c df / f^2] / [c / f]^5 / [e^(h f / (k T)) - 1]
= Int 8 pi h df f^3 / c^3 / [e^(h f / (k T)) - 1]

So when the peak wavelength is found by taking the greatest value for dP / dw, or 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1], we find the greatest value for that, the peak wavelength taken in these terms, always occurs at a wavelength of w_peak = (2.8977685×10^(−3) m·K) / T. The peak frequency is taken in terms of the peak value for dP / df, or 8 pi h f^3 / c^3 / [e^(h f / (k T)) - 1] using the frequency formula, giving a peak frequency of (5.879×10^10 Hz/K) * T. However, these peak values do not correlate with each other quite as they should in reference to the same peak of the same graph. For instance, when multiplied together, the peak wavelength and the corresponding peak frequency should give the speed of light, but they don't. They give only .5682 c regardless.

Now, when looking at the equations, I can understand where these slightly off definitions might come from. One sees dP = 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1] and would naturally determine that dP / dw = 8 pi h c / w^5 / [e^(h c / (k T w)) - 1] and figures they have determined a method for finding the peak wavelength in absolute terms by dividing out the derivatives, and probably not much else can directly be seen that can be done to determine this in any other way. However, when writing that into a computer program to integrate intervals of pressure using small intervals of wavelengths, I found that is not necessarily the case.

The simplest way to integrate it in a computer program is to just keep adding the intervals of pressure for very small intervals of dw, so each iteration climbs with w' = w + dw, and we have P = Int 8 pi h c (w' - w) / w^5 / [e^(h c / (k T w)) - 1] = Int 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1]. Finding frequency in the same way gives f' = f - df where P = Int 8 pi h (f - f') f^3 / c^3 / [e^(h f / (k T)) - 1] = Int 8 pi h df f^3 / c^3 / [e^(h f / (k T)) - 1].

But that is extremely slow, so a better or at least faster way is to multiply the wavelengths by some constant near unity each time, like z = 1.0001, so that w' = z w and therefore dw = w' - w = z w - w. That gives

P = Int 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1]
= Int 8 pi h c (z w - w) / w^5 / [e^(h c / (k T w)) - 1]
= Int 8 pi h c (z - 1) / w^4 / [e^(h c / (k T w)) - 1]

For frequency, where f' = f/z and so dw becomes c (f - f') / (f f') = c (f (1 - 1/z)) / (f (f/z)) = c (z - 1) / f, and we get

P = Int 8 pi h c dw / w^5 / [e^(h c / (k T w)) - 1]
= Int 8 pi h c [c (z - 1) / f] / (c / f)^5 / [e^(h c / (k T w)) - 1]
= Int 8 pi h (z - 1) f^4 / c^3 / [e^(h f / (k T)) - 1]

Then each of the peak values, instead of being found with dP / dw and dP / df, are instead found with dP / (z - 1) in both cases, giving 8 pi h c / w^4 / [e^(h c / (k T w)) - 1] and 8 pi h f^4 / c^3 / [e^(h f / (k T)) - 1], providing exactly corresponding values for dP with both as they are plotted out in a graph and the peak values for the wavelength and frequency correspond also where w_peak * f_peak = c. Basically it amount to the same thing as taking the greatest value for (dp / P) / (dw / w) and (dP / P) / (df / f) instead of just the derivatives alone, in order to find the peak values for w and f accordingly.

grav
2009-Nov-19, 01:48 AM
No, it's precise, for a blackbody of a single T.

The position of the Wien peak depends entirely on what units are being considered, and there are many.

F_nu vs. wavenumber, k = 2pi/lambda
F_nu vs. nu (frequency)
F_lambda vs. lambda
F_lambda vs. nu
nuF_nu vs. nu
nuF_nu vs. wavenumber
photon flux per unit whatever interval vs. whatever....