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Just a quick question here about red/blueshift in special relativity: if a source is emitting blackbody radiation in its rest frame S, is the spectrum as observed in another inertial frame S?€? also a blackbody spectrum?

Yes, a blackbody spectrum transforms to another blackbody spectrum. The conversion factor for the absolute temperature assocated with the observered spectrum is the same number as the relativistic Doppler frequency shift: ??(1+??cos??).
(Where ?? is v/c, ?? is sqrt(1 - ????), and ?? is the angle between the velocity vector v and the position of the light source, as seen in the rest frame).

Grant

granthutchison wrote:Yes, a blackbody spectrum transforms to another blackbody spectrum. The conversion factor for the absolute temperature assocated with the observered spectrum is the same number as the relativistic Doppler frequency shift: ??(1+??cos??).
(Where ?? is v/c, ?? is sqrt(1 - ????), and ?? is the angle between the velocity vector v and the position of the light source, as seen in the rest frame).

Grant

...and if Chaos now asks why this is so?

Bye Fridger

t00fri wrote:...and if Chaos now asks why this is so?

Then I'll answer him.

Grant

Thanks. I'd got the transformation law for an individual frequency from transforming the 4-vector (?‰/c,k) but I hadn't managed to convince myself that a blackbody distribution mapped to another blackbody distribution, though it seemed like that would be the case. Temperature change would then come from combining Wien's displacement law and the transformation of the peak frequency.

chaos syndrome wrote:... but I hadn't managed to convince myself that a blackbody distribution mapped to another blackbody distribution, though it seemed like that would be the case.

One way to convince yourself without wading through the maths is to look at a log-log plot of black body spectra for various temperatures.
It's clear that all the curves are same shape, and differ only in their horizontal and vertical displacement. Since it's a log plot, we can see immediately that applying a constant multiple to all the frequencies in a given spectrum (as we do with Doppler) will simply move the curve to a new horizontal location without changing its shape. Let's call the relativistic Doppler multiplier for frequency ?·. If we multiply frequencies throughout by ?·, then the peak frequency will move from a location corresponding to the rest-frame temperature T, to a location corresponding (according to Wien) to a temperature of ?·.T.
But what about the necessary vertical scaling? This is proportional to the total flux, of course, which is in turn proportional to T^4. If our new apparent temperature under relativistic Doppler is to be ?·.T, then our flux F needs to be transformed to ?·^4.F. And it is: each photon is received in the moving frame with an energy ?· times its rest frame energy; such photons are received at ?· times the rest-frame rate; and the effect of relativistic stellar aberration alters the angular area of a radiant source by a factor of ?·^2. So the apparent flux per steradian is ?·^4 times the rest-frame flux.
We've therefore consistently accounted for the preservation of curve shape, the shift in position against the frequency axis and the shift in position against the flux axis: all the appearances in the moving frame are of a black body with temperature ?·.T.

If you do want to do the underlying maths, you'll find it's possible to recast Planck's equation for the black body spectrum (by inserting Stefan-Boltzmann and then moving terms around) so that the shape of the curve is defined by an expression in (h??/kT), and its height depends on T^4: playing with ?? and T by plugging in the ?·'s then reproduces the behaviour of the graphs in my link.

Grant