Doppler shifted blackbody spectrum

[E92M3]

You can know the temperature of a star by fitting a black body spectrum. BUt what if the star is moving with some radial velocity v? I worked out that:

$$I(\lambda_0,T)=\frac{8\pi h c}{\lambda_0^5}\frac{1}{e^{\frac{hc}{\lambda_0kT}}-1}$$
$$\lambda=\lambda_0\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$
$$I(\lambda,T)=I(\lambda_0,T)\frac{d\lambda_0}{d\lambda}$$

$$I(\lambda,T)=\frac{8\pi h c}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT} \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}}-1}$$
$$=I(\lambda,T')$$

where $$T'=T\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}}$$

Am I correct here? If so, how can we actually tell the temperature of stars?

 

[George Jones]

What is $T/T'$ for typical star speeds relative to us?

[edit]Also, can't spectroscopic analysis give the amount of the Doppler shift?[/edit]

 

[Dale]

You can know the temperature of a star by fitting a black body spectrum. BUt what if the star is moving with some radial velocity v? I worked out that:

$$I(\lambda_0,T)=\frac{8\pi h c}{\lambda_0^5}\frac{1}{e^{\frac{hc}{\lambda_0kT}}-1}$$
$$\lambda=\lambda_0\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$
$$I(\lambda,T)=I(\lambda_0,T)\frac{d\lambda_0}{d\lambda}$$

$$I(\lambda,T)=\frac{8\pi h c}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT} \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}}-1}$$
$$=I(\lambda,T')$$

where $$T'=T\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}}$$

That is an interesting idea. I don't think your third and fourth steps are correct, but I wonder if your basic premise is correct that a Doppler-shifted blackbody spectrum is itself a blackbody spectrum. It might be right.

Am I correct here? If so, how can we actually tell the temperature of stars?

Stars are not perfect blackbodies, they have characteristic emission spectra peaks on top of the blackbody spectrum. We can use the emission peaks to determine the Doppler shift. Then we can look at the Doppler-corrected blackbody portion of the spectrum to determine the temperature.

 

[twofish-quant]

That is an interesting idea. I don't think your third and fourth steps are correct, but I wonder if your basic premise is correct that a Doppler-shifted blackbody spectrum is itself a blackbody spectrum. It might be right.

The wikipedia article for "Black body" mentions that you have to do a solid angle correction.

Doppler-shifted blackbody spectrums are blackbodies, that's how we can talk about 3 kelvin background radiation. I vaguely remember a thermodynamic argument why a blackbody in one reference frame must be a black body in all reference frames.

Then we can look at the Doppler-corrected blackbody portion of the spectrum to determine the temperature.

To get precision measurements of stellar temperatures, people don't fit black body curves. What people do is to look at the strength of specific spectral lines and those change in very strong ways with respect to temperature.

 

[Dale]

To get precision measurements of stellar temperatures, people don't fit black body curves. What people do is to look at the strength of specific spectral lines and those change in very strong ways with respect to temperature.

That is interesting and it makes sense. The sharp spectral peaks are always going to be more reliable to measure than the rather broad blackbody spectrum. I was only thinking about measuring their frequency, but there is certainly additional information in their amplitude too.

Do you know if there is any phase information, or are the spectral lines incoherent?

 

[twofish-quant]

I was only thinking about measuring their frequency, but there is certainly additional information in their amplitude too.

There's *tons* of information in spectral lines.

Do you know if there is any phase information, or are the spectral lines incoherent?

I know that there is phase information in the general output of stars which is important with things like supernova. I don't know off hand if people use this for spectral lines.