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As you know, Doppler effect can change the wavelength of the light.

Suppose there are two stars at same distance and same temperature. If one star is receding and the other star is approaching, their colors would look different.

What I want to know is, how probable is it that two stars show different colors because of Doppler shift? In other words, if two stars at same distance have different colors, say blue and red, then is it safe to assume that two stars have different surface temperature?

Actually, this was the question from the final exam for Wave & Heat.

Q. Stars emit radiation whose spectrum is very similar to that of an ideal blackbody. Two stars, identical in size, are at the same distance from us. One of the stars appears reddish in color while the other one looks very blue. Choose the correct statement.

(a) The red star is hotter.
(b) The blue star is hotter.
(C) They could both have the same surface temperature.

I read that some stars near the galactic center moves very fast. http://en.wikipedia.org/wiki/S2_(star)

If we only consider Planck distribution, only (b) is possible. I'm wondering whether (c) is also a realistic case, or it is only conceptually possible.

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It is conceptually possible for a velocity of around $0.25c$(red star turns blue). Whether it is observed or not, it's a different story altogether. –  Cheeku Dec 29 '13 at 14:04
    
I'd say(b) is the correct answer. For Doppler shift to make a significant effect (as described), the relative velocity of the two stars has to be close to that of light. This is much larger than any obvsered, even for S2. –  Walter Dec 29 '13 at 16:18
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1 Answer

up vote 4 down vote accepted

As you correctly mention, there are rare stars which move particularly fast. Check also hypervelocity stars section here, where a stellar population moving at around $10^3 \textrm{km/s}$ is described (by the way, these velocities were most likely produced in the galactic center). Note also, that these stars are too fast to stay in the disk for long times. As another sidenote, galactic center in not accessible in visual due to dust extinction.

Redshifts are calculated particularly easy: $$z=\dfrac{v}{c}$$, where $v$ is the velocity along the line of sight, $c$ is the speed of light, $z=\dfrac{\delta\lambda}{\lambda}$ and $(v/c)^2\approx 0$ (which holds even for hypervelocity stars).

Because hypervelocity stars are rare, they most likely can be found only close to regular stars. Yet, being very optimistic, take $v=3\cdot 10^3 \textrm{km/s}$, and you obtain that $\dfrac{\delta\lambda}{\lambda}=10^{-2}$ under very favourable conditions. In general it is not small, but when it comes to colors, the change is negligible.

Say, green light at $\lambda=555 \textrm{nm}$ would now have $\lambda=550 \textrm{nm}$ or $560 \textrm{nm}$ in wavelegth, i.e. would still be very green. So Doppler effect has no bearing on your problem.

What has possible bearing, however, is dust reddening. One of the two stars might be obscured by the dust, unlike the other one, and look reddish. So, both b) and c) are possible answers.

P.S. And actually, if a star "appears reddish", it is most likely dust extinction. Black body temperature colors are either blue, bluish, white, yelowish, yellow, orange or red.

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