Speed of a star in Special Relativity

[RafaPhysics]

Homework Statement

I've been trying to solve this problem but I'm pretty sure there's something more that I could do, but I can't see it.

Relevant Equations

$$v_r=u sin\frac{2\pi}{T}\left(t-\frac{R}{c}+\frac{Rv_r}{c^2}\right)$$

Hey, I have this problem from the Special Relativity by AP. French . Exercise 3.3, Chapter 3.
The figure shows a double-star system with two stars, A and B, in circular orbits of the same period T about their center of mass. The earth is in the plane definied by these orbits at a distance R of many light-years. Let the speed of A in its orbit be u; then at any instant it has a velocity $v_r(=ucos\theta)$ along the line from the double-star system to the earth. When light emitted from A reaches the earth, its observed Doppler shift (change of wavelength of characteristic spectral lines) reveals the value of vr at the instant of emission.
If the speed of light from A to the earth were modified by the motion of A, so as to be equal to $c+v_r$, show that the value of $v_r$, as inferred from spectroscopic observations on earth, would appear to be varying with time in accordance with the following equation if $u\ll c$: It's the equation above.
I've tried with the doppler efect, but I can't still see when It turns to $sin$, there's something more but simply I can't see it. Please I need help.

 

[PeroK]

I guess only AP French knows what laws of physics apply and which do not in this case! I'd assumed he means using the classical Doppler shift, without time dilation. I.e. assuming Newtonian physics with no invariant speed. Is that what you did?

 

[RafaPhysics]

I guess only AP French knows what laws of physics apply and which do not in this case! I'd assumed he means using the classical Doppler shift, without time dilation. I.e. assuming Newtonian physics with no invariant speed. Is that what you did?

Yes, I did it. But I only come to the expression of the Doppler efect, but from that point to the expression above there's still a world of difference.

 

[PeroK]

Yes, I did it. But I only come to the expression of the Doppler efect, but from that point to the expression above there's still a world of difference.

Why did you choose the AP French book?

I'm must admit, I'm not sure what he's assuming. Perhaps, if we assume that $f_s$ is the frequency at the source, then the crests of a light wave would be emitted at time $\frac 1 {f_s}$ apart. And, in that time, $v_r$ would increase by a tiny amount. The next crest would catch up with the one before it. That would explain why the distance $R$ gets into the equation. If $R$ was too large, then the light wave crests would arrive out of sequence! The assumption $u \ll c$ might be there to ensure that the crests arrive in the correct order?! Or, maybe not! Maybe the observation of adjacent crests coincide?!

If you are not too committed to French, I'd try another book. In any case, I'd be tempted to skip this question.

 

[PeroK]

PS and it's not clear what the spectroscopists assume. Do they assume the relativistic Doppler shift? Is that the idea?

 

[RafaPhysics]

PS and it's not clear what the spectroscopists assume. Do they assume the relativistic Doppler shift? Is that the idea?

I think they assume that the velocity of ligth depends of source motion, like It was Galilean, I had an idea, I have to take an infinitesimal time interval, after that take an infinitesimal distance interval, and then divide it by the already calculated time interval, but it still doesn't give the result.

 

[PeroK]

I think they assume that the velocity of ligth depends of source motion, like It was Galilean, I had an idea, I have to take an infinitesimal time interval, after that take an infinitesimal distance interval, and then divide it by the already calculated time interval, but it still doesn't give the result.

It's not to do with infinitesimals For the emission event:
$$\frac{dv_r}{dt} = \frac{d}{dt}(u\cos \theta) = -u(\sin \theta)\frac{d\theta}{dt} = -u(\frac{2\pi}{T})\sin \theta$$$$\Delta v_r = \frac{dv_r}{dt}\Delta t = \frac{dv_r}{dt}\frac 1 {f_s} = $$For the observation event:
$$t_1 = \frac{R}{c + v_r}, \ t_2 = \frac{1}{f_s} + \frac{R}{c +v_r + \Delta v_r}$$$$f_o = \frac{1}{t_2 - t_1}$$And so on ...