0

I am trying to determine the frequency redshift when a signal is sent between two spacecraft orbiting a star at different radii. I have not been able to find a formula for this, I can only find formula for a signal emitted at a particular radius from a star and received at infinity. Both spacecraft are well outside the Schwartzchild radius, so I think I only need the Newtonian limit.

I think what I need is:

Fr = Fs/(1+(Pr-Ps)/c^2)

Fr = Frequency at receiver Fs = Frequency at source Pr = Gravitational Potential at receiver Ps = Gravitational Potential at source c = speed of light

I determine Gravitational Potential as: P = -GM/r

Is it correct to use the difference in Potential in this way? Is the rest of the equation correct?

I suspect there may be something wrong because if the receiving ship immediately repeats the signal back to the sending ship, the above formula does not result in the final signal having the same frequency as the original transmission, i.e. the red shift in one direction does not cancel the blue shift in the other direction. I would think it should.

Ron Smith
  • 3
  • You can relate them both to infinity, right? So what is the problem? – dmckee --- ex-moderator kitten Mar 14 '13 at 19:18
  • Sorry, I don't know what you mean by that. – Ron Smith Mar 14 '13 at 19:35
  • I mean the same thing that John did. You have asked how A is related to B, but you already know how each of A and B are related to C. As a result, the problem should be straight forward. You want to make a habit of noticing those kinds of chains of relationship because they are one of the core tools you use in solving physics questions (both textbook types and in the real world). – dmckee --- ex-moderator kitten Mar 14 '13 at 19:42

1 Answers1

1

You need to calculate the time dilation using the Schwarzschild metric. Fortunately this gives the nice simple result that the time, $t_r$, experienced by an observer at distance $r$ is related to the time, $t_\infty$, experienced by an observer at infinity by:

$$ t_r = t_\infty \sqrt{1 - \frac{2GM}{rc^2}} $$

This assumes that the spaceship isn't orbiting fast compared to the speed of light. The frequency measured by the observer at infinity is just:

$$ f_\infty = f_r\frac{t_r}{t_\infty} = f_r \sqrt{1 - \frac{2GM}{rc^2}} $$

Note that as $r$ tends towards the event horizon, $2GM/c^2$, the frequency at infinity tends to zero. This is what makes black holes black!

Between two spaceships at $r_1$ and $r_2$ the frequency shift is just the ratio of the time dilations:

$$ f_1 = f_2 \frac{t_2}{t_1} = f_2 \sqrt{ \frac{1 - \frac{2GM}{r_2c^2}}{1 - \frac{2GM}{r_1c^2}}} $$

and vice versa.

John Rennie
  • 355,118
  • Oops, I've noticed you did say you only needed the Newtonian limit. However, the full relativistic calculation is still pretty simple. – John Rennie Mar 14 '13 at 19:12
  • OK, I think I get most of this, but why is the frequency shift the ratio of the time dilations? – Ron Smith Mar 14 '13 at 20:00
  • Because frequency is just the number of events per unit time. Suppose the observer at infinity and the spaceship have something emitting pulses at 1Hz, and suppose because of time dilation $t_r$ = 0.5$t_\infty$. That means while the observer at infinity counts 10 seconds the spaceship counts only 5 seconds and emits 5 pulses. The observer at infinity counts 5 pulses in 10 seconds so the frequency has been shifted to 0.5Hz. – John Rennie Mar 14 '13 at 20:11
  • OK I get it now. I need to take this one step further. If there is another large gravitating body, e.g. a gas giant, how would I also include the effects of it? would I need to calculate the time dilation for the star and the gas giant separately then multiply, or do they need to be added somehow in the Schwartzchild metric? I suspect done separately then multiplied, but I am not sure. – Ron Smith Mar 14 '13 at 20:30
  • I see now that multiplication is wrong. I am now adding the Grav Potent of each body inside the Shwartzchild metric, i.e. replacing -GM/R with appropriate sum of potentials. I am pretty sure this is correct. – Ron Smith Mar 14 '13 at 23:31