Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Voyager I, as an example, taking account gravity
and setting aside effects of speed as cause of time dilation.

If it is very far away from earth and sun, so then there must be a difference in the spacetime curvature there in the ship compared with here in earth, It means a detectable difference between our local clocks and its onboard clocks.

Imagine a signal transmision was designed to be at 1 byte for second at local Voyager clock

Should we receive it at higher and higher rates ?

( because of solar system gravity decrease as it(Voyager) moves away and it will keep decreasing while it doesn't reach a middle point between another massive object)

Or the middle Voyager-Earth light path would compensate the effect, making the high rates generation from low curvature zones being delayed enough for us to receive it at same rate that it was generated?

share|improve this question
add comment

1 Answer

There is in principle a "gravitational blueshift" for signals traveling from Voyager to us. The data rate we receive will be higher than the data rate transmitted by a factor $(1+\Delta\Phi/c^2)$, where $\Delta\Phi$ is the difference in Newtonian gravitational potential between the locations.

(Of course, this formula is only valid in weak gravitational fields, where it makes sense to talk in terms of Newtonian gravitational potentials.)

If I'm not mistaken, $\Delta\Phi/c^2\approx 10^{-8}$ for this sort of system, so the shift is quite small in practice. In particular, it's much smaller than than the ordinary Doppler effect due to the fact that both Voyager and Earth are moving. A motion at a speed of just 3 m/s would cause a Doppler shift as large as this gravitational shift, and both bodies are moving much faster than that.

share|improve this answer
1  
For the purposes of plugging into the above, taking infinity to be the zero of gravitational potential, the surface of the earth is around 53 km/s or $\Delta \Phi = 1300\text{ (km/s)}^2$ down the well. –  dmckee May 16 '11 at 19:39
4  
And that is exactly the source of the blue shift. Light falls down the well, but instead of accelerating (it cannot since $c$ is constant) it picks up energy in the form of frequency. –  ja72 May 16 '11 at 19:44
    
I have some questions. Is that equation an aproximation? (beyond that all are), I mean, it doesn't model gravity for all solar system (much less all universe), it must be for a spherical mass, perhaps it's not easy to calculate the potential, it could be more complex, then any Vogayer or similar traveler that send us periodic signals is a gravity probe, and if suddenly the signals come quickly we could suspect there is some -gravitational potential valley-, is this feasible? –  HDE May 17 '11 at 14:21
2  
It's a weak-field approximation. That is, it applies only when gravity is, in some appropriate sense, "nearly Newtonian." I don't immediately remember all the details of what has to be assumed, but it's something like (1) all masses moving nonrelativistically, (2) $\Phi/c^2\ll 1$ over the region of interest. So it does work for the entire solar system, for example. So yes, if you can measure this effect, you're probing the gravitational potential. In practice it's not a great way to probe gravity, though, because the effect is small compared to other signals such as Doppler shifts. –  Ted Bunn May 17 '11 at 17:03
1  
I'm sorry, but I don't understand your last comment. Anyway, here's what I'm saying. If the distribution of mass is complicated, then the function $\Phi({\bf r})$ is complicated, but it's still true that the time dilation factor is $1+\Delta\Phi/c^2$. The validity of this depends on the fields being weak and the matter being nonrelativistic, but not on any assumption about the geometry being simple or symmetric or anything like that. The proof of this result is found in general relativity textbooks. I like the one by Schutz for this sort of thing. –  Ted Bunn May 17 '11 at 20:22
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.