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My question is very simple, and most likely a stupid one:

One observer is at a point in space were the gravitational force form massive bodies (or a single massive body) cancel each-other out. The second observer is in another hypothetical situation where there exists no massive bodies and therefore no gravity. My question is this, is there a relativistic time difference between the two?

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Think in terms of potentials rather than force fields. –  leftaroundabout Jul 5 '11 at 23:11
It is not a stupid question at all! A free falling or an orbiting observer about a point mass feels no force too, but his clock is anyway slower. It is the gravitational potential in the expression of the metric what matters, and an equilibrium point between two planets is a local maximum. Tipically, an extense object in that place suffers tidal forces. It is a very interesting question! –  Eduardo Guerras Valera Nov 6 '12 at 2:03
Sorry, I rather mean a saddle point, not a maximum. –  Eduardo Guerras Valera Nov 6 '12 at 7:41

4 Answers 4

up vote 12 down vote accepted

In the "weak field limit" where the graviational forces are small (such as anything in the solar system, and basically anything not right next to a black hole), the time dilation relative to a distant observer is:

$\Delta T/\Delta T_0 = 1 - \Phi/c^2$.

here $\Delta T_0$ is the time elapsed for an observer at infinity, $\Delta T$ is that time elapsed at some point in the system, and $\Phi$ is the gravitational potential at that point. As @leftaroundabout stated correctly the important factor is the gravitational potential not gravitational forces. Since potentials add, instead of cancelling like the forces, we get twice the time dilation with two planets than we get with one, as @Jim Graber said.

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In this context it's probably worth adding that this formula assumes units where the speed of light $c=1$. In other systems of units, replace $\Phi$ by $\Phi/c^2$. –  Ted Bunn Jul 6 '11 at 14:10
@Ted Bunn: Thanks for pointing that out. I changed it so the factor of $c$ is there explicitly. –  BebopButUnsteady Jul 6 '11 at 16:50
And $\Phi$ is set to be 0 at the observer at infinity. –  Leos Ondra Nov 6 '12 at 12:22

The gravitational force may be zero at the mid-point, but the spacetime itself is still stretched out pretty strongly - twice as strongly in fact - and that's all that matters.

So, yes, time is still skewed.

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Yes. Good example: First observer right in the middle between two identical massive bodies. Second observer far away. In the weak field limit, the redshift is twice as much as if there were only one massive body and otherwise the same set up.

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I haven't taken physics since high school, is there a formal solution to this problem? –  brysgo Jul 5 '11 at 23:24
I was really confused by your answer, and I think it's because you say "Yes" when you should say, and your answer supports, "No". The question asks if 2 gravitational fields can cancel each other out, but we only know of gravity that is additive and your answer is 2x time the redshift, not 0x as predicted by the question. –  AlanSE Jul 6 '11 at 2:03

A good place to look is Wikipedia.

According to that article, if we suppose that this metric field is stationary then the gravity redshift is

$$1 + z = \biggl[\frac{g_{tt}(\text{receiver})}{g_{tt}(\text{sender})}\biggr]^{\frac{1}{2}} = \frac{f(\text{sender})}{f(\text{receiver})}$$

So there is no redshift under the conditions described because

$$\frac{g_{tt}(\text{receiver})}{g_{tt}(\text{sender})} = 1$$

Note, that the formulas are different for cosmological redshifts where space as a whole has accelerating expansion. The above problem is for the peculiar positions inside two local regions of space neglecting the cosmological expansion as a small effect.

See http://www.physics.uq.edu.au/download/tamarad/.

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The question refers to gravitational fields (in the Newtonian sense) cancelling, but nothing in this answer has anything to do with gravitational fields. –  Ben Crowell Aug 28 '13 at 5:53

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