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.

The theory of relativity makes very precise predictions about how an object's motion through space-time affects the passage of time for both the object and observers in other frames of reference.

I am curious, however, whether this motion is equivalent to changing / fluctuating gravitational fields? What if instead of an object moving through space/time we instead had the same object exposed to shifting gravitational fields, similar to what it would experience if it were actually moving relative to other masses? Can it be shown that the same set of effects would still exist from the object's frame of reference (eg time dialation, length-contraction, etc)?

share|improve this question
    
How can you replace a free motion with an accelerated motion (in a gravitational field) for the sake of "the same" time dilation? How about the energy-momentum evolution? –  Vladimir Kalitvianski May 15 '11 at 19:19
add comment

1 Answer 1

Absolute motion is a very, very tricky concept in General Relativity. You can always change coordinates to a coordinate system where an arbitrary path is 'stationary'${}^{1}$, thereby embedding the dynamics into the spacetime, and removing them from the matter of interest.

Furthermore, since measurable quantities are usually either scalars formed by contracting $4-D$ quantities, or coordinate invariant integrals, the machinery of differential geometry will guarantee that all of the same effects will happen in any coordinate system. What you do, then, is choose a set of coordinates to make the calculation that you're doing look simpler. So, for instance, if we're studying the expansion of a flat universe, we can look at it in the so-called "comoving coordinates":

$$ds^{2} = -dt^{2} + a(t)^{2}d^{3}{\vec x}$$

where $d^{3}{\vec x}$ is the three-metric of flat $3-D$ space. Here, you see the explicit expansion and contraction of the $3-D$ space by multiplying by the factor of $a(t)$, which governs the expansion and contraction of the space. Also, it is readily apparent in this system that every person in the universe can measure their clocks in the same way--apparently an expanding spatially flat universe has a universal time coordinate. It looks like space is expanding. But, you can also look at this spacetime using a different radial coordinate: $R=a(t)r$. If you do this, then the metric becomes:

$$ds^{2} = -\left(1-\left(\frac{\dot a}{a}\right)^{2}r^{2}\right)dt^{2} - 2\,dt\,dR \left(\frac{\dot a}{a}\right)r + d^{3}{\vec x}$$

Now, it looks like space isn't expanding at all--there is no time dependence in the spacelike 3-metric! But what have we changed? Well, we've introduced terms that look like a motion into our metric--the factor $\left(1-\left(\frac{\dot a}{a}\right)^{2}r^{2}\right)dt^{2}$ is associated with a time dialation effect, while the factor $2\,dt\,dR \left(\frac{\dot a}{a}\right)r$ will give you a doppler effect--in this coordiante system, we can describe the distant galaxies as moving away from us on top of a fixed $3-D$ metric.

Which picture is right? Both are. Any quantities that we calculate will give us the exact same answers, thanks to the rules of differential geometry (which I won't get into here). All we're doing is using different variables to describe the effects--and depending on which variables you use, different effect become more obvious.

${}^{1}$The easiest way to do this is by finding the path of the particle of interest, and finding a set of "parallel" curves, and then choosing their proper time as your time coordinate, but there are other ways, certainly.

share|improve this answer
    
By the way, the time dilation effect cited above is precisely the redshift of distant galaxies. –  Jerry Schirmer May 18 '11 at 15:34
add 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.