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Question about time dilation in the schwarzchild metric?

by zeromodz
Tags: dilation, metric, schwarzchild, time
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zeromodz
#1
Apr11-11, 09:42 AM
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My first question is the following. Does the radial component of the schwarzchild metric account for just the radius of the body in study or is it the distance between the body and the observer, where the body is treated as a singularity (Point mass particle)?

My second question is about how the metric expresses time dilation and length contraction.

c^2dτ^2 = -(1 - Rs / R)c^2dt^2 + ((1 - Rs / R)^-1)dR^2 + R^2(θ^2 + sinθ^2dΦ^2)

(1 - Rs / R) < 1, so multiplying anything times it will give you a smaller number. So I understand the length contraction part of it, where R is the distance seen by the observer due to gravitational length contraction, but the proper length is longer so we divide to get a larger number. What about time though, shouldn't dt < dτ? because time slows down due to gravity. Why isn't (1 - Rs / R) being divided by time? Thanks
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bcrowell
#2
Apr11-11, 10:50 AM
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Hi, zeromodz,

In any metric, when you write down the line element, τ can be interpreted as the proper time of an observer (assuming the world-line we're talking about is timelike). For an observer low down in a gravity well, we expect clocks to run slowly relative to those outside, so we expect dτ to be smaller for them. This is how the Schwarzschild metric, written in the Schwarzschild coordinates, behaves.

You have to watch out because in GR, typically coordinates have no built-in physical meaning whatsoever. The Schwarzschild t coordinate can be interpreted as the proper time of an observer far away and at rest, as you can verify by plugging in a large r and dr=0 in the metric. This is the only reason it's valid to compare t and τ.

Ben


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