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cragar

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and the t in v=gt would be the time in my free-fall frame. Then once I knew v , I could use the standard time dilation equation from SR.

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- Thread starter cragar
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cragar

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and the t in v=gt would be the time in my free-fall frame. Then once I knew v , I could use the standard time dilation equation from SR.

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A.T.

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No. Gravitational time dilation is different from movement time dilation. Here is the formula:If I wanted to figure out the gravitational time dilation at different points in a the field around earth, could I just use v=gt. and find the speed that I would be moving if I went into free-fall from point x to point y. To figure out the time difference between x and y?

http://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere

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cragar

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PAllen

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Because both curvature and motion contribute to clock difference between a source and receiver. Imagine a receiver on hovering high above earth. Imagine a bunch of clocks at some place on earth: sitting on the ground, thrown upwards, kicked sidway, etc. The receiver will see a different rate for each of the source clocks, influenced by their difference in motion and by the curvature of spacetime between them and the receiver.

In a situation like this (essentially static gravity), we call the difference observed for the clock sitting on the ground gravitational time dilation. Then the different clocks moving relative to this can be analyzed as an additional motion effect.

There is a way to treat all the cases in a uniform way a as a function of source and target world line and the intervening geometry. However, I don't know if you have the background for it: do you know what parallel transport of 4-vectors in curved spacetime is?

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cragar

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PAllen

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If you want to use something from SR for gravitational time dilation, what you should use is SR doppler rather than SR time dilation. Then, the parallel transport method I alluded to applies - if you compute the tangent vector (4-velocity) of a ground stationary observer, and parallel transport along a light path (null geodesic) to a higher stationary observer world line, you find the two vectors (transported tangent; and tangent of higher observer world line) are not parallel, implying a local relative motion that is equivalent to the gravitational effect. Then, if you apply the SR doppler formula, using this local relative velocity of the source along with the local light propagation vector, you would get a correct figure for gravitational redshift (= what is called gravitational time dilation).

- #7

grav-universe

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cragar said:I was just wondering if I Could find the speed one would be at if he was in free-fall until he got to the other point, And use that speed to find the time dilation using SR equations.

Actually, if you start with a scalar speed v of a particle as measured by a local static observer at r and freefall, regardless of the path, to radius s with resulting locally measured scalar speed u, the ratio of the kinetic time dilations will equal the ratio of the gravitational time dilations, so that

sqrt(1 - (v/c)^2) / sqrt(1 - (u/c)^2) = sqrt(1 - 2 m / r) / sqrt(1 - 2 m / s)

This comes from conservation of energy, whereby the quantity E_r z_r = m c^2 sqrt(1 - 2 m / r) / sqrt(1 - (v/c)^2) is conserved for a particular freefalling particle, where z_r is the local gravitational time dilation. The same is true for frequencies of light, although massless, giving E_r z_r = h f_r z_r. So the smaller the gravitational time dilation, the greater the locally measured frequency of light.

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