View Full Version : length contraction in gravitational field

2009-Aug-24, 05:13 PM
Okay, so according to a distant observer, the speed of light as it travels tangent to a gravitational field is c sqrt(1-r/R) and it travels radially at c (1 - r/R). If a distant observer were to send pulses of light to an observer on the planet with the frequency of f, so a time between pulses of 1/f, if the first pulse is sent at T=0, the light will take a time of t to reach the observer, so reach him at T1=t according to the distant observer. A second pulse is sent at T=1/f, and travels the same path as the first in a time of t, so reaches the observer at T2=1/f + t. According to the distant observer, then the pulses are still reaching the observer with a time between pulses of T2-T1 = (1/f + t) - t = 1/f, so with the same frequency f. The observer's clock, however, is time dilated by sqrt(1 - r / R), so the frequency according to the observer on the planet is that much faster, so f' = f / sqrt(1 - r/R).

Now when the light is travelling radially, the emitting observer measures the same frequency for the light as it reaches the planet observer but with a reduced speed for the light itself, so w_r = (c (1 - r/R)) / f = (1 - r/R) w. The planet observer measures a local speed for the light of c, but a greater frequency, so w' = c / (f / sqrt(1 - r/R)) = sqrt(1 - r/R) w. That means that the same distance between pulses that the distant observer measures of (1 - r/R) w is measured as 1 / sqrt(1- r/R) greater by the planet observer, which means the planet observer measures distances as elongated by 1/ sqrt(1 - r/R) radially.

In the tangent direction, the frequency as it is received by the planet observer will be the same according to the distant observer and greater by 1/sqrt(1 - r/R) according to the planet observer due to the time dilation. The wavelength as it is received is, according to the distant observer, w_r = (c sqrt(1 - r/R)) / f = sqrt(1 - r/R) w. According to the planet observer, it is w' = c / (f / sqrt(1 - r/R)) = sqrt(1 - r/R) w. So distances are measured the same tangentially.

So distances are measured as 1 / sqrt(1 - r/R) greater radially and the same tangentially by the planet observer than the distant observer. Does this mean that the planet observer will see his own planet as having the same width but enlogated from where he is standing to the other side of the planet through the center of gravity, producing a sort of oval shape that will change as he moves from one spot to another on the planet, or will he still see his own planet as spherical, but the distant observer will see it flattened along the line of sight, which changes as the distant observer changes position?

2009-Aug-25, 12:29 AM
I'm thinking that the planet should remain spherical to the distant observer, because otherwise the distribution of mass would make the gravity of the planet as found by integrating point masses different than just that coming from the center of the planet, and that is the same according to GR as far as I know. There is the precession, but a planet that is flattened according to a distant observer will produce less gravity, not more like precession does, and precession is a variable of the speed of the observer, so there would be no difference for the stationary distant observer we are considering in that respect anyway. So even if the planet is spherical according to the distant observer and oval shaped according to the one in the gravitational field, point masses will still distribute themselves evenly around the planet because the gravity is still the same for each of the point masses wherever they happen to be.