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Unformatted text preview: PHYS5523 Theory of Relativity Test 1: Solutions September 20, 2011 Problem 1: Circular Motion and Time Dilation. In spherical polar coordinates the four acceleration and four velocity of this obect must take on a particular simple form, given that we are dealing with circular motion in a plane. Thus the velocity in the radial and directions must be zero and we are left with vectoru = ( u t , , ,u ) (1) Similarly the object feels an acceleration in the radial direction, but it otherwise under the impression that it is stationary, thus in its own frame vectora = (0 ,g, , 0) (2) We know that vectoru vectoru = u u = 1 and in our coordinates = diag ( 1 , 1 ,r 2 ,r 2 sin 2 ) (3) Thus u t 2 = 1 + r 2 sin 2 u 2 (4) In order to compare u with g we can simply note that vectora = dvectoru d (5) Looking at only the radial part of this expression, and recalling that there is no motion in the radial direction or out of the plane of motion, hence r = = r = = 0 and cos = 0 since we are in the equatorial plane a r = g = r sin 2 (6) and provided we are careful to differeniate with respect to proper time, = u and so u = g r sin 2 (7) Thus u t 2 = 1 + r 2 sin 2 ( g r sin 2 ) = 1 gr (8) 1 Therefore, and converting to conventional units u t = dt d = radicalbig 1 gr/c 2 (9) Two clocks in this frame of reference which are rotating together rigidly and therefore not moving with respect to each other in that rotating frame should have the same coordinate time. Therefore the ratio of their proper times, if one is at radius R and the other is at the center (r=0) is d R d o = 1 / radicalbig 1 gR/c 2 (10) Now consider an outside inertial observer looking at these two clocks. One is not moving (the center one) the other is moving with linear speed v at any given instant (only the direction of the motion changes over time). Thus thisgiven instant (only the direction of the motion changes over time)....
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