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The Schwarzschild Solution
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To obtain an equation for the trajectory of a light ray, we differentiate with respect to a second time to get rid of the constant e2 . This gives u u + u 3mu 2 = 0 Dividing through by u , we obtain the nal result u + u = 3mu 2 (10.54)
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The standard procedure is to solve this equation using perturbation methods. First we set = 3m and then try a solution of the form u = u 0 + u 1 + O 2 Ignoring the higher order terms, we have u = u 0 + u 1 u = u 0 + u 1 Now, ignoring terms that are second order and higher, we have 3mu 2 = u 2 u 2 . Inserting these results into (10.54), we obtain 0 u 0 + u 1 + u 0 + u 1 = u 2 0 We now equate terms by their order in . We start with u0 + u0 = 0 The solution of this equation is given by u 0 = A sin + B cos . Without loss of generality, we can choose our initial conditions so that B = 0 and u 0 = A sin . This equation represents straight-line motion since r = 1/u, and using y = r sin in polar coordinates, we can write this as 1/A = r sin = y. Therefore, the constant A represents the distance of closest approach to the origin. Next, we equate terms that are rst order in . u 1 + u 1 = u 2 = A2 sin2 0 The homogeneous equation is the same as we had before u1 + u1 = 0 (10.55)
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Fig. 10-3. Lowest order representation of the trajectory of a light ray. The light ray follows the straight path represented by the dashed line. The distance r0 is the distance of closest approach to the origin.
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with solution u H = B sin + C cos . Without loss of generality, we take 1 B = 0. For the particular solution, we guess that u p = D sin2 + E cos2 ; therefore, differentiating we have u p = 2D sin cos 2E cos sin u p = 2D cos2 2D sin2 2E cos2 + 2E sin2 and so, we have u p + u p = D sin2 + 4D sin2 = 3D sin2 u p + u p = 2D cos2 2D sin2 2E cos2 + 2E sin2 +D sin2 + E cos2 = 2D cos2 D sin2 E cos2 + 2E sin2 Now the particular solution must satisfy u p + u p = A2 sin2 and this only be true can if 2D E = 0, leaving us with D sin2 + 4D sin2 = 3D sin2 Using u p + u p = A2 sin2 , we conclude that D = A2 /3. And so the particular solution becomes u 1 = D sin2 + E cos2 =
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A2 2 A2 sin + 2 cos2 3 3
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A2 A2 1 cos2 + 2 cos2 3 3 A2 A2 + cos2 = 3 3 The complete rst-order solution is then u1 = A2 1 + K cos + cos2 3 (10.56)
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where K is another integration constant. Putting everything together and using = 3m, the complete solution (which is an approximation) for the trajectory of a light ray is u = u 0 + u 1 = A sin + mA2 1 + K cos + cos2 (10.57)
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The mA2 will cause a de ection of the trajectory from a straight-line path. In an astrophysics situation, a light ray originating from a distant star approaches the Sun along an asymptotic straight line, is de ected a small amount by the Sun s gravitational eld, and then heads off into the distance along another asymptotic straight line. The asymptotes correspond to u = 0 and are parallel to the x-axis. We can take = (where is some small angle) and use the small angle approximation sin , cos 1 to write (10.57) as u A + mA2 (2 + K ) We can de ne a new constant = 2 + K , and setting this equal to zero, we nd (see Fig. 10-4) = m A The minus sign indicates that the de ection is towards the Sun. Recalling that the constant A is inversely related to the straight-line distance we found above, we can write m = r0 The total de ection is given by = 4m r0
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