Derivation of the Geodesic Equation and De¯ning the Christo®el Symbols Dr. Russell L. Herman March 13, 2008 We begin with the line element 2 ® ¯ ds = g®¯dx dx (1) where g®¯ is the metric with ®; ¯ = 0; 1; 2; 3. Also, we are using the Einstein summation convention in which we sum over repeated indices which occur as a subscript and superscript pair. In order to ¯nd the geodesic equation, we use the variational principle which states that freely falling test particles follow a path between two ¯xed points in spacetime which extremizes the proper time, ¿. The proper time is de¯ned by d¿ 2 = ¡ds2: (We are assuming that c = 1.) So, formally, we have Z B p Z B q 2 ® ¯ ¿AB = ¡ds = ¡g®¯dx dx : A A In order to write this as an integral that we can compute, we consider a parametrized worldline, x® = x®(σ); where the parameter σ = 0 at point A and σ = 1 at point B. Then, we write Z 1 · ® ¯ ¸1=2 Z 1 · ® ¸ dx dx dx ® ¿AB = ¡g®¯ dσ ´ L ; x dσ: (2) 0 dσ dσ 0 dσ £ ¤ dx® ® Here we have introduced the Lagrangian, L dσ ; x : We note also that d¿ L = : dσ Therefore, for functions f = f(¿(σ)), we have df df d¿ df = = L : dσ d¿ dσ d¿ We will use this later to change derivatives with respect to our arbitrary pa- rameter σ to derivatives with respect to the proper time, ¿: 1 Using variational methods as seen in classical dynamics, we obtain the Euler- Lagrange equations in the form µ ¶ d @L @L ¡ + = 0: (3) dσ @(dxγ /dσ) @xγ We carefully compute these derivatives for the general metric. First we ¯nd @L 1 @g dx® dx¯ = ¡ ®¯ @xγ 2L @xγ dσ dσ L @g dx® dx¯ = ¡ ®¯ : (4) 2 @xγ d¿ d¿ The σ derivatives have been converted to ¿ derivatives. Now we compute µ ¶ @L 1 dx¯ dx® = ¡ g ± + ± @(dxγ /dσ) 2L ®¯ dσ αγ dσ ¯γ µ ¶ 1 dx¯ dx® = ¡ g + g 2L γ¯ dσ αγ dσ 1 dx® = ¡ g : (5) L αγ dσ In the last step we used the symmetry of the metric and the fact that ® and ¯ are dummy indices. We di®erentiate the last result to obtain µ ¶ µ ¶ d @L d 1 dx® ¡ = g dσ @(dxγ /dσ) dσ L αγ dσ µ ¶ d dx® = L g d¿ αγ d¿ · ¸ d2x® dg dx® = L g + αγ αγ d¿ 2 d¿ d¿ · ¸ d2x® @g dx¯ dx® = L g + αγ αγ d¿ 2 @x¯ d¿ d¿ · µ ¶ ¸ d2x® 1 @g @g dx® dx¯ = L g + αγ + γ¯ : (6) αγ d¿ 2 2 @x¯ @x® d¿ d¿ Again, we have used the symmetry of the metric and the reindexing of repeated indices. Also, we have eliminated appearances of L by changing to derivatives with respect to the proper time. So far we have found that µ ¶ d @L @L 0 = ¡ + dσ @(dxγ /dσ) @xγ · µ ¶ ¸ d2x® 1 @g @g dx® dx¯ L @g dx® dx¯ = L g + αγ + γ¯ ¡ ®¯ : (7) αγ d¿ 2 2 @x¯ @x® d¿ d¿ 2 @xγ d¿ d¿ 2 Rearranging the terms on the right hand side and changing the dummy index ® to ±; we have µ ¶ d2x® 1 @g dx® dx¯ 1 @g @g dx® dx¯ g = ®¯ ¡ αγ + γ¯ αγ d¿ 2 2 @xγ d¿ d¿ 2 @x¯ @x® d¿ d¿ · ¸ 1 @g @g @g dx® dx¯ = ¡ αγ + γ¯ ¡ ®¯ 2 @x¯ @x® @xγ d¿ d¿ · ¸ 1 @g @g @g dx± dx¯ = ¡ ±γ + γ¯ ¡ ±¯ 2 @x¯ @x± @xγ d¿ d¿ dx± dx¯ ´ ¡g ¡® : (8) αγ ±¯ d¿ d¿ We have found the geodesic equation, d2x® dx± dx¯ = ¡¡® (9) d¿ 2 ±¯ d¿ d¿ where the Christo®el symbols satisfy · ¸ 1 @g @g @g g ¡® = γ± + γ¯ ¡ ±¯ : (10) αγ ±¯ 2 @x¯ @x± @xγ This is a linear system of equations for the Christo®el symbols. If the metric is diagonal in the coordinate system, then the computation is relatively simple as there is only one term on the left side of Equation (10). In general, one needs to use the matric inverse of g®¯: Also, you should note that the Christo®el symbol is symmetric in the lower indices, ® ® ¡±¯ = ¡¯± We can solve for the Christo®el symbols by introducing the inverse of the metric, gµγ ; satisfying µγ ¹ g gαγ = ±®: (11) ¹ Here, ±® is the Kronecker delta, which vanishes for ¹ 6= ® and is one otherwise. Then, µγ ® ¹ ® ¹ g gαγ ¡±¯ = ±®¡±¯ = ¡±¯: (12) Therefore, · ¸ 1 @g @g @g ¡¹ = gµγ γ± + γ¯ ¡ ±¯ : (13) ±¯ 2 @x¯ @x± @xγ 3.