Derivation of the Geodesic Equation and Defining the Christoffel Symbols

Derivation of the Geodesic Equation and Deﬁning the Christoﬀel Symbols

Dr. Russell L. Herman March 13, 2008

We begin with the line element

2 α β ds = gαβdx dx (1) where gαβ is the metric with α, β = 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 find the geodesic equation, we use the variational principle which states that freely falling test particles follow a path between two fixed points in spacetime which extremizes the proper time, τ. The proper time is defined by dτ 2 = −ds2. (We are assuming that c = 1.) So, formally, we have

Z B p Z B q 2 α β τAB = −ds = −gαβdx dx . A A In order to write this as an integral that we can compute, we consider a parametrized worldline, xα = xα(σ), where the parameter σ = 0 at point A and σ = 1 at point B. Then, we write

Z 1 · α β ¸1/2 Z 1 · α ¸ dx dx dx α τAB = −gαβ dσ ≡ L , x dσ. (2) 0 dσ dσ 0 dσ £ ¤ dxα α Here we have introduced the Lagrangian, L dσ , x . We note also that dτ L = . dσ Therefore, for functions f = f(τ(σ)), we have

df df dτ df = = L . dσ dτ dσ dτ We will use this later to change derivatives with respect to our arbitrary parameter σ to derivatives with respect to the proper time, τ.

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 ﬁnd

∂L 1 ∂g dxα dxβ = − αβ ∂xγ 2L ∂xγ dσ dσ L ∂g dxα dxβ = − αβ . (4) 2 ∂xγ dτ dτ The σ derivatives have been converted to τ derivatives. Now we compute µ ¶ ∂L 1 dxβ dxα = − g δ + δ ∂(dxγ /dσ) 2L αβ dσ αγ dσ βγ µ ¶ 1 dxβ dxα = − g + g 2L γβ dσ αγ dσ 1 dxα = − g . (5) L αγ dσ In the last step we used the symmetry of the metric and the fact that α and β are dummy indices. We diﬀerentiate the last result to obtain µ ¶ µ ¶ d ∂L d 1 dxα − = g dσ ∂(dxγ /dσ) dσ L αγ dσ µ ¶ d dxα = L g dτ αγ dτ · ¸ d2xα dg dxα = L g + αγ αγ dτ 2 dτ dτ · ¸ d2xα ∂g dxβ dxα = L g + αγ αγ dτ 2 ∂xβ dτ dτ · µ ¶ ¸ d2xα 1 ∂g ∂g dxα dxβ = L g + αγ + γβ . (6) αγ dτ 2 2 ∂xβ ∂xα dτ dτ 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 ¶ ¸ d2xα 1 ∂g ∂g dxα dxβ L ∂g dxα dxβ = L g + αγ + γβ − αβ . (7) αγ dτ 2 2 ∂xβ ∂xα dτ dτ 2 ∂xγ dτ dτ

2 Rearranging the terms on the right hand side and changing the dummy index α to δ, we have µ ¶ d2xα 1 ∂g dxα dxβ 1 ∂g ∂g dxα dxβ g = αβ − αγ + γβ αγ dτ 2 2 ∂xγ dτ dτ 2 ∂xβ ∂xα dτ dτ · ¸ 1 ∂g ∂g ∂g dxα dxβ = − αγ + γβ − αβ 2 ∂xβ ∂xα ∂xγ dτ dτ · ¸ 1 ∂g ∂g ∂g dxδ dxβ = − δγ + γβ − δβ 2 ∂xβ ∂xδ ∂xγ dτ dτ dxδ dxβ ≡ −g Γα . (8) αγ δβ dτ dτ We have found the geodesic equation,

d2xα dxδ dxβ = −Γα (9) dτ 2 δβ dτ dτ where the Christoﬀel symbols satisfy · ¸ 1 ∂g ∂g ∂g g Γα = γδ + γβ − δβ . (10) αγ δβ 2 ∂xβ ∂xδ ∂xγ

This is a linear system of equations for the Christoﬀel 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 gαβ. Also, you should note that the Christoﬀel symbol is symmetric in the lower indices,

α α Γδβ = Γβδ

We can solve for the Christoﬀel symbols by introducing the inverse of the metric, gµγ , satisfying µγ µ g gαγ = δα. (11) µ Here, δα is the Kronecker delta, which vanishes for µ 6= α and is one otherwise. Then, µγ α µ α µ g gαγ Γδβ = δαΓδβ = Γδβ. (12) Therefore, · ¸ 1 ∂g ∂g ∂g Γµ = gµγ γδ + γβ − δβ . (13) δβ 2 ∂xβ ∂xδ ∂xγ