Derivation of geodesic deviation equation

Question: Let $x^{a}$ be a geodesic, and $x^{a} + \epsilon^{a}$ be a neighbouring geodesic such that $\epsilon^{a}$ is infinitesimal. Prove the so-called geodesic deviation equation:

$\frac{D^{2} \epsilon^{a}}{D \tau^{2}} = \, – R^{a}_{\: bcd} \epsilon^{c} \frac{dx^{b}}{d \tau} \frac{dx^{d}}{d \tau}$


Relevant Geodesic equations:

$\frac{d^{2} x^{a}}{d \tau^{2}} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \frac{d x^{c}}{d \tau} = 0$        — Eqn 1

$\frac{d^{2} \left( x^{a} + \epsilon^{a} \right)}{d \tau^{2}} + \Gamma^{a}_{\: bc} \frac{d \left(x^{b} + \epsilon^{b} \right)}{d \tau} \frac{d \left( x^{c} + \epsilon^{c} \right)}{d \tau} = 0$       — Eqn 2


We know that:

$\frac{D \epsilon^{a}}{D \tau} = V^{a} = U^{b}\nabla_{b} \epsilon^{a}$, where $U^{b} = \frac{d x^{b}}{d \tau}$    — Eqn 3

$\frac{D^{2} \epsilon^{a}}{D \tau^{2}} = W^{a} = U^{b}\nabla_{b} v^{a}$     — Eqn 4


From equation 4,

$$\begin{aligned} V^{a} &= U^{b} \nabla_{b} \epsilon^{a} \\
&= \frac{d \epsilon^{a}}{d \tau} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \epsilon^{c} \end{aligned}$$

$$\begin{aligned} W^{a} &= U^{b} \nabla_{b} V^{a} \\ &= \frac{d V^{a}}{d \tau} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} V^{c} \end{aligned}$$

Sub. $V^{a}$ from equation 3 into the above equation, (Note that you will need to change the index of V to the dummy index)

$$\begin{aligned} W^{a} &= \frac{d}{d \tau} \left( \frac{d \epsilon^{a}}{d \tau} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \epsilon^{c} \right) + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \left( \frac{d \epsilon^{c}}{d \tau} + \Gamma^{c}_{\: ef} \frac{dx^{e}}{d \tau} \epsilon^{f} \right) \\ &= \frac{d^{2} \epsilon^{a}}{d \tau^{2}} + \Gamma^{a}_{\: bc} \frac{d^{2} x^{b}}{d \tau^{2}} \epsilon^{c} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \frac{d \epsilon^{c}}{d \tau} + \frac{d \Gamma^{a}_{\: bc}}{d \tau} \frac{d x^{b}}{d \tau} \epsilon^{c} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \frac{d \epsilon^{c}}{d \tau} + \Gamma^{a}_{\: bc} \Gamma^{c}_{\: ef} \frac{dx^{b}}{d \tau} \frac{dx^{c}}{d \tau} \epsilon^{f}  \end{aligned}$$

Using equation 1 to get rid of $\frac{d^{2} x^{a}}{d \tau^{2}}$ and chain rule to change $\frac{d \Gamma^{a}_{\: bc}}{d \tau}$ to $\frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \frac{dx^{e}}{d \tau} \epsilon^{c}$,

$$W^{a} = \frac{d^{2} \epsilon^{a}}{d \tau^{2}} + 2 \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \frac{d \epsilon^{c}}{d \tau} – \Gamma^{a}_{\: bc} \Gamma^{b}_{\: ef} \frac{d x^{e}}{d \tau} \frac{d x^{f}}{d \tau} \epsilon^{c} + \Gamma^{a}_{\: bc} \Gamma^{c}_{\: ef} \frac{dx^{b}}{d \tau} \frac{dx^{e}}{d \tau} \epsilon^{f} + \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \frac{dx^{e}}{d \tau} \frac{dx^{b}}{d \tau} \epsilon^{c}$$

To get rid of the first two terms in the above equation, we go back to the geodesic equation – Eqn 2.

Expanding equation 2, (Since $\Gamma^{a}_{\: bc}$ contains epsilon, we have $\Gamma^{a}_{\: bc} + \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \epsilon^{e} + \, … $)

$$\frac{d^{2}}{d \tau^{2}} + \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \frac{dx^{c}}{d \tau} + \frac{d^{2} \epsilon^{a}}{d \tau^{2}} + \Gamma^{a}_{\: bc} \frac{d x^{b}}{d \tau} \frac{d \epsilon^{c}}{d \tau} + \Gamma^{a}_{\: bc} \frac{d x^{c}}{d \tau} \frac{d \epsilon^{b}}{d \tau} + \frac{ \partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \epsilon^{e} \frac{dx^{b}}{d \tau} \frac{dx^{c}}{d \tau} = 0$$

Note: Terms containing $\epsilon^{a} \frac{d \epsilon^{b}}{d \tau}$ and $\frac{d \epsilon^{b}}{d \tau} \frac{d \epsilon^{c}}{d \tau}$ are thrown away since $\epsilon$ is small.

The first two terms are 0 from equation 1, the original geodesic. And note that $\Gamma^{a}_{bc} = \Gamma^{a}_{cb}$. Hence, equation 2 becomes:

$$\frac{d^{2} \epsilon^{a}}{d \tau^{2}} + 2 \Gamma^{a}_{\: bc} \frac{dx^{b}}{d \tau} \frac{d \epsilon^{c}}{d \tau} + \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \frac{dx^{b}}{d \tau} \frac{dx^{c}}{d \tau} \epsilon^{e} = 0$$

Substituting the above equation into the final equation for $W^{a}$

$$W^{a} = \, – \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \frac{dx^{b}}{d \tau} \frac{dx^{c}}{d \tau} \epsilon^{e} + \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{e}} \frac{dx^{e}}{d \tau} \frac{dx^{b}}{d \tau} \epsilon^{c} – \Gamma^{a}_{\: bc} \Gamma^{b}_{\: ef} \frac{d x^{e}}{d \tau} \frac{d x^{f}}{d \tau} \epsilon^{c} + \Gamma^{a}_{\: bc} \Gamma^{c}_{\: ef} \frac{dx^{b}}{d \tau} \frac{dx^{e}}{d \tau} \epsilon^{f}$$

Relabelling the terms, we obtain: (Note: I’ve switched the indices to force it to be similar to the answer)

$$\begin{aligned} W^{a} &= \, – \frac{\partial \Gamma^{a}_{\: bd}}{\partial x^{c}} \frac{dx^{b}}{d \tau} \frac{dx^{d}}{d \tau} \epsilon^{c} + \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{d}} \frac{dx^{b}}{d \tau} \frac{dx^{d}}{d \tau} \epsilon^{c} – \Gamma^{a}_{\: fc} \Gamma^{f}_{\: bd} \frac{d x^{b}}{d \tau} \frac{d x^{d}}{d \tau} \epsilon^{c} + \Gamma^{a}_{\: df} \Gamma^{f}_{\: bc} \frac{dx^{b}}{d \tau} \frac{dx^{d}}{d \tau}\epsilon^{c} \\ &= \, \left( – \frac{\partial \Gamma^{a}_{\: bd}}{\partial x^{c}} + \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{d}} – \Gamma^{a}_{\: fc} \Gamma^{f}_{\: bd}  + \Gamma^{a}_{\: df} \Gamma^{f}_{\: bc} \right) \frac{dx^{b}}{d \tau} \frac{dx^{d}}{d \tau} \epsilon^{c} \end{aligned}$$

Since $\Gamma^{a}_{\: bcd} = \frac{\partial \Gamma^{a}_{\: bd}}{\partial x^{c}} – \frac{\partial \Gamma^{a}_{\: bc}}{\partial x^{d}} + \Gamma^{f}_{\: bd} \Gamma^{a}_{\: cf} – \Gamma^{f}_{\: bc} \Gamma^{a}_{\: df}$,

$$W^{a} = \, – R^{a}_{\: bcd} \epsilon^{c} \frac{dx^{b}}{d \tau} \frac{dx^{d}}{d \tau}$$



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1 thought on “Derivation of geodesic deviation equation”

  1. sir I find difficult to undestand expansion of equation 2above. Can you please explain step by step expansion of equation 2 ?


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