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I'm stuck trying to follow Foster and Nightingale's derivation of the geodesic equation from two neighbouring geodesics $x^{a}\left(u\right)$ and $\tilde{x}^{a}\left(u\right)$ joined by a connecting vector $\xi\left(u\right)$. My problem may be that I'm unsure what “first order” means in the context of this derivation. And there again it may not.

We know that $$\tilde{x}^{a}=x^{a}+\xi^{a}.$$ And, to first order$$\tilde{\Gamma}_{bc}^{a}=\Gamma_{bc}^{a}+\partial_{d}\Gamma_{bc}^{a}\xi^{d}.$$

The two geodesic equations are:

$$\frac{d^{2}\tilde{x}^{a}}{du^{2}}+\tilde{\Gamma}_{bc}^{a}\frac{d\tilde{x}^{b}}{du}\frac{d\tilde{x}^{c}}{du}=0$$

and$$\frac{d^{2}x^{a}}{du^{2}}+\Gamma_{bc}^{a}\frac{dx^{b}}{du}\frac{dx^{c}}{du}=0.$$

Subtract the second geodesic equation from the first geodesic equation to get

$$\frac{d^{2}\xi^{a}}{du^{2}}+\tilde{\Gamma}_{bc}^{a}\frac{d\tilde{x}^{b}}{du}\frac{d\tilde{x}^{c}}{du}-\Gamma_{bc}^{a}\frac{dx^{b}}{du}\frac{dx^{c}}{du}=0.$$ Substituting the above equations for $\tilde{x}^{a}$ and $\tilde{\Gamma}_{bc}^{a}$ into this and I end up with

$$\frac{d^{2}\xi^{a}}{du^{2}}+\Gamma_{bc}^{a}\frac{dx^{b}}{du}\frac{d\xi^{c}}{du}+\Gamma_{bc}^{a}\frac{d\xi^{b}}{du}\frac{dx^{c}}{du}+\partial_{d}\Gamma_{bc}^{a}\xi^{d}\frac{dx^{b}}{du}\frac{dx^{c}}{du}+\partial_{d}\Gamma_{bc}^{a}\xi^{d}\frac{dx^{b}}{du}\frac{d\xi^{c}}{du}+\partial_{d}\Gamma_{bc}^{a}\xi^{d}\frac{d\xi^{b}}{du}\frac{dx^{c}}{du}=0.$$ This is correct, but only if I can omit the last two terms $\left(\partial_{d}\Gamma_{bc}^{a}\xi^{d}\frac{dx^{b}}{du}\frac{d\xi^{c}}{du}\right)$ and $\left(\partial_{d}\Gamma_{bc}^{a}\xi^{d}\frac{d\xi^{b}}{du}\frac{dx^{c}}{du}\right).$ Foster and Nightingale say “only first order [in $\xi^{a}$ ] terms have been retained” . But why are these two terms second order? Does $\xi^{d}\frac{d\xi^{c}}{du}$ count as a second order term in $\xi^{a}$? Thanks

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    $\begingroup$ Questions where the answer can be "yes" are generally frowned upon...but the answer is yes. $\endgroup$ – Ryan Unger Aug 28 '15 at 16:46
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yep, think of $ \xi $ as a unit vector and replace all instances of it with $ \epsilon \xi$ where $ \epsilon $ is some small number. Then you will see that those two terms are second order in $ \epsilon $.

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    $\begingroup$ Thanks, but I still don't see why something multiplied by the derivative of that something should be second order. How do we know that $\frac{d\xi^{c}}{du}$ is a small number? I'm assuming second order in this derivation means something squared, not a second order differential equation. $\endgroup$ – Peter4075 Aug 28 '15 at 17:33
  • $\begingroup$ Just think of it as a taylor series expansion (which is exactly what it is). No matter how big $ \frac{d\xi}{du} $ is, for a small enough $ \epsilon $, $\epsilon^2 \xi \frac{d\xi}{du} $ will be order $ \epsilon $ smaller than either $ \epsilon \xi $ or $ \epsilon \frac{d\xi}{du} $ and therefore negligible. Remember that the expression above is only valid to first order in $ \epsilon \xi $ in the first place (look at the definition of $ \bar{\Gamma} $) $\endgroup$ – Graham Reid Aug 28 '15 at 19:37

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