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I have been studying the basics of general relativity with Hartle's Gravity. He presents the geodesic equation as

$$ \frac{d^2x^{{\mu}}}{ds^2}=-{\Gamma}^{{\mu}}_{{\alpha}{\beta}}\frac{dx^{{\alpha}}}{ds}\frac{dx^{{\beta}}}{ds} $$

However, in reading Padmanabhan's Gravitation, he says that the equation

$$ 0=k^{{\alpha}}{\nabla}_{{\alpha}}k_{{\beta}} $$

is the geodesic equation for a wave vector $k$ defined as $k_{{\alpha}}={\nabla}_{{\alpha}}{\psi}$, where ${\psi}$ is just a scalar function.

How do these two definitions of the geodesic equation represent the same thing? They do not look at all alike. In fact, if I try to work it out, I get

$$ 0=k^{{\alpha}}\,\bigg(\frac{{\partial}k_{{\beta}}}{{\partial}x^{{\alpha}}}-{\Gamma}^{{\delta}}_{{\beta}{\gamma}}k_{{\delta}}\bigg) $$

$$ 0=k^{{\alpha}}\,\frac{{\partial}k_{{\beta}}}{{\partial}x^{{\alpha}}}-{\Gamma}^{{\delta}}_{{\beta}{\gamma}}k_{{\delta}}k^{{\alpha}} $$

For one thing, the $x^{{\alpha}}$ is not supposed to be on the bottom of the derivative! If $k_{{\alpha}}$ is supposed to satisfy the geodesic equation, I expected this to look like the first equation I wrote.

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1 Answer 1

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In short, the first equation you wrote is in component form where curves are parametrized as $x^{\mu}(s)$, while the second equation is more general. Now let's flesh out the details.

What does it mean for a curve to be a geodesic? Intuitively it has to be straightest curve possible in curved spacetime. The way you do that is to propagate the tangent vector $T^{\alpha}$ of a curve C along itself! That yields the coordinate independent geodesic equation: $$T^{\alpha}\nabla_{\alpha}T^{\beta} = 0$$

Then you can write this tensor equation in component form, by choosing a coordinate system and its associated Christoffel symbol. The covariant derivative is expressed as: $$\nabla_{\alpha}T^{\beta} = \partial_{\alpha}T^{\beta}+\Gamma^{\beta}_{\alpha\gamma}T^{\gamma}$$ Now if you plug this equation into the coordinate independent geodesic equation, you get: $$ T^{\alpha}\partial_{\alpha}T^{\beta}+T^{\alpha}\Gamma^{\beta}_{\alpha\gamma}T^{\gamma} = 0$$ Now if you parametrize the curve C as $x^{\mu}(s)$, then the tangent vector field becomes:$T^{\mu} = \frac{dx^{\mu}}{ds}$, and $T^{\alpha}\partial_{\alpha}$ represents the derivative operator $\frac{d}{ds}$. Therefore, the previous equation turns into: $$\frac{d}{ds}\frac{dx^{\beta}}{ds}+\Gamma^{\beta}_{\alpha\gamma}\frac{dx^{\alpha}}{ds}\frac{dx^{\gamma}}{ds} = 0$$ $$\frac{d^2x^{\beta}}{ds^2}+\Gamma^{\beta}_{\alpha\gamma}\frac{dx^{\alpha}}{ds}\frac{dx^{\gamma}}{ds} = 0$$ Which is precisely the equation that you started with.

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    $\begingroup$ Good answer. Just replace t with s so there is no confusion. $\endgroup$
    – Bob Bee
    Jul 7, 2016 at 3:09
  • $\begingroup$ Edits look good $\endgroup$
    – Bob Bee
    Jul 7, 2016 at 4:10

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