# Alternative approaches to formulating geodesic equations on Riemannian manifolds and proof of their equivalence

A geodesic can be defined as an extremal path between two points on a manifold in the sense that it minimises or maximises some criterion of interest (e.g., minimises distance travelled, maximises proper time, etc). Such a path will satisfy some geodesic equations equivalent to the Euler-Lagrange equations of the calculus of variations. A geodesic can also be defined in a conceptually different way as the `straightest’ possible path between two points on a manifold. In this case the path will satisfy geodesic equations derived by requiring parallel transport of a tangent vector along the path. Although these are conceptually different ways of defining geodesics, they are mathematically equivalent. In the present note I want to explore the derivation of geodesic equations in these two different ways and prove their mathematical equivalence.

Now, in the calculus of variations we typically define a system’s action $S$ to be the time-integral of a Lagrangian $L$:

$S \equiv \int^{t_B}_{t_A} L(q_i, \dot{q_i}) dt$

where $L(q_i, \dot{q_i})$ says that the Lagrangian is a function of position coordinates $q_i$ and velocities $\dot{q_i}$ (and $i$ ranges over however many coordinates there are). We find the trajectory that yields a desired extremal value of the action $S$ as the one that satisfies the Euler-Lagrange equations

$0 = \frac{d}{dt} \big(\frac{\partial L}{\partial \dot{q_i}} \big) - \frac{\partial L}{\partial q_i}$

Let us now suppose that we are facing an exactly analogous situation in which there are two points on the manifold, $A$ and $B$, and we are considering possible paths between them to try to find the extremal one. We can describe any path between $A$ and $B$ by specifying the coordinates of the points along it as functions of a parameter $\sigma$ that goes from a value of $0$ at $A$ to a value of $1$ at $B$, i.e., by specifying the functions $x^{\mu}(\sigma)$. Noting that the line element can be written as

$ds^2 = g_{\mu \gamma} dx^{\mu} dx^{\gamma}$

we can write the length of a particular path as

$s = \int \sqrt{ds^2} = \int^1_0 \sqrt{g_{\mu \gamma} \frac{dx^{\mu}}{d \sigma} \frac{dx^{\gamma}}{d \sigma}} d \sigma$

Note that the metric is a function of the coordinates of points along the path, which in turn are functions of the parameter $\sigma$, i.e., $g_{\mu \gamma} = g_{\mu \gamma}(x^{\alpha}(\sigma))$. This situation is exactly analogous to the usual calculus of variations scenario because, writing $\dot{x}^{\alpha} \equiv d x^{\alpha}/d \sigma$, we see that we have a Lagrangian function

$L(x^{\alpha}, \dot{x}^{\alpha}) = \sqrt{g_{\mu \gamma} \ \dot{x}^{\mu} \ \dot{x}^{\gamma}}$

and we hope to find the path $x^{\mu}(\sigma)$ that makes the integral of the Lagrangian extreme. This will be the path that satisfies the Euler-Lagrange equations

$0 = \frac{d}{d \sigma} \big(\frac{\partial L}{\partial \dot{x}^{\alpha}}\big) - \frac{\partial L}{\partial x^{\alpha}}$

This corresponds to $N$ separate differential equations in an $N$-dimensional manifold, one equation for each value of the index $\alpha$.

We can manipulate the Euler-Lagrange equations to get geodesic equations which are easier to use in particular contexts. First, note that

$\frac{\partial L}{\partial \dot{x}^{\alpha}} = \frac{\partial}{\partial \dot{x}^{\alpha}}\sqrt{g_{\mu \gamma} \ \dot{x}^{\mu} \ \dot{x}^{\gamma}}$

$= \frac{1}{2L} (g_{\mu \gamma} \ \delta^{\mu}_{\alpha} \ \dot{x}^{\gamma} + g_{\mu \gamma} \ \dot{x}^{\mu} \ \delta^{\ \gamma}_{\alpha})$

because, for example, $\partial \dot{x}^{\mu}/\partial \dot{x}^{\alpha} = \delta^{\mu}_{\alpha}$. Also note that the metric is treated as a constant as it depends on $x^{\alpha}$ not on $\dot{x}^{\alpha}$. Doing the sums over the Kronecker deltas we get

$\frac{\partial L}{\partial \dot{x}^{\alpha}} = \frac{1}{2L}(g_{\alpha \gamma} \ \dot{x}^{\gamma} + g_{\mu \alpha} \ \dot{x}^{\mu})$

$= \frac{1}{2L}(g_{\alpha \mu} \ \dot{x}^{\mu} + g_{\alpha \mu} \ \dot{x}^{\mu})$

$= \frac{1}{L} g_{\alpha \mu} \ \dot{x}^{\mu}$

But notice that since

$s = \int L d \sigma$

we have

$\frac{ds}{d \sigma} = L$

so

$\frac{1}{L} = \frac{d \sigma}{ds}$

and we can write

$\frac{\partial L}{\partial \dot{x}^{\alpha}} = \frac{1}{L} g_{\alpha \mu} \frac{d x^{\mu}}{d \sigma}$

$= g_{\alpha \mu} \frac{d x^{\mu}}{d \sigma} \frac{d \sigma}{ds}$

$= g_{\alpha \mu} \frac{d x^{\mu}}{ds}$

Next, we have

$\frac{\partial L}{\partial x^{\alpha}} = \frac{\partial}{\partial x^{\alpha}}\sqrt{g_{\mu \gamma} \ \dot{x}^{\mu} \ \dot{x}^{\gamma}}$

$= \frac{1}{2L} \frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \dot{x}^{\mu} \dot{x}^{\gamma}$

$= \frac{1}{2} \frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{d \sigma} \frac{d x^{\gamma}}{d \sigma} \frac{d \sigma}{ds}$

$= \frac{1}{2} \frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{d \sigma}$

Putting these results into the Euler-Lagrange equations we get

$0 = \frac{d}{d \sigma} \big(g_{\alpha \mu} \frac{d x^{\mu}}{ds} \big) - \frac{1}{2}\frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{d \sigma}$

Finally, multiplying through by $d \sigma/ds$ we get

$0 = \frac{d}{ds} \big(g_{\alpha \beta} \frac{d x^{\beta}}{ds} \big) - \frac{1}{2}\frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{ds}$

where I have also renamed $\mu \rightarrow \beta$ in the first term to make it clearer that the Einstein summations in the first and second terms are independent. This is the first version of the geodesic equations, derived by requiring that the path between the points $A$ and $B$ should be extremal in the sense of satisfying the Euler-Lagrange equations of the calculus of variations.

We will now derive a second version of the geodesic equations by requiring the geodesic to be a path that is locally straight. In differential geometry a path is defined as straight if it parallel transports its own tangent vector, i.e., if the tangent vector does not change as we move an infinitesimal step along the path. If we take an arbitrary point on the path to be $x^{\mu} \ e_{\mu}$ and we take $ds$ to be an infinitesimal displacement along the path, then a tangent vector to the path is

$\frac{d x^{\mu}}{d \sigma} e_{\mu}$

and we want

$\frac{d}{ds}\big(\frac{d x^{\mu}}{d \sigma}e_{\mu} \big) = \frac{d^2 x^{\mu}}{ds d \sigma} e_{\mu} + \frac{d x^{\mu}}{d \sigma} \frac{d e_{\mu}}{ds} = 0$

Multiplying through by $d \sigma/ds$ this gives

$\frac{d^2 x^{\mu}}{ds^2} e_{\mu} + \frac{d x^{\mu}}{ds} \frac{d e_{\mu}}{ds} = 0$

But

$\frac{d e_{\mu}}{ds} = \frac{\partial e_{\mu}}{\partial x^{\gamma}} \frac{d x^{\gamma}}{d s}$

$= \frac{d x^{\gamma}}{ds} \Gamma^{\alpha}_{\hphantom{\alpha} \mu \gamma} e_{\alpha}$

Putting this into the equation gives

$\frac{d^2 x^{\mu}}{ds^2} e_{\mu} + \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{ds} \Gamma^{\alpha}_{\hphantom{\alpha} \mu \gamma} e_{\alpha} = 0$

To enable us to factor out the basis vector we can rename the indices in the second term as $\mu \rightarrow \alpha$ and $\gamma \rightarrow \beta$ to get

$\frac{d^2 x^{\mu}}{ds^2} e_{\mu} + \frac{d x^{\alpha}}{ds} \frac{d x^{\beta}}{ds} \Gamma^{\mu}_{\hphantom{\mu} \alpha \beta} e_{\mu} = 0$

$\iff$

$\big[\frac{d^2 x^{\mu}}{ds^2} + \frac{d x^{\alpha}}{ds} \frac{d x^{\beta}}{ds} \Gamma^{\mu}_{\hphantom{\mu} \alpha \beta} \big] e_{\mu} = 0$

$\implies$

$\frac{d^2 x^{\mu}}{ds^2} + \frac{d x^{\alpha}}{ds} \frac{d x^{\beta}}{ds} \Gamma^{\mu}_{\hphantom{\mu} \alpha \beta} = 0$

This is the second version of the geodesic equations, derived by assuming that the path between the two points on the manifold is locally straight.

We now have two seemingly different versions of the geodesic equations, namely

$0 = \frac{d}{ds} \big(g_{\alpha \beta} \frac{d x^{\beta}}{ds} \big) - \frac{1}{2}\frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{ds}$

and

$0 = \frac{d^2 x^{\mu}}{ds^2} + \frac{d x^{\alpha}}{ds} \frac{d x^{\beta}}{ds} \Gamma^{\mu}_{\hphantom{\mu} \alpha \beta}$

We will next show that they are in fact mathematically equivalent. Starting from the first version, we can expand out the brackets to get

$0 = \frac{\partial g_{\alpha \beta}}{\partial x^{\sigma}}\frac{dx^{\sigma}}{ds} \frac{dx^{\beta}}{ds} + g_{\alpha \beta} \frac{d^2 x^{\beta}}{ds^2} - \frac{1}{2}\frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{ds}$

$\iff$

$0 = \frac{1}{2}\frac{\partial g_{\alpha \beta}}{\partial x^{\sigma}}\frac{dx^{\sigma}}{ds} \frac{dx^{\beta}}{ds} + \frac{1}{2}\frac{\partial g_{\alpha \beta}}{\partial x^{\sigma}}\frac{dx^{\sigma}}{ds} \frac{dx^{\beta}}{ds} + g_{\alpha \beta} \frac{d^2 x^{\beta}}{ds^2} - \frac{1}{2}\frac{\partial g_{\mu \gamma}}{\partial x^{\alpha}} \frac{d x^{\mu}}{ds} \frac{d x^{\gamma}}{ds}$

Now we rename the indices as follows: $\sigma \rightarrow \alpha$ in the first term; $\sigma \rightarrow \beta$ in the second term; $\beta \rightarrow \mu$ and $\alpha \rightarrow \sigma$ in the third term; and $\alpha \rightarrow \sigma$, $\mu \rightarrow \alpha$, $\gamma \rightarrow \beta$ in the fourth term. We get

$0 = \frac{1}{2}\frac{\partial g_{\sigma \beta}}{\partial x^{\alpha}}\frac{dx^{\alpha}}{ds} \frac{dx^{\beta}}{ds} + \frac{1}{2}\frac{\partial g_{\sigma \alpha}}{\partial x^{\beta}} \frac{dx^{\alpha}}{ds} \frac{dx^{\beta}}{ds} + g_{\sigma \mu} \frac{d^2 x^{\mu}}{ds^2} - \frac{1}{2}\frac{\partial g_{\alpha \beta}}{\partial x^{\sigma}} \frac{d x^{\alpha}}{ds} \frac{d x^{\beta}}{ds}$

We can write this as

$0 = \frac{dx^{\alpha}}{ds} \frac{dx^{\beta}}{ds} \frac{1}{2} [\partial_{\alpha} \ g_{\beta \sigma} + \partial_{\beta} \ g_{\sigma \alpha} - \partial_{\sigma} \ g_{\alpha \beta}] + g_{\sigma \mu} \frac{d^2 x^{\mu}}{ds^2}$

Finally, multiplying through by $g^{\mu \sigma}$ and using the facts that

$g^{\mu \sigma} \ g_{\sigma \mu} = 1$

and

$\Gamma^{\mu}_{\hphantom{\mu} \alpha \beta} = \frac{1}{2} g^{\mu \sigma} [\partial_{\alpha} \ g_{\beta \sigma} + \partial_{\beta} \ g_{\sigma \alpha} - \partial_{\sigma} \ g_{\alpha \beta}]$

we get

$0 = \frac{d x^{\alpha}}{ds} \frac{d x^{\beta}}{ds} \Gamma^{\mu}_{\hphantom{\mu} \alpha \beta} + \frac{d^2 x^{\mu}}{ds^2}$

which is the second version of the geodesic equation. Thus, the two versions are equivalent as claimed.