The Geodesic Equation for a Metric

From the given metric we can obtain the matrix form of covariant metric tensor which is ${ g }_{ ij }$ = $\begin{bmatrix} \left( 1+{ y }^{ 2 } \right) & 0 \\ 0 & \left( 1+{ x }^{ 2 } \right) \end{bmatrix}$ . Also we can compute the inverse of the metric tensor which is known as the contravariant metric tensor and its of the form ${ g }^{ ij }$ = $\begin{bmatrix} \frac { 1 }{ \left( 1+{ y }^{ 2 } \right) } & 0 \\ 0 & \frac { 1 }{ \left( 1+{ x }^{ 2 } \right) } \end{bmatrix}$

And the remaining thing which is to be done is to find the Christoffel symbols for $x$ component $\Gamma^x_{ij}$ (here $i$ and $j$ are dummy variables and they can take either values of $x$ or $y$ ) for this we make use of the equation ${ \Gamma }_{ ij }^{ x }=\frac { 1 }{ 2 } { g }^{ xl }\left( \frac { \partial { g }^{ jl } }{ \partial i } +\frac { \partial { g }^{ il } }{ \partial j } -\frac { \partial { g }^{ ij } }{ \partial l } \right)$ .when $i = x$ and $j = y$ or vice versa then $\Gamma^x_{xy}$ = $\Gamma^x_{yx}$ = $\frac { 1 }{ 2 } { g }^{ xx }\left( \frac { \partial { g }_{ xx } }{ \partial y } \right)$ = $\frac { y }{ 1+{ y }^{ 2 } }$ and when $i = y$ and $j = y$ then $\Gamma^x_{yy}$ = $-\frac { 1 }{ 2 } { g }^{ xx }\left( \frac { \partial { g }_{ yy } }{ \partial x } \right)$ = $\frac { -x }{ 1+{ y }^{ 2 } }$ . [[Note: ${ g }^{ xx } = \frac { 1 }{ 1+{ y }^{ 2 } }$ and ${ g }_{ xx }=\left( 1+{ y }^{ 2 } \right)$ like wise ${ g }^{ yy }=\frac { 1 }{ 1+{ x }^{ 2 } }$ and ${ g }_{ yy }=\left( 1+{ x }^{ 2 } \right)$ . well the value for ${ \Gamma }_{ xx }^{ x } = 0$ ]]

On substituting these values in the geodesic equation $\frac{d^2 x}{d\tau^2} + \Gamma^x_{ij} \frac{dx^i}{d\tau} \frac{dx^j}{d\tau} = 0.$ we obtain the answer for the $x$ -component of the geodesic equation for that given metric.

The relevant Christoffel symbols are the $\Gamma^x_{ij}$ . The nonzero ones can be computed to be:

$\begin{aligned} \Gamma^x_{yx} &= \Gamma^x_{xy} = \frac{y}{1+y^2}\\ \Gamma^x_{yy} &= \frac{-x}{1+y^2} \end{aligned}$

Plugging in to the geodesic equation: $\frac{d^2 x}{d\tau^2} + \Gamma^x_{ij} \frac{dx^i}{d\tau} \frac{dx^j}{d\tau} = 0.$ obtains the answer.