# Christoffel symbols in sinusoidal coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Problem P17.6.

We are now in a position to revisit the system with sinusoidal coordinates. To review, we had a 2-d system with coordinates ${u}$ and ${w}$ defined in terms of the usual rectangular coordinates ${x}$ and ${y}$ by

 $\displaystyle u$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (1)$ $\displaystyle w$ $\displaystyle =$ $\displaystyle y-A\sin\left(bx\right) \ \ \ \ \ (2)$

The metric for this system is

$\displaystyle g_{ij}=\left[\begin{array}{cc} 1+\left[Ab\cos\left(bu\right)\right]^{2} & Ab\cos\left(bu\right)\\ Ab\cos\left(bu\right) & 1 \end{array}\right] \ \ \ \ \ (3)$

We looked at an object with a velocity given by ${\mathbf{v}=\left[v,0\right]}$ where ${v}$ is a constant. Clearly the acceleration of the object is zero, but if we calculate the velocity components in the ${uw}$ system, we get

 $\displaystyle v^{u}$ $\displaystyle =$ $\displaystyle v\ \ \ \ \ (4)$ $\displaystyle v^{w}$ $\displaystyle =$ $\displaystyle -Ab\cos\left(bu\right)v \ \ \ \ \ (5)$

so although ${dv^{u}/dt=0}$, ${dv^{w}/dt\ne0}$ since ${u=x=vt}$ varies with time.

To get the ‘true’ acceleration, we need to find the actual differential ${d\mathbf{v}}$ and divide this by ${dt}$. We’ve seen how to do this when we defined the Christoffel symbols:

$\displaystyle d\mathbf{v}=\left[\frac{\partial v^{k}}{\partial x^{j}}+v^{i}\Gamma_{\; ij}^{k}\right]\mathbf{e}_{k}dx^{j} \ \ \ \ \ (6)$

To calculate ${d\mathbf{v}}$, we need the ${\Gamma_{\; ij}^{k}}$ for the ${uw}$ system, which we can calculate in the usual way using the geodesic equation. We start with

$\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (7)$

Since ${g_{ij}}$ is independent of ${w}$ only derivatives with respect to ${u}$ are non-zero. Consider first ${a=u}$; then we get

 $\displaystyle \left[1+\left[Ab\cos\left(bu\right)\right]^{2}\right]\ddot{u}+\left(Ab\cos bu\right)\ddot{w}-A^{2}b^{3}\cos bu\sin bu\dot{u}^{2}+\left[-Ab^{2}\sin bu-\left(-Ab^{2}\sin bu\right)\right]\dot{u}\dot{v}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \left[1+\left[Ab\cos\left(bu\right)\right]^{2}\right]\ddot{u}+\left(Ab\cos bu\right)\ddot{w}-A^{2}b^{3}\cos bu\sin bu\dot{u}^{2}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (9)$

Now for ${a=w}$:

$\displaystyle Ab\cos bu\ddot{u}+\ddot{w}-Ab^{2}\sin bu\dot{u}^{2}=0 \ \ \ \ \ (10)$

We would like to compare these equations with the equation involving the Christoffel symbols:

$\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (11)$

However, because the metric here is not diagonal, we get second derivatives of more than one coordinate in each equation. We can eliminate ${\ddot{w}}$ by multiplying 10 by ${Ab\cos bu}$ and subtracting it from 9. This gives the convenient result:

$\displaystyle \ddot{u}=0 \ \ \ \ \ (12)$

From this we conclude that

$\displaystyle \Gamma_{\; ij}^{u}=\left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (13)$

We can substitute 12 into 10 to get

$\displaystyle \ddot{w}-Ab^{2}\sin bu\dot{u}^{2}=0 \ \ \ \ \ (14)$

from which we conclude

$\displaystyle \Gamma_{\; ij}^{w}=\left[\begin{array}{cc} -Ab^{2}\sin bu & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (15)$

We can now evaluate 6 to find:

 $\displaystyle dv^{u}$ $\displaystyle =$ $\displaystyle \left[\frac{\partial v^{u}}{\partial x^{j}}+v^{i}\Gamma_{\; ij}^{u}\right]dx^{j}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (17)$ $\displaystyle dv^{w}$ $\displaystyle =$ $\displaystyle \left[\frac{\partial v^{w}}{\partial x^{j}}+v^{i}\Gamma_{\; ij}^{w}\right]dx^{j}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[Ab^{2}v\sin bu-vAb^{2}\sin bu\right]du\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (20)$

Thus by taking the proper derivative using the Christoffel symbols, the differentials of both components of velocity are zero, so the acceleration is zero in the ${uw}$ system as well.