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.

Leave a Reply

Your email address will not be published. Required fields are marked *