Geodesic equation and four-velocity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 8; Problem 8.4.

The geodesic equation is

\displaystyle  \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\partial_{a}g_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (1)

We can write this in terms of the four-velocity {u^{i}\equiv dx^{i}/d\tau} if we expand the first derivative:

\displaystyle   \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right) \displaystyle  = \displaystyle  \partial_{i}g_{aj}\frac{\partial x^{i}}{\partial\tau}\frac{\partial x^{j}}{\partial\tau}+g_{aj}\frac{d}{d\tau}\left(\frac{dx^{j}}{d\tau}\right)\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \partial_{i}g_{aj}u^{i}u^{j}+g_{aj}\frac{du^{j}}{d\tau} \ \ \ \ \ (3)

Therefore, the geodesic equation is

\displaystyle  \partial_{i}g_{aj}u^{i}u^{j}+g_{aj}\frac{du^{j}}{d\tau}-\frac{1}{2}\partial_{a}g_{ij}u^{i}u^{j}=0 \ \ \ \ \ (4)

We can express this in a different form by multiplying by {u^{a}} and summing:

\displaystyle   u^{a}\partial_{i}g_{aj}u^{i}u^{j}+g_{aj}u^{a}\frac{du^{j}}{d\tau}-\frac{1}{2}u^{a}\partial_{a}g_{ij}u^{i}u^{j} \displaystyle  = \displaystyle  g_{aj}u^{a}\frac{du^{j}}{d\tau}+\frac{1}{2}u^{a}u^{i}u^{j}\partial_{a}g_{ij}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  g_{aj}u^{a}\frac{du^{j}}{d\tau}+\frac{1}{2}u^{i}u^{j}\frac{dg_{ij}}{d\tau}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}\frac{d}{d\tau}\left(g_{ij}u^{i}u^{j}\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}\frac{d}{d\tau}\left(\mathbf{u}\cdot\mathbf{u}\right) \ \ \ \ \ (8)

The geodesic equation thus confirms that {\frac{d}{d\tau}\left(\mathbf{u}\cdot\mathbf{u}\right)=0} along a geodesic, which we knew beforehand, since {\mathbf{u}\cdot\mathbf{u}=-1} is an invariant.

7 thoughts on “Geodesic equation and four-velocity

  1. Pingback: Geodesic equation: 2-d space-time | Physics tutorials

  2. Pingback: Geodesic equation in 2-d: exponential metric | Physics tutorials

  3. Dan

    It seems like you’ve used “j” as a replacement index for Thomas Moore’s “beta” and “mu.” It looks like an additional index needs to be added to your geodesic equation to make it work.

    Reply
    1. gwrowe Post author

      Equation 1 is correct. Remember that repeated indices are just dummy indices since they are summed, so it doesn’t matter what you call them. In my previous posts I was lazy and used Roman letters for indices since they are easier to type than Greek letters, which Moore uses. [Actually about half the textbooks on general relativity use Roman indices anyway, so I’m not that far out of line.] Moore uses {\beta} for my {j} in the first term, and {\mu} for my {i} and {\nu} for my {j} in the second term, though he could have used {\nu} instead of {\beta} in the first term as well and still have a correct equation. The only index that isn’t summed is {a} (Moore uses {\alpha}).

      Reply

Leave a Reply

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