# Existence of affine parameters of null geodesics

1. Oct 25, 2014

### center o bass

We have a general spacetime interval $ds^2 = g_{\mu \nu} dx^\mu dx^\nu$.
One way to define an affine parameter is to define it to be any parameter $u$ which is related to the path length $s$ by $u = as + b$ for two constants $a,b$. One can show that for the tangent vector $u^\alpha = \frac{dx^\alpha}{du}$ to a geodesic this implies a geodesic equation
$$\nabla_{u} u^\alpha = 0$$
in comparison to the more general
$$\nabla_{u} u^\alpha = f(x) u^\alpha.$$
However, if the geodesic is a null-geodesic, we can not use parameters $u$ which is related to the pathlength by $u=as + b$ since $s = 0$ along any null curve. However, it is still claimed that one can find parameters for which
$$\nabla_{u} u^\alpha = 0$$
holds also for null-geodesics. This thus defines a more general class of "affine parameters".

Now, my question is -- how do we know that there exists such parameters also for null-geodesics?

2. Oct 25, 2014

### Staff: Mentor

This is backwards. The affine parameter is defined as one which obeys the geodesic equation; then you can show that, for a non-null geodesic, path length is a valid affine parameter. You're right that this doesn't work for a null geodesic (because the path length is identically zero), but that only affects the second part (showing that path length works as an affine parameter), not the first part (finding parameters that obey the geodesic equation).

By showing that null geodesics obey the geodesic equation in the simpler form $\nabla_u u = 0$. The simplest way to do that is to start from the more general form $\nabla_u u = f(x) u$ and then show that one can always transform the parameter so that the equation assumes the simpler form. See, for example, Eric Poisson's A Relativist's Toolkit:

3. Oct 27, 2014

### stevendaryl

Staff Emeritus
I had the same confusion about affine parameters. Here's the way I understand it now:

Suppose you have a path through spacetime, parametrized in some coordinate system as $X^\mu(s)$. The corresponding "velocity" 4-vector has components $U^\mu(s) = \frac{dX^\mu}{ds}$.

What makes such a path a geodesic? Basically, if it's a geodesic, that means the "velocity" is constant along the path. But we have to think about what that means in curved spacetime with an arbitrary parametrization. First of all, in curved spacetime, vectors can only be compared if they are measured at the same spacetime point. So to see if $U^\mu(s)$ is constant, we need to do the following: Let $\tilde{U}^\mu(s)$ be the result of taking the vector $U^\mu(s)$, defined at the point $X^\mu(s)$ and "parallel-transporting" it back to the point $X^\mu(0)$. (In terms of connection coefficients $\Gamma^\mu_{\nu \lambda}$, $\tilde{U}^\mu(s) = U^\mu(0) + (\frac{dU^\mu}{ds} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda) s + \ldots$, where $\ldots$ represents higher-order terms in $s$.) If $X^\mu(s)$ is a geodesic, then (in the limit of infinitesimal $s$), $\tilde{U}^\mu(s)$ and $U^\mu(0)$ must represent the same 4-velocity, physically.

The second thing you need to realize is that the parametrization is not physically meaningful. Changing parameters from $s$ to $s'$ will change the 4-velocity from $U^\mu(s)$ to $U'^\mu = U^\mu \frac{ds}{ds'}$. Since $\frac{ds}{ds'}$ is absolutely arbitrary, that means that any scalar multiple of a 4-velocity (multiplying every component by the same real number) produces a physically equivalent 4-velocity.

So in terms of what it means for the path $X^\mu(s)$ to be a geodesic, we don't want to require that:

$\tilde{U}^\mu(s) = U^\mu(0)$

$\tilde{U}^\mu(s) = f(s) U^\mu(0)$

where $f(s)$ is an arbitrary real function of $s$. So as you move along the path $X^\mu(s)$, the overall scale of the 4-velocity $U^\mu(s)$ might change, but if it's a geodesic, then the velocity stays parallel to the original velocity.

We say that $s$ is affine if the scale doesn't change. That is, if $f(s) = 1$. For an affine parameter $s$,

$\tilde{U}^\mu(s) = U^\mu(0)$

which, in terms of connection coefficients means that the additional term is zero, so:

$\frac{dU^\mu}{ds} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda = 0$

Absolutely nothing in this discussion depends on whether the geodesic is spacelike, timelike or lightlike. The only thing that is different for a timelike geodesic (the path followed by a massive object) is that the requirement that the scale not change along the path can be stated more succinctly as follows:

$U^\mu(s) U_\mu(s) =$ a constant along the path.

If $U^\mu(s)$ is lightlike, then that equation becomes $0 = 0$, and it doesn't tell you anything about the scaling.

There is one other difference with lightlike geodesics, which is this:

For a timelike geodesic, the geodesic equation can be obtained by extremizing the proper time:

$\int \sqrt{g_{\mu \nu} \frac{dX^\mu}{ds} \frac{dX^\nu}{ds}} ds$

But for a lightlike geodesic, any null path (geodesic or not) gives the same value 0, so it's a little complicated to see how a null geodesic comes out of the action integral.

4. Oct 28, 2014

### center o bass

I agree with you that there is no problem in defining what we mean by an affine parameter independent of whether the curve is timelike or null according to your prescription. However, defining something does not mean that it exist!

The advantage of the path-length parameter is that it definitely exist, and we can construct a whole range of affine parameters from it. The existence of affine parameter for null curves require a proof.

5. Oct 28, 2014

### stevendaryl

Staff Emeritus
Well, for an arbitrary parameter $s$, the geodesic equation has the form:

$\dfrac{d^2 X^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \dfrac{dX^\nu}{ds} \dfrac{dX^\lambda}{ds} = f(s) \dfrac{dX^\mu}{ds}$

with an arbitrary function $f(s)$

If you change parameters from $s$ to $q$, then the equation becomes:

$(\dfrac{d^2 X^\mu}{dq^2} + \Gamma^\mu_{\nu \lambda} \dfrac{dX^\nu}{dq} \dfrac{dX^\lambda}{dq})(\frac{dq}{ds})^2 = (f(s) \frac{dq}{ds} - \frac{d^2q}{ds^2}) \dfrac{dX^\mu}{dq}$

So to get the right-hand side to equal 0, you have to solve the differential equation:
$f(s) \frac{dq}{ds} - \frac{d^2q}{ds^2} = 0$

(which you solve to get the affine parameter $q$)

So when you ask for a proof that an affine parameter exists, are you just asking for a proof that that differential equation has a solution, or a proof that the solution gives an affine parametrization?

6. Oct 28, 2014

### WannabeNewton

The 4-momentum $p^{\mu} = \frac{dx^{\mu}}{d\lambda}$ is a physical observable i.e. measurable quantity. Furthermore the (parametrization independent) worldline of a coherent beam of light is also a quantity that can be constructed from observation. Obtaining these quantities one can then construct a (gauge dependent) $\lambda$ such that $x^{\mu}(\lambda)$ reproduces the observables; having constructed such a $\lambda$ from the measurable quantities it is a simple matter to define a reparametrization $\bar{\lambda} = \bar{\lambda}(\lambda)$ such that $p^{\nu}\nabla_{\nu}p^{\mu} = 0$. It is ad-hoc but that's the best one can do when one does not have a comoving clock so as to simply use the clock time as the worldline parameter.

Last edited: Oct 28, 2014
7. Oct 29, 2014

### center o bass

I'm satisfied with the fact that an arbitrary parameter can be turned into an affine one. Thank you! :)