# Affine parameterization of a light ray

Hello,

Is this parameterization correct? -

##r(t) = R = \mbox{const}##
##\theta(t) = 0 = \mbox{const}##
##z(t) = ct##
##t = t##

This is supposed to be the null geodesic curve in the case of a light ray, emitted at point {##r=R,\theta=0,z=0,t=0##} parallel to the ##z-##axis in flat spacetime. I'm posting because I am unsure if the time ##t## can serve as an affine parameter.

Are there better parameterization choices here?

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PeterDonis
Mentor
2019 Award
I'm posting because I am unsure if the time ##t## can serve as an affine parameter.
It can.

Are there better parameterization choices here?
Not really "better", but another equally good choice in this case would be ##z##; the only difference is the factor of ##c##, and if you use "natural" units in which ##c = 1##, that difference goes away.

• kkz23691
It seemed to me that the "coordinate time" ##t## would appear to be a reasonable parameter, because I understand it to be the reading of a stationary clock; while the "proper time" ##\tau## would be the reading of a clock traveling with the ray, and it would not be convenient to use ##\tau## as a parameter because it would show the same reading at any point of the curve.

Matterwave
Gold Member
It seemed to me that the "coordinate time" ##t## would appear to be a reasonable parameter, because I understand it to be the reading of a stationary clock; while the "proper time" ##\tau## would be the reading of a clock traveling with the ray, and it would not be convenient to use ##\tau## as a parameter because it would show the same reading at any point of the curve.
There can be no clock which travels along with the light ray. There is no ##\tau## for a light ray, only an interval ##s=0## along the whole light ray. And you are right, this is definitely not a good affine (or really any other kind as well) parameter. Null curves can not be parametrized by their arc length.

• kkz23691
stevendaryl
Staff Emeritus
Something that always struck me as mysterious is that you can derive the geodesic equation

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

by extremizing proper time: $\tau = \int \sqrt{g_{\mu \nu} U^\mu U^\nu} ds$ (where $U^\mu = \frac{dx^\mu}{ds}$)*

but the equation works unchanged even for lightlike geodesics. The original integral is identically zero for lightlike geodesics, so extremizing it doesn't give you any equations of motion.

* Actually, extremizing proper time gives something like:

$\frac{dU^\mu}{ds} + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda = U^\mu \frac{d}{ds} log(\mathcal{L})$, where $\mathcal{L} = \frac{d \tau}{ds}$. By choosing $s$ to be affine, you can make $\mathcal{L} =$ a constant, so the right-hand side vanishes.

bcrowell
Staff Emeritus
Gold Member
The original integral is identically zero for lightlike geodesics, so extremizing it doesn't give you any equations of motion.
We don't consider the set of all geodesics from A to B and then extremize proper time to pick out one geodesic. We consider the set of all curves from A to B and extremize proper time, and that gives us one curve (or in some cases more than one), which is a geodesic. Did you mean to say "The original integral is identically zero for lightlike *curves*?"

stevendaryl
Staff Emeritus
We don't consider the set of all geodesics from A to B and then extremize proper time to pick out one geodesic. We consider the set of all curves from A to B and extremize proper time, and that gives us one curve (or in some cases more than one), which is a geodesic. Did you mean to say "The original integral is identically zero for lightlike *cuves*?"
Yes, that's what I meant.

bcrowell
Staff Emeritus