# Affine parameters in GR

In describing the length of a curve in spacetime it is necessary to parametrize it and in GR one comes across the notion of affine parameters.
One definition of affine parameters u are parameters which are related 'affinely' to the length of the curve s trough u = as + b where a and b are constants. These parameters has the property that the tangent vector to the curve remain constant in length - however spacetime is a psaudo-Riemannian manifold, so for null curves which have zero length (and are crucial in the description of photons) one can not use affine parameters to parametrize these curves since the length does not vary along them.

Generally a geodisic can be defined as a curve for which the tangent vector remain parallel to itself i.e.

$$\frac{d \vec t}{du} = \lambda(u) \vec t$$

and another definition of affine parameters is to parametrize the curve such that

$$\frac{d \vec t}{du} = 0.$$

But this implies that also the length of the tangent vector remain constant, so I worry that this definition of an affine parameter is equivalent with the old one which could not be used for null curves. Is it possible to find parameters which are not related by to the length of the curve, but which still make the length of the tangent vector constant along the curve? If so does anyone have an example of such a parameter?

Or to state my question more clearly: Does one have to use an affine parameter in order to satisfy the geodisic equation

$$\frac{d \vec t}{du} = 0.$$

If not is this somehow obvious? Any examples?

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Dale
Mentor
These parameters has the property that the tangent vector to the curve remain constant in length - however spacetime is a psaudo-Riemannian manifold, so for null curves which have zero length (and are crucial in the description of photons) one can not use affine parameters to parametrize these curves since the length does not vary along them.
Hmm, are you sure about this. I thought that you could use affine parameters on a null path, you just cannot use the spacetime interval.

Hmm, are you sure about this. I thought that you could use affine parameters on a null path, you just cannot use the spacetime interval.
From the first definition which defines an affine parameter to be related to the length, i.e. the spacetime interval. So if one takes this as a definition one can not use an affine parameter.

However from the second definiton which defines an affine parameter to have the property that ##\lambda(u) =0## it might be possible. My worry is just that these definitions might be equivalent and that u then has to be related to the spacetime interval since it implies a constant tangent vector.

I guess it does not have to, but I would like to know why.

WannabeNewton
Hmm, are you sure about this. I thought that you could use affine parameters on a null path, you just cannot use the spacetime interval.
What DaleSpam says is correct regarding affine parameters and null geodesics (see below).

Let ##\gamma :J\subseteq \mathbb{R}\rightarrow M## be a regular curve (we say a curve is regular if it is a parametrized ##C^{\infty}## function into the codomain). Here ##J## is an interval (so that it is connected) and ##(M,g_{ab})## is the space-time.

The usual definition of a geodesic we see in GR is the following: ##\gamma## is a geodesic iff ##\nabla_{\xi}\xi = 0## where ##\xi## is the tangent to the curve or, in the usual GR notation, ##\xi^{a}\nabla_{a}\xi^{b} = 0##. However, we can relax this condition to just ##\xi^{a}\nabla_{a}\xi^{b} = \alpha \xi^{b}## for some scalar function ##\alpha = \alpha(t)## where ##t## is the curve parameter. We can do this because the intuitive notion of a geodesic as being "as straight as possible" only requires that the tangent vector be parallel transported not that its length remain constant, which is what the usual definition demands on top.

Now, define a new curve parameter by ##t' = \beta (t)##. Then, ##\xi^{a}\nabla_{a}\xi^{b} = \dot{\beta}^{2}\xi'^{a}\nabla_{a}\xi
'^{b} + \dot{\beta}\xi'^{a}\xi'^{b}\nabla_{a}\dot{\beta} = \alpha \dot{\beta} \xi'^{b}##. Note that if we then choose ##\beta## so that ## \alpha = \xi'^{a}\nabla_{a}\dot{\beta} ## i.e. ## \alpha\dot{\beta} = \xi^{a}\nabla_{a}\dot{\beta} = \ddot{\beta}## then ##\xi'^{a}\nabla_{a}\xi'^{b} = 0## so we can always re-parametrize our curve so that it takes the usual form anyways. The above is an ordinary differential equation for ##\dot{\beta}## that can be solved using integrating factors (by the way, this is problem 3.5 of Wald). Note in particular that if ##\xi^{a}## is the tangent to a null geodesic i.e. ##\xi^{a}\xi_{a} = 0## then ##\xi'^{a}\xi'_{a} = 0## as well, where ##\xi^{a} = \beta \xi'^{a}## as above, so the re-parametrization will keep the geodesic null. Thus we can always parametrize null geodesics by an affine parameter.

If our geodesic ##\gamma## is such that ##\nabla_{\xi}\xi = 0## then we say it is affinely parametrized. It is easy to show that any other re-parametrization of ##\gamma## such that ##\nabla_{\xi}\xi = 0## still holds after the re-parametrization must be of the form ##t \mapsto at + b, a,b\in \mathbb{R}## i.e. all affine parameters of ##\gamma## are linearly related.

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pervect
Staff Emeritus
I agree, null geodesics have affine parameterizations.

Wald has a discusssion of this on pg 41, and if you want a practical example:

Consider the Schwarzschild metric. Define a more or less random null tangent vector ##u^a##= [1-2m/r, 0, 0, 1].

Wald's remarks boil down to:

an affinely parameterized geodesic satisfied ## u^a \nabla_a u^b = 0##, a non-affine paramaterization yields the weaker condition ## u^a \nabla_a u^b = \alpha u^b ##

Wald also remarks that: "It is easy to show that given a curve which satisfies (the second condition), we can always reparameterize it so that it satisfies (the first)."

I haven't done this "easy exercise" personally, Wald mentions problem 5.

A note on notation: ##\nabla_a## is the covariant derivative. Some older texts use semi-colon noation, in that notation ## \nabla_a u^b = u^a{}_{;b}##

Back to our example

Check that ##u^a##= [1-2m/r, 0, 0, 1].is a null vector. Then check to see whether or no it's affinely parameterized. Compute

## u^a \nabla_a u^b ##. Find that it is NOT affinely parameterized. Not too surprising, since we picked a semi-random null vector.

If you want to finish convincing yourself, you'll have to do what Wald suggested and demonstrate a reparameterization of this to make it affinely parameterized.

I'd suggest solving for the vector field so that ## u^a \nabla_a u^b = 0 ##, then demonstrating that they both generated by the same radial infall curve or "orbit".

I haven't done this part, I'm trusting my textbook on this one,.

As an aside, I know some people do not like the idea of defining parallel transport in the manner Wald does. They insist (with good reason) that $\nabla_a$ has to be defined for a congruence of curves.

I'm not sure how they deal with the issue of null geodesics.

pervect
Staff Emeritus
Just in case it helps:

For the example ##u^a## = [1-2m/r, 0, 0, 1]

##
\nabla_a u^b = \left[ \begin {array}
{cccc} {\frac {m}{{r}^{2}}}&0&0&{\frac { \left( r-2\,m \right) m}{{r}^
{3}}}\\0&{\frac {r-2\,m}{{r}^{2}}}&0&0
\\0&0&{\frac {r-2\,m}{{r}^{2}}}&0
\\{\frac {m}{ \left( r-2\,m \right) r}}&0&0&{\frac {
m}{{r}^{2}}}\end {array} \right]
##

using GRTensor

WannabeNewton
I haven't done this "easy exercise" personally, Wald mentions problem 5.
I did it in my above post, if you're interested.

pervect
Staff Emeritus