For time-like geodesic, the affine parameter is the proper time $\tau$ or its linear transform, and the geodesic equation is

$$\frac{\mathrm d^{2}x^{\mu}}{\mathrm d\tau^{2}}+\Gamma_{\rho\sigma}^{\mu}\frac{\mathrm dx^{\rho}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}=0. $$

But proper time $\Delta\tau=0$ for null paths, so what the physical meaning of is the affine parameter for null geodesic?

  • 2
    $\begingroup$ Spacetime is locally flat, and in any flat space you have parallelism. This structure of parallels is completely independent of whether you think of the points as representing points in relativistic spacetime, points in Newtonian spacetime, or points in Euclidean space. It could even be a space such as a graph of temperature versus time. Once you've got a notion of parallelism, you are automatically able to construct a system of measurement along any given line. You can see the construction worked out here: lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.1 $\endgroup$
    – user4552
    Commented Nov 27, 2011 at 15:00

5 Answers 5


If you forget about the affine-ness for a moment: you can parametrize a null geodesic in any way you want. Actually, you can parametrize any geodesic (heck, even any curve) in any way you want; all you need is a monotonic function that maps points on the geodesic to unique values of the parameter. But for timelike geodesics, you almost always use the proper time because it's a nice, sensible physical quantity that also happens to work as a parameter.

With null geodesics, you don't have the proper time as an option because the proper time mapping assigns the same value to all points on the geodesic. So you have to pick some other parametrization. In principle, again, it can be any monotonic function that maps points on the geodesic to unique values of the parameter.

However, it's possible to pick a way to parametrize the null geodesic in a way that is "sensible" in the same way that proper time is "sensible" for a timelike geodesic. This is called an affine parameter. In particular, one way to define an affine parameter is that it satisfies the geodesic equation. (Note: the geodesic equation does not work for just any arbitrary parametrization of a geodesic. You have to use an affine parameter.) Another way is to say that iff the parametrization is affine, parallel transport preserves the tangent vector, as Wikipedia does. Another way is to say that the acceleration is perpendicular to the velocity given an affine parameter, as Ron did. All these definitions are equivalent.

It turns out, although I don't know the details of a proof, that there is a unique affine parameter for any geodesic, up to transformations of the form $t \to at+b$.

  • 12
    $\begingroup$ The property you mentioned in the last paragraph is what affine means in mathematics. $\endgroup$
    – timur
    Commented Oct 31, 2014 at 13:31
  • $\begingroup$ The condition that the acceleration is perpendicular to the velocity, if I understand it correctly, cannot be applied to null geodesics. Even with non-affine parametrization, the acceleration $a^{\mu}=k^{\nu}\nabla_{\nu} k^{\mu}=\kappa k^{\mu}$ is orthogonal to $k^{\mu}$ as $k^{\mu}k_{\mu}=0$. $\endgroup$
    – Curiosa
    Commented Sep 24, 2016 at 2:13
  • $\begingroup$ So, to answer the OP's question: none. $\endgroup$
    – tparker
    Commented Apr 13, 2017 at 5:07
  • $\begingroup$ @tparker Yeah, you could say that ;-) $\endgroup$
    – David Z
    Commented Apr 13, 2017 at 5:10
  • $\begingroup$ +1: This answer goes into first principles as to "why" we do what we do. It seems affine parameter is unique up to t->at+b vs proper time is only unique up to t->t+b, so you lose only a "little" in terms of "the natural parameter". $\endgroup$ Commented Apr 1, 2020 at 21:54

Following David Z's answer, the proof for the last paragraph is:

  • since $t$ is an affine parameter it satisfies: \begin{equation} \frac{\mathrm d^2x^a}{\mathrm dt^2}-\Gamma^a_{bc}\frac{\mathrm dx^b}{\mathrm dt}\frac{\mathrm dx^c}{\mathrm dt}=0 \tag1 \end{equation}

  • the parameter $t'$ must be related in some way to $t$, that is: $$t'=t'(t) \tag2$$

  • use the chain rule to get:

$$\frac{\mathrm dt'}{\mathrm dt}\frac{\mathrm d}{\mathrm dt'}\left(\frac{\mathrm dt'}{\mathrm dt}\frac{\mathrm dx^a}{\mathrm dt'}\right)-\Gamma^a_{bc}\left(\frac{\mathrm dt'}{\mathrm dt}\frac{\mathrm dx^b}{\mathrm dt'}\right)\left(\frac{\mathrm dt'}{\mathrm dt}\frac{\mathrm dx^c}{\mathrm dt'}\right)=0 \\ \Leftrightarrow \left(\frac{\mathrm dt'}{\mathrm dt}\right)^2\frac{\mathrm d^2x^a}{\mathrm dt'^2}+\frac{\mathrm d^2t'}{\mathrm dt^2}\frac{\mathrm dx^a}{\mathrm dt'}-\Gamma^a_{bc}\left(\frac{\mathrm dt'}{\mathrm dt}\right)^2\frac{\mathrm dx^b}{\mathrm dt'}\frac{\mathrm dx^c}{\mathrm dt'}=0 \tag3 $$

  • the affine parameter $t'$ must also satisfy: $$\frac{\mathrm d^2x^a}{\mathrm dt'^2}-\Gamma^a_{bc}\frac{\mathrm dx^b}{\mathrm dt'}\frac{\mathrm dx^c}{\mathrm dt'}=0 \tag4$$

  • comparing $(3)$ and $(4)$ we get the conditions that: $$ \left(\frac{\mathrm dt'}{\mathrm dt}\right)^2\neq0 ~~\wedge~~ \frac{\mathrm d^2t'}{\mathrm dt^2}=0 \tag5 $$

  • which yields $t'=\alpha t+\beta$ with $\alpha \neq0$


An affine parameter is one which makes the acceleration perpendicular to the velocity. It is the limit of proper time as you take the limit that the geodesic becomes null, rescaled to make the total length stay finite.

  • $\begingroup$ An affine parameter makes the inner product (defined with the metric) of the tangent vector (velocity) and its derivative vanish. In the case of a null geodesic, this means that the derivative of the tangent vector is a multiple of the tangent vector (so both are parallel). $\endgroup$
    – ThomasS
    Commented Nov 6, 2022 at 6:52

A useful property of the affine parametrization is the following. At any point along the geodesic we can define the momentum $k^\mu = \dot X^\mu $ where the dot is the derivative with respect to affine time. This momentum is defined up to an overall rescaling, due to the possibility of rescaling the parameter. Now the useful thing is that the same derivative, with respect to the same affine parameter, but at a different point along the geodesic gives now the momentum at a different point. If you had a massless particle with the initial momentum at the first point, then this final momentum gives you the redshifted, or blue shifted momentum that the particle would have along the geodesic.

This implies that the affine parameter can be viewed as proportional to the phase of the quantum wavefunction, or phase of the corresponding wave equation.


According to Lasenby et al. (General relativity An introduction to Physicists)[section 3.16] given a curve $x^{\mu}(u)$, $u$ is an affine parameter if the tangent vector with respect to $u$ is parallel transported along the curve.

  • $\begingroup$ Is this also true for null geodesics, as requested? $\endgroup$
    – Kyle Kanos
    Commented Dec 28, 2019 at 19:23
  • $\begingroup$ Yes! But you have to make sure if you choose a suitable parameter for this kind of geodesic. $\endgroup$ Commented Jan 9, 2020 at 22:08

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