Let $(M,g)$ be the Schwarzschild spacetime. The usual metric tensor in Schwarzschild coordinates reads: $$g=-f(r)dt^2+f(r)^{-1}dr^2+r^2d\Omega^2\quad f(r)=1-\frac{2M}{r}.$$

Now consider radial null geodesics. One can show that $r$ works as an affine parameter along such curves, which can be written as

$$t(\lambda)=\pm r_\ast(\lambda)+C,\quad r(\lambda)=\lambda,\quad \theta(\lambda)=\theta_0,\quad \phi(\lambda)=\phi_0.$$

Where $r_\ast(r)$ is the function of $r$ defined by $$r_\ast(r)=r+2M\ln \frac{|r-2M|}{2M}.$$

So we have one fact here: $r$ works as an affine parameter for radial null geodesics, be incoming or outgoing.

We can further use this to construct the Eddington-Finkelstein coordinate systems $(u,r,\theta,\phi)$ and $(v,r,\theta,\phi)$ on which these curves become coordinate lines. Of course $r$ is still an affine parameter.

On the other hand we have the Kruskal-Szekeres null coordinates $U,V$. These are defined by $$U=-e^{-u/4M},\quad V=e^{v/4M}.$$

I've read that they are affine parameters on top of the radial null geodesics. But why is that? I can't understand this.

My naive guess was that one had to switch to $(u,v,\theta,\phi)$ coordinates and work out the relation between $u,v$ and the affine parameter.

So take one outgoing radial null geodesic. We have $(u,\theta,\phi)$ constant. We further have


Thus $$v=u_0+2r+4M\ln \frac{r-2M}{2M}.$$

Now perform a change of affine parameter by $r\mapsto \lambda+2M$. We would have (upon redefining $u_0$)


If we could discard $u_0+2\lambda$, this would give

$$v=4M\ln (\lambda/2M)\Longrightarrow \lambda = 2M e^{v/4M}$$

But it doesn't seem that we can discard $u_0+2\lambda$, even with further affine reparameterization.

So what is going on here? Why people say that $U,V$ are affine parameters? Or they are somehow affine parameters just on the horizon? In that case $\lambda =0$. But then again, we would have $\ln 0$ and this wouldn't make sense, so this is clearly wrong.


2 Answers 2


After working this out I'm confident that I have found out the answer to my question. I shall post it here. Corrections are highly appreciated in the event something is wrong. The main point seems to be:

In Schwarzschild spacetime there are three categories of radial null geodesics: the ingoing, outgoing and the horizon generators.

This has been also a subject of another question of mine which I answered giving the details of this construction.

The ingoing/outgoing radial null geodesics have $r$ as the affine parameter and they have a simple characterization in terms of the Eddington-Finkelstein coordinate systems.

Indeed in the ingoing Eddington-Finkelstein coordinates $(v,r,\theta,\phi)$ the ingoing radial null geodesics are simply the coordinate lines of $r$, parameterized as $(v_0,r,\theta_0,\phi_0)$ by $r$ which is an affine parameter.

In the outgoing Eddington-Finkelstein coordinates $(u,r,\theta,\phi)$ the outgoing radial null geodesics are simply the coordinate lines of $r$ as well, with parameterization $(u_0,r,\theta_0,\phi_0)$ by $r$ which is again an affine parameter.

Regarding these two classes of radial null geodesics, using the Eddington-Finkelstein coordinates is merely a convenient choice of coordinates making them coordinate lines. They can also be written down and studied in Schwarzschild coordinates as I show in the other answer.

Now we come to the horizon generators. These are radial null geodesics with $r$ constant. They are obviously a separate class since the ingoing/outgoing are parameterized by $r$ which can't be turned into a constant simply by reparemeterization.

It can be shown - and the details are reffered again to the linked question - that this class only has solutions when the constant $r$ is $r = 2M$. These therefore can't be studied with Schwarzschild coordinates simply because this region is not covered by that chart.

It has two parts, however, each covered by one of the Eddington-Finkelstein coordinates. Working in the ingoing system, it can be seen that such radial null geodesics are parameterized by $$(v,r,\theta,\phi)=(4M\ln \lambda,2M,\theta_0,\phi_0)$$ where $\lambda$ is an affine parameter. In the same way, in the outgoing Eddington-Fineklstein system, these radial null geodesics are parameterized by $$(u,r,\theta,\phi)=(-4M\ln\lambda,2M,\theta_0,\phi_0)$$

It can be shown that the surface $r = 2M$ in each of these systems is a null surface and these geodesics are precisely the generators of the surface.

Finally, if one pass to the double-null Eddington-Finkelstein coordinates $(u,v,\theta,\phi)$ a drawback occurs. The surface $r = 2M$ is pushed to infinity of the coordinates. One therefore would like to introduce coordinates pulling it back to finite coordinate value.

A nice and geometrically motivated way is to peform a coordinate transformation satisfying two conditions:

  1. We leave the angular part alone and transform $U = U(u)$ and $V = V(v)$. This ensures that $v$ constant is the same as $V$ constant and $u$ constant is the same as $U$ constant. Thus in such a system, the ingoing/outgoing radial null geodesics are still coordinate lines. This changes nothing about their affine parameter being $r$.

  2. The coordinates $U,V$ should, when evaluated along the $r = 2M$ generators give the affine parameter. This makes then well defined and finite at $r = 2M$ by definition.

Inspection of the parameterizations of the horizon generators suggest $U = - e^{-u/4M}$ and $V = e^{v/4M}$. They satisfy the two conditions, the minus sign being a convenience, which doesn't change the fact that $U$ computes an affine parameter along horizon generators.

Finally, it can be seen that along the outgoing/ingoing null geodesics, $U$ and $V$ can't be affine parameters. This is because their coordinate lines, which are the outgoing/ingoing null geodesics, if parameterized by $U$ and $V$ do not satisfy the autoparallel equation in general. This would demand $\Gamma_{UU}^U =0$ and $\Gamma_{VV}^V=0$. But this answer shows these connection coefficients which can be seen to be zero just on the horizon.

So the conclusions are:

  1. The Kruskal Coordinates $U$ and $V$ have as coordinate lines all radial null geodesics. The ingoing ones are $(V,\theta,\phi) = (V_0,\theta_0,\phi_0)$ with $V_0 \neq 0$. These are affinely parameterized by $r$, and when parameterized by $U$ are not affinely parameterized since $\Gamma_{UU}^U\neq 0$ here.

    The outgoing ones are $(U,\theta,\phi)=(U_0,\theta_0,\phi_0)$ with $U_0 \neq 0$. These are again affinely parameterized by $r$ and when reparameterized by $V$ are not affinely parameterized since $\Gamma_{VV}^V\neq 0$ here.

  2. The horizon generators are also coordinate lines. The future horizon generators are $(U_0,\theta_0,\phi_0)$ with $U_0 = 0$ and they are affinely parameterized by $V$. The past horizon generators are $(V_0,\theta_0,\phi_0)$ with $V_0 =0$ and are affinely parameterized by $U$. Obviously $r$ can't be a parameter here since it is constant by definition.

So $(U,V)$ and $r$ are affine parameters of radial null geodesics, but each for certain subcategories of said geodesics.


From Wald, consider the constant of motion $E = -g_{ab}k^a(\partial/\partial t)^b$ where $k^a$ is tangent to the null geodesic (and $(\partial/\partial t)^b$ is a timelike Killing field). Write the metric as $$ ds^2 = -\frac{2Me^{-r/2m}}{r}e^{(v-u)/4M}\, du\,dv$$ where the $r$-dependent factor is an implicit function of $u$ and $v$. The constant of motion can be written $E =e^{(v-u)/4M}\,dt/d\lambda$, where $\lambda$ is an affine parameter along the geodesic. Then solve the integral: $$\lambda = \frac{1}{2E}\int e^{(v-u)/4M}\,dv$$ where $u$ is being held constant, giving $V = e^{v/4m}$. $U$ is derived similarly.

  • $\begingroup$ Thanks for the answer. But that's not $E$ for Rindler? For Schwarzschild I got, in $(u,v)$ coordinates $$E=g(\partial_t,\gamma')=g_{uv}(\dot{u}+\dot{v})= -\frac{2Me^{-r/2m}}{r}e^{(v-u)/4M}(\dot{u}+\dot{v}).$$ For $u$ constant, the equation would be $$E= -\frac{2Me^{-r/2m}}{r}e^{(v-u)/4M}\dfrac{dv}{d\lambda}$$. This doesn't seem to be easily integrable as the Rindler case because of the $e^{-r/2M}/r$ part. Is there some workaround? I tried using the information that $r$ is the affine parameter to proceed but didn't get $U,V$. $\endgroup$
    – Gold
    Commented Mar 8, 2019 at 2:38
  • $\begingroup$ Can’t we perform the $v$ integration while holding $r$ constant? $\endgroup$
    – bapowell
    Commented Mar 8, 2019 at 3:23
  • $\begingroup$ But the problem is that $v$ has been defined in terms of $r$ right? In other words $$v = t+r_\ast = t+r+2M\ln \frac{r-2M}{2M}.$$ In fact $$r_\ast = \frac{v-u}{2}$$ defines $r$ implicitly as a function of $(u,v)$. So even if $u$ is constant, it is stil a function of $v$. Please correct me if I'm wrong. $\endgroup$
    – Gold
    Commented Mar 8, 2019 at 3:27
  • $\begingroup$ I haven't worked it out, if you can't tell ;), but my suspicion is that if you write the $d\lambda$ integral in terms of $r$, you'll get $\lambda \propto \exp(r_*/2M)$. $\endgroup$
    – bapowell
    Commented Mar 8, 2019 at 18:20
  • $\begingroup$ I concluded that $U,V$ are not afine parameters on top of ingoing/outgoing null radial geodesics of Schwarzschild. They are affine parameters on top of horizon generators. If they were affine parameters on said geodesics, for $r > 2M$ we would have $\Gamma_{UU}^U$ and $\Gamma_{VV}^V$ zero by the geodesic equation in affine parameter form. These Christoffel symbols are computed (physics.stackexchange.com/questions/404611/…) and nonzero for $r > 2M$. I added one answer. Corrections are highly appreciated. Thanks for the help! $\endgroup$
    – Gold
    Commented Mar 15, 2019 at 14:50

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