I'm trying to understand which part of the following metric determines whether photons travel on a "straight" line (thinking of $(t,r,\theta,\phi)$ as a flat background), the metric I'm considering is:

$$ds^2 = -F(r)dt^2 + F(r)^{-1}dr^2 + r^2 d\Omega^2$$

In the Minkowski case ($F = 1$) it's easy to see that the geodesics travel on straight lines because we can transform to cartesian coordinates and see this immediately.

My difficulty is in understanding the distinction between "de Sitter" space ($F(r) = 1 - r^2$) and "Schwarzschild" ($F(r) = 1 - \frac{1}{r}$). In both these cases the $\phi$ equation of motion is the same:

$$2r\dot{r}\dot{\phi} + r^2 \ddot{\phi} = 0$$

Then by uniqueness of solutions to ODEs it seems that if $\phi = 0$ and $\dot{\phi}=0$ initially then $\phi=0$ for all s (s being the geodesic parameter). (This is after all the same usual argument for restricting to the equatorial plane when our metric is spherically symmetric - right?)

So now I'm struggling to understand how the nature of $F$ tells us whether the null geodesics are straight - in the De Sitter case I understand that photons should travel an on straight lines and in the Schwarzschild case clearly not (e.g. gravitational lensing).

  • $\begingroup$ You need to solve for the orbit to determine if an object moves "in a straight line". For the case you specify (zero initial velocity in the $\phi$ direction), you will get straight line motion anyway (right though $r=0$). $\endgroup$
    – levitopher
    Commented May 8, 2016 at 19:55
  • $\begingroup$ possible answer? physics.stackexchange.com/q/205125 $\endgroup$
    – levitopher
    Commented May 8, 2016 at 20:00
  • $\begingroup$ @levitopher Thanks for your comment - sure, in that case they will go straight through the origin - I was just writing down that equation really as an example that these two metrics have the same equations of motion (except in the r direction) and where is the difference that leads to non-straight orbits. The linked answer doesn't really answer the question - I'd really like to see an argument coming directly from the geodesic equations. I'm aware of the conformal diagram for de Sitter space. Thanks $\endgroup$
    – Wooster
    Commented May 8, 2016 at 20:52
  • $\begingroup$ Ok, but the answer comes from the differences in the radial equation of motion. I think the best you can do is solve the geodesic equations for a generic function $F$, and notice that you will need to take derivatives of that function. Different functions, different derivatives. In fact, the wikipedia page almost shows you exactly what I mean: en.wikipedia.org/wiki/Schwarzschild_geodesics, go down to "Mathematical derivation of the orbit equation" and compare with what you have. $\endgroup$
    – levitopher
    Commented May 10, 2016 at 2:49
  • 1
    $\begingroup$ What is the definition of straight line here? $\endgroup$
    – Prahar
    Commented May 10, 2016 at 21:23

2 Answers 2


We can write down the Lagrangian for this problem with general $F$: (restricted to the equatorial plane)

$$L = -F(r)\dot{t}^2 + F(r)^{-1}\dot{r}^2 + r^2 \dot{\phi}^2$$

Now using the conserved quantities (energy and angular momentum), and using the convenient substitution $u = \frac{1}{r}$ and rewriting the problem as a differential equation for $u = u(\phi)$ we find the following:

$$\frac{d^2u}{d\phi^2} + \frac{u^2}{2}F'(u) + uF(u) = 0$$

Now by rotational symmetry we need only consider straight line solutions $u = k \cos \phi$ for some constant k and substituting this into our problem we find that we have straight line solutions if and only if $F$ satisfies the differential equation:

$$F'(u) + 2F(u) - 2 = 0$$

In particular this is true for the De Sitter case but not the Schwarzschild case. (We of course also have the restriction that $F$ must be such that the metric solves Einstein's equation and I believe these are the only two non-trivial cases)


Calculating the geodesic equations in something like De Sitter is not hard, but it's already been done so I'll just link them. For instance, I found a thesis by someone name Chris Ripkin:


Go to chapter 3, "Geodesics". Since De Sitter is maximally symmetric, the geodesics will have constant angular coordinates. You can check out the radial coordinate in the link, but these are straight, radial paths.

I guess the problem with what you're trying to do (determine how the nature of $F$ characterizes the geodesics) is that you're not considering the reason $F$ has the specific form that it does. The Einstein equations for DeSitter are $$R_{\mu\nu}=\Lambda g_{\mu\nu}$$ (vacuum universe with cosmological constant). In addition, when we find the metric, we set the integration constant $M=0$. For Schwarzschild, we assume zero cosmological constant and non-zero integration constant $M$. Ok we started with the same form for the metric, but these two choices give you two very different geometries.

  • $\begingroup$ So because it is maximally symmetric then showing that the geodesic through the origin is straight is enough? $\endgroup$
    – Wooster
    Commented May 10, 2016 at 22:15

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