Light bending in Schwarzschild spacetime

In 1919 Sir Arthur Eddington observed a solar eclipse, and made some very important observations. He looked at light from stars passing close to the sun, and found that the stars were out of place. Of course, they only looked out of place. Their light was being bent by the sun. This was one of the earliest confirmations of Einstein's General Theory of Relativity. How exactly does general relativity make this prediction? Let's look at the calculation.

As always in physics, we start by writing down the action for the system. The action for a relativistic point particle is given by

We want to investigate light bending around the sun, a spherical object, so the best metric to use is the Schwarzschild metric,

where we set .

Let us look at the action. It is reparameterization invariant; we can replace and the action will remain the same. That's good — physics should not depend on the coordinates. It also vanishes when . That's not good, since we are looking at light beams. The way out is to introduce an extra parameter and work with this action instead:

This extra parameter is called an “einbein.” In order for the action to still be invariant under reparameterizations, we need to transform as , as you can easily check. Varying the action with respect to gives

When you plug this into the new action, you get back the old familiar action. Incidentally, from the equation for , we see that it is proportional to the square root of the metric on the particle's worldline. This shows us that the arbitrary choice of parameterization of the worldline is equivalent to an arbitrary choice of . Taking an affine parameterization , where is the proper time, then, since is a constant, we can treat as independent of .

Let's now consider massless particles, . In this case we find that implies that . This shows us that light rays travel on null geodesics. We still have the freedom to choose . Choosing to be constant is again called affine parameterization. Let us choose . The Lagrangian becomes:

where . Note that we cannot call the proper time now — proper time is not defined for massless particles. It is simply a parameterization of the worldline for the light ray.

The Euler-Lagrange equation for gives

Without loss of generality, we can choose coordinates such that and initially. It should be easy to see why this can be done for . You can see why it can be done for as well if you think of the particle's velocity vector initially. It's an arrow, pointing in a certain direction. You can then rotate the coordinate system so that this vector is tangent to a line of constant . This gives initially. Now look at the equation of motion for again. With the initial conditions, initially, giving initially. You can differentiate the equation of motion times if you want, and you will always find that initially. This means that with the initial conditions we have chosen, will be for all time. So orbits are restricted to a plane, as they are in classical mechanics.

That simplifies things. Let's look for more conserved quantities. Notice that the action does not depend on time. Using the Euler-Lagrange equation for , we get that , where is a constant. If you use Noether's theorem you can show that this is the conserved charge corresponding to the time-independence of the action, and so it corresponds to the energy. In the exact same way, we can solve the Euler-Lagrange equation for and get , and is the angular momentum.

We can now use that light rays travel on null geodesics to get another equation of motion:

Here we have used that and substituted in the constants and . Let us now define a dimensionless variable , where is some length scale. We also define and . We plug all this stuff in and find, after the dust settles, that

where .

We now have a more useful equation for the path that a light ray will take as it bends around the sun. For one thing, we have eliminated the arbitrary parameter and only have as a function of . That is all we need to find the angle by which the path of the light ray bends.

Let us assume that . This means that we are assuming the light ray passes far from the sun, relative to its mass. Now we make an approximation. We will try to solve for order by order. Let . To zeroth order,

It is insightful to consider this solution geometrically. Converting back to gives . But this is precisely the equation of a straight line a distance from the origin (in polar coordinates.) The line comes in from infinity at , is at the minimum distance when , and goes to infinity again at . In our case, the sun is at the origin. The line is straight, so no light bending occurs at this order. We'll have to go to the next order.

To first order, we use and plug it into the equation of motion.

Using the solution for and simplifying,

The solution for is

for a constant . This is the deviation from the straight line.

We can now find how much the deviation is exactly. A straight line comes from infinity at (at the right in the figure) and goes to infinity at (at the left in the figure.) The actual trajectory is a slight perturbation of this — it comes in at on the right and leaves at on the left, with .

When , , so we solve this equation for the solutions near to get :

For we get (after linearizing)

and for

We can add these two equations together, and get that the total deflection is . This deflection has been measured experimentally.