The Geodesic Deviation Equation

John Baez

Start with two clocks next to each other, out in the wilderness of empty space! Now drag one a foot away from the other and do your best to leave it at rest relative to the first. Let them float out there in the wilderness of space. They are in free fall, so they trace out geodesics. These geodesics may converge or diverge, and the rate of this "geodesic deviation" may be measured by shining a laser from one to the other and measuring the redshift.

But what does geodesic deviation have to do with curvature, exactly? Let's look at a spacetime diagram of the situation. L et's use "v" to denote the velocity vector of the first clock at time zero, and let "w" denote the vector from the first clock to the second":

```   |       |
|       |
v^       ^
| w     |
P->-----Q

```
I've labelled the initial positions of the two clocks at rest by P and Q. Note that the velocity vector of the clock at Q is just the result of PARALLEL TRANSPORTING v in the direction w.

Now, suppose we let each clock wait a second. They now have new positions (in spacetime) P' and Q'.

```   P'->----Q'
|       |
|       |
v^       ^
| w     |
P->-----Q

```
Now, what's the velocity vector of the clock at Q'? Well, think how we got it: first we parallel transported v over to Q along w. Then we parallel tranported the result over to Q', since the curve from Q to Q' is a geodesic, which means its velocity vector is parallel translated along itself.

Now for the big question! We want to know if the clock at Q' is moving away from the clock at P'. To answer this, we compare its velocity vector to the following vector: what we get by first parallel translating v along itself over to P', and then over to Q'. That's the velocity the clock at Q' would have it it were at rest relative to the clock at P'.

Note that when we do this, we are taking the vector v and parallel translating it two different ways from P to Q' and getting two slightly different answers... then we compare these answers. If the answers were the same, the second clock would remain at rest relative to the first. But in fact they are not, and the difference tells us how the second one begins accelerating away from the first.

Now remember how curvature works: the result of

dragging v from P to Q to Q'

minus the result of

dragging v from P to P' to Q'

is going to be

-R(w,v,v)

where R is the Riemann tensor. But this is just the same as

R(v,w,v)

since the Riemann tensor is defined so that it's skew-symmetric in the first two slots.

In short, the GEODESIC DEVIATION EQUATION says the following:

Two initially comoving particles in free fall will accelerate relative to one another in a manner determined by the curvature of space. Suppose the velocity of one particle is v, and the initial displacement from it to the second is small, so that it may be represented as a vector w. Then the acceleration A of the second relative to the first is given by R(v,w)v. Or if you like indices,

Aa = Rabcd vb wc vd