## General Relativity Tutorial -

## 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,

A^a = R^a_{bcd} v^b w^c v^d