General Relativity Tutorial - Torsion

John Baez

Oz asks: "Why do we assume [the connection] in GR is torsion free? A short couple of non-sarcastic information-rich understandable sentences would suffice. Enough to place it in a footnote of the brain."

Non-sarcastic, eh? You must think I'm really mean and nasty, when actually I am the sweetest, nicest guy you could ever meet.

Relatively few people understand why in GR we assume the connection --- the gadget we use to do parallel translation --- is torsion-free. It's often presented as a technical assumption not worth trying to understand. But here's a nice way understanding what torsion-free-ness means (in conjunction with our other assumptions: that parallel translation be linear and metric-compatible).

Take a tangent vector v at P. Parallel translate it along a very short curve from P to Q, a curve of length epsilon. We get a new tangent vector w at Q. Now let two particles free-fall with velocities v and w starting at the points P and Q. They trace out two geodesics. Let me try to draw this:

    |      |
    |      |
    |      |
    ^v     ^w
    |      | 

Remember, this is a picture in spaceTIME. Here I've drawn what it might look like in flat Minkowski spacetime, where the geodesics are boring old straight lines, and I've drawn everything very rectilinearly, since ASCII is so bad at drawing curves. Okay. Now, let's call our two geodesics C(t) and D(t), respectively. Here we use as the parameter t the proper time: the time ticked out by stopwatches falling along the geodesics. (We set the stopwatches to zero at the points P and Q, respectively.)

Now we ask: what's the time derivative of the distance between C(t) and D(t)? Note this "distance" makes sense because C(t) and D(t) are really close, so we can define the distance between them to be the arclength along the shortest geodesic between them.

If our connection is torsion-free, then no matter how we choose P and Q and v, the time derivative of the distance between C(t) and D(t) at t = 0 is ZERO, up to terms proportional to epsilon2! (One can derive this from the definition of torsion, assuming our recipe for parallel transport is metric preserving.)

If v got "rotated" a bit when we dragged it over to Q, and things looked like this:

    |           /
    |          /
    |         /
    ^v       ^w
    |       / 

then the time derivative of the distance would not be zero (it'd be proportional to epsilon). In this case the torsion would not be zero.

In GR we assume this kind of "rotation" effect doesn't happen. In some other theories of gravity there is torsion. But there's no experimental evidence for torsion, so most people stick with GR.

You'll notice I gave a necessary condition for a connection to be torsion free, but not a sufficient one. So, I haven't fully explained torsion-freeness: it actually says a bit more. Here's what it says.

Work using coordinates so that we can freely identify vectors with points. Consider a small square of size epsilon in the xy plane, where x and y are names for the first two our coordinates. Imagine this as the usual xy plane you know and love. Suppose the bottom left corner of our square is the origin.

Now, we can take the vector (epsilon,0,0,...) at the origin --- it's pointing in the x direction --- and we can parallel transport it in the y direction to the top left corner of our square, getting a vector v.

Alternatively, we can take the vector (0,epsilon,0,0,...) at the origin --- it's pointing in the y direction --- and we can parallel transport it in the x directin to the bottom right corner of our square, getting a vector w.

The condition that the connection be torsion free says that the tips of v and w touch --- up to terms of order epsilon^3. The idea is that neither v nor w has rotated. If this is true in every coordinate system, our connection has no torsion.