General Relativity Tutorial - Tensors

John Baez

Now what's a tensor? Well, there are a million ways to think of it, but a good way is to think of it as a machine that eats a list of say, 3 tangent vectors, and spits out a number, for example, or maybe a tangent vector. (This isn't quite the most general sort of tensor but it's good enough for starters.) We require that the output depend in a linear way on each of the inputs.

So for example a while back I discussed the Riemann tensor R^a_{bcd}. This was a thing that ate 3 tangent vectors and spit out a tangent vector... which is why there are 3 subscripts and one superscript.

Namely, if we parallel translate a tangent vector u around the little parallelogram of size epsilon whose edges point in the directions of the tangent vectors v and w, it changes by a little bit. Namely, it changes by the tangent vector whose component in the a direction is

- epsilon^2 R^a_{bcd} v^b w^c u^d + terms on the order of epsilon^3

Here we sum over b, c, and d. The thing "v^b" is the component of the vector v in the b direction... in whatever the hell coordinate system we happen to be using. And remember, indices like a,b,c,d range from 0 to 3 if we are working in 4d spacetime.

Oz says: "OK, I bet in reality it's not quite as simple as this however. "

Well, that was a pretty precise definition a large class of tensors, but not of the most general kind.

Here it is again, more formally so you will feel the suffering normally associated with education:

A tensor of "rank (0,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a number. The output must depend linearly on each input.

A tensor of "rank (1,k)" at a point p of spacetime is a function that takes as input a list of k tangent vectors at the point p and returns as output a tangent vector at the point p. The output must depend linearly on each input.

I am avoiding defining tensors of rank (n,k) for other values of n, because there is actually a fair amount of physics I can do without dragging them in. Eventually I would need to explain them, and you'd see they weren't much worse.

Oz remarks: "I take it that "linear way" is a fundamental property of a Tensor."

Indeed!!!!!!!!!! It's a branch of Linear Algebra.

Oz comments: "In general I assume there are an infinity of possible tangent vector directions like v,w above defining some parallelogram of size epsilon (which some nasty person will presumably tend to zero). I presume this is operational for a well behaved space over which tangent vectors are definable."

Sure, there are infinitely many tangent vectors at a point. This is not so bad. For example, note that the output

R^a_{bcd} v^b w^c u^d

(let's ignore that epsilon junk and the niggly minus sign) depends linearly on the inputs u, v, and w, so we don't need to know it for *infinitely* many choices of u, v, and w to figure out what it will be for all possible choices. Linearity keeps life simple.

Oz notes: "Somebody has sneakily brought in some co-ordinates whilst nobody was looking (a,b,c ....). Now telling me they are local just won't do and nor will telling me they are 'whatever co-ordinates you desire'. I have (in GR contexts) been bludgeoned into realising that I need to reconsider the concept of 'co-ordinates' altogether."

Good. There's nothing like a good bludgeoning now and then. Well, certainly they are local coordinates, because you would be hard pressed to flatten out a whole pumpkin and impose *global* coordinates on it, rendering it a mere plane. And certainly the coordinates above are indeed "whatever coordinates you desire". But you are starting to sound like a mathematician --- high praise in my book --- with your complaint about the unpleasant appearance of coordinates. So to reward you, I will explain how it works without coordinates. You'll see it's much simpler.

The Riemann tensor is a tensor of rank (1,3) at each point of spacetime. Thus it takes three tangent vectors, say u, v, and w as inputs, and outputs 1 tangent vector, say R(u,v,w). As usual, the output depends linearly on each input. The Riemann tensor is defined like this:

Take the vector w, and parallel transport it around a wee parallelogram whose two edges are the vectors epsilon u and epsilon v , where epsilon is a tiny number longing to approach zero. The vector w comes back a bit changed by its journey; it is now a new vector w'. We then have

w' - w = -epsilon^2 R(u,v,w) + terms of order epsilon^3

Note: I am now saying just what I said before, but without those yucky coordinates! If you insist on using coordinates, I will say "Go ahead! Pick any that you like! I don't care which!" The only effect will be to turn the above elegant equation into the grungier but sometimes more practical one:

w'^a - w^a = - epsilon^2 R^a_{bcd} v^b w^c u^d + terms of order epsilon^3