General Relativity Tutorial - The Metric

John Baez

Oz says: "I feel that the concept of 'metric' has been rather glossed over. I mean, it has a proper name that almost means something. Something to do with 'measurement'. Could you possibly expand a little on the above, please?"

The "metric" is no more than the generalization from space to spacetime of the dot product you know and love. It "measures" lengths and angles.

Oz asks: "So do I have it right? The metric is merely (!) a prescription or mechanism to extract 'lengths' and 'angles' from a pair of tangent vectors. We could chose all sorts of mechanisms, most of which would be 'unphysical', but for GR we use the g one as in g(v,w)."

Yes. For each pair of tangent vectors v,w based at the same point, g(v,w) is a kind of "dot product" of them. If we were in boring flat old Minkowski space... the land of *special* relativity... the vectors v and w would look like this:

v = (t,x,y,z)

w = (t',x',y',z')

and the metric would be given by:

g(v,w) = - tt' + xx' + yy' + zz'

Just the usual dot product you learn in college, with a minus sign thrown in front of the time part to make time be different from space. Alternatively, we could throw minus signs in front of all the space parts. General relativists like to do it the way I do above, so that the geometry of space stays as much like it used to be as possible. Particle physicists like to do it the other way.)

Oz wonders: "Presumably there are other metrics (not g(v,w)) that apply to other strange and wonderful constructs in mathematics that we need not consider here. Probably luckily."

Well, in the world of general relativity, "metric" means no more and no less than a symmetric nondegenerate tensor of rank (0,2), or if you prefer, a dot product thingie.

In other realms, "metric" means other things, and then to be specific we say that the metric in GR is a "Lorentzian metric" if it has that minus sign in front of the time part, or a "Riemannian metric" if it's just for space. But let's not worry about those other realms just now, eh?

Keith Ramsay elaborates: "The metric on space-time consists of the information describing lengths and times. (For those who have time for a book, Geroch's "Relativity from A to B" gives what seems a good explanation of the physical significance of the metric.)

In Minkowski space, one has a metric dt2-dx2-dy2-dz2. This takes two vectors (x1,y1,z1,t1) and (x2,y2,z2,t2) and gives back the number, t1t2-x1x2-y1y2-z1z2. Here I'm setting the units so that the speed c of light is 1. I'm also adopting the "+---" convention. The "-+++" convention (change all the signs) is also used.

Suppose you are in Minkowski space-time, and that your path can be described by the parametrized curve (x(s),y(s),z(s),t(s)). The elapsed time on your clock between s=0 and whenever you are at a given value of s is a function of s. Let's call it f(s). f'(s) is the rate at which time is passing per unit of s (whatever s may be).

From special relativity, one knows that in this model, the rate of elapsed time for you, relative to t, is sqrt(1-v2/c2). Since I'm setting c=1, this is just sqrt(1-v2). To find the rate of elapsed time w.r.t. s, multiply by dt/ds: sqrt(1-v2)dt/ds. Then of course v2=(dx/dt)2+(dy/dt)2+(dz/dt)2, so one gets

sqrt((dt/ds)2 (1-(dx/dt)2-(dy/dt)2-(dz/dt)2)) =

sqrt((dt/ds)2 - (dx/ds)2 - (dy/ds)2 - (dz/ds)2)

using the chain rule from calculus. Now the square of this

(dt/ds)2 - (dx/ds)2 - (dy/ds)2 - (dz/ds)2

is what you get by applying the metric, above, to the vector (dx/ds,dy/ds,dz/ds,dt/ds), the tangent vector to your parametrization of your path (world-line) through space-time.

This is directly analogous to the way that one can compute the length of a parametrized curve (x(s),y(s),z(s)) in Euclidean space by integrating

sqrt((dx/ds)2+(dy/ds)2+(dz/ds)2)

over the portion of the curve in question. The metric in space-time "measures" elapsed time, roughly the way the metric in Euclidean space "measures" distance.

It also measures distances, although the interpretation seems a little less direct. If you have a tiny rigid rod moving along a path in space-time, the points in space-time occupied by its ends, which are simultaneous in the locally inertial frame for the rod, are a distance apart equal to the length of the rod. Here, you can compute distance using the metric with the sign changed:

sqrt((x1-x2)2+(y1-y2)2+(z1-z2)2-(t1-t2)2),

where (x1,y1,z1,t1) and (x2,y2,z2,t2) are the coordinates of the points of space-time in question.

Now, all that was special relativity (!). In general relativity, one deals with other space-times. The metric, however, will do the same for you. Some astronaut has had an amazing career in which he's flown close to event horizons of black holes and things like that. Once he's retired and died of old age, it's up to you to write an obituary. One thing you want to work out: how old was he, after all that? Sure he was born in 2025 and died only in 2211, but what with all the time-dilation he certainly wasn't 186 years old. So draw his world-line in space-time, and parametrize it somehow. Then use the metric to determine the rate of elapsed personal time for him, per unit of parameter. (It's sometimes nice to make elapsed time itself the parameter.) Apply the metric tensor to the tangent vector v at a point and v. (It's a 2-tensor, which takes two tangent vectors and gives back a number. Put the same vector in both

arguments.) Take the square root of the result. That's the rate of elapsed time per unit of parameter.

Angles come out of the metric in a way similar to the way angles can be figured out from distances in ordinary geometry. The metric is directly analogous to the "dot product", v.w=|v||w|cos(angle).

The metric is assumed to be one which at any one point can be transformed by a change of coordinates into a Minkowski metric. So there is always an implicit "+---" (or "-+++") "signature". This makes the geometry used in general relativity different from much of metric differential geometry, where the signature is positive. One manifestation of this feature is that in some tangent directions to space-time, the metric is 0. These are "null" directions, and massless particles (such as light, presumedly) travel along them. A common tactic is to employ some mathematical tool to convert the problem into one which involves a positive metric (using "imaginary time")."