# What objects, in General Relativity, carry units?

1. Mar 12, 2006

### pensano

Hi, maybe someone could answer this for me, or at least confirm my answer. (It's not homework.)

What objects, in General Relativity, carry units?

My thinking is that coordinates, $$x^i$$, on manifold patches have no units. And the parameter, $$t$$, for a path $$x^i(t)$$ has no units. So velocities with respect to that parameter have no units. But proper time does have units, seconds say, $$[\tau]=s$$, and is the integral of the velocity magnitude
$$\Delta \tau = \int dt \sqrt{v^i v^j g_{ij}}$$
So the metric needs to have units, $$[g_{ij}]=s^2$$.

Does this seem right? Or does one actually ascribe units to manifold coordinates?
My feeling is one could do that, but those units would be meaningless and be swallowed by the metric, which is the only physical object giving meaning to distance and carrying units.

Thanks!

2. Mar 12, 2006

### Cexy

I believe that the v_i have units of length over time and the metric is dimensionless, so that the overall dimension of the gunf under the integral sign is

$$[T]\left([L]^2 [T]^{-2}\right)^{1/2} = [L]$$

But we know that the action is dimensionless, so we multiply by the appropriate combination of c, G and M in order to cancel the one factor of length.

Last edited: Mar 12, 2006
3. Mar 12, 2006

### pensano

If $$v_i=\frac{dx^i}{dt}$$ has units of length over time, then the coordinates of a path on the manifold, $$x^i$$, need to have units of length, and the parameter, $$t$$, needs to have units of time. But that doesn't make sense to me, since manifold coordinates just come from charts -- maps from manifold points to $$\Re^4$$. Unless I'm just being dumb about something?

4. Mar 13, 2006

### Cexy

The v_i, living in the tangent space of the manifold, are often taken to be velocities. So I guess that means that the x_i have units of length.

I'm afraid that I don't have a philosophical reason as to 'why' this works out, other than the fact that it 'just works'.

Edit: You could try thinking about it this way:

Take v to be a vector and e_i a basis for the tangent space (at some point p), so that the components of v in the basis are v^i, and we have

$$v = v^i e_i$$

We want the magnitude squared of v to have units of length squared over time squared, but we define |v|^2 as

$$|v|^2 = g(v,v) = g(v^ie_i, v^je_j) = v^iv^jg(e_i,e_j) = v^iv^jg_{ij}$$

The only way for this to be sensible is to give units of length over time to the v_i, and no units to the g_ij.

Last edited: Mar 13, 2006