# Curved spacetime is described by a manifold

1. Mar 10, 2006

### pensano

In general relativity, curved spacetime is described by a manifold and a metric or frame on top of it.

Can the manifold coordinates carry units of, say, meters and seconds, or do the metric components have those units?

2. Mar 10, 2006

### chroot

Staff Emeritus
All of the "machinery" for measuring distances and angles is "built into" the metric, so the unit system you choose affects the components of the metric. The manifold itself involves no notion of distance.

If you choose different unit systems, you're effectively just applying a linear transformation to the metric.

- Warren

Last edited: Mar 10, 2006
3. Mar 10, 2006

### pensano

So... a velocity has no units?

But the maginitude of a velocity does?

4. Mar 10, 2006

### pervect

Staff Emeritus
A velocity has no units when it is expressed in geometric units. This is the simplest choice. However, as chroot indicated, it is a choice.

If you chose to, you can chose to use 4-vectors with standard units.

If you do chose standard units, 4 vectors have units of

(time, space, space, space)

i.e. the "time" component of the 4 vector has units of time, and the "space" component has units of distance.

Since a 4-velocity is the rate of change of time of a 4-vector with respect to proper time, its unit are just (1/time) * (units of 4-vector). Thus a 4-velocity has units of

(dimensionless, velocity, velocity, velocity)'

The magnitude of a 4-vector will be a velocity - for example, calling the components of the 4-vector (x0,x1,x2,x3) we see that the magnitude in Minkowski space is

sqrt(c^2*dimensionless^2 - velocity^2 - velocity^2 - velocity^2)

this generalizes to higher dimensional spaces.

5. Mar 10, 2006

### pensano

Let me clarify my question.

So, say you're on a part of the manifold covered by a chart with coordinates $$x^i$$. And there's some path, $$x^i(\tau)$$, through that chart, parametrized by $$\tau$$. The velocity components along the path are
$$v^i = \frac{d x^i}{d \tau}$$
And the magnitude of the velocity is
$$|v| = \sqrt{v^i v^j g_{ij}}$$

My question is: what units, such as meters and seconds, can be associated with these quantities? (Presuming I choose not to set c=1 or G=1)

Can the coordinates have units, such as $$[x^0]=s,[x^1]=m$$?

If not, does the parameter have units, $$[\tau]=s$$?

How about the metric, $$[g_{ij}]=?$$

And what're the units of $$|v|$$?

It's fine with me if some components have no units, like $$[v^0]=1$$ and $$[v^1]=m/s$$ (as pervect is suggesting?) but then, for consistancy, don't the coordinants need units?

Is chroot suggesting that only the metric should have units?

(I'm not sure if this tex will work, but I'm giving it a try)

Last edited: Mar 10, 2006
6. Mar 10, 2006

### pervect

Staff Emeritus
I guess my notation wasn't clear.

The 4-vector $x^i$ has units of time when x=0, and units of space when x > 0. Assuming that the ordering is the usual ordering.

The 4-velocity, $$v^i = \frac{d x^i}{d \tau}$$, is dimensionless when x=0, and has units of velocity when i>0.

The magnitude of the 4 velocity is a velocity, and all its terms must have units of a velocity. Thus one can infer that $g_{00}$, for instance, must have units of of a velocity, while $g_{11}$, for instance, must be dimensionless.

7. Mar 10, 2006

### pensano

So it's OK then to associate units, like seconds and meters, to the coordinates of a manifold?

That won't freak out the mathematicians?

8. Mar 10, 2006

### pensano

pervect, if the magnitude of the velocity is $$[|v|]=\frac{m}{s}$$ then when you integrate to get the proper time elapsed along the path
$$\Delta \tau = \int d\tau |v|$$
you'll get $$[\Delta \tau]=m$$. But it should be $$s$$, so I think the maginitude of velocity needs to be unitless, yes?

9. Mar 10, 2006

### pensano

And if $$\tau$$ is the proper time along a path, doesn't $$|v|=1$$?

10. Mar 11, 2006

### pervect

Staff Emeritus
The Lorentz interval

$$ds^2 = c dt^2 - dx^2 - dy^2 - dz^2$$

has units of distance. To convert it to units of time, you need to divide the above by 'c'.

See for instance

http://farside.ph.utexas.edu/teaching/jk1/lectures/node14.html

This can be re-written to give

$$d\tau = \sqrt{1-(\frac{v}{c})^2} \, dt$$

where v is the magnitude of the 3 velocity (NOT the magnitude of the 4-velocity, which is always equal to c).

I don't know where you got your formulas from.