Does the curvature of spacetime have a unit?

[JF131]

Does the curvature of spacetime have a unit?

 

[Bill_K]

Does the curvature of spacetime have a unit?

The Riemann curvature tensor has dimensions (length)^-2.

 

[bcrowell]

The Riemann curvature tensor has dimensions (length)^-2.

Only if you use coordinates that have units of length, which is not necessary. Take the Schwarzschild spacetime described in Schwarzschild coordinates $(t,r,\theta,\phi)$. Then, e.g., we have $R_{\phi\phi r r}=m\sin^2\theta/(r-2m)$, which is unitless.

And of course there are other measures of curvature, which can have different units.

 

[Bill_K]

Only if you use coordinates that have units of length, which is not necessary. Take the Schwarzschild spacetime described in Schwarzschild coordinates $(t,r,\theta,\phi)$. Then, e.g., we have $R_{\phi\phi r r}=m\sin^2\theta/(r-2m)$, which is unitless.

No, I disagree. Just because you can use different coordinates does not mean that the concept of units becomes ambiguous! One might just as well say the same thing about the ordinary 3-momentum -- that its units were ambiguous because of the possibility of writing it in polar coordinates.

You might note that under a coordinate transformation the covariant and contravariant components may acquire different units, but the tetrad components remain the same.

And of course there are other measures of curvature, which can have different units.

What other measures of curvature have different units?? The Riemann, Weyl and Ricci tensors, as well as the Ricci scalar, all have the same units.

 

[pervect]

I think it's good to emphasize that most measures of curvature are tensors, in particular the curvature of a 4-d space-time takes more than one number to describe.

I'm not sure what to say about the units question. If we take momentum as an example, in my opinion we do get different units for momentum depending on the whether we take polar or rectangular coordinates. But the momenta we get in this manner are generally regarded as different concepts - i.e. angular momentum and linear momentum - as well as having different units.

Sidestepping this issue for a bit as not being helpful to the layperson, it might be helpful to talk about sectional curvature, http://en.wikipedia.org/wiki/Sectional_curvature, which can be described as the curvature of a plane slice This is a single number and it has units of 1/distance^2. Furthermore, I seem to recall that if you know the sectional curvature of all possible plane slices, you know the Riemann curvature. It gives a somewhat intuitive idea of why there is more than 1 number that represents curvature of a 4-d space-time (the existence of multiple plane slices. By itself the notion of sectional cuvature doesn't do a lot to tell you how many numbers are needed to completely descibe the curvature of space-time, but emphasizing that it's "more than 1" is a good start).

And sectional curvature is specific enough that it has a specific unit.

 

[BruceW]

In SI units, I think we can agree that curvature has dimensions of (length)^-2 but what I think bcrowell meant to say, is that we don't have to use SI units, and physicists don't always use SI units, since there are more natural systems of units to use. So for example, in geometrized units, mass and length have the same dimensions. and in the full Planck units, everything is dimensionless.

 

[Rena Cray]

Assume bcrowell's example of the Schwarzschild curvature,

$R_{\phi\phi r r}=m\sin^2\theta/(r-2m)$.

This element is dimensionless according to bcrowell, but what of the entire tensor?

For example $R_{\phi\phi r r} d\phi d\phi dr dr$
would have dimension $[D^2]$.

 

[Dale]

Only if you use coordinates that have units of length, which is not necessary. Take the Schwarzschild spacetime described in Schwarzschild coordinates $(t,r,\theta,\phi)$. Then, e.g., we have $R_{\phi\phi r r}=m\sin^2\theta/(r-2m)$, which is unitless.

I tend to like to consider coordinates to be unitless, i.e. They are just ordered 4-tuples of numbers. That puts all of the units in the other tensors and ensures that they are self-consistent.

I don't know if that is a standard approach, so I won't try to promote it, but considerations like this are what got me thinking along those lines.

 

[bcrowell]

What other measures of curvature have different units?? The Riemann, Weyl and Ricci tensors, as well as the Ricci scalar, all have the same units.

For example, the Kretschmann scalar has units of L^-4 (and because it's a scalar, this is independent of the choice of coordinates, even if one takes the point of view I expressed in #3).

 

[bcrowell]

I tend to like to consider coordinates to be unitless, i.e. They are just ordered 4-tuples of numbers. That puts all of the units in the other tensors and ensures that they are self-consistent.

It's definitely odd to think of a vector as having components whose units don't agree with one another -- but that's how I think about it.

 

[Rena Cray]

I disagree. The units of tensor components should not match, but compliment the basis. Otherwise it leads to a poor fit between nature and mathematical representation.

A good example among many is the action density pseudo scalar. This should not be represented as a scalar. It fails to be a true tensor because it is presumed to have scalar property. In addition it has to be pseudo-ed to wedge it into place.

We would better represent action density as

[itex]L = L_{\mu\nu\rho\sigma} [S/D^3T] dx^\mu dx^\nu dx^\rho dx^\sigma [D^3T][/itex].

Here, S means units of action.

Now we have a true tensor, without the need to normalize with $\sqrt{-|g|}$.