# Coordinates in general relativity

1. Feb 14, 2014

### GarageDweller

Hello fellow PF go-ers
I am having trouble with coordinates in curved space time lately, allow me to demonstrate my issue.
Take the metric of flat space in spherical coordinates for example, a diagonal metric with values 1,r^2 and r^2sinΘ. It appears to me that only when we know that the Θ and ψ variables are dimensionless, can we transform this back into the usual flat spacetime in cartesian coordinates.
However what if I made an abitrary transformation of coordinates and did not reveal to you the dimensions of the various coordinates, how would one know whether the metric is flat or not?

The problem seems to be that the cartesian system of coordinates is held on a pedestal, and I think this should not be.

2. Feb 14, 2014

### Staff: Mentor

Hi GarageDweller,

I think that there are a few different conventions possible. The bottom line requirement that everything has to meet is that $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ must have units of length squared. However, how you split that up into dx or g doesn't really matter, and you can do it separately for each coordinate or over the whole tensor. My personal preference is to think of dx as always being unitless and g as always having units of length squared so that everything is consistent from coordinate system to coordinate system and from component to component, but I know it is not a universal preference.

As long as you meet the requirement you are OK.

As far as flatness goes, I don't think that the units are an issue. If the components of the curvature tensor are 0 then they will be 0 regardless of the units.

3. Feb 14, 2014

### Bill_K

Tensor analysis does not know or care about dimensions. The metric gμν transforms as a tensor. In the new coordinates, always

gμ'ν' = ∂xμ/∂xμ' ∂xν/∂xν' gμν.

Dimensions don't play a role in this at all. The new coordinates can be ANY functions of the old ones, even functions that seem to be dimensionally inconsistent, such as x = r + Θ.

I would calculate the Riemann tensor.

4. Feb 14, 2014

### Staff: Mentor

This actually is one of the reasons that I like to think of coordinates as just numbers. Then there is no hesitation to use functions like that.

5. Feb 14, 2014

### bcrowell

Staff Emeritus
Yeah, I prefer the opposite convention, so r has units of distance, and Θ is unitless. This has the advantage that quantities like r and Θ have the units you expect, but the disadvantage that different components of the same tensor can have different units. But as DaleSpam says, this is just a matter of personal preference.

6. Feb 14, 2014

### Staff: Mentor

But even with that disadvantage, it still works out that $ds^2$ has units of length squared. The different components of $dx$ and the different components of $g$ combine together to give you the right units.