LTB Metric: Dimensions Explained

[micomaco86572]

I was reading some papers about Lemaitre-Tolman-Bondi model these days, and was confused about the dimension of this metric.
As we know, the parabolic LTB line element takes the form:$ ds^{2}=-c^{2}dt^{2}+(R')^{2}dr^{2}+R^{2}d\Omega^{2}$.
In my GR lessons I was told that the metric is dimensionless. But here something seems to be paradoxical. If the coefficient of the second term is dimensionless, then we can deduce that R must has a dimension of length, which would conflict with the fact that the coefficient fo the thrid term, R, is required to be dimensionless. And vice versa.Forgive my poor English. lol.

 

[tiny-tim]

Hi micomaco86572!

A metric isn't dimensionless … it has dimensions of length-squared.

c R' and Ω are dimensionless, and everything else is length-squared.

 

[micomaco86572]

Hi micomaco86572!

A metric isn't dimensionless … it has dimensions of length-squared.

c R' and Ω are dimensionless, and everything else is length-squared.

Thx for your reply!

I may put it in a wrong way and didn't make it clear. I actually meant the component of the metric tensor is dimensionless. Of course the line element has a dimension of length. But why didi u say c is dimensionless? It should has the dimension of lenght/time, shouldn't it? And I am still not very sure about whether the \Omega has a dimension or R. Could u show me some proof or evidence?

Thx again for your reply!

 

[tiny-tim]

Hi micomaco86572!

(just got up … :zzz:)

(oh, and have an omega: Ω and try using the X² tag just above the Reply box )

… the component of the metric tensor is dimensionless. Of course the line element has a dimension of length. But why didi u say c is dimensionless? It should has the dimension of lenght/time, shouldn't it? And I am still not very sure about whether the \Omega has a dimension or R. Could u show me some proof or evidence?

c is dimensionless because length and time are the same dimension (just think about dt² - dx² ).

Ω is dimensionless because it's area/radius² (similarly, ordinary angle, = arc-length/radius, is dimensionless).

And R' is dimensionless because it's ∂R/∂r … see http://en.wikipedia.org/wiki/Lemaitre–Tolman_metric"

 

[DrGreg]

c is dimensionless because length and time are the same dimension (just think about dt² - dx² ).

With respect, I think that's a slightly weird way of thinking of it, in the context of the equation being discussed in this thread. It is true that relativists often use the convention that c = 1, by measuring time and distance in appropriate units (e.g. years and light-years). Under that convention, you can regard "time" and "distance" as being dimensionally the same. But under that convention, the letter c doesn't appear in any equations, as it is 1.

If you have an equation with an explicit c in it, then you have to regard time and distance as being dimensionally different, otherwise why would you bother writing the c? So I don't really buy "c is dimensionless" but I do accept "1 is dimensionless" (when c = 1).

 

[tiny-tim]

… If you have an equation with an explicit c in it, then you have to regard time and distance as being dimensionally different, otherwise why would you bother writing the c? So I don't really buy "c is dimensionless" but I do accept "1 is dimensionless" (when c = 1).

Hi DrGreg!

Yes, I always use c = 1, so I get confused when it isn't.

hmm … let's think …

although my reasoning was a bit iffy (in particular, I should have written "c²dt² - dx²" ),

I'm still going to maintain that length and time have the same dimensions, and that c is a dimensionless constant, like the 12 for converting feet to inches, or like the 4π for converting from some cgs units to SI units.

What do other people think?

 

[micomaco86572]

I don't think c is dimensionless. As DrGreg said, 1 is dimensionless, but c is not. In natural unit c is set to be 1, so it is dimensionless, but in the SI units, it has to have a dimension like the gravitational constant G or some other constants.