- #1
Radiohannah
- 49
- 0
Hello!
May I ask a few questions, I am trying to understand how the metric tensors work in general relativity. In particular, why you would want to make either the radial or transverse part look Euclidean? In what situations is that useful?
I am happy that in a very general form, the metric tensor looks like this:
[tex]c^{2} d \tau^{2} = c^{2} d t^{2} - R^{2} (t) [a^{2}(r) dr^{2} + b^{2} (r) [d \theta^{2} + sin^{2} \theta d\phi^{2}]][/tex]
In which I am assuming that the radial part is the
[tex] R^{2} (t) a^{2}(r) dr^{2} [/tex]
And the transverse part is the
[tex] R^{2} (t) b^{2} (r) [d \theta^{2} + sin^{2} \theta d\phi^{2}] [/tex]
Now I am getting confused as to how you can chose the functions a(t) or b(t) to make something "look as Euclidean as possible"..?
You can divide the radial term by " 1 - kr^{2}" to make the transverse part look Euclidean, so you would be left with...
[tex] R^{2} (t) [ \frac{dr^{2}}{1 - kr^{2}} + r^{2} d \phi^{2}] [/tex]
So the transverse part looks Euclidean because...you are left with just the r squared d phi squared term? I don't understand why this would be useful?
Similarly, by making the radial part look Euclidean, you have the "dr^{2}" term by itself and b(r) is some sin^{2}r term or sinh^2{r} term.
I don't see how from that could you deduce whether "r" is an angular distance or just a comoving distance? I'm getting a bit confused and would appreciate any help :-)
Thank you
Hannah :-)
May I ask a few questions, I am trying to understand how the metric tensors work in general relativity. In particular, why you would want to make either the radial or transverse part look Euclidean? In what situations is that useful?
I am happy that in a very general form, the metric tensor looks like this:
[tex]c^{2} d \tau^{2} = c^{2} d t^{2} - R^{2} (t) [a^{2}(r) dr^{2} + b^{2} (r) [d \theta^{2} + sin^{2} \theta d\phi^{2}]][/tex]
In which I am assuming that the radial part is the
[tex] R^{2} (t) a^{2}(r) dr^{2} [/tex]
And the transverse part is the
[tex] R^{2} (t) b^{2} (r) [d \theta^{2} + sin^{2} \theta d\phi^{2}] [/tex]
Now I am getting confused as to how you can chose the functions a(t) or b(t) to make something "look as Euclidean as possible"..?
You can divide the radial term by " 1 - kr^{2}" to make the transverse part look Euclidean, so you would be left with...
[tex] R^{2} (t) [ \frac{dr^{2}}{1 - kr^{2}} + r^{2} d \phi^{2}] [/tex]
So the transverse part looks Euclidean because...you are left with just the r squared d phi squared term? I don't understand why this would be useful?
Similarly, by making the radial part look Euclidean, you have the "dr^{2}" term by itself and b(r) is some sin^{2}r term or sinh^2{r} term.
I don't see how from that could you deduce whether "r" is an angular distance or just a comoving distance? I'm getting a bit confused and would appreciate any help :-)
Thank you
Hannah :-)