Hello!May I ask a few questions, I am trying to understand how the

  • #1
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 :-)
 

Answers and Replies

  • #2
5,428
291


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]
you haven't mentioned this, but I assume you're aware that this is a general form of a spacetime where the spatial part is expanding/contracting, written in spherical polar coordinates.

In particular, why you would want to make either the radial or transverse part look Euclidean? In what situations is that useful?
The motivation for this comes from the desire for the metric to describe a realistic cosmos, in which spatial (hyper-)slices are isotropic and homogenous (maximally symmetric -> constant curvature).

This metric

[tex]
c^{2} d \tau^{2} = c^{2} d t^{2} - R^{2} (t) [dr^{2}/(1-k\ r^2) +r^2\ d \theta^{2} + r^2\ \sin^{2} \theta d\phi^{2}]
[/tex]

satisfies that condition. For the case k = 0 the spatial part is Euclidean which is the simplest case of the FRW spacetime.

I'm not sure if this answers your question but I should point out you've dropped a term from this

[tex]
R^{2} (t) [ \frac{dr^{2}}{1 - kr^{2}} + r^{2} d \phi^{2}]
[/tex]

which should read

[tex]
R^{2} (t) [ \frac{dr^{2}}{1 - kr^{2}} + r^{2} d \theta^{2} + r^2\ \sin(\theta)^2\ d\phi^2]
[/tex]
 
  • #3


Brilliant thank you :-)! woops
 
  • #4


HOWEVER, may I ask

I don't see how you could then interpretate distances from this metric? ie from a Schwarzschild metric?

I know that for a stationary observer, the radial and the angular parts would equate to 0. And I know that in general the distance between 2 points separated by dr would NOT = dr. Does this mean that you cannot interpretate distances from a Schwarzschild metric in the frame? Or is that the invariant part?

I hope that makes sense.....

Cheers!

Hannah
 
  • #5
5,428
291


I don't understand your questions.

Defining distance is tricky in GR and is a subject in its own right. But the interval defined by the metric is the proper interval and is invariant. If it is integrated along a worldline all observers will agree on the calculation and the observations.

If you ask the question 'How far is it between points p1 and p2' there is no definate answer without making further assumptions about the means of measurement and the distances involved.
 
  • #6


Sorry, I'm finding it difficult to grasp some of the basics

From a Schwarzschild metric

You can get the proper and coordinate times, and the time measured by an observer at some distance from the source.

I was wondering if you could similarly get a 'distance' from the metric, measured by a stationary observer?

When you say "'there is no definate answer without making further assumptions about the means of measurement and the distances involved",... by these assumptions do you mean, for instance, working in the frame of a stationary observer? So taking the angular and radial parts of the metric, dr, d theta and d phi, to be 0 and solving for a distance with those conditions?

The distance between two points separated by an interval "dr" is not dr in the metric and so I was trying to figure out how you could go about calculating a distance measured by an observer, if it was possible.

I hope that makes more sense, I'm really struggling with this topic, sorry :-(
 
  • #7
5,428
291


Suppose we have two observers held fixed at the coordinates [tex](r_1, \theta_1,\phi_1)[/tex] and [tex](r_2, \theta_2,\phi_2)[/tex]. No meaningful distance can be found by plugging those coordinates into the metric, which I think is what you said.

Instead we have to consider two cases,

1. an observer who measures a distance between two events on their own worldline

2. an observer who measures the distance between themselves and another worldline.

In the first case the observer will use local coordinates, which define ruler distance as coordinate distance.

The second case is more difficult, but clearly whatever definition is used, the distance measured depends on the particulars of the worldlines.

As I mentioned, distance is GR is a big topic, and I quote two abstracts to give you a taste,


Measurement of Distance in General Relativity

Newman, E.; Goldberg, J. N.

Physical Review, vol. 114, Issue 6, pp. 1391-1395

An instrinsic geometrical definition of distance is presented in terms of the geodesic deviation of null rays. This definition is applied to the Schwarzschild solution and the homogeneous and isotropic cosmological solutions of the Einstein field equations. In the former case the distance is proportional to the radial coordinate of the standard metric and in the latter case it yields the usual cosmological distance. However, since the geodesic deviation is a scalar, the results are independent of the particular metric used. The relationship of this definition to observation is discussed and it is concluded that it agrees with the astronomical definition of luminosity distance.
This is a link to an early paper by Walker,

http://qjmath.oxfordjournals.org/content/os-4/1/71.extract
 

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