Questions about the FLRW metric

  • Thread starter Mishra
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Hello,

I have two questions regarding the FLRW metric, it is more about its interpretation.

The metric reads:

##dl²=dt²-a²(\frac{dR²}{1-kR²}+R²d\Omega²)## where ##a## is the radius of the 3-sphere (universe), and ##R=r/a## a normalized radial coord.

What I don't understand is this statement: "##k=+1## then the universe is closed".
I understand that when ##k=+1## then if ##R>1## the metric changes its signature which does not make sense in GR. Therefore ##R<1## if ##k=+1##.

My question is then:
If ##R=r/a##, then ##R<1## means that ##r<a##. Which just means you just don't observe things outside the universe of radius ##a##. I don't see why the radius of the universe ##a## could not be infinite.

The same problem the other way around: When ##k=-1##, ##r## can take an arbitrary value. Meaning it can be bigger than ##a##. How can the radial coord be bigger than the radius of the universe ?

How does the radial coord. tells us anything about the finiteness of the universe?




Also that FLRW metric seems like a huge regression from the Schwarchild's metric which has been built to keep the signature. Why not just use the Schwarchild's metric or something more general like a simple spherical metric (which is used to find the Schwarchild's metric) ?

I am obviously missing something. Maybe the
Schwarchild's metric does not fit the problems addressed by Friedmann and Lemaître but I don't see how...

Thank you and have a great weekend!
 

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  • #2
PeterDonis
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where ##a## is the radius of the 3-sphere (universe)
Only in the ##k = 1## case. For the ##k = 0## and ##k = -1## cases, the universe is spatially infinite, so the scale factor ##a## does not have the same straightforward interpretation.

I don't see why the radius of the universe ##a## could not be infinite.
Because if it were, the line element would be undefined (an infinite value ##a## would be multiplying the entire spatial part).

When ##k=-1##, ##r## can take an arbitrary value. Meaning it can be bigger than ##a##. How can the radial coord be bigger than the radius of the universe
In the ##k = -1## or ##k = 0## case, the universe is spatially infinite, and ##a## is not "the radius of the universe". See above.
 
  • #3
PeterDonis
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Also that FLRW metric seems like a huge regression from the Schwarchild's metric which has been built to keep the signature.
This is confused in at least two ways. First, the FLRW metric and the Schwarzschild metric have nothing to do with each other; they are two different solutions to the Einstein Field Equation, which model two different physical situations--the FLRW metric models an expanding universe, and the Schwarzschild metric models the vacuum spacetime around a single isolated gravitating body which is static (i.e., nothing changes with time). There's no point in trying to compare the two, since they're modeling two different scenarios.

Second, it is not really correct to say that the Schwarzschild metric is "built to keep the signature". Any metric in GR must have a Lorentzian signature. Sometimes that means you have to impose constraints on the ranges of coordinates (as in the case of ##R## in the FLRW metric as you have written it), but that's because of the way the coordinates are defined; it has nothing to do with the intrinsic properties of the metric.
 
  • #4
PeterDonis
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If ##R=r/a##, then ##R<1##. Which just means you just don't observe things outside the universe of radius ##a##.
No, it means that the universe does not exist outside of radius ##a##; that is, the universe has a finite volume. But this finite volume is unbounded; it is a 3-sphere, just as the surface of the Earth is a 2-sphere, a finite area but unbounded--no edge.
 
  • #5
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Thanks for you answers!



This is confused in at least two ways. First, the FLRW metric and the Schwarzschild metric have nothing to do with each other; they are two different solutions to the Einstein Field Equation, which model two different physical situations--the FLRW metric models an expanding universe, and the Schwarzschild metric models the vacuum spacetime around a single isolated gravitating body which is static (i.e., nothing changes with time). There's no point in trying to compare the two, since they're modeling two different scenarios.
Got it, it was stupid of me, they indeed have nothing in common! It also seems that the Schw. metric is not homogenous and isotropic. I'll investigate more...


Only in the k=1k = 1 case. For the k=0k = 0 and k=−1k = -1 cases, the universe is spatially infinite, so the scale factor aa does not have the same straightforward interpretation.
Well this solves all the problems. But I don't get it. What is ##a## then ? In my notes it is defined as the radius of the 3-sphere. How is it not the radius of the universe?
 
  • #6
PeterDonis
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It also seems that the Schw. metric is not homogenous and isotropic.
Correct. It is spherically symmetric, but that's all.

What is ##a## then ? In my notes it is defined as the radius of the 3-sphere.
That's true for the ##k = 1## case. It is not true for the ##k = 0## or ##k = -1## cases, because the universe is not spatially a 3-sphere in those cases; it is spatially infinite. The scale factor ##a## in these cases is simply a measure of "how much the universe has expanded", heuristically speaking; it converts coordinate distances (in terms of what you are calling ##R##) into proper distances (in terms of what you are calling ##r##) at a given instant of coordinate time. That still has meaning even if the universe is spatially infinite.
 

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