Tod & Hughston GR Intro: FRW Metric Derivation w/ R=6k or R=3k?

If so, how does this affect the overall solution?It's possible, but I'm not sure how it would affect the overall solution. It could be worth looking into further, possibly by consulting other sources or reaching out to the authors for clarification.
  • #1
binbagsss
1,259
11
I'm looking at Tod and Hughston Introduction to GR and writing the metric in the two forms;

[1]##ds^{2}=dt^{2}-R^{2}(t)(\frac{dr^{2}}{1-kr^{2}}+r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}))##

[2] ##ds^{2}=dt^{2}-R^{2}(t)g_{ij}dx^{i}dx^{j}##

where

##g_{ij}dx^{i}dx^{j}=d\chi^{2}+\chi^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}) ## for ##k=0##
##=d\chi^{2}+sin^{2}\chi{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}) ## for ##k=1##
##=d\chi^{2}+sinh^{2}\chi{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}) ## for ##k=-1##

Now in solving for the form [2] Tod computes the Ricci scalar of ##ds^{2}=d\chi^{2}+f^{2}(\chi)(d\theta^{2}+sin^{2}\theta d\phi^{2})## and finds ##R=-(2\frac{f''}{f}-\frac{1}{f^{2}}+\frac{(f')^{2}}{f^{2}}## then integrates, uses ##R=3k## and solves for all 3 cases ##k=0,\pm 1##.

My question

##R=3k## doesn't seem right to me, since in 3-d space we can write ##R_{ab}=2kg_{ab}##. Of course you could just define a constant ##K=2k##, but it uses the constant ##k## in the FRW metric of the form [1] not ##k##, comparing to Introduction to GR lecture notes by sean M.Caroll,I thought that this should be ##R=6k##

...In Caroll's notes he uses ##R_{ab}=2kg_{ab}## in the derivation and gives the FRW metric the same as in form [1] with small ##k##. So it doesn't look as though Tod has used ##K=2k##.
Can anyone help explain how Tod uses ##R=3k##?

Thanks.
 
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  • #2
binbagsss said:
I thought that this should be ##R=6k##

That's what I get when I compute the Ricci scalar for the spatial part of [1] (i.e., the part inside the parentheses after ##R^2(t)## ).

binbagsss said:
Tod computes the Ricci scalar of ##ds^{2}=d\chi^{2}+f^{2}(\chi)(d\theta^{2}+sin^{2}\theta d\phi^{2})## and finds ##R=-(2\frac{f''}{f}-\frac{1}{f^{2}}+\frac{(f')^{2}}{f^{2}}##

When I compute the Ricci scalar for this, I get twice the answer you are giving from Tod. So it looks to me like Tod is dividing by 2 somewhere in order to equate his answer for [2] with R = 3k for [1].
 
  • #3
PeterDonis said:
That's what I get when I compute the Ricci scalar for the spatial part of [1] (i.e., the part inside the parentheses after ##R^2(t)## ).
When I compute the Ricci scalar for this, I get twice the answer you are giving from Tod. So it looks to me like Tod is dividing by 2 somewhere in order to equate his answer for [2] with R = 3k for [1].

Thanks for your reply. I can't see at all where Tod is dividing by 2 though.
Could the answer possibly lie in the fact that simulatenously rescaling:
##k \to k/| k/##
##r \to \sqrt{| k|} r ##
##a \to a/\sqrt{|k|} ##
The FRW metric is unchanged?
If it could however, I'm unsure how to justify explicitly.
Thanks.
 
  • #4
binbagsss said:
Could the answer possibly lie in the fact that simulatenously rescaling:
##k \to k/| k/##
##r \to \sqrt{| k|} r##
##a \to a/\sqrt{|k|}##
The FRW metric is unchanged?

I don't see how that would account for the factor of 2. I don't have Tod and Hughston, so I can't say where the discrepancy arises.
 
  • #5
I don't suppose anyone else has any ideas on where this extra factor of 2 is coming from?
Is it possible that ##\chi## could have been re-scaled such that solving the case ##R=3k## is the same as solving for ##R=6k##.
 
Last edited:

Related to Tod & Hughston GR Intro: FRW Metric Derivation w/ R=6k or R=3k?

1. What is the significance of the Tod & Hughston GR Intro?

The Tod & Hughston GR Intro is a mathematical framework that describes the evolution of the universe according to Einstein's theory of General Relativity. It helps to explain the behavior of matter and energy on a large scale, and is a fundamental tool in modern cosmology.

2. What is the FRW metric in this context?

The FRW (Friedmann–Lemaître–Robertson–Walker) metric is a mathematical formula that describes the geometry of the universe in the context of General Relativity. It is based on the assumption that the universe is homogeneous and isotropic, meaning that it looks the same in all directions and at all points in space.

3. How is the FRW metric derived with R=6k or R=3k?

The FRW metric derivation with R=6k or R=3k is a mathematical process that involves applying the Einstein field equations to a universe with a constant curvature. The value of R, which represents the radius of curvature, can be set to either 6k or 3k to produce different solutions for the evolution of the universe.

4. What is the difference between R=6k and R=3k in the FRW metric derivation?

The difference between R=6k and R=3k in the FRW metric derivation lies in the curvature of the universe. R=6k corresponds to a closed universe, where space is finite and curves back on itself. R=3k corresponds to an open universe, where space is infinite and has a flat curvature.

5. How does the FRW metric derivation contribute to our understanding of the universe?

The FRW metric derivation is a key tool in modern cosmology, as it allows us to make predictions about the behavior of the universe based on its initial conditions. It has helped us to understand the expansion of the universe, the formation of galaxies and other large-scale structures, and the distribution of matter and energy in the universe.

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