I'm confused about the different ways of writing the Friedmann-Robertson-Walker (FRW) metric using the normalised and non-normalised scale factor. Peacock, for example, (see equation 3.13) gives $$c^{2}\,\mathrm d\tau^{2}=c^{2}\,\mathrm dt^{2}-R^{2}\left(t\right)\left(\frac{\mathrm dr^{2}}{1-kr^{2}}+r^{2}\,\mathrm d\psi^{2}\right),$$ where $R\left(t\right)$ is the non-normalised scale factor and $k=0,\,\pm1$.

He then gives an alternative form using $a$, the dimensionless, normalised scale factor, where $a\left(t\right)\equiv R\left(t\right)/R_{0}$ so $$c^{2}\,\mathrm d\tau^{2}=c^{2}\,\mathrm dt^{2}-a^{2}\left(t\right)\left(\frac{\mathrm dr^{2}}{1-k\left(Ar\right)^{2}}+r^{2}\,\mathrm d\psi^{2}\right),$$ where $A=1/R_{0}$. This surely means that the $kA^2$ term no longer equals 0, +1 or -1. However, and this is what loses me, I've also seen (equation 1, here: https://ned.ipac.caltech.edu/level5/Sept02/Reid/Reid2.html) the metric as $$c^{2}\,\mathrm d\tau^{2}=c^{2}\,\mathrm dt^{2}-a^{2}\left(t\right)\left(\frac{\mathrm dr^{2}}{1-kr^{2}}+r^{2}\,\mathrm d\psi^{2}\right),$$ where $a(t)$ is dimensionless and $k$ again equals 0, +1 or -1. What am I misunderstanding? Thank you.


I can get from the first to the second metric by making the substitutions $R\rightarrow R/R_{0}$, $r\rightarrow rR_{0}$, $k\rightarrow k/R_{0}^{2}$. But I can't see how I can get to the third metric, where not only does (presumably - it's described as being "dimensionless") $a\left(t\right)\equiv R\left(t\right)/R_{0}$ but also $k$ still equals 0, +1 or -1.

  • $\begingroup$ The $a$ in the second form is the same as the $R$ before. Don't pay too much attention to the names of things. $\endgroup$
    – Javier
    Jun 29, 2018 at 13:44
  • $\begingroup$ @Javier - how can they be the same if they're different equations, and if they are the same how can the third equation be true? $\endgroup$
    – Peter4075
    Jun 29, 2018 at 17:10
  • $\begingroup$ You might be interested in this answer of mine: physics.stackexchange.com/questions/397279/…. What matters is the form of the equations, not the names of the variables. $\endgroup$
    – Javier
    Jun 29, 2018 at 18:50
  • $\begingroup$ @Javier - OK, thanks for that. So are you saying that in order for the third metric I give to be correct we need to rescale the radial coordinate $r$? $\endgroup$
    – Peter4075
    Jun 29, 2018 at 19:17
  • $\begingroup$ It really depends. You can't tell just from the form of the metric, because there is still some coordinate freedom. If you want $k \in \{-1,0,1\}$, then $r$ must be adimensional and hence rescaled, and $a$ will be a length. But you can have $k$ be just any real number, and $r$ will be a length. $\endgroup$
    – Javier
    Jun 29, 2018 at 19:34

1 Answer 1


In the first expression:

  • $R$ has dimensions of length
  • $r$ and $k$ have no dimensions
  • $k$ is scaled so it's $-1$, $0$, or $1$

To go from the first expression to the second expression, we define $$r' = r R_0, \quad k' = k / R_0^2, \quad a(t) = R(t) / R_0$$ where I'm denoting the new variables with a prime just to be extra clear. Then

  • $r'$ has dimensions of length and $k'$ has dimensions of inverse length squared
  • $a$ has no dimensions

These are the two main conventions for the FRW metric. Your third example is a bit unusual, because it tries to set $k'$ to $-1$, $0$, or $1$, which doesn't make any sense because $k'$ has dimensions.

However, $k'$ is a constant, so we can use it to define our unit of length. Taking positive curvature for concreteness, this means that rather than measuring lengths in meters, we measure in a unit system where the numeric value of $k'$ is $1 \, (\text{length unit})^{-2}$. Another way of saying this is that this example is setting $k' = 1$ in the same sense that you may set $c = 1$ in special relativity. You can always do this by defining your unit system appropriately, but then naive dimensional analysis will stop working, and you'll have to put the $k'$'s and $c$'s back in at the end to get valid numbers out.

  • $\begingroup$ Defining $k' = 1$ - sneaky! That's the most underhand, duplicitous thing I've heard since the Brexit campaign promised that the EU need us more than we need them. But it makes sense (defining $k' = 1$, that is). Thanks. $\endgroup$
    – Peter4075
    Jul 1, 2018 at 11:40

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