# Is the scale factor a scalar?

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Is the scale factor a scalar? Assuming so, gets a weird result. I think.
Is the scale factor a scalar?
I think that the answer is no but I want to check because god (or the universe) has been playing tricks on me...
At Sean Carroll's invitation I wanted to check that the tensor$$K_{\mu\nu}=a^2\left(g_{\mu\nu}+U_\mu U_\nu\right)$$was a Killing tensor. ##U^\mu=\left(1,0,0,0\right)## is the four velocity of all comoving observers. The FLRW metric in use is given by $${ds}^2=-{dt}^2+a^2\left(t\right)\left[\frac{{\rm dr}^2}{1-\kappa r^2}+r^2{d\theta}^2+r^2\sin^2{\theta}{d\phi}^2\right]$$ For that to be so we need ##\nabla_{(\sigma}K_{\mu\nu)}=0## and the first step is to calculate the components of ##\nabla_\sigma K_{\mu\nu}##. I had the Christoffel symbols to hand and started to compute as follows$$\nabla_\sigma K_{\mu\nu}=\left(g_{\mu\nu}+U_\mu U_\nu\right)\nabla_\sigma\left(a^2\right)+a^2\nabla_\sigma\left(g_{\mu\nu}+U_\mu U_\nu\right)$$then ##\nabla_\sigma\left(a^2\right)## became ##2a\partial_\sigma a## which all looked very nice. Unfortunately it produces exactly the right answer except for a wrong sign in the ##\nabla_iK_{0j}## components (##i,j=123##). WTF. Of course I might have made a mistake*, thus my question: Is the scale factor a scalar? It it's not then ##\ \nabla_\sigma\left(a^2\right)## is nonsense.

The correct result comes from computing ##\nabla_\sigma K_{\mu\nu}=\mathrm{\partial}_\sigma K_{\mu\nu}-\Gamma_{\sigma\mu}^\lambda K_{\lambda\nu}-\Gamma_{\sigma\nu}^\lambda K_{\mu\lambda}## I now know.
* Edit. I did make a mistake. See below.

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Mentor
Is the scale factor a scalar?

It should be. It has a single numerical value at every event in spacetime, and its gradient is well-defined.

Unfortunately it produces exactly the right answer except for a wrong sign in the ##\nabla_iK_{0j}## components (##i,j=123##).

Are you taking into account that lowering the index from ##U^0## to ##U_0## introduces a minus sign (from the sign of ##g_{00}##)?

• George Keeling and vanhees71
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then ##\nabla_\sigma\left(a^2\right)## became ##2a\partial_\sigma a##

Shouldn't this be ##\nabla_\sigma \left( a^2 \right) = 2 a \nabla_\sigma a##?

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Shouldn't this be ##\nabla_\sigma \left( a^2 \right) = 2 a \nabla_\sigma a##
If ##a## scalar then ##\nabla_\sigma \left( a^2 \right) = 2 a \nabla_\sigma a= 2 a \partial_\sigma a##
and I think I did take into account that ##U_\mu=(-1,0,0,0)## I will check my calculations again tomorrow.

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• vanhees71
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If ##a## scalar then ##\nabla_\sigma \left( a^2 \right) = 2 a \nabla_\sigma a= 2 a \partial_\sigma a##

Ah, yes, got it.

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I'm a bit puzzled by the statement ##a## in the FLRW metric being a scalar as it is just an element in the metric components in a specific frame (the comoving fundamental frame)
$$\mathrm{d} s^2=\mathrm{d} t^2 - a^2(t) \left [\frac{\mathrm{d} r^2}{1-K r^2} + r^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2) \right].$$
From that I'm not so sure that ##K_{\mu \nu}## are really tensor components. Where did you find this? I've not found it in Carroll's GR lecture notes.

• George Keeling
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Ouch! I did start off by forgetting that ##U_0=-1## and corrected it - but only three times out of four. Having spotted the fourth I get the right answer. Thank you Mr Donis.

I now believe that the scale factor is a scalar (it satisfies ##\nabla_\sigma a= \partial_\sigma a##).

It might also be interesting that using$$\nabla_\sigma K_{\mu\nu}=\left(g_{\mu\nu}+U_\mu U_\nu\right)\nabla_\sigma\left(a^2\right)+a^2\nabla_\sigma\left(g_{\mu\nu}+U_\mu U_\nu\right)$$is much more efficient than using$$\nabla_\sigma K_{\mu\nu}=\mathrm{\partial}_\sigma K_{\mu\nu}-\Gamma_{\sigma\mu}^\lambda K_{\lambda\nu}-\Gamma_{\sigma\nu}^\lambda K_{\mu\lambda}$$(as long as you don't make a careless mistake!)

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I still don't understand, why ##a## should be a scalar :-(.

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I'm a bit puzzled by the statement ##a## in the FLRW metric being a scalar as it is just an element in the metric components in a specific frame
I agree! Just because it's in the formula for the line element doesn't make it a scalar. However, the formula$$K_{\mu\nu}=a^2\left(g_{\mu\nu}+U_\mu U_\nu\right)$$ is 8.98 in section 8.5 on redshifts and distances in Carroll's book. He goes on to use it to work out things about redshifts and distances. I haven't quite got that far yet. The RHS contains ##a## and a bunch of tensors. Must ##a## be a (rank 0) tensor if the LHS is to be a tensor?

The fact that both expansions of ##\nabla_\sigma K_{\mu\nu}## work, and the former because ##a## behaves like a scalar, is evidence. The latter expansion does not assume that ##a## is a scalar. If ##a## looks, acts, smells like a scalar, it is a scalar. Surely?

Mind you, if ##a## was not a scalar, making ##K## not a tensor then the second expansion would also be invalid! The whole project falls to pieces. I will read on.

Mentor
I'm a bit puzzled by the statement ##a## in the FLRW metric being a scalar as it is just an element in the metric components in a specific frame

That doesn't mean it can't be a scalar. The Schwarzschild ##r## coordinate is a scalar, even though it is "just an element in the metric components in a specific frame".

The question is whether there is an invariant way to assign a value for ##a## to every event in spacetime. Obviously there is, since ##a## is a function of FRW coordinate time, and FRW coordinate time labels the invariant family of spacelike hypersurfaces that are homogeneous and isotropic. So each such hypersurface can be labeled with its value of ##a## and that labeling will be invariant. In any other coordinates besides FRW coordinates, this invariant will not be a function of just one coordinate, but it will still be an invariant. That is sufficient to make ##a## a scalar.

• vanhees71 and George Keeling
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I didn't claim it isn't a scalar. I only said that I don't understand why it is one. I also don't understand why you say a coordinate is a scalar since if anything changes under coordinate transformations its the coordinates, right?

So can you explicitly express somehow ##a## with some tensor contractions? Then it would be obvious for me that it's a scalar.

Mentor
I didn't claim it isn't a scalar. I only said that I don't understand why it is one.

I explained why.

I also don't understand why you say a coordinate is a scalar

I didn't say any coordinate is a scalar, I said the Schwarzschild ##r## coordinate is a scalar. It's a scalar because it's defined to be the "areal radius", i.e., ##r = \sqrt{A / 4 \pi}## on a 2-sphere of area ##A##. That makes it an invariant, since the area ##A## of a 2-sphere containing a given event in a spherically symmetric spacetime is an invariant.

can you explicitly express somehow ##a## with some tensor contractions?

Perhaps. The expansion scalar for the congruence of comoving observers in FRW spacetime, which is the contraction ##\nabla_a u^a## for a 4-velocity field ##u^a##, is ##3 \dot{a} / a##. Since the FRW ##t## coordinate is the same as proper time for comoving observers, I think we can rewrite ##\dot{a}## as ##da / d \tau##, and the equation

$$\theta = \frac{3}{a} \frac{da}{d\tau}$$

is sufficient to show that ##a## is a scalar.

• vanhees71