# General expression of the redshift: explanation?

In some papers, authors put the following formula for the cosmological redshift $z$ :

$1+z=\frac{\left(g_{\mu\nu}k^{\mu}u^{\nu}\right)_{S}}{\left(g_{\mu\nu}k^{\mu}u^{\nu}\right)_{O}}$

where :

• $S$ is the source and $O$ the observer

• $g_{\mu\nu}$ is the metric

• $k^{\mu}=\frac{dx^{\mu}}{d\lambda}$ is the coordinate derivative regarding the affine parameter $\lambda$

• $u^{\nu}$ is the 4-velocity of the cosmic fluid

My first question is : according to the Einstein summation, is it a fraction of sums $\left(\frac{\sum_{\mu\nu}X}{\sum_{\mu\nu}Y}\right)$ or a sum of fractions $\left(\sum_{\mu\nu}\frac{X}{Y}\right)$ ?

My second question (and more important) is : where does this formula come from ? Where can I find a "demonstration"/"derivation"/"explanation" of this ?

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Please tell us where you found this, so we have some context. The only reasonable thing I can imagine the "cosmic fluid" would refer to would be the rest frame of the CMB or the Hubble flow. Let's consider the simplest case, where both source and observer are at rest relative to the CMB. Then the velocity four-vectors $u$ are both of the form (1,0,0,0) in their respective frames S and O. The equation becomes $1+z=f_S/f_O$, which simply reads as a definition of $z$ in terms of the Doppler shift. – Ben Crowell Jun 17 '13 at 17:54
You can also consider the case where S and O are separated by a small distance, so cosmological expansion is negligible. Then the same Lorentz frame can cover both of them, and I assume (haven't worked it out) that you then recover the SR Doppler shift. – Ben Crowell Jun 17 '13 at 17:57
The $u^\lambda$ is presumably the 4-velocity of an observer. The inner product is simple the frequency (up to a constant) the observer with that 4-velocity measures. The identity is quite trivial to expect to see a proof in any textbook. – MBN Jun 18 '13 at 12:07
@MBN: It may be a trivial identity in SR, but he's saying it's valid for GR, in cosmological spacetimes. That seems extremely nontrivial to me. If it's cosmologically valid, can you point us to a textbook where it's discussed? – Ben Crowell Jun 18 '13 at 14:58
@BenCrowell: I thought it followed directly from the equivalence principle. It is a local statement, the frequency that an observer measures. I have seen it in Wald where he derives the red shift for Schwarzschield and FRWL spacetime. But he only states it. – MBN Jun 18 '13 at 15:34

For the first question, it is the same quantity, the Minkowski dot product of the four vectors $k$ and $u$ that you may call $A_{O}$ and $A_{S}$, in general

$$g_{\mu\nu}k^{\mu}u^{\nu}=k_{\nu}u^{\nu}\equiv A$$

computed for the source and computed for the observer. So you have

$$1+z=\frac{A_{S}}{A_{O}}$$

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So this is a fraction of sums and not a sum a fractions ? – Vincent Jun 17 '13 at 15:22
Yes. The proof goes along the lines that the energy of the photon ($E=h\nu$) is precisely that dot product, but I can't give a detailed explanation – Jorge Jun 17 '13 at 15:26
This is fine as far as the purely notational issues, but it doesn't explain why the inner product $k_\nu u^\nu$ is of interest, or why the "cosmic fluid" is relevant. – Ben Crowell Jun 17 '13 at 17:46
@BenCrowell fair enough, I just answered half of the OP question, and I claim that I can't give a full explanation on why is that quanitity relevant. – Jorge Jun 17 '13 at 18:16
@BenCrowell: $k_\nu u^\nu$ is just the energy, and the "cosmic fluid" comes in because we're dealing with source and observer that are comoving with the Hubble flow – Christoph Jun 17 at 18:39

The energy $h\nu$ of a photon is simply the contraction $$h\nu = g_{\alpha\beta} k^\alpha u^\beta$$ of its momentum $k^\mu$ with the frame's velocity $u^\mu$.

The frequency shift is then given by $$1 + z = \frac{\nu_S}{\nu_O} = \frac{(g_{\alpha\beta} k^\alpha u^\beta)_S}{(g_{\sigma\rho} k^\sigma u^\rho)_O}$$ where we denote the velocities of both source and observer with $u^\mu$ as we're dealing with the special case of cosmological redshift, where source and observer are supposed to be comoving with the Hubble flow.

The momentum $(k^\mu)_S$ and $(k^\mu)_O$ of the photon at times of emission and absorption are related by parallel transport along the photon worldline, and if the worldline is not affinely parametrized, solving the parallel transport equation is a way to compute the frequency shift that works for arbitrary spacetimes.

This is actually nothing but the generalization of the special relativistic Doppler effect, and in Minkowski spacetime (or even just normal coordinates if we transport the source velocity along the photon geodesic), it reduces to the Doppler factor.

I've taken the time to write this up in more detail.

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