# Relativistic Redshift and understanding it's approximation

I was reading an article, and I saw the expression $$1+z=\frac{(g_{\mu\nu}k^{\mu}u^{\nu})_e}{(g_{\mu\nu}k^{\mu}u^{\nu})_o},$$ where $$e$$ represents the emitter frame, $$o$$ the observer frame, $$g_{\mu\nu},$$ is the metric, $$k^{\mu}$$ is the photon four-momentum and $$u^{\nu}$$ is the four-velocity of the source or observer.

Has anyone seen this expression before? I want to understand how we can obtain the usual cosmological redshift equation, that is $$1+z=\frac{a_o}{a_e} \equiv \frac{a(t_o)}{a(t_e)}$$

from this expression and understand the metric potentials etc. Any reference would be appreciated. Thanks.

• Could you define what is meant by $a_o$ and $a_e$? I'm a bit more used to the first formula, but not with the second Feb 7, 2022 at 11:55
• It's the well known cosmological redshift Feb 7, 2022 at 12:00

This answer is pretty much based on Wald's General Relativity, particularly Sec. 5.3.

For completeness, let me begin by saying the first equation you wrote is a more direct definition of redshift. Notice that $$g_{\mu\nu} k^{\mu} u^{\nu}$$ is exactly the frequency an observer with four-velocity $$u^{\mu}$$ measures the photon to have. Hence what the expression says is simply that the redshift $$z$$ is such that $$1 + z = \frac{\omega_{\text{emitted}}}{\omega_{\text{detected}}},$$ pretty much what we expected the word "redshift" to mean. It is also applicable when studying the gravitational redshift due to a black hole, for example, so it is more general than the cosmological expression (although I believe I've never seen someone using $$z$$ when talking about black holes, the RHS is pretty common).

To derive the expression, I'll follow the arguments Wald gives on pp. 103–104. Let us write the FLRW metric as $$\mathrm{d}s^2 = -\mathrm{d}\tau^2 + a(\tau)^2 \mathrm{d}\Sigma^2, \tag{1}$$ so $$\tau$$ is cosmological time. $$\Sigma$$ is the spatial section of spacetime, taken to be orthogonal to $$\left(\frac{\partial}{\partial \tau}\right)^\mu$$. We'll be using an observer with four-velocity $$u^\mu = \left(\frac{\partial}{\partial \tau}\right)^\mu$$, so I'll write simply $$u^\mu$$ from now on.

Now consider $$k^\mu$$. Let $$\xi^\mu$$ be its projection onto $$\Sigma$$, so that we can write $$k^\mu = -k^\nu u_\nu u^\mu + \xi^\mu.$$

Since $$k^\mu$$ is null, it must hold that its projection into $$\Sigma$$ and its projection into $$u^\mu$$ are actually of the same magnitude. Essentially, its spacelike part and its timelike part are of the same magnitude, so that the final vector ends up being null. Hence, $$k_\mu u^\mu = - \frac{k_\mu \xi^\mu}{\sqrt{\xi_\nu \xi^\nu}}, \tag{2}$$ where the denominator on the RHS is a normalization factor, since $$\xi^\mu\xi_\mu$$ need not be one according to our definition.

Using these formulae, we see that your first expression now can be written as $$1 + z = \left(\frac{k_\mu \xi^\mu}{\sqrt{\xi_\nu \xi^\nu}}\right)_{e} \left(\frac{k_\mu \xi^\mu}{\sqrt{\xi_\nu \xi^\nu}}\right)_{o}^{-1},$$ where I used Eq. (2).

However, notice that since $$\xi^\mu$$ lies in $$\Sigma$$, its magnitude should scale with $$a(\tau)$$. In other words, we know that $$\frac{\sqrt{\xi^\mu \xi_\mu}|_o}{\sqrt{\xi^\mu \xi_\mu}|_e} = \frac{a_o}{a_e},$$ which allows us to conclude that $$1 + z = \frac{a_o}{a_e} \frac{k_\mu \xi^\mu |_e}{k_\mu \xi^\mu |_o}.$$

If we can prove that $$k_\mu \xi^\mu |_e = k_\mu \xi^\mu |_o$$, we're done. The trick now is noticing that $$\xi^\mu$$ is a Killing field, for $$\Sigma$$ is maximally symmetric. In other words, since $$\Sigma$$ is homogeneous and isotropic, $$\xi^\mu$$ must be a Killing field. However, given any Killing field $$\chi^\mu$$ and any four-velocity $$U^\mu$$ of a geodesic, it is well-known that $$\chi^\mu U_\mu$$ is a constant along the geodesic. Hence, $$k_\mu \xi^\mu |_e = k_\mu \xi^\mu |_o$$ does hold, and we get to the result $$1 + z = \frac{a_o}{a_e}.$$

## Topics Differential Geometry Raised in the Comments

### What does $$u^\mu = \left(\frac{\partial}{\partial \tau}\right)^\mu$$ means?

In Differential Geometry, and hence also in General Relativity, vectors are defined as directional derivatives (or in a different way that is equivalent to this). Hence, in GR it is common to sometimes write vectors in terms of the derivatives they are built of. These derivatives can be thought of as the basis of the vector space.

In this particular case, $$u^\mu = \left(\frac{\partial}{\partial \tau}\right)^\mu$$ means the four-velocity $$u^\mu$$ is given, in terms of the coordinates I used to write the metric in Eq. (1), as $$u^\mu = (1,0,0,0)^\intercal$$. In other words, I'm taking the observer to be always in the reference frame that keeps space isotropic.

### Killing Vector Fields

Killing fields are vector fields that keep the metric unchanged as you move along them. They are a mathematical way of stating more precisely what one means by homogeneity, isotropy, rotational symmetry, time-translation symmetry, and so on. They are the key objects encoding the notion of symmetry in GR.

The result I mentioned of how $$U^\mu \chi_\mu$$ doesn't changed can be thought of as a very different way of stating momentum conservation, for example. Since the spacetime is homogeneous and isotropic, there is always a symmetry in the direction of $$\xi^\mu$$ in space, and as a consequence momentum is conserved in that direction by means of Noether's theorem. Since $$u^\mu \xi_\mu$$ is the momentum of the photon as measured by the observer with four-velocity $$u^\mu$$, it doesn't change as one follows the trajectory of the observer along spacetime.

### Which assumptions (isotropy, homogeneity, flatness, etc) are we making?

I'm assuming a FLRW metric, i.e., the spatial sections are homogeneous and isotropic. These hypothesis enter on the form of the metric, and also on the piece of the argument depending on $$\xi^\mu$$ being a Killing field. I'm not sure if the formula you quoted holds for more general cases, but my guess is that it will likely get more complicated.

Also, I'm assuming an observer in the reference frame that keeps the Universe isotropic (notice that isotropy always requires a favorite reference frame, since boosting from an isotropic frame will pick up a favorite direction). This comes into play when I write $$k^\mu$$ in terms of $$u^\mu$$ and $$\xi^\mu$$ and when I argue that the magnitudes of both projections should be equal. The result can be generalized to other observers by simply computing the relativistic Doppler shift at the point of emission or detection, for example.

For example, suppose the emission frequency is measured by an observer that keeps the Universe isotropic, but the detection frequency is measured by a moving observer. One can first compute the cosmological redshift in the way I described, then make the correction from an isotropic to a moving observer afterwards.

• This is reallly helpful, but feeling a bit lost due to my level of knowledge. I am not really familier with the killing vectors...For instance what it means to have $u^{\mu}=(\frac{\partial}{\partial \tau})^{\mu}$. Is there way to explain it without using the killing vectors ? Feb 7, 2022 at 15:03
• What is the main assumption behind this calculation. Assuming CP/FLRW metric or something else ? Feb 7, 2022 at 15:15
• @Neptune Please check the updated version. If there's anything that's still unclear, please let me know =) Feb 7, 2022 at 15:24
• Oh CP is the Cosmological Principle Feb 7, 2022 at 16:35
• @Neptune Oops, actually that expression should be contracted on both sides. I just fixed the answer, thanks for pointing it out! And you're welcome! =D Feb 7, 2022 at 17:12