I Redshift and the Friedmann metric

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1. Jun 30, 2018

redtree

My discussion of the Friedmann metric comes from the derivation presented in section 4.2.1 of the reference: https://www1.maths.leeds.ac.uk/~serguei/teaching/cosmology.pdf

I have a couple of simple questions on the derivation. The are placed at points during the derivation.

I note the following for the Friedmann metric for $k=0$:

\begin{split}

\partial \textbf{s}^2 &= -\partial t^2 + a^2(t) \left[ \partial dr^2 + r^2 \left( \partial \theta^2 + \sin^2{\theta}\partial \phi^2 \right)\right]

\end{split}

Which I rewrite as follows:

\begin{split}

\partial \textbf{s}^2 &= -\partial t^2 + a^2(t) \partial \vec{r}^2

\end{split}

For a zero rest-mass object $\partial \textbf{s}^2=0$, such that:

\begin{split}

\partial t^2 &= a^2(t) \partial \vec{r}^2

\end{split}

Thus:

\begin{split}

\partial t &= a(t) \partial \vec{r}

\end{split}

Such that:

\begin{split}

\frac{\partial t}{a(t)} &= \partial \vec{r}

\end{split}

Thus, where $t_E$ denotes time at emission, $t_O$ denotes time at observation and $R_E$ denotes radial distance at emission:

\begin{split}

\int_{t_E}^{t_O}\frac{\partial t}{a(t)} &= \int_{0}^{R_E}\partial \vec{r}

\end{split}

QUESTION: Why use $R_E$? Isn't the distance travelled by the photon $R_O$, where $R_0 = R_E + \partial \vec{r}$?

The derivation continues as follows for another photon emitted at $t_E + dt_E$ and observed at $t_O + dt_O$, such that:

\begin{split}

\int_{t_E+dt_E}^{t_O+dt_O}\frac{\partial t}{a(t)} &= \int_{0}^{R_E}\partial \vec{r}

\end{split}

QUESTION: Again, why use $R_E$? Isn't the distance travelled for this photon $R_O + \partial \vec{r}_O + \partial \vec{r}_E?$ If $R_O >> \partial \vec{r}_O + \partial \vec{r}_E$, then $R_O + \partial \vec{r}_O + \partial \vec{r}_E \approx R_O$. I assume this is the logic.

Continuing the derivation:

\begin{split}

\int_{t_E+dt_E}^{t_O+dt_O}\frac{\partial t}{a(t)} - \int_{t_E}^{t_O}\frac{\partial t}{a(t)}&= 0

\end{split}

Where:

\begin{split}

\int_{t_E+dt_E}^{t_O+dt_O}f(t) \partial t &= -f(t_E) \partial t_E+ \int_{t_E}^{t_O+dt_O}f(t) \partial t

\\

&=+f(t_O)\partial t_O - f(t_E) \partial t_E+ \int_{t_E}^{t_O}f(t) \partial t

\end{split}

Assuming $f(t) = \frac{1}{a(t)}$:

\begin{split}

\frac{\partial t_O}{a(t_O)} - \frac{\partial t_E}{a(t_E)} &=0

\end{split}

Such that:

\begin{split}

\frac{\partial t_O}{\partial t_E} &= \frac{a(t_O)}{a(t_E)}

\end{split}

Assuming $T_E = \partial t_E$ and $T_O = \partial t_O$:

\begin{split}

\frac{T_O}{T_E} &= \frac{a(t_O)}{a(t_E)}

\end{split}

Given:

\begin{split}

1+z &= \frac{\lambda_O}{\lambda_E}

\\

&=\frac{T_O}{T_E}

\end{split}

Thus:

\begin{split}

1+z &= \frac{a(t_O)}{a(t_E)}

\end{split}

2. Jun 30, 2018

Orodruin

Staff Emeritus
First of all, you should not be using $\partial$ to denote the differentials. There are many possible notations that are more or less standard, but that is not one of them.

To answer your questions, $R_O = R_E + dr$ makes very little sense. Note that $R_E$ is the comoving distance to the emitter, not a physical distance.

Edit: You can also find a different way to do the derivation (as well as the standard way) in my PF Insight.

3. Jul 17, 2018

redtree

I take your point regarding $\partial$.

Given $R_E$ denotes the comoving distance, then it would be constant with the cosmological expansion, whereas the proper distance increases with the expansion. I don't understand why the integration of $d\vec{r}$ is over the interval of the comoving distance and not the proper distance. Isn't $d\vec{r}$ a small slice of the proper distance interval?

4. Jul 17, 2018

Orodruin

Staff Emeritus
No. It is an infinitesimal change in the comoving distance.

5. Jul 17, 2018

redtree

Such that $a(t) d\vec{r}$ is the change in proper distance. Got it.

Last edited: Jul 17, 2018
6. Jul 17, 2018

redtree

I have one other question regarding the equation. Why the assumption that the cosmic expansion affects space but not time? Why not assume that it affects time and not space? Or both equally?

7. Jul 18, 2018

Orodruin

Staff Emeritus
The FLRW metric is based on the assumption of space being homogeneous and isotropic and its scale depending on the time, that is what the scale factor is. You could reparametrise it using different coordinates but then your time would not be the proper time of a comoving observer.

8. Jul 18, 2018

redtree

It just seems both arbitrary and classical to assume that spacetime would expand and even inflate only in its spacial components. GR and SR affect both space and time. Why wouldn’t the spacetime expansion affect both as well? I’m not expecting an answer. I’m just commenting.

9. Jul 18, 2018

Orodruin

Staff Emeritus
As I said in my previous post, it is just a coordinate choice (and a quite reasonable one at that) to use the proper time of comoving observers as the time coordinate. It is perfectly possible to make a different choice - it will not affect the physics - you will just spend more time interpreting your results than you would do otherwise. Hence, it is not arbitrary.

Even if it was arbitrary, it would not be strange in my opinion. The scale factor changes with time, but you don’t get more change in time per change in time.

10. Jul 19, 2018

redtree

Another assumption is that $a(t)$ is a scale factor associated with expansion. That is also an assumption. The math can be done without assuming a physical interpretation of $a(t)$.