- #1

- 218

- 3

## Main Question or Discussion Point

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{equation}

\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}

\end{equation}

Which I rewrite as follows:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Thus:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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{equation}

\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}

\end{equation}

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{equation}

\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}

\end{equation}

Where:

\begin{equation}

\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}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Given:

\begin{equation}

\begin{split}

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

\\

&=\frac{T_O}{T_E}

\end{split}

\end{equation}

Thus:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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{equation}

\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}

\end{equation}

Which I rewrite as follows:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Thus:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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{equation}

\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}

\end{equation}

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{equation}

\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}

\end{equation}

Where:

\begin{equation}

\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}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Given:

\begin{equation}

\begin{split}

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

\\

&=\frac{T_O}{T_E}

\end{split}

\end{equation}

Thus:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}