Gravitational redshift

Before I talk about gravitational redshift, let me just briefly talk about doppler shift.

Doppler shift is a high school concept that most of us have learnt at some point or the other in our lives. If you haven’t learnt it, safe to say you’ve surely experienced it. Say you’re on a platform and a train is approaching you. The train is sounding a horn, to catch the attention of passengers on the platform. The increase in the pitch of the horn as the train is approaching you and the decrease in pitch as the train is moving away is the phenomenon of doppler shift.

The same can happen to light. Say a source of light is moving towards you, the frequency you perceive the light to be at is not the same as the frequency of light emitted by the source. This is because of the fact that a clock in the frame of the source ticks slower than your clock, which is at rest. This is the phenomenon of time dilation.

Venturing further now, the Schwarzschild metric describes the spherically symmetric curved space time around a mass. It is the most general vaccum solution. The metric is as follows

$$g_{\mu\nu} =diag((1-\frac{2GM}{r}),\frac{1}{(1-\frac{2GM}{r})},r^2, r^2\sin^2\theta)$$

Euclidean space in cartesian coordinate system has a metric of the form diag$(1,1,1)$. A matric describes space or spacetime of a manifold, in a specific coordinate system.

Now, let’s say that there is a source at $r = r_E$ who is emitting radiation and there is an observer at $r = r_R$ who is receiving the radiation. i.e they are separated radially. Now let’s say that the source emits two pulses of light at $t_E$ and $t_E+\Delta t_E$ which the observer perceives at $t_R$ and $t_R+\Delta t_R$

Given the metric, we can write

$$c^2dt^2(1-\frac{2GM}{r}) = \frac{dr^2}{(1-\frac{2GM}{r})}$$
$$dt = \frac{dr}{c(1-\frac{2GM}{r})}$$
$$\int^{t_R}_{t_E}dt = \int^{r_R}_{r_E}\frac{dr}{c(1-\frac{2GM}{r})}$$
$$t_R-t_E=f(r_E,r_R)$$

Similarly for the second pulse

$$\int^{t_R+\Delta t_R}_{t_E+\Delta t_R}dt = \int^{r_R}_{r_E}\frac{dr}{c(1-\frac{2GM}{r})}$$
$$t_R-t_E + \Delta t_R- \Delta t_E =f(r_E,r_R)$$
$$\Delta t_R- \Delta t_E = 0$$
$$\Delta t_R = \Delta t_E$$

meaning that for an observer at r = $\inf$, the time difference between the two events at the source and at the observer are the same.

Now, let’s talk about how clocks at $r_E$ and $r_R$ tick i.e $\tau$

$$ds^2 = c^2d\tau^2 = c^2dt^2(1-\frac{2GM}{r})$$
$$d\tau = dt(1-\frac{2GM}{r})^{1/2}$$

$$\Delta\tau_E = \Delta t_E(1-\frac{2GM}{r_E})^{1/2}$$
$$\Delta\tau_R = \Delta t_R(1-\frac{2GM}{r_R})^{1/2}$$

$$\frac{\Delta\tau_E}{\Delta\tau_R} = \frac{\Delta t_E(1-\frac{2GM}{r_E})^{1/2}}{\Delta t_R(1-\frac{2GM}{r_R})^{1/2}}$$

$$\frac{\Delta\tau_E}{\Delta\tau_R} = \frac{(1-\frac{2GM}{r_E})^{1/2}}{(1-\frac{2GM}{r_R})^{1/2}}$$

we can convert the time difference to frequency and therefore energy, therefore changing the above equation to

$$\frac{E_R}{E_E} = \Bigg(\frac{1-\frac{2GM}{r_E}}{1-\frac{2GM}{r_R}}\Bigg)^{1/2}$$

So, for the case $r_E < r_R$, it can be seen that $E_E > E_R$ i.e for a photon emitted at $r_R$, by the time it reaches $r_R$, it would have lost a certain amount of energy.

I would like to remind you again that this is curved spacetime and not flat spacetime. In qualitative terms, in a central gravitational potential, a photon will lose energy as it climbs out of the potential.

An interesting nugget of information is the physical implication of clocks in curved spacetime. The spacetime around earth is curved and clocks in space i.e on satellites run a bit slower than a clock on earth. And this difference is crucial to calibrate GPS signals, without which the accuracy with which one can pin point their location becomes abysmal.

This article was written using StackEdit1.

---

1. StackEdit is a full-featured, open-source Markdown editor based on PageDown, the Markdown library used by Stack Overflow and the other Stack Exchange sites. ↩