# Schwarzschild metric: gravitational redshift

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapters 9; Problem 9.1.

The relation between the proper time interval and Schwarzschild time interval for an object at rest is

$\displaystyle \Delta\tau=\left(1-\frac{2GM}{r}\right)^{1/2}\Delta t \ \ \ \ \ (1)$

We can use this relation to derive a formula for the gravitational redshift. The key to this is that the wavelength of light is ${\lambda=c\Delta\tau}$, where ${c=1}$ is the speed of light and ${\Delta\tau}$ is the time interval as measured by an observer required for a single wavelength to be emitted or received. Why ${\Delta\tau}$ instead of ${\Delta t}$? As far as I understand it, this is because ${\tau}$ is the only correct measure of time for an object at rest. The Scwarzschild ${t}$ coordinate, although it’s called the ‘time coordinate’, isn’t really a measure of time directly.

The redshift arises because if we emit a light beam at ${r=r_{E}}$ in a direction radially outwards from the mass ${M}$ and receive the light beam at ${r=r_{R}>r_{E}}$, then the interval ${\Delta t}$ required for the passage of a single wavelength must be the same at both the emitter and the receiver. Why? Because the metric doesn’t depend on ${t}$.

The proper time interval, however, is not the same at the two points because of the relation above. Plugging in the values, we get

 $\displaystyle \frac{\lambda_{R}}{\lambda_{E}}$ $\displaystyle =$ $\displaystyle \frac{\Delta\tau_{R}}{\Delta\tau_{E}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-2GM/r_{R}}{1-2GM/r_{E}}} \ \ \ \ \ (3)$

This formula could be used, for example, to calculate the redshift due to a star when observed from Earth.

If both distances are large compared to ${2GM}$, we can expand the formula in a series up to first order:

 $\displaystyle \frac{\lambda_{R}}{\lambda_{E}}$ $\displaystyle \approx$ $\displaystyle \left(1-\frac{GM}{r_{R}}\right)\left(1+\frac{GM}{r_{E}}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+GM\left(\frac{1}{r_{E}}-\frac{1}{r_{R}}\right)+\ldots \ \ \ \ \ (5)$

As a further approximation, if the distance ${h=r_{R}-r_{E}\ll r_{E}}$, that is, the distance between emission and reception is small compared with the radial coordinate, then we can write

 $\displaystyle \frac{\lambda_{R}}{\lambda_{E}}$ $\displaystyle \approx$ $\displaystyle 1+GM\left(\frac{1}{r_{E}}-\frac{1}{r_{R}}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+GM\left(\frac{h}{r_{E}r_{R}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1+\frac{GM}{r^{2}}h \ \ \ \ \ (8)$

where ${r}$ in the last line can be taken as the average of ${r_{R}}$ and ${r_{E}}$. In this limit, we’d expect Newton’s law of gravitation to apply, and a particle a distance ${r}$ from a mass ${M}$ experiences an acceleration ${g=GM/r^{2}}$, so we have

$\displaystyle \frac{\lambda_{R}}{\lambda_{E}}\approx1+gh \ \ \ \ \ (9)$

As an example, suppose we have a neutron star with mass ${M=3\times10^{30}\mbox{ kg}}$ and Schwarzschild radial coordinate at the surface of ${r_{E}=1.2\times10^{4}\mbox{ m}}$. The redshift observed by a satellite orbiting the star at a radius ${r_{R}=1.7\times10^{4}\mbox{ m}}$ can be calculated using the approximation formula. We need to express ${G}$ in relativistic units (that is, where ${c=1}$ so that ${GM}$ has the units of length). Since the units of ${G}$ are ${\mbox{m}^{3}\mbox{kg}^{-1}\mbox{s}^{-2}}$, we need to eliminate the reference to seconds which we can do by dividing by ${c^{2}=9\times10^{16}\mbox{m}^{2}\mbox{s}^{-2}}$. That is

 $\displaystyle G$ $\displaystyle =$ $\displaystyle \frac{6.67\times10^{-11}}{9\times10^{16}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 7.41\times10^{-28}\mbox{m kg}^{-1} \ \ \ \ \ (11)$

For the neutron star,

 $\displaystyle GM$ $\displaystyle =$ $\displaystyle \left(7.41\times10^{-28}\right)\left(3\times10^{30}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.223\times10^{3}\mbox{ m} \ \ \ \ \ (13)$

We take

 $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(r_{E}+r_{R}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.45\times10^{4}\mbox{ m}\ \ \ \ \ (15)$ $\displaystyle g$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{2}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2.223\times10^{3}}{\left(1.45\times10^{4}\right)^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.06\times10^{-5}\mbox{ m}^{-1} \ \ \ \ \ (18)$

Incidentally, this is a massive acceleration compared to that on the Earth’s surface. In SI units, this comes out to ${\left(1.06\times10^{-5}\right)\left(9\times10^{16}\right)=9.54\times10^{11}\mbox{m s}^{-2}}$.

The fractional redshift is

 $\displaystyle \frac{\lambda_{R}-\lambda_{E}}{\lambda_{E}}$ $\displaystyle \approx$ $\displaystyle g\left(r_{R}-r_{E}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1.06\times10^{-5}\right)\left(5\times10^{3}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0529 \ \ \ \ \ (21)$

The exact value is

 $\displaystyle \frac{\lambda_{R}-\lambda_{E}}{\lambda_{E}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-2GM/r_{R}}{1-2GM/r_{E}}}-1\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0831 \ \ \ \ \ (23)$

This is the gravitational redshift formula. For ${r_{R}\rightarrow\infty}$, the formula reduces to

$\displaystyle \frac{\lambda_{R}}{\lambda_{E}}=\frac{1}{\sqrt{1-2GM/r_{E}}} \ \ \ \ \ (24)$

This formula could be used, for example, to calculate the redshift due to a star when observed from Earth.

If both distances are large compared to ${2GM}$, we can expand the formula in a series up to first order:

 $\displaystyle \frac{\lambda_{R}}{\lambda_{E}}$ $\displaystyle \approx$ $\displaystyle \left(1-\frac{GM}{r_{R}}\right)\left(1+\frac{GM}{r_{E}}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+GM\left(\frac{1}{r_{E}}-\frac{1}{r_{R}}\right)+\ldots \ \ \ \ \ (26)$

As a further approximation, if the distance ${h=r_{R}-r_{E}\ll r_{E}}$, that is, the distance between emission and reception is small compared the radial coordinate, then we can write

 $\displaystyle \frac{\lambda_{R}}{\lambda_{E}}$ $\displaystyle \approx$ $\displaystyle 1+GM\left(\frac{1}{r_{E}}-\frac{1}{r_{R}}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+GM\left(\frac{h}{r_{E}r_{R}}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 1+\frac{GM}{r^{2}}h \ \ \ \ \ (29)$

where ${r}$ in the last line can be taken as the average of ${r_{R}}$ and ${r_{E}}$. In this limit, we’d expect Newton’s law of gravitation to apply, and a particle a distance ${r}$ from a mass ${M}$ experiences an acceleration ${g=GM/r^{2}}$, so we have

$\displaystyle \frac{\lambda_{R}}{\lambda_{E}}\approx1+gh \ \ \ \ \ (30)$

As an example, suppose we have a neutron star with mass ${M=3\times10^{30}\mbox{ kg}}$ and Schwarzschild radial coordinate at the surface of ${r_{E}=1.2\times10^{4}\mbox{ m}}$. The redshift observed by a satellite orbiting the star at a radius ${r_{R}=1.7\times10^{4}\mbox{ m}}$ can be calculated using the approximation formula. We need to express ${G}$ in relativistic units (that is, where ${c=1}$ so that ${GM}$ has the units of length). Since the units of ${G}$ are ${\mbox{m}^{3}\mbox{kg}^{-1}\mbox{s}^{-2}}$, we need to eliminate the reference to seconds which we can do by dividing by ${c^{2}=9\times10^{16}\mbox{m}^{2}\mbox{s}^{-2}}$. That is

 $\displaystyle G$ $\displaystyle =$ $\displaystyle \frac{6.67\times10^{-11}}{9\times10^{16}}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 7.41\times10^{-28}\mbox{m kg}^{-1} \ \ \ \ \ (32)$

For the neutron star,

 $\displaystyle GM$ $\displaystyle =$ $\displaystyle \left(7.41\times10^{-28}\right)\left(3\times10^{30}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.223\times10^{3}\mbox{ m} \ \ \ \ \ (34)$

We take

 $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(r_{E}+r_{R}\right)\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.45\times10^{4}\mbox{ m}\ \ \ \ \ (36)$ $\displaystyle g$ $\displaystyle =$ $\displaystyle \frac{GM}{r^{2}}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2.223\times10^{3}}{\left(1.45\times10^{4}\right)^{2}}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.06\times10^{-5}\mbox{ m}^{-1} \ \ \ \ \ (39)$

Incidentally, this is a massive acceleration compared to that on the Earth’s surface. In SI units, this comes out to ${\left(1.06\times10^{-5}\right)\left(9\times10^{16}\right)=9.54\times10^{11}\mbox{m s}^{-2}}$.

The fractional redshift is

 $\displaystyle \frac{\lambda_{R}-\lambda_{E}}{\lambda_{E}}$ $\displaystyle \approx$ $\displaystyle g\left(r_{R}-r_{E}\right)\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1.06\times10^{-5}\right)\left(5\times10^{3}\right)\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0529 \ \ \ \ \ (42)$

The exact value is

 $\displaystyle \frac{\lambda_{R}-\lambda_{E}}{\lambda_{E}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-2GM/r_{R}}{1-2GM/r_{E}}}-1\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0831 \ \ \ \ \ (44)$

## 5 thoughts on “Schwarzschild metric: gravitational redshift”

1. Harald

When r_R comes close to the Schwarzschild radius, it looks like the wave length goes close to zero and the frequency and thereby the energy of the photon becomes very large. But this is true only for the observer sitting at r_R and watching, right?

1. growescience

I think that’s right, although I’m not sure I really understand this argument myself. I’m also not sure that a photon emitted at the Schwarzschild radius would actually escape at all.