What I know

I am told that in an inertial frame, the 4-momentum of a photon is:

$$p^{\mu}=\frac{E}{c}\left(1, \frac{\vec{p}}{|\vec{p}|}\right).\tag{1}$$

I can also show with the geodesic equation, that for a stationary metric, $p_0$ is constant.

Physical Setup

Lets have a photon emitted from the surface of a gravitating object. It is received by an observer at infinity. I am interested in how much the photon has redshifted.

To get the energy of the photon observed by the observer, I do, using the above equation:

$$E_{\text{Obs}}=p^0_{\text{Obs}}c=g^{00}_{\text{Obs}} p_0c\tag{2}$$

Where I have raised the index with the metric: $p^0=g^{00}p_0$.

Let's have on observer where the photon was emitted. It'll detect an energy for it:

$$E_{\text{Emit}}=p^0_{\text{Emit}}c=g^{00}_{\text{Emit}} p_0c\tag{3}$$.

To get ratio of frequencies, I divide the energies. $p_0$ and $c$ is the same for both observers. I obtain:

$$\frac{E_{\text{Obs}}}{E_{\text{Emit}}}=\frac{g^{00}_{\text{Obs}}}{g^{00}_{\text{Emit}}}=\frac{g_{00 \text{Emit}}}{g_{00 \text{Obs}}}\tag{4}$$

The problem

When I compare with my lecture notes, I find out that I am missing a square root & I should have obtained:

$$\frac{E_{\text{Obs}}}{E_{\text{Emit}}}=\sqrt{\frac{g_{00 \text{Emit}}}{g_{00 \text{Obs}}}}\tag{5}$$

What is wrong with my above calculation?

  • 1
    $\begingroup$ This is a bit more complicated than that. To measure the frequency of a photon one needs an observer and its 4-velocity (which must be properly normalized with the metric!). The measured frequency is then the scalar product of this velocity and the photon 4-momentum. $\endgroup$
    – Nikodem
    Jan 28, 2021 at 22:07

1 Answer 1


In order to measure the frequency of a photon one needs an observer and its 4-velocity which must be properly normalized with the metric. A static observer will have $v^\mu = (v^0, 0, 0, 0)$ normalized with $g_{\mu\nu} v^\mu v^\nu = g_{00} (v^0)^2 = 1$ which gives $v^0 = 1/\sqrt{g_{00}}$.

If the spacetime is stationary it has a Killing-Vector $\xi^\mu = (1,0,0,0)$. Let the photon has a 4-velocity $u^\mu$ normalized to zero $u_\mu u^\mu = 0$ along its geodesics. Then it has a constant of motion $\omega = u_\mu \xi^\mu = u_0$ which is its energy.

The frequency (energy) measured by the observer $v^\mu$ is $E_{obs} = u_\mu v^\mu = v^0 = 1/\sqrt{g_{00}}$. That is why the ratio of two frequencies measured at different positions is given by the ratio of the inverse(!) square roots of the metric component $g_{00}$: $$ \frac{E_2}{E_1} = \frac{\sqrt{g_{00}(x_1)}}{\sqrt{g_{00}(x_2)}} $$ This is the basic formula giving the well-known redshift in the Schwarzschild metric.

Referring to your calculation: I guess you got the normalization of $v^\mu$ wrong. Neither $v_0$ nor $v^0$ is constant and independent of the metric which is what you probably assumed. (I set $c=1$ for convenience but this should not be an issue to complement it.)


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