In general, the redshift of a photon emitted from one static observer at point $A$ to another static observer located at point $B$ is of the form

$$ z_{AB} = \frac{\omega_A}{\omega_B} - 1 $$

where $\omega_A$ is the frequency of the photon observed in observer A's frame of reference. I understand how, in the presence of a weak gravitational field, we may take

$$ \frac{\omega_A}{\omega_B} = \sqrt{\frac{g_{00}(R + L)}{g_{00}(R)}} \approx \sqrt{\frac{1 + 2\Phi(R + L)}{1 + 2\Phi(R)}} $$

where $g_{00}$ is the time-time component of the metric, $\Phi(r) = -\frac{GM}{r}$ is the classical Newtonian potential, $R$ is the radius of the Earth, and $L$ is the distance from Earth's surface to the static observer at point B. Furthermore, I know what the redshift should be:

$$ z_{AB} = \frac{gL}{c^2} $$

where $g = \frac{GM}{R^2}$ is the acceleration due to gravity and we've divided by $c^2$ to get the correct units. We may easily verify that

$$ z_{AB} \approx \Phi(R + L) - \Phi(R) $$

reproduces the correct result. However, the problem I am facing is in demonstrating that this follows from the expression I have above for $\frac{\omega_A}{\omega_B}$. Specifically, this is what my professor did:

\begin{eqnarray} \sqrt{\frac{1 + 2\Phi(R + L)}{1 + 2\Phi(R)}} &\approx \sqrt{1 + 2[\Phi(R + L) - \Phi(R)]}\\ &\approx 1 + [\Phi(R + L) - \Phi(R)]. \end{eqnarray}

The second approximation is not mysterious. We're simply using a Taylor expansion of the form

$$ \sqrt{1 + x} \approx 1 + \frac{1}{2}x + \cdots $$ but I have no idea where the first approximation comes from. If it's a Taylor expansion of some sort, I don't know how to formulate it, let alone verify it. I'd really appreciate it if someone could fill in this gap in my understanding of what is otherwise a very neat derivation. Thanks!


1 Answer 1


It's the same idea, except now the expansion is $$\frac{1}{1+x} \approx 1 - x$$

  • $\begingroup$ Yeah, I figured it out xD. You treat the term in the numerator as a constant and take the Taylor expansion of the denominator. I was confused because I was assuming that you had to expand using both expressions involving Phi. I was wrong. $\endgroup$ Nov 28, 2021 at 20:03

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