# Taylor Expansions in Derivation of Gravitational Redshift in weak field limit

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!

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