# Expansion for gravitational time dilation

In the section on gravitational time dilation in Prof. David Tong's lecture notes on general relativity, he performs the following expansion:

$$t\sqrt{1-\frac{2GM}{r_{A}c^{2}}+\frac{2GM{\Delta}r}{r^{2}_{A}c^{2}}}\\{\approx}t\sqrt{1-\frac{2GM}{r_{A}c^{2}}}\left(1+\frac{GM{\Delta}r}{r^{2}_{A}c^2}\right)$$

I was hoping someone could fill in the steps between the lines here, as I'm having some trouble.

• It's a binomial expansion to first order, $(1+x)^n \approx 1 + nx$, where you can let the $1$ be $\sqrt{1 - 2GM/(r_A c^2)}$, and your $x = 2GM\Delta r/(r^2_A c^2)$. Then, with $n = 1/2$, you can see why the two gets cancelled on our $x$-value. Mar 1, 2022 at 15:28
• @Cassem02 If I expand $(c+x)^{\frac{1}{2}}$ I get $\sqrt{c}+\frac{x}{2 \sqrt{c}}-...$ Mar 1, 2022 at 16:06

I called the function f(x), so $$x$$ instead of $$\Delta r$$ and all divided by $$t$$.

$$f(x) = \sqrt{1-\frac{2GM}{r_Ac^2}+\frac{2GM x}{r_A^2c^2}}$$

$$f'(x) = \frac{1}{2} \frac{1}{\sqrt{1-\frac{2GM}{r_Ac^2}+\frac{2GM x}{r_A^2c^2}}} \frac{2GM }{r_A^2c^2}$$

With normal Taylor expansion around $$x_0 = 0$$ to first order:

$$f(x) \approx \sqrt{1-\frac{2GM}{r_Ac^2}} + \frac{1}{ \sqrt{1-\frac{2GM}{r_Ac^2}}}\frac{GM x}{r_A^2c^2}$$

$$= \sqrt{1-\frac{2GM}{r_Ac^2}} (1+\frac{1}{1-\frac{2GM}{r_Ac^2}} \frac{GM x}{r_A^2c^2})$$

If you say $$\frac{1}{1-\frac{2GM}{r_Ac^2}} \approx 1$$ you recover your result. But as for now, doing so does not make sense to me.

• I think you're correct as he does mention making that approximation just below. Thanks! Mar 1, 2022 at 22:04