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


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

  • $\begingroup$ 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. $\endgroup$
    – MathZilla
    Mar 1, 2022 at 15:28
  • $\begingroup$ @Cassem02 If I expand $(c+x)^{\frac{1}{2}}$ I get $\sqrt{c}+\frac{x}{2 \sqrt{c}}-...$ $\endgroup$
    – Martin
    Mar 1, 2022 at 16:06

1 Answer 1


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.

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

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