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Suppose the line element is $$ds^2 = -A(t,r)^2dt^2+B^2(t,r)dr^2+C^2(t,r)d\theta^2+C^2\sin^2\theta d\phi^2.$$
Since the metric is diagonal, to find the determinant I can multiply the diagonal entries, $$\det g_{ab} = g = -A^2B^2C^4 \sin^2\theta.$$ I have a few questions about this.

  1. First off, why do we call the metric determinant $g$?
  2. Why isn't it true that $g = g_{ab} g^{ab} = 4$? Isn't that how $g$ is defined?
  3. When will it be true that $g = 1$?
knzhou
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user2129579
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    It's just notation. We define $g = \det g_{ab}$. In some texts, they use the definition $g = | \det g_{ab} |$ instead. – Prahar Oct 18 '17 at 15:28
  • @Prahar Okay. But what if I write $g_{ab}$$g^{ab}$=$g$, is this correct? – user2129579 Oct 18 '17 at 15:32
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    $g_{ab} g^{ab}$ has nothing to do with the metric determinant. Think about it in terms of matrices; then $g_{ab} g^{ab} = \text{tr} (g^T g^{-1})$, which is something completely different. – knzhou Oct 18 '17 at 15:38
  • @user2129579 - No. $g_{ab} g^{ab}$ has nothing to do with the determinant of $g$. – Prahar Oct 18 '17 at 15:42
  • If you want to use index notation, the determinant of $g_{ab}$ can be written as $g\equiv\text{det}(g_{ab})=\epsilon_{i_0 i_1\cdots i_n} g_{0,i_0}g_{1,i_1}\cdots g_{n,i_n}$, where I have assumed the indices $a$ and $b$ run from $0$ to $n$. See https://en.wikipedia.org/wiki/Determinant for more details. – L. Werneck Oct 26 '17 at 15:20

2 Answers2

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  1. First off, why do we call the metric determinant $g$?

Because we define the notation that way.

  1. Why isn't it true that $g = g_{ab} g^{ab} = 4$? Isn't that how $g$ is defined?

If the determinant of the metric could be written using abstract index notation, without resorting to non-tensorial objects like the Levi-Civita tensor, then it would be an observable quantity that was a property of space at a particular point. There are such quantities: they're measures of curvature or related quantities involving derivatives of the curvature. But the metric itself (as opposed to its derivatives) doesn't let you compute a curvature, and therefore any such quantity that doesn't involve a derivative cannot be an observable.

What this tells you is that the determinant of the metric isn't a property of space, it's a property of the coordinates you've chosen. For example, if you use coordinates in which the basis vectors aren't orthogonal, then the determinant of the metric will be smaller.

  1. When will it be true that $g = 1$?

Never, in 3+1 dimensions, because $g$ will be negative. If you want to know when $g=-1$, then the answer is whenever you choose coordinates that make that true.

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I will add something to 2. First of all,

$$g \neq g_{ab}g^{ab}$$

but

$$ \delta g = gg^{ab} \delta g_{ba} $$

which you can prove from

$$ \partial_\mu(\det(A)) = \det(A)tr(A^{-1}\partial_\mu A) $$

which you can prove from

$$ \det \left( e^{B} \right) = e^{\mathrm{tr}(B)}$$

Ege Özmeral
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