# How does the determinant of the metric transform

1. Oct 18, 2017

### Milsomonk

1. The problem statement, all variables and given/known data
In special relativity the metric is invariant under lorentz transformations and therefore so is the determinant of the metric. How does the metric determinant transform under a more general transformation $$x^{a\prime}=J^{a\prime}_{\quad a}x^{a}$$ where $$J^{a\prime}_{\quad a}$$ may not satisfy the conditions of the lorentz group.

2. Relevant equations
$$x^{a\prime}=J^{a\prime}_{\quad a}x^{a}$$

3. The attempt at a solution

So I need to show how the determinant transforms, but in general I thought that the determinant was a scalar and thus did not transform, clearly this isn't correct so my next thought was to perform the above transformation on the metric as follows:
$$\eta_{a^{\prime}b^{\prime}}=J_{a\prime}^{\quad a}J_{b\prime}^{\quad b}\eta_{ab}$$
Then take the determinant of the result. But this doesn't appear to get me anywhere and doesn't make use of the fact that J is in general not a lorentz transform, any guidance as to where I should go next or whether i'm barking up the wrong tree entirely would be much appreciated :)

2. Oct 18, 2017

### Orodruin

Staff Emeritus
The metric determinant is not a scalar, but a scalar density of weight $\pm 2$ (with sign depending on weight convention). You should be able to show this by expressing the metric determinant in terms of the permutation symbols and the metric. (Note that the permutation symbols are tensor densities.)

3. Oct 19, 2017

### Milsomonk

Aha I'll give that a go, thanks very much :)