Varying metric determinant

[Themetricsystem]

Let h = det h_{alpha beta}. The number of dimensions is not necessarily four. Show that
$$\[ \delta h = -h h_{\alpha \beta} \delta h^{\alpha \beta} \, ; \]$$

delta h is the variation in h.

Not sure how to start.

 

[dextercioby]

HINT:

$$\mbox{det} A=\exp\mbox{Tr} \ \ln A$$

Daniel.

 

[tacman]

To begin, we can rewrite the equation given as:

\[
\delta h = -h \, \mathrm{det}(h_{\alpha \beta}) \, \delta h^{\alpha \beta}
\]

We can see that the determinant term, h, is being multiplied by the variation in the inverse metric, $\delta h^{\alpha \beta}$. This makes sense, as the determinant is used to calculate the inverse metric.

To show that this equation holds, we can start by considering the definition of the determinant of a matrix. In this case, the matrix is the metric tensor $h_{\alpha \beta}$.

The determinant of a matrix is defined as the product of its eigenvalues, or in the case of a symmetric matrix like $h_{\alpha \beta}$, it is equal to the product of the diagonal elements.

So, we can write the determinant of $h_{\alpha \beta}$ as:

\[
\mathrm{det}(h_{\alpha \beta}) = h_{11}h_{22}...h_{nn}
\]

Where n is the number of dimensions. This shows that the determinant is not necessarily four, as it depends on the number of dimensions.

Now, let's consider the variation in the determinant, $\delta h$. This can be written as:

\[
\delta h = h_{11}h_{22}...h_{nn} \, \delta h^{\alpha \beta}
\]

We can see that this is similar to the original equation given, with the determinant term being multiplied by the variation in the inverse metric. However, there is an additional factor of $h_{\alpha \beta}$ in the original equation.

To account for this, we can rewrite the determinant as:

\[
\mathrm{det}(h_{\alpha \beta}) = h_{11}h_{22}...h_{nn} \, h_{\alpha \beta}
\]

Substituting this into our equation for $\delta h$, we get:

\[
\delta h = -h \, \mathrm{det}(h_{\alpha \beta}) \, \delta h^{\alpha \beta}
\]

Which is the same as the original equation given. This shows that the equation holds, and that the number of dimensions is not necessarily four.