A Solving Jackson's 3rd Ed. Equations Involving A, L and g

[anbhadane]

TL;DR Summary

I am reading Jackson's book, where on (p.545) I got some doubts.

In Jackson, (3rd edition p. 545), there are equations they are given as,
$$A = e^L $$
$$det A = det(e^L) = e^{Tr L}$$
$$g\widetilde{A}g = A^{-1} $$
$$ A = e^L , g\widetilde{A}g = e^{{g\widetilde{L}g}} , A^{-1} = e^{-L}$$
$$ g\widetilde{L}g = -L $$

I have several doubts.

1) $$det(e^L) = e^{TrL}$$ How determinant is equal to RHS of equation?, does here are we assuming there is special type of L ?

2) Now in $$ g\widetilde{A}g = e^{{g\widetilde{L}g}}$$, How is it possible to have $$g = e^{g}?$$

 

[martinbn]

For 1) https://en.wikipedia.org/wiki/Matrix_exponential#The_determinant_of_the_matrix_exponential
For 2) not sure what the question is

p.s. You wrote doubts. Did you mean questions?

 

[anbhadane]

yes doubts are questions.
Oh, I come to know about the first part but still for second question how is it possible to have metric tensor equal to exponential of metric tensor?

 

[George Jones]

but still for second question how is it possible to have metric tensor

Is $g$ the matrix for the metric tensor for Minkowski spacetime in Cartesian coordinates? If so, then $g^2$ = $I$ and $g = g^{-1}$. Consequently,
$$\begin{align}
e^{g \tilde L g} &= e^{g \tilde L g^{-1}} \\
&= I + g \tilde L g^{-1} + \frac{1}{2!} \left( g \tilde L g^{-1} \right)^2 + \frac{1}{3!} \left( g \tilde L g^{-1} \right)^3 + \ldots \\
&= I + g \tilde L g^{-1} + \frac{1}{2!} g \tilde L g^{-1} g \tilde L g^{-1} + \frac{1}{3!} g \tilde L g^{-1} g \tilde L g^{-1} g \tilde L g^{-1} + \ldots \\
&= g I g^{-1} + g \tilde L g^{-1} + \frac{1}{2!} g \tilde L^2 g^{-1} + + \frac{1}{3!} g \tilde L^3 g^{-1} +\ldots \\
&= g \left( \tilde L + \frac{1}{2!} \tilde L^2 + \frac{1}{3!} \tilde L^3 + \dots \right) g^{-1} \\
&= g e^{\tilde L} g^{-1} .
\end{align}$$

 

[martinbn]

Oh, I come to know about the first part but still for second question how is it possible to have metric tensor equal to exponential of metric tensor?

He has that $A=e^L$, here $A$ isn't a metric, neither is $L$.

 

[anbhadane]

Now I got it. I was thinking in very different way. First one was easy, and second one is essentially taylor expansion. But here we are just taking approximation, thank you all of you.

 

[martinbn]

Now I got it. I was thinking in very different way. First one was easy, and second one is essentially taylor expansion. But here we are just taking approximation, thank you all of you.

There are no approximations. What George Jones wrote is exact.

 

[anbhadane]

Oh,sorry , got it.

 

[anbhadane]

Oh, similar way of expansion, calculations on 547, are done. In Jackson's (p.547) equation such as (11.94) , (11.96), (11.98) are expanded by Taylor expansion.