# Problem understanding these concepts in Jackson

• A

## Summary:

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

## Main Question or Discussion Point

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 $$g = e^{g}?$$

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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
Staff Emeritus
Gold Member
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##.

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