Re: Jackson's derivation of the Lorentz Matrix

On Apr 16, 5:13 am, pmb <pm...@xxxxxxxxxxx> wrote:
On Apr 14, 11:19 am, "Dirk Van de moortel"



<dirkvandemoor...@xxxxxxxxxxxxxxxxxx> wrote:
pmb<pm...@xxxxxxxxxxx> wrote in message

  ba982bd3-d508-4211-9146-c13f7d730...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx

On Apr 13, 3:37 pm, "Dirk Van de moortel"
<dirkvandemoor...@xxxxxxxxxxxxxxxxxx> wrote:
blackhead <larryhar...@xxxxxxxxxxxx> wrote in message

c6c2f773-f274-435f-acba-3027e569f...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx

I'm looking at the 3rd edition of Jackson's Electrodynamics, pages 544
and 545:

^T is the transpose operator, I is the identitiy matrix, A the Lorentz
matrix

g = 1 0 0 0
0 -1 0 0
0 0 -1 0
0 0 0 -1

A must satisfy A^T g A = g (11.86)

He then starts the explicit construction of A by making the guess,
where L is a 4 x 4 matrix:

A = exp(L) (11.87)

11.86 can be rewritten as:

g A^T g = A^(-1) (11.88)

From the definition (11.87) and the fact that g^2 = I we have:

A^T = exp(L^T), g A^T g = exp(g L^T g), A^(-1) = exp (-L)

How does he get A^(-1) = exp (-L) ?

Standard property:
exp(X) exp(-X) = exp(O) = I
so
exp(-X) = [exp(X)]^(-1) = A^(-1)

It's not as easy as that since A and exp(L) are both matrices.

It is one of the standard maxtrix exponentals properties.
It is as "easy" as that.

If you want to confuse him it's that easy. If I hadn't pointed out
what I did then it'd be very easy for him to assume that exp(A)exp(B)
= exp(A+B) in the future because you claimed that "it is one of the
standard matrix exponentials properties" - Its standard *only* when
the arguments are numbers. It's invalid in general when the arguments
are matrices. You shouldn't complain when someone adds words of
caution when you post a poor explanation like you did here. Or is it
the way I phrased it, i.e. are you whining about me phrasing it as
"It's not as easy as that"?  ... Never mind. It's obvious that you
don't like people having to clarify your misleading response.

This newsgroup is not a free education in mathematics, Pete.
.

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