- #1

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Not being a mathematician, what I'd really like is a matrix representation

and a rule for getting the covariant derivative in the event of a broken symmetry.

I'd be much obliged if any one can give me any information.

M.

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- Thread starter Mentz114
- Start date

- #1

- 5,428

- 292

Not being a mathematician, what I'd really like is a matrix representation

and a rule for getting the covariant derivative in the event of a broken symmetry.

I'd be much obliged if any one can give me any information.

M.

- #2

HallsofIvy

Science Advisor

Homework Helper

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[tex]\left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right)\left(\begin{array}{c}x \\ y \\ z \\ 1\end{array}\right)= \left(\begin{array}{c}x+ u \\ y+ v \\ z+ w \\ 1\end{array}\right)[/tex]

- #3

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HallsofIvy, thank you very much. This looks promising. Seems to form a group under multiplication, with an identity if u=v=w=0.

[tex] \left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right)\left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right) = \left(\begin{array}{ccccccc}1 && 0 && 0 && 2u \\ 0 && 1 && 0 && 2v\\ 0 && 0 && 1 && 2w \\ 0 && 0 && 0&& 1\end{array}\right)

[/tex]

[tex] \left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right)\left(\begin{array}{ccccccc}1 && 0 && 0 && u \\ 0 && 1 && 0 && v\\ 0 && 0 && 1 && w \\ 0 && 0 && 0&& 1\end{array}\right) = \left(\begin{array}{ccccccc}1 && 0 && 0 && 2u \\ 0 && 1 && 0 && 2v\\ 0 && 0 && 1 && 2w \\ 0 && 0 && 0&& 1\end{array}\right)

[/tex]

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