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garrett
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#29
Jun27-06, 07:53 PM
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Quote Quote by garrett
[tex]
\xi^-_i{}^B
= \delta_{iB} \frac{\sin(r)\cos(r)}{r} + x^i x^B ( \frac{1}{r^2} - \frac{\sin(r)\cos(r)}{r^3} ) + \epsilon_{ikB} x^k \frac{\sin^2(r)}{r^2}
[/tex]
And now I have to go figure out what the inverse of that is...
[tex]
\xi_B{}^i
= \delta_{Bi} \frac{r \cos(r)}{\sin(r)} + x^B x^i ( \frac{1}{r^2} - \frac{\cos(r)}{r \sin(r)} ) + \epsilon_{Bik} x^k
[/tex]

:)

By the way, if you're trying to do this yourself by hand, I calculated the inverse by making the ansatz:
[tex]
\xi_B{}^i
= \delta_{Bi} A + x^B x^i B + \epsilon_{Bik} x^k C
[/tex]
and solving for the three coefficients.

Now I'm going for a bike ride, then coming back to do rotations.