0
$\begingroup$

I'm very confused by the notation used in General Relativity. For example, how are indices changed for the Lorentz transformation matrix $\Lambda$ with its components $\Lambda_\mu^{\text{ }\nu}$ or $\Lambda_\nu^{\text{ }\mu}$? [Actually, does LaTeX provide embedded means for typing this notation?] Is $(\Lambda_\mu^{\text{ }\nu})^{T}=\Lambda_\nu^{\text{ }\mu}$? Is $(\Lambda_\mu^{\text{ }\nu})^{-1}=\Lambda_{\text{ }\mu}^{\text{}\nu}$?

Can someone please explain (with briefly clarifying why such and such indices are used) how to derive the following?

$$ds^2 = \eta'_{\rho \sigma}\tilde{\Lambda}_{\text{ }\mu}^{\text{}\rho} dx^\mu\tilde{\Lambda}_{\text{ }\nu}^{\text{}\sigma}dx^\nu$$

I would appreciate if someone could please suggest a straightforward source on this notation, since we are not given ones in this course.

$\endgroup$
1
$\begingroup$

Remember that indices are just a bookkeeping technique for summation, with the added benefit that upper and lower indices indicate how the object transforms under coordinate transformations (i.e. covariance and contravariance). The exact symbol used in the index only matters when it is multiplied by another tensor, which implies summation under the Einstein summation notation. So $\Lambda_\mu^\nu V_\nu = \Lambda_\nu^\mu V_\nu$ only if $\Lambda$ is symmetric.

One way of easily seeing this is by viewing the two-index tensors as matrices. In the first term, $\Lambda_\mu^\nu V_\nu$, you are taking the standard matrix multiplication of a vector, while in the second term, $\Lambda_\nu^\mu V_\nu$ you are essentially taking the transpose matrix multiplication of a vector.

As for your second expression, assuming standard special relativity conventions, is saying the space-time interval, $ds^2 = \eta_{\nu\mu} dx^\mu dx^\nu$, is invariant under Lorentz transformations, $\Lambda^\mu_\rho$. This isn't something that can be "derived" from first principles, as Lorentz invariance is usually assumed, although it can be tested, most notably by looking for particles moving faster than the speed of light.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.