0
$\begingroup$

I just got Sean Carroll's Spacetime and Geometry: An Introduction to General Relativity a couple of weeks ago, and I have resolved to go through the entire book. In the first chapter, he prompts the reader to check that $\omega_\mu V^\mu$ is Lorentz invariant by using the transformation laws of vectors and their dual covectors.

When I read this, I assume he wants the reader to show that $${\Lambda^{\upsilon'}}_\mu \omega_\mu V^\mu = \omega_{\upsilon'} V^{\upsilon'}$$ after a Lorentz transformation has been applied to the left side. However I am confused when applying a Lorentz transformation to the LHS, because in the book Carroll shows that covectors transform according to the inverse of the transformation used on vectors, so I have 2 questions:

  1. How do I obtain $\omega_{\upsilon'}$ from ${\Lambda^{\upsilon'}}_\mu \omega_\mu$?

  2. If question 1 is meaningless or not well posed, my main question is how do I show that the action of a covector on a vector is Lorentz invariant, using only the transformation properties of vectors and covectors?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

$s=\omega_\mu V^\mu$ is scalar. It has "no" indices; the $\mu$ that looks like an index is internal to the definition of $s$ (it is summed over), in the same way that $c=\int_0^5x\,dx$ is a constant with no dependence on the variable $x.$

There are two ways to transform $\omega_\mu V^\mu$ by a Lorentz transform $\Lambda.$ One way is to transform the quantity as a whole. The rule is, for every free index $\mu$ in the quantity, to add a factor of ${\Lambda^{\mu'}}_\mu$ (or ${\Lambda_{\mu'}}^\mu$). But $\omega_\mu V^\mu$ has no free indices, so it doesn't change at all: $\omega_\mu V^\mu$ transformed by $\Lambda$ is $\omega_\mu V^\mu.$ I.e. "scalars are invariant."

The other way is to transform each of its parts. $V^\mu$ has one up index, so transforms (by the above generic rule!) to ${\Lambda^{\mu'}}_\mu V^\mu.$ $\omega_\mu$ has one down index so transforms to ${\Lambda_{\mu'}}^\mu\omega_\mu.$ I will rename $\mu$ to $\nu$ in the previous sum to avoid clashing, and then substitute both transformations into $s$ to find that $\omega_\mu V^\mu$ transformed by $\Lambda$ is also ${\Lambda^{\mu'}}_\mu{\Lambda_{\mu'}}^\nu\omega_\nu V^\mu.$

Your task is to show these two versions of the transformation to be consistent (that applying the vector and covector transformation laws agrees with the scalar (lack of) transformation law): $${\Lambda^{\mu'}}_\mu{\Lambda_{\mu'}}^\nu\omega_\nu V^\mu=\omega_\mu V^\mu$$

$\endgroup$
3
  • $\begingroup$ So the "rule" just adds the convenient form of the Lorentz transform to whatever it is transforming? I thought the transform was some kind of function that did something to vectors and covectors. Assuming this rule, is ${\Lambda^{\upsilon'}}_\mu \omega_\mu$ a meaningless operation? $\endgroup$
    – Chidi
    Apr 30, 2022 at 5:38
  • $\begingroup$ The transform is a function that "does something" to by adding the "convenient form" of the Lorentz transform. They not mutually exclusive concepts. If you would like it explicitly, you could consider $f^\mu(V)={\Lambda^\mu}_\nu V^\nu$ and $f_\mu(\omega)={\Lambda_\mu}^\nu\omega_\nu$ to define the transformation of $V$ to $f(V)$ and $\omega$ to $f(\omega).$ I don't think anyone bothers with this notation. Yes, ${\Lambda^{\nu'}}_\mu\omega_\mu$ is meaningless, since $\mu$ appears twice, both down. An index in a term must appear either as a pair (one up, one down) or just once. $\endgroup$
    – HTNW
    Apr 30, 2022 at 5:49
  • $\begingroup$ Alright, that cleared it up for me, thank you. $\endgroup$
    – Chidi
    Apr 30, 2022 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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