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?


1 Answer 1


$s=\omega_\mu V^\mu$ 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): $${\Lambda^{\mu'}}_\mu{\Lambda_{\mu'}}^\nu\omega_\nu V^\mu=\omega_\mu V^\mu$$

  • $\begingroup$ 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

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