Proving that the product of contravariant and covariant vectors is Lorentz-invariant

Question

I have to prove that $A_{\mu} B^{\mu}$ is Lorentz invariant, and I'd like to check my understanding with you if you don't mind.

My first question is about the definition of Lorentz invariance. Does it mean the following:

$A'_{\mu} B'^{\mu} = A_{\mu} B^{\mu}$ ?

If so, I think I have an idea how to prove it but I might get confused with the indexes. I would express $A'_{\mu}$ and $B'^{\mu}$ as

$A'_{\mu} = a_{\mu}^{\phantom{\mu} \nu} A_{\nu}$ and

$B'^{\mu} = a^{\mu}_{\phantom{\mu} \nu} B^{\nu}$.

Is that correct? Moreover, if I was to calculate the product above, can I use the same indices for both terms? That is:

$A'_{\mu} B'^{\mu} = a_{\mu}^{\phantom{\mu} \nu} A_{\nu} a^{\mu}_{\phantom{\mu} \nu} B^{\nu}$?

If so, I can rewrite the following as:

$A'_{\mu} B'^{\mu} = a_{\mu}^{\phantom{\mu} \nu} A_{\nu} a^{\mu}_{\phantom{\mu} \nu} B^{\nu} = \underbrace{a_{\mu}^{\phantom{\mu} \nu} a^{\mu}_{\phantom{\mu} \nu}}_{= \delta_{\nu}^{\nu}} A_{\nu} B^{\nu} = 1 \cdot A_{\mu} B^{\mu} = A_{\mu} B^{\mu}$.

Does all of that makes sense? I'm still trying to get a feel of this new notation for me.

Thank you very much in advance for your comments.

Julien.

Answer 1

Lorentz invariance means that the value does not change under Lorentz transformations, so we need to show that $A'_\mu B'^\mu=A_\mu B^\mu$. Let the transformation be given by the matrix $a^\nu_\mu$$~$ ($v'^\nu=a^\nu_\mu v^\mu$ for any vector $v^\mu$).

Then $$A'_\mu \,B'^\mu=a^\nu_\mu \, A_\nu \, a^\mu_\sigma\, B^\sigma =A_\nu \,a^\nu_\mu\, a^\mu_\sigma\, B^\sigma = A_\nu \, \delta^\nu_\sigma B^\sigma = A_\nu \, B^\nu.$$

So in fact, your approach was almost correct, but you need to introduce new indices for each transformation.

Regarding your comment: The objects $a^\mu_\nu$ denote the components of the matrix (they are numbers!), therefore their order can always be exchanged. The order of the tensors is encoded in the summation indices.

For example, let $(A)^i_j=a^i_j$, $(B)^i_j=b^i_j$, $(C)^i_j=c^i_j$, then $(ABC)^i_j= a^i_k b^k_l c^l_j$ whereas $(ACB)^i_j=a^i_k c^k_l b^l_j = a^i_k b^l_j c^k_l$.