Lorentz boost matrix

Question

In my textbook, there is a proof that the dot product of 2 four-vectors is invariant under a Lorentz transformation. While I understood most of the derivation (I am a beginner and we haven't done any math regarding this notation), there is one step which I do not understand:

$$ (\Lambda^μ_α)(\Lambda_{μβ})x^αy^β =(\Lambda^μ_α)(\Lambda^β_μ)x^αy_β.$$

As you can see the $\beta$ subscript on the $\Lambda$ becomes a superscript, while the $\beta$ superscript on $y$ vector becomes a subscript. I know for a fact that you can change the index position on the $\Lambda$ but the index also changes, meaning we use a new character when the index goes up or down after we multiply $\Lambda$ with the metric tensor. So how does the book do it here?

Answer 1

If you have a quantity with a down index, say $A_\mu$, you can raise the index using the metric (in this case the Minkowski metric $\eta$) as $A_\mu=A^\nu\eta_{\nu\mu}$. In an analogous way, we can lower the index of $B^\mu$ as $B^\mu=B_\nu\eta^{\nu\mu}$.

Now suppose we have a term as in your example where there is a quantity with a down-index next to one with an up-index, and they are contracted (i.e. summed over / named the same) as in $A_\mu B^\mu$. We can raise the index of $A_\mu$ and lower the one of $B^\mu$. It's always good to introduce a new letter when raising or lowering an index if you are not sure. So we would do $A_\mu=A^\nu\eta_{\nu\mu}$ just as before, and $B^\mu=B_\sigma\eta^{\sigma\mu}$ (where I have introduced a new letter $\sigma$ instead of $\nu$). With this we have

\begin{align} A_\mu B^\mu&=A^\nu\eta_{\nu\mu}B_\sigma\eta^{\sigma\mu}\\ &=A^\nu B_\sigma\eta_{\nu\mu}\eta^{\sigma\mu}. \end{align}

If we think of the metric $\eta_{\mu\nu}$ as a matrix, notice that the product $\eta_{\nu\mu}\eta^{\sigma\mu}$ is the way to write the matrix multiplication of a matrix times its inverse in the Einstein convention. A matrix times its inverse gives the identity matrix, whose components are given by a Kronecker delta $\delta_\nu^\sigma$. So if we write $\eta_{\nu\mu}\eta^{\sigma\mu}=\delta_\nu^\sigma$, we are left with

$$A_\mu B^\mu=A^\nu B_\sigma\delta_\nu^\sigma.$$

This last expression is summed over $\nu$ and over $\sigma$, but only the terms with $\sigma=\nu$ survive the sum because of the Kronecker delta: $A^\nu B_\sigma\delta_\nu^\sigma=A^\nu B_\nu$. Renaming now the index $\nu$ as $\mu$ (we can do that, it's a dummy index) we end up with

$$A_\mu B^\mu=A^\mu B_\mu.$$

So with a little more practice you'll be able to identify the change $A_\mu B^\mu\to A^\mu B_\mu$ as something trivial.

Answer 2

If I understand OP's question correctly, he does not understand why $\sum_\beta \Lambda_{\mu\beta} y^\beta = \sum_\beta\Lambda_{\mu}^{\;\;\beta} y_\beta$. This is how it works in detail $$ \sum_\beta \Lambda_{\mu\beta} y^\beta =\sum_{\beta,\sigma} \Lambda_\mu^{\;\;\sigma} \eta_{\sigma\beta} y^\beta = \sum_\sigma \Lambda_\mu^{\;\;\sigma} y_\sigma = \sum_\beta \Lambda_\mu^{\;\;\beta} y_\beta $$ I have written out the sum symbol explicitly, but in these expression repeated indices are understood to be summed over.

Note also that I write $\Lambda_\mu^{\;\;\sigma}$ and not $\Lambda_\mu^{\sigma}$ as it is important to know which of the two indices is raised.