How to manipulate this equation with the metric tensor? (index notation) [closed]

Question

I try to follow a derivation in special relativity using index notation. According to the solution, from this equation:

\begin{equation} \eta^{\mu\tau}v'_{\tau}=\Lambda^{\mu}_{\;\sigma}\eta^{\sigma\nu}v_{\nu} \label{eq:1} \end{equation}

I should get this:

\begin{equation} v'_{\mu} = \eta_{\mu\tau}\Lambda^{\tau}_{\;\sigma}\eta^{\sigma \nu}v_{\nu} \label{eq:2} \end{equation}

But I do not know how.

To get the right hand side of the first equation, I multiply by $\delta^{\mu}_{\tau}\eta_{\mu\tau}$, so I have

\begin{equation} \eta_{\mu\tau}\eta^{\mu\tau}v'_{\mu} = \eta_{\mu\tau}\Lambda^{\tau}_{\;\sigma}\eta^{\sigma \nu}v_{\nu} \end{equation}

But $\eta_{\mu\tau}\eta^{\mu\tau} = 4$ (because it is the trace of the four dimensional identity matrix).

Am I wrong somewhere or is the derivation wrong?

Answer 1

We begin with $$ \eta^{\mu\tau}v'_{\tau} = \Lambda^{\mu}_{\ \ \sigma} \eta^{\sigma\nu}v_\nu $$

We can use the fact that the metric contracted with its inverse gives a Kronecker delta $ \eta_{\alpha\mu}\eta^{\mu\tau} = \delta_{\alpha}^{\tau} $ and multiply both sides by $\eta_{\alpha\mu}$.

We are not allowed to contract both indices $\mu,\tau$ of $\eta$, as you attempted, because when one index $\tau$ is contracted already with a bottom index of $v'_\tau$ as it does, you can't also contract it with another tensor. The result you mention that $\eta_{\mu\tau}\eta^{\mu\tau}=4$ is true but will not be relevant here because there is a contravariant tensor on the LHS (one free upper index $\mu$). In other words we may define $V^{^\prime \mu} = \eta^{\mu\tau}v'_\tau$ and then you can clearly see why the only way to combine $\eta$ with $V^{^\prime \mu}$ is one which will give another vector, rather than a number, hence at least one index must remain free when we multiply the LHS by $\eta$.

So, the LHS then becomes:

$$\eta_{\alpha\mu}\eta^{\mu\tau}v'_{\tau} = \delta_{\alpha}^{\tau}v'_{\tau} = v^{\ \prime}_\alpha $$ So that the equation is now: $$ v^{\ \prime}_\alpha = \eta_{\alpha\mu}\Lambda^{\mu}_{\ \ \sigma} \eta^{\sigma\nu}v_\nu \tag{$*$}$$ Which is in fact identical, via renaming of indices to the result you want to obtain: $$ v'_{\mu} = \eta_{\mu\tau}\Lambda^{\tau}_{\;\sigma}\eta^{\sigma \nu}v_{\nu}$$

To see how, simply rename indices in $(*)$ as follows: $ \mu \rightarrow \tau $, $\alpha \rightarrow \mu$

This is perfectly allowable because first we rename $\mu$ which is a repeated index that's summed over. The symbol for such indices don't matter as long as they don't collide with another already used symbol, which $\tau$ isn't in $(*)$. After that, we can also rename the free index $\alpha$ that appears on both sides to $\mu$, and the equality is established.