# Lorentz Transformations in Index Notation

So I was refreshing my special relativity knowledge for a review I need to write next year and was really confused by the following:

I was thinking about a frame $S$ with another frame $S'$ boosted along the $x$ direction of $S$. I know that a general Lorentz Transform can be written in index notation as:

$x'^\mu=L_{\nu}^{\mu}x^\nu$

Of course the interval between two events as observed from $S$ and $S'$ will be the same and this can be given by:

$\eta_{\alpha\beta}x^{\alpha}x^{\beta}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}$

then using the first equation we can write:

$\eta_{\alpha\beta}x^{\alpha}x^{\beta}=\eta_{\mu\nu}(L^{\mu}_{\alpha}x^{\alpha})(L^{\nu}_{\beta}x^{\beta})=\eta_{\mu\nu}L^{\mu}_{\alpha}L^{\nu}_{\beta}x^{\alpha}x^{\beta}$

then we can simply rearrange this to write:

$(\eta_{\alpha\beta}-\eta_{\mu\nu}L^{\mu}_{\alpha}L^{\nu}_{\beta})x^{\alpha}x^{\beta}=0$

and so finally we can write

$K_{\alpha\beta}x^{\alpha}x^{\beta}=0$ where $K_{\alpha\beta}=\eta_{\alpha\beta}-\eta_{\mu\nu}L^{\mu}_{\alpha}L^{\nu}_{\beta}$

This is all basic. The book I was reading then went on to say that because this equation must hold for all coordinates then this implied that:

$K_{\alpha\beta}+K_{\beta\alpha}=0$

and further to this $K_{\alpha\beta}$ must be symmetric so that from the above equation we find that in fact $K_{\alpha\beta}=0$. I do not understand why this should be the case, any guidance would be hugely appreciated.

First I'll answer your question mathematically using no physical intuition. Let's start from the equation that you wrote down:

$$K_{\alpha\beta}x^\alpha x^\beta=0$$

Exchanging dummy indices:

$$K_{\beta\alpha} x^\beta x^\alpha=0$$

Commuting the two components of $x$:

$$K_{\beta\alpha}x^\alpha x^\beta =0$$

Now let's add this to the first equation I wrote down above:

$$(K_{\alpha\beta}+K_{\beta\alpha})x^\alpha x^\beta=0$$

Now I use the fact that you refer to in your original question. Since this equation must hold for any $x$, we can eliminate the factors of $x$ such that:

$$(K_{\alpha\beta}+K_{\beta\alpha})=0$$

By the definition of $K$ which you state in the problem statement, we see that $K$ is symmetric by considering $K$ in matrix notation:

$$K=\eta-L^T \eta L$$

Since $\eta$ is symmetric, it thus follows that $K$ is symmetric. Therefore, we can rewrite $(K_{\alpha\beta}+K_{\beta\alpha})=0$ as

$$K_{\alpha \beta} + K_{\alpha \beta}=0$$ $$2K_{\alpha \beta}=0$$ $$K_{\alpha \beta}=0$$

Q.E.D.

As far as the physical intuition here, return to the definition of $K$. $K$ takes in a spacetime event $x$ and spits out the difference between its interval from the origin in $S$ and its interval from the origin in $S'$. This interval is invariant; therefore, this difference must always be equal to zero, i.e. $K_{\alpha \beta}x^\alpha x^\beta =0$ for all $x$. Since $K=0$ for all $x$, it must be the case that $K=0$.

• Great, thanks so much for taking the time to answer. Brilliant, informative answer. I was just not getting the fact that K is symmetric because Eta is symmetric. Makes more sense seeing it matrix notation like that as L is just a constant defining the transformation so of course the symmetry of K would depend on the symmetry of Eta. Thanks again. – Newtonian Aug 1 '16 at 20:13

In general one has $$K_{\alpha\beta} = \frac{1}{2}K_{\alpha\beta} + \frac{1}{2}K_{\alpha\beta} + \frac{1}{2}K_{\beta\alpha} - \frac{1}{2}K_{\beta\alpha} = \frac{1}{2}K_{\left\{\alpha, \beta\right\}} + \frac{1}{2}K_{\left[\alpha, \beta\right]}$$ where $\left\{\cdot,\cdot\right\}$ denotes the symmetric permutation of the indeces and $\left[\cdot,\cdot\right]$ the antisymmetric one. Plugging the above into the original equation we have $$\frac{1}{2}K_{\left\{\alpha, \beta\right\}}x^{\alpha}x^{\beta} + \frac{1}{2}K_{\left[\alpha, \beta\right]}x^{\alpha}x^{\beta} = 0.$$ The second contribution above vanishes, being the product of an antisymemtric term by a symmetric one (just work it out), hence we are left with $$K_{\left\{\alpha, \beta\right\}}x^{\alpha}x^{\beta} = 0$$ that must imply $K_{\left\{\alpha, \beta\right\}} = 0$ if you want it to hold true however you choose $(x^{\alpha}, x^{\beta})$. It is easy to see, from its definition, that $K_{\alpha\beta} = K_{\beta\alpha}$; as such $K_{\left\{\alpha, \beta\right\}} = K_{\alpha\beta} + K_{\beta\alpha} = 2 K_{\alpha\beta} = 0$, that completes the proof.