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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.

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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 $K=0$.

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  • $\begingroup$ 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. $\endgroup$ – Newtonian Aug 1 '16 at 20:13
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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.

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