# Special relativity: how to prove that $g = L^t g L$?

We have $$X^\textrm{t}gX = 0 \iff X^\textrm{t}L^\textrm{t}gLX = 0,$$ where $X$ is a column vector of length four, $L$ is a non-singular $4 \times 4$ matrix, 't' denotes matrix transpose, and $$g = \left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}\right)\,.$$ doesn't it immediately follow that $g = L^\textrm{t}gL,$ since $$X^\textrm{t}gX = 0 \iff X^\textrm{t}\left(L^\textrm{t}gL\right)X = 0?$$ Why or why not? I ask because the proof in my book takes up an entire page, so I have a feeling that this argument is not sound.

It shouldn't matter for the question, but $X$ and $L$ come from the following equations: $$X = \left(\begin{matrix} ct_2 \\ x_2 \\ y_2 \\ z_2 \end{matrix}\right) - \left(\begin{matrix} ct_1 \\ x_1 \\ y_z \\ z_1 \end{matrix}\right)\,,$$ where $t_1$, $x_1$, $y_1$, $z_1$ and $t_2$, $x_2$, $y_2$, $z_2$ are the inertial coordinates of two events, and $$\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = L \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right) + C,$$ which gives the Lorentz transformation from the primed to the unprimed inertial coordinate system.

doesn't it immediately follow that $g=L^tgL$ ?
Consider, for example, $L = aI$ where a is a nonzero real number, and $I$ is the $4\times 4$ identity matrix. This matrix $L$ is nonsingular, and it has the property that \begin{align} X^tgX = 0\,\quad\text{if and only if}\quad X^tL^tgLX = 0 \end{align} for all $X\in \mathbb R^4$, but notice that \begin{align} L^t gL = a^2 g\neq g. \end{align}
Intuition. By the way, you might be wondering how I came up with such a slick counterexample (if I may indulge in a bit of self-flattery). Well it came from some physical intuition. Note that the condition $X^tgX = 0$ simply says that $X$ is a null vector, so if $X^tgX = 0$ if and only if $X^tL^tgLX = 0$, then this just means that $g$ and $L^tgL$ agree in their action in the light cone. Then I thought, "oh, but the light cone is scale invariant, so metrics related by scaling can still agree on the light cone." Voila.
• Thanks. I guess it would only follow if $X^\textrm{t}gX = X^\textrm{t}L^\textrm{t}gLX$. – Randy Randerson Mar 15 '14 at 5:43
• @fctaylor25 Yes, provided there is an implicit universal quantifier "for all $X$" there (or some sufficiently large class of $X$ at least). I also added a section on intuition. – joshphysics Mar 15 '14 at 5:50