3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

You ask

doesn't it immediately follow that $g=L^tgL$ ?

Nope.

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.

$\endgroup$
  • 1
    $\begingroup$ Thanks. I guess it would only follow if $X^\textrm{t}gX = X^\textrm{t}L^\textrm{t}gLX$. $\endgroup$ – Randy Randerson Mar 15 '14 at 5:43
  • 2
    $\begingroup$ @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. $\endgroup$ – joshphysics Mar 15 '14 at 5:50
  • $\begingroup$ I'm glad you added the section on intuition. I was wondering how you thought of that. $\endgroup$ – Randy Randerson Mar 15 '14 at 6:15
  • $\begingroup$ @fctaylor25 I'm glad you're glad. Happy to help. $\endgroup$ – joshphysics Mar 15 '14 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.