Invariance of $ds^2$ from invariance of all null intervals

Question

Is this linear algebra statement true?

Let $\eta= \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}$.

If $x^T (\Lambda^T\eta\Lambda) x$=0 for all $x$ such that $x^T \eta x=0$, then $\Lambda^T\eta\Lambda=a\eta$ for some $a \in \mathbb{R}$.

If so how does one prove it? Is a stronger statement true (e.g. $a>0$)?

Motivation: I can't figure out why this statement from the Wikipedia article is true using the above mathematical language:

Since if $ds^{2}=0$, then the interval will be null in any other system (second postulate), and since $ds^{2}$ and $ds'^{2}$ are infinitesimals of the same order, they must be proportional to each other,

$ds^{2}=ads'^{2}$.

The translation of the above is that $x^T (\Lambda^T\eta\Lambda) x$ must be proportional to $x^T \eta x$ for all $x$. Why?

Answer 1

Proposition. In a space $V\cong \mathbb{R}^n$ of dimension $n=p+q$ let there be given an indefinite metric tensor $$\eta~=~\begin{pmatrix} \mathbf{1}_{p\times p}& \mathbf{0}_{p\times q} \cr \mathbf{0}_{q\times p} & -\mathbf{1}_{q\times q} \end{pmatrix}_{n\times n}~=~{\rm diag}(\underbrace{+1,\ldots,+1}_{p\text{ times}},\underbrace{-1,\ldots,-1}_{q\text{ times}})\tag{1}$$ of signature $(p,q)$, and a (possibly degenerate & indefinite) metric tensor $g$. Assume that all null-vectors for $\eta$ are also null-vectors for $g$: $$\forall v\in V :~~v^t\eta v~=~0~~\Rightarrow~~v^tg v~=~0.\tag{2}$$ Then $g$ is proportional to $\eta$: $$\exists \lambda\in \mathbb{R}:~~g~=~\lambda \eta.\tag{3}$$

Sketched proof of proposition:

1. Let $e_i=(0,\ldots, 0,1,0,\ldots, 0)^t$ be the $i$th unit-vector (of $\eta$-length-square $\pm 1$). Write the metric tensor $$g~=~ \begin{pmatrix} a& b^t \cr b & c \end{pmatrix} \tag{4}$$ in terms of a symmetric $p\times p$ matrix $a$, a symmetric $q\times q$ matrix $c$, and a rectangular $q\times p$ matrix $b$.

2. Use the following "polarization trick" to show that the $b$-block vanishes: $$b~=~0.\tag{5}$$ If $g_{ij}=e_i^tge_j$ corresponds to a matrix element in the $b$-block, then $v_{\pm}:=e_i\pm e_j$ are null-vectors, so $$4g_{ij}~=~4e_i^tge_j~=~(v_++v_-)^tg(v_+-v_-)~=~v_+^tgv_+ -v_-^tgv_-~\stackrel{(2)}{=}~0+0~=~0. \tag{6}$$

3. We can diagonalize the symmetric $a$ and $c$ blocks by orthogonal matrices while keeping $\eta$ invariant. In other words, we may assume w.l.o.g. that $$g\text{ is diagonal}.\tag{7}$$

4. Finally, by considering null-vectors of the form $v:=e_i+ e_j$, it becomes clear that $$g_{ii}+g_{jj}~=~e_i^tge_i+e_j^tge_j~\stackrel{(7)}{=}~v^tgv~\stackrel{(2)}{=}~0.\tag{8}$$ This implies that both $a$ and $c$ are proportional to an identity matrix. The sought-for eq. (3) follows. $\Box$

Answer 2

Prelude: 1+2D

In 1+2D we have a matrix $$\Lambda = \begin{bmatrix}a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} $$ which is being used to generate a symmetric matrix $\eta' = \Lambda^T \eta \Lambda$. We can rotate the two spatial dimensions into each other to find a family of null vectors $v(\theta) = [1, \cos\theta,\sin\theta]^T$ such that $v^T \eta v = 0$ and you want to consider only the $\Lambda$ such that $v^T \eta' v =0$ too, for all $\theta$. This would mean that $$(a + b\cos\theta + c\sin\theta)^2 = (d+e\cos\theta + f \sin\theta)^2 + (g + h\cos\theta + i\sin\theta)^2.$$ So we have 6 degrees of freedom (symmetric 3x3 matrix $\eta'$) but presumably we have 5 equations here: terms in $\theta$ proportional to $1,$ $\cos\theta,$ $\sin\theta,$ $\cos(2\theta),$ and $\sin(2\theta):$ $$\begin{align} 2a^2 + b^2 + c^2 &= 2d^2 + e^2 + f^2 + 2g^2 + h^2 + i^2\\ ab &= de + gh\\ ac &= df + gi\\ b^2 - c^2 &= e^2 - f^2 + h^2 - i^2\\ bc &= ef + hi \end{align}$$The three "small" equations above set all of the off-diagonal elements to be 0 in the resulting matrix. The first "big" equation can be reduced to $a^2 + b^2 = d^2 + e^2 + g^2 + h^2$ in light of the second, thus we would have $$\begin{align}a^2 - d^2 - g^2 &= - b^2 + e^2 + h^2 &:= K\\ b^2 - e^2 - h^2 &= c^2 - f^2 - i^2 &= -K \end{align}$$ proving the form $$\begin{bmatrix}K&0&0\\0&-K&0\\0&0&-K\end{bmatrix}$$ for some $K$ which likewise does not have to be positive.

Extending to 1+N dimensions

Now let's just do the same thing as before, but probe $\eta'$ in 1+N dimensions with some basic null vectors that comprise the unit vector in the time direction $\hat w$ and the unit vector in some arbitrary space dimension $\hat x$, e.g. $$(\hat w \pm \hat x)^T\eta'(\hat w \pm \hat x) = 0.$$Since $\eta'$ is symmetric one gets results like $$\hat w^T \eta' \hat w ~\pm~ 2 \hat w^T\eta'\hat x ~+~ \hat x^T\eta'\hat x ~=~ 0$$ and this then argues that these off-diagonal elements $\hat w^T\eta'\hat x = 0$ directly.

The above rotational argument from 1+2D gives the same for the $\hat x^T\eta'\hat y$ terms if we just do a rotation from any spatial coordinate into any other, call them $\hat x$ and $\hat y$: we have even that $$(\hat w + \hat x \cos\theta + \hat y \sin\theta)^T \eta' (\hat w + \hat x \cos\theta + \hat y \sin\theta) = 0$$ and the $\sin(2\theta)$ component of that equation comes exclusively from $2 \hat x^T \eta' \hat y \cos\theta \sin\theta$ and this can only be zero if $\hat x^T \eta' \hat y = 0.$

So we've proven that all off-diagonal elements must be zero and then we can just probe with those first null vectors again, so if we use $\hat w + \hat x$ then we determine that the $(w, w)$ diagonal element must be the negative of the $(x, x)$ diagonal element, but since we chose $\hat x$ arbitrarily this must apply to all of the other diagonal elements: it must have the form $\operatorname{diag}(K, -K, -K, \dots)$.

Answer 3

[converted from a comment into an answer]

The WP article isn't claiming that it's true on mathematical grounds, it's just saying that it needs to be true on physical grounds. It also isn't claiming that it holds for any Λ, but only for a Λ that represents a change of coordinates to a new set of Minkowski coordinates.

I don't think your claim, without any conditions on Λ, holds. Let $\Lambda=\operatorname{diag}(2,2)$. This is just rescaling the coordinates by a factor of 2. For this $\Lambda$, we have $\Lambda^T\eta\Lambda=4\eta$. This is a counterexample to your conjecture that "If $x^T (\Lambda^T\eta\Lambda) x$=0 for all $x$ such that $x^T \eta x=0$, then $\Lambda^T\eta\Lambda=\eta$. "

Also, what is your definition of "a set of Minkowski coordinates"?

Minkowski coordinates are coordinates in which the metric has the form $\operatorname{diag}(1,-1,-1,-1)$.

So I think what you were interpreting in the WP article as a mathematical theorem is actually a combination of a physical argument (second postulate) with a definition (defining Minkowski coordinates as above).

A straightforward way to see that your conjecture shouldn't be expected to hold without any condtions on $\Lambda$ is that it's written in a way that assumes that we can do a certain coordinate transformation and the corresponding inverse transformation using the matrix $\Lambda$ and its transpose. This is not generally true for coordinate transformations. It's a mor special property that happens to hold for Lorentz boosts from one set of Minkowski coordinates to another.