Invariance of spacetime interval by Schutz

Question

I am reading the book on General Relativity by Bernard Schutz. In it he proves the invariance of the interval in special relativity using the following argument.

$S^2=0$ for all light-like paths. This directly follows from the postulates.

After this, he makes a coordinate transformation. Knowing that coordinate transformations are linear, he can say that the transformed interval is $S'^2=t'-r'=M_{\alpha\beta}x^\alpha x^\beta$.

Then he goes to argue that $M_{\alpha\beta}=\eta_{\alpha\beta}$, the minkowski metric, if $S^2=S'^2=0$. This is then taken as proof that $S^2$ is invariant between any two events.

My problem with this is that this seems circular and insufficient. If it is assumed that $S^2=0$ for light-like paths then $M_{\alpha\beta}=\eta_{\alpha\beta}$ by assumption. Also, it has not been proven that $S^2$ works for any event, only for light-like events.

What am I missing?

Here is the material for this proof in case I understood the proof wrong:

He then goes to argue that $\phi(v)=1$ but that is not relevant here.

Answer 1

The claim is that if the relationship between coordinates in the two inertial frames is linear and if $\Delta S^2 = 0$ necessarily implies that $\Delta S'^2 = 0$, then in general we must have that $\Delta S^2 \propto \Delta S'^2$.

To understand this, consider the 1+1-dimensional case. Note that in general, $$\Delta S'^2 = \Delta t'^2 - \Delta x'^2 = M_{00} \Delta t^2 + (M_{01}+M_{10}) \Delta t \Delta x + M_{11} \Delta x^2 $$ Let $\Delta S^2 = \Delta t^2 - \Delta x^2= 0$. Plugging this in to $\Delta S'^2$ yields $$\Delta S'^2 = (M_{00}+M_{11} \pm[M_{01}+M_{10}])\Delta x^2$$ where we have used that $\Delta t = \pm \Delta x$. If we now demand that $\Delta S^2 = 0 \implies \Delta S'^2 = 0$ regardless of whether $\Delta t= \pm \Delta x$, then it follows that both $M_{00}+M_{11}=0$ and $M_{01}+M_{10}=0$. As a result, we can let $M_{11} = -M_{00}$ and so we find that in general, $$\Delta S'^2 = M_{00}(\Delta t^2 - \Delta x^2) = M_{00} \Delta S^2$$

---

The extension to 3+1-dimensions is straightforward: $$\Delta S'^2 = M_{00} \Delta t^2 + (M_{0 i} + M_{i 0}) \Delta t \Delta r^i + M_{ij} \Delta r^i \Delta r^j$$ where $i,j$ run from $1$ to $3$. Letting $\Delta S^2 = \Delta t^2 - \delta_{ij} \Delta r^i \Delta r^j=0$, we find that $$\Delta S'^2 = \big(M_{00}\delta_{ij} + M_{ij} \big)\Delta r^i\Delta r^j \pm(M_{0i}+M_{i0})\Vert \Delta r\Vert \Delta r^i$$ Demanding that this vanish regardless of whether $\Delta t = \pm \Vert \Delta r \Vert$ implies that $M_{ij}=-M_{00} \delta _{ij}$ and that $(M_{0i}+M_{i0})\Delta r^i=0$, and so $$\Delta S'^2 = M_{00} \big(\Delta t^2 - \delta_{ij} \Delta r^i \Delta r^j) = M_{00} \Delta S^2$$

Answer 2

I would suggest you read Landau & Lifshitz argument for invariance of $ds^2$, particularly the Wikipedia link because it states clearly the theorem which is being proved, and it fills in a few steps which Schutz skips/leaves as an exercise.

"If it is assumed $S^2=0$ for light-like paths then $M_{ab}=\eta_{ab}$" by assumption.

No, that's not the definition, nor the assumption. We're starting with two Lorentzian inner products, let us denote them as $g$ and $h$ (i.e they are bilinear forms; they eat two vectors $v,w$ as input and output real numbers $g(v,w)$ and $h(v,w)$ respectively; and they do this in a bilinear fashion... also they have "Lorentz signature"). Consider the following two statements:

• If $g=Ch$ for some $C\in\Bbb{R}\setminus\{0\}$ then the quadratic forms of $g$ and $h$ have the same zero set (i.e $g(v,v)=0$ if and only if $h(v,v)=0$).
• If the quadratic forms of $g$ and $h$ have the same zero set, then $g$ and $h$ are proportional by a non-zero scalar.

The first statement is absolutely trivial from the definitions, simply because $C\neq 0$, and as a result, $g(v,v)= Ch(v,v)=0\iff h(v,v)=0$. However, the statement Schutz (and L&L in the other link) are making is the second statement (the converse), and this is not an obvious assertion (well to L&L it is).

Just to emphasize why it is not an obvious thing, imagine two single-variable functions $f_1,f_2:\Bbb{R}\to\Bbb{R}$. Let us say $f_1(x)=x^3$ and $f_2(x)=x^4$. Then, both these functions have the same zero set (i.e $f_1^{-1}(\{0\})=f_2^{-1}(\{0\})=\{0\}$, i.e $f_1(x)=f_2(x)=0\iff x=0$; in words the origin is the only place these functions vanish). But obviously $f_1$ is not a scalar multiple of $f_2$. This shows you already that there is something very specific about bilinearity (of the Lorentz inner products $g$ and $h$) that one has to exploit in order to show that equality of the zero level set (i.e equality of their respective light cones) implies proportionality.