0
$\begingroup$

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:

enter image description here

enter image description here

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

$\endgroup$

2 Answers 2

3
$\begingroup$

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 $$\Delta S'^2 = M_{00} \big(\Delta t^2 - \delta_{ij} \Delta r^i \Delta r^j) = M_{00} \Delta S^2$$

$\endgroup$
3
  • $\begingroup$ First of all, thank you for answer. I think I know what you are trying to get at. If these M's don't depened on the relative speeds of the coordinate systems, then proving that $S'^2=S^2=0$ results in $S'^2=M_{00}S^2$ in all coordinate systems. However I also think that the invariance of speed for $M_{01}=M_{10}$ and $M_{11}$ needs to be established as well. That was not done in the book, only for $M_{00}$. Although, I think that the argument for their invariance is identical to that of $M_{00}$. $\endgroup$ May 10 at 10:49
  • $\begingroup$ @JoelJärnefelt Even if they do depend on the relative velocity of the coordinate frames the point is that they don’t depend on the coordinates $\Delta t$ or $\Delta r$ (otherwise the transformation would be nonlinear). Where do you think the argument breaks down? It’s quite explicit, so which step in my answer do you think is invalid? $\endgroup$
    – J. Murray
    May 10 at 12:09
  • $\begingroup$ I actually took a long walk and thought this through. The argument works perfectly. The reason is that the M's only depend on the relative differences of the coordinate systems not on the distances that are measured. Those are only used to help determine the M's. $\endgroup$ May 10 at 17:45
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you for the answer. I wonder why in L&L the second statement is "obvious" since pretty much the entire proof of invariance depends on it. $\endgroup$ May 10 at 10:56

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.