Bernard Schutz proof of the invariance of the interval

[kmm]

TL;DR Summary

Bernard Schutz's proof of the invariance of the interval, appears to me to apply to a special case. I don't understand how his proof applies to the general case.

I've been going through Bernard Schutz's A First Course in General Relativity, and I'm hung up on his "proof" of the invariance of the interval. At the beginning of section 1.6, he claims that he will prove the invariance of the interval, and after a few lines shows that the universality of the speed of light implies that the interval of the two frames $\overline {\mathbf{O}}$ and $\mathbf{O}$ satisfy the relation, $$ \Delta \overline {s}^2 = \phi (\mathbf{v}) \Delta s^2$$
He then states that he will show that $\phi (\mathbf{v})=1$, which is the statement that the interval is independent of the observer. I'm hung up on the first of the two parts of the proof. The first part of the proof is in showing that $\phi (\mathbf{v})$ depends only on $|\mathbf{v}|$. He has us consider a rod at rest in $\mathbf{O}$ and oriented perpendicular to the velocity $\mathbf{v}$ of $\overline {\mathbf{O}}$. He goes on to show that the interval in each frame of reference, is equal to the square of the length of the rod. So we now have, $$(\text {length of rod in} \: \overline {\mathbf{O}})^2 = \phi (\mathbf{v})(\text{length of rod in} \: \mathbf{O})^2$$
He then goes on to say that the length of the rod cannot depend on the direction of the velocity, because the rod is perpendicular to it and there are no preferred directions of motion. At this point he claims to have finished the first part of the proof showing that $$\phi (\mathbf{v}) = \phi (| \mathbf{v} |)$$
The second part of the proof starts here and he goes on to show that $\phi (\mathbf{v})=1$, and so it has been proven that the interval is independent of the observer. Given that $\phi (\mathbf{v}) = \phi (| \mathbf{v} |)$, the second part of the proof seems straightforward to me. But this hinges on the first part, which was found from a very specific case. Given that one reference frame moves perpendicular to a rod at rest in another reference frame, it's clear to me that they can move along any direction and the length of the rod won't change. But how can we extend this case to the general case of when they are not moving perpendicular, so that we can always say $\phi (\mathbf{v}) = \phi (| \mathbf{v} |)$? With this rod example, it seems to me that all that has been shown is that $\phi (\mathbf{v})$ is only dependent on $|\mathbf{v}|$ when one frame moves perpendicular to a rod that is at rest in another. I appreciate any help!

 

[Martin Scholtz]

I saw this kind of proof of invariance of the spacetime interval in Landau-Lifshitz course on teoretical physics. I know Bernard Schutz from conferences but actually haven't read any of his book (I think he has a book on differential geometry too).

Regarding your question: you are right. You cannot prove it, you must assume it. But it is a natural assumption, at least in special relativity. Look, in principle, function $\phi$ you mention could depend on everything: on the position $\mathbf{x}=(x,y,z)$, on time $t$ and velocity $\mathbf{v} = (v^x, v^y, v^z)$. Why does it not? It is because of the symmetries of the space and time. Now forget about general relativity, I will focus on flat spacetime without gravitation.

• Time homogeneity
  • All time instants are equivalent. It doesn't matter when you perform the experiment, if the external conditions are the same. Hence, also the function $\phi$ cannot depend on time because it would mean that the spacetime interval transforms differently on Monday and Thursday. This symmetry ultimately leads, via Noether theorem, to the conservation of energy.
• Space homogeneity
  • The same is true for the places in space: all points are equivalent and experiment does not depend on place, provided that all external conditions are the same. Hence, $\phi$ cannot depend on position, otherwise special relativity would look differently in different parts of the universe. This property implies the conservation of momentum.
• Space isotropy
  • Thus, $\phi$ cannot depend neither on time nor on position and what remains is just a velocity. But again, all directions of the space are equivalent. Imagine you have a laboratory in empty space and perform an experiment. Now you arbitrarily rotate your lab about arbitrary axis. The outcome of the experiment must be exactly the same, otherwise you could distnguish a special direction in an empty space. Hence, factor $\phi$ cannot depend on the absolute orientation in space only on the magnitude of relative velocity of the two systems. Noether theorem implies the conservation of angular momentum in this case.

To summarize the proof briefly as I understand it (I have no idea how much it is similar to Schutz's proof). I appologize if will produce some typos or bigger mistakes, let me know because I will do the proof "on fly" withut deeper thinking.

Let $S$ and $S'$ be two inertial frames. That is, each observer assigns to any event a 4-tuple of coordinates $x^\mu = (c t, x, y,z)$ and $x'^\mu = (c t',x',y',z')$.

Consider two infinitesimally close events $A$ and $B$ (infinitesimal, because it's closer to geometrical point of view and consequently to general relativity) and construct quantities

1. $d s^2 = - c^2 \, dt^2 + d x^2+d y^2 +d z^2$
2. $d s'^2 = - c^2 \, dt'^2 + d x'^2+d y'^2 +d z'^2$

Differentials transform always linearly, even if the coordinate transformation $x^\mu \mapsto x'^\mu$ is non-linear:
$d x'^\mu = \dfrac{\partial x'^\mu}{\partial x^\nu} d x^\nu$

(the fraction is the Jacobi matrix - it is made of partial derivatives but it is a matrix and hence represents a linear transformation). Hence, the space time intervals must be related by a quadratic form $g_{\mu\nu}$ as follows:

$d s'^2 = \sum\limits_{\mu,\nu=0}^3 g_{\mu\nu} dx^\mu\,d x^\nu = g_{00} c^2 d t^2 + 2 \sum\limits_{i=1}^3 g_{0i} c dt\,d x^i + \sum\limits_{i,j=1}^3 g_{ij}\,dx^i\,d x^j$

If both frames $S,S'$ are inertial, the non-diagonal elements of $g_{\mu\nu}$ must vanish,

$g_{\mu\nu} = \begin{pmatrix} g_{00} & 0 & 0 & 0 \\ 0 & g_{11} & 0 & 0 \\ 0 & 0 & g_{22} & 0 \\ 0 & 0 & 0 & g_{33} \end{pmatrix}$
because in the spacetime interval there are no $dt\,d x$ nor $d x\, dy$ terms. Thus, we have

$ds'2 = g_{00}\,c^2\,dt^2 + g_{11} dx^2 + g_{22}dy^2 + g_{33}dz^2$

Now suppose that the event $A$ is an emission of a light signal and $B$ is the absorption of this signal. Since $c$ is the same in both frames, the distance traveled by the signal is

$dr = c\, dt$ in $S$ and $dr' = c\,dt'$ in $S'$, where $r^2 = dx^2 +dy^2+dz^2$ and similarly for $dr'$. Rewriting the intervals in inertial frames in the form
$ds^2 = -c^2 \,dt^2 + d r^2$ and $ds'^2 =-c^2\,dt'^2 + dr'^2$ we immediately see that for events connected with the light signal we have $ds^2 = ds'^2 = 0$.

Now the symmetries enter the game for the first time. In general,
$dr'^2 = \sum\limits_{i,j=1}^3 g_{ij} dx^i\, dx^j$
But since the space is isotropic, no direction is preferred, all directions in space must contribute to the value of $dr$ in the same way, with the same weight. Mathematically, if we rotate just the spatial part of $g_{\mu\nu}$, i.e. only the $g_{ij}$ part, we must get the same $g_{ij}$. This is possible ony if all $g_{ii}$ are the same. This is a crucial assumption and cannot be derived from more basic principles, but physically it is perfectly reasonable. Let us denote $f = g_{ii}$ for $i=1,2,3$, so that

$g_{\mu\nu} = \begin{pmatrix} g_{00} & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & f & 0 \\ 0 & 0 & 0 & f \end{pmatrix}$

and

$ds'^2 = g_{00} c^2\,dt^2+ f^2\,dr^2 =-g_{00} c^2\,dt^2 + f^2 (dx^+dy^2+dz^2)$.

We already know that if the two events are connected with a light ray, $ds'^2 = ds^2=0=-c^2 dt^2 + dr^2$. Therefore, we have $g_{00} = -f$. and we can write
$ds'^2 = f \, ds^2$
(Notice that the transformation rule doesn't depend on the events we observe, so if there is a proportionality for light propagation, it will be the same relation for any two events).

Now we return to your original question. By homogeneity of space and time, function $f$ does not depend on coordinates, it can depend on the relative velocity only. By isotropy of space, it cannot depend on the direction of the velocity, otherwise it would be possible to select a preferred direction in space. Forget about Schutz's example with the perpendicular rod. Imagine two inertial frames moving relative to each other. You will measure some transformation of the spacetime interval. Now you rotate entire system by some angle. What changed? Nothing! You cannot distinguish experimentally these two situations, therefore $f$ can depend only on the magnitude of the velocity: $f =f(|\mathbf{v}|)$. In addition, if the two systems are not moving, spacetime interval must be the same, $f(0) = 1$.

Now suppose that you transform $d s^2$ from $A$ to $B$ ($ds'^2$) and then back to $A$ ($ds'' = ds$). We have
$ds''^2 = f(|\mathbf{-v}|) ds'^2 = f(|\mathbf{-v}|) \,f(|\mathbf{v}|) ds^2 = f(\mathbf{v})^2 d s^2 = ds^2$
i.e. $f(\mathbf{-v})^2 = 1$, implying $f^2=1$.

OK, I tried to be as rigorous as possible in the online discussion, let me know if you find a mistake.

 

[kmm]

Thank you very much for your thorough response! Your explanation is much clearer to me. It makes sense now that it is a natural assumption to make that $\phi$ be independent of velocity. I can see now what Schutz was trying to get across, but I think that Schutz not stating the assumptions and using the rod example as part of this "proof" were more of a distraction to me than anything.

 

[kent davidge]

hey, good explanation, but I think there's a problem

you concluded that the interval is given by

Thus, we have
$ds'2 = g_{00}\,c^2\,dt^2 + g_{11} dx^2 + g_{22}dy^2 + g_{33}dz^2$

but then, for light, you considered

Rewriting the intervals in inertial frames in the form
$ds^2 = -c^2 \,dt^2 + d r^2$ and $ds'^2 =-c^2\,dt'^2 + dr'^2$

and this form contradicts you earlier version that I quoted before

I mean, you should write it in the form given before, which would not lead you to conclude that $ds^2 = 0$ for light unless you put in the symmetry requirements you gave just later.

 

[Martin Scholtz]

Thank you very much for your thorough response! Your explanation is much clearer to me. It makes sense now that it is a natural assumption to make that $\phi$ be independent of velocity. I can see now what Schutz was trying to get across, but I think that Schutz not stating the assumptions and using the rod example as part of this "proof" were more of a distraction to me than anything.

Thank you, I'm glad if my answer clarified anything. Maybe Schutz wanted to motivate the assumption but he should tell it more explicitly. Good luck with further study of gravity!

 

[Martin Scholtz]

hey, good explanation, but I think there's a problem

you concluded that the interval is given by


but then, for light, you considered

and this form contradicts you earlier version that I quoted before

I mean, you should write it in the form given before, which would not lead you to conclude that $ds^2 = 0$ for light unless you put in the symmetry requirements you gave just later.

Hello, thank you for the warning, yes, my explanation is potentially confusing. Nevertheless, we know from the beginning that intervals in inertial frames have isotropic diagonal for, cf. eqs. 1. and 2. at the beginning of my explanation. The purpose of symmetries etc. is to relate the general transformation of the metric to the result 2. we want to get. The symmetry arguments don't apply in curved spacetimes or even for nonlinear coordinate transformations.

I won't change my post but I will mention your objection. Thank you once more.

 

[Martin Scholtz]

I was kindly warned in the post by Kent David about a potential confusion of my answer. Please, see his post and my reaction.