Derivation of Equation (1.6) in Schutz: A First Course in GR

[msbell1]

Homework Statement

From pages 10--11 in "A First Course in General Relativity" (Second Edition) by Bernard Schutz:

Given

$$\Delta\overline{s}^2 = \phi\left(\textbf{v}\right)\Delta s^2,$$

where $\Delta \overline{s}^2$ is the interval measured between two events in frame $O'$, which is moving in the positive $x$-direction with constant velocity $\textbf{v}$ relative to frame $O$, and $\Delta s^2$ is the interval between the same two events as measured in $O$, show that $$\phi\left(\textbf{v}\right) = \phi\left(\left| \textbf{v}\right|\right).$$ (That is, show that the direction of the relative velocity between the two frames is irrelevant. Note: this is part of a two-part proof in which Schutz shows that $\phi\left(\textbf{v}\right) = 1$.)

Homework Equations

Schutz does this by considering frame $O'$ to be moving in the $+x$-direction relative to frame $O$, with relative velocity $\textbf{v}$, as described above. He then imagines a rod lying along the $y$-axis, and shows that the two events corresponding to the ends of the rod at $x=z=t=0, y=a,b$ (which are simultaneous in frame $O$) are also simultaneous in frame $O'$. I can follow his argument for that part.

He then states that

(length of rod in $O'$)$^2 = \phi\left(\textbf{v}\right)$ (length of rod in $O$)$^2$, which I can also understand.

The Attempt at a Solution

Then he says "On the other hand, 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 (the principle of relativity). Hence the first part of the proof concludes that $$\phi\left(\textbf{v}\right) = \phi\left(\left|\textbf{v}\right|\right)$$." I am really confused by that statement. Yes, the velocity is perpendicular to the rod. But what's the deal with the "and there is no preferred directions of motion" part? That statement on its own makes sense, but I don't know what ties the two parts of his statement together in a way that proves what he says it proves.

 

[andrewkirk]

The point is that
$$\phi\left(\textbf{v}\right)
= \left(\frac{\textrm{length of rod in }O'}{\textrm{length of rod in }O}\right)^2$$
and the ratio cannot vary with the direction of $\mathbf v$.If it did, there would be one or more directions in which the ratio was maximised and one or more in which it was minimised. That would establish a 'preference of direction' of the former over the latter, or vice versa.

If you accept that, then we know that the ratio depends on $\mathbf v$ which is fully specified by the spherical coordinates $v,\psi,\theta$ (normally in spherical coordinates one uses $\phi$ where I've written $\psi$, but Schutz has already used that for something else - his function) and it can't depend on the last two since they specify direction. So it can depend only on $v$, which is another name for $|\mathbf v|$.

Schutz then says
$$
\phi\left(\textbf{v}\right) = \phi\left(\left|\textbf{v}\right|\right)
$$
which is an abuse of notation, and quite wrong. When reading Schutz it's important to bear in mind that he's a physicist, not a mathematician, so his use of maths is often sloppy.

What he wishes to convey is that the length ratio depends only on $v$, not on $\psi$ or $\theta$. Strictly, what he is doing is writing
$$
\phi\left(\textbf{v}\right) = \zeta\left(\left|\textbf{v}\right|\right) = \zeta(v)
$$
for some unknown function $\zeta$.

 

[msbell1]

Thanks for the reply. I'm still a bit unsettled by this. I think my main hang-up is that this relationship between the lengths was obtained by assuming the rod was oriented perpendicular to the velocity of frame $O'$ relative to frame $O$. I can see how the argument holds for any other direction in a plane that is perpendicular to the rod, but what about when the relative velocity is not perpendicular to the rod? I guess the generalization to that case is what is confusing me. I don't see how it follows from the argument in the textbook.

 

[andrewkirk]

Schutz requires that the orientation of the rod is chosen to be perpendicular to the direction of $\mathbf v$. If that were not the case, the ratio would depend on direction. At the extreme, if the rod is aligned with the direction of motion, the ratio will be the usual factor $\sqrt{1-(v/c)^2}$.

The order of development is: we are given two inertial frames with a relative velocity of $\mathbf v$ and we define $\phi(\mathbf v)=-M_{00}$ in Schutz's notation. We want to work out what the function $\phi$ is. To do that, we consider a rod oriented perpendicular to $\mathbf v$, that is stationary in one of the frames, and do some calculations about the ratio of its lengths in the two frames.

 

[msbell1]

Ok, another follow up question. So I now know that the length of a rod as measured by inertial observers doesn't depend on the direction of $\textbf{v}$ as long as $\textbf{v}$ is perpendicular to the rod. If the purpose of the proof is to show in general that the spacetime interval between two events is invariant as measured in all inertial frames, how does that result follow from the special case of measuring the length of a rod when $\textbf{v}$ is perpendicular to the rod? (I'm ok with the second part of the proof where Shutz shows that $\phi\left(v\right)=1$, but the jump in the first part of the proof from considering a special case in which $\textbf{v}$ is perpendicular to a rod, to the more general case of any spacetime interval is still confusing to me.) Thanks.