- #1

msbell1

- 25

- 0

## 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.