#### kmm

Gold Member

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- 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!

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!