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I'm trying to understand the proof that spacetime interval is invariant under for any two inertial observers. I know it's easy to arrive at the result using Lorentz transformation but I'm trying to understand geometrical way of doing it, specifically as done in Schutz. The proof starts in page 9 in the above link. Going into the details I couldn't understand the following part.

Once we arrive at the result $\Delta \bar s^2 = \phi(v) \Delta s^2 $, we set on to prove first that $\phi(v) = \phi(|v|)$ and then $\phi(v) = 1 $. To prove the first part, we choose a particular class of observer pair $O$ and $\bar O$ where relative velocity $v$ is perpendicular to the length of rod on $y$-axis (essentially $y=\bar y$ and $z= \bar z$) and then show that $\phi(v) = \phi(|v|)$.

Next part claims that this character of $\phi(v)$ is true for any general class of observer pair(for example $O$ and $\bar O$ where relative velocity $v$ such that $y \neq \bar y$ and $z \neq \bar z$). I couldn't understand this extension. I'm looking for some explanation at this step. Thanks, any help is appreciated.

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  • $\begingroup$ I changed the link, to google books. $\endgroup$
    – levitt
    Apr 10, 2015 at 9:53
  • $\begingroup$ I am also totally baffled by this step in the logic. $\endgroup$
    – Mike Bell
    Oct 14, 2017 at 17:02
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    $\begingroup$ I see that no one has answered your question, but I was wondering if you've been able to answer it for yourself. I am really confused by the jump in the logic of the proof from $\phi(\mathbf{v}) = \phi(\left|\mathbf{v}\right|)$. Do you get it (now that you've had a couple of years to think about it :) )? $\endgroup$
    – Mike Bell
    Oct 18, 2017 at 14:27
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    $\begingroup$ Hi, I don't remember if I have been able to answer it to myself. It has been really long time since I last worked on relativity. But I will spend some time in the next few days to see if I am able to understand it although I can't gaurentee if will be able to resolve this. I will let you know if I anything to say. If you are stuck at this point, I suggest that you leave this part alone, and proceed further. $\endgroup$
    – levitt
    Oct 19, 2017 at 8:21

2 Answers 2

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Yes, the logic provided in schutz is wrong though the correct logic would be that space and time are homogeneous and isotropic. Actually, ϕ can be a function of velocity and of coordinates of spacetime but due to homogeneity of space,for a moment we can say function is dependent on velocity only as homogeneity of space says every point in space is equivalent therefore whatever O' coordinate frame we choose can be shifted to origin of original frame. Now by isotropy of space we can say that wherever O' frame moves it doesn't matter in which direction, it would give same result of the some experiment(which we call event) perfomed.Thus function only depends only on speed and further arguments given by schutz can be applied to get ϕ(v)=1.

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Let two inertial systems $\;\mathrm S_1,\mathrm S_2\;$ with common axis $\;x_1\equiv x_2\;$ and parallel axes $\;y_1\parallel y_2\;$ and $\;z_1\parallel z_2$. The system $\;\mathrm S_2\;$ is moving along the common axis $\;x_1\equiv x_2\;$ with constant velocity vector $\;\boldsymbol{\upsilon}\;$ as in Figure. For the space-time intervals
\begin{equation} \mathrm ds^2_1 \boldsymbol{=}\mathrm dt^2_1\!\boldsymbol{-}\!\mathrm dx^2_1\!\boldsymbol{-}\!\mathrm dy^2_1\!\boldsymbol{-}\!\mathrm dz^2_1\,, \quad \mathrm ds^2_2 \boldsymbol{=}\mathrm dt^2_2\!\boldsymbol{-}\!\mathrm dx^2_2\!\boldsymbol{-}\!\mathrm dy^2_2\!\boldsymbol{-}\!\mathrm dz^2_2 \tag{01}\label{eq01} \end{equation} we have \begin{equation} \mathrm ds^2_2 \boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)\mathrm ds^2_1 \tag{02}\label{eq02} \end{equation} where $\;\phi\left(\boldsymbol{\upsilon}\right)\;$ a function of the vector variable $\;\boldsymbol{\upsilon}\;$ to be determined.

Now, suppose a third system $\;\mathrm S_3\;$ at rest with respect to $\;\mathrm S_2\;$ with common axis $\;y_2\equiv y_3\;$ and anti-parallel axes $\;x_3\parallel \boldsymbol{-} x_2\;$ and $\;z_3\parallel \boldsymbol{-} z_2\;$. Then \begin{equation} \mathrm ds^2_3 \boldsymbol{=}\mathrm ds^2_2 \stackrel{\eqref{eq02}}{\boldsymbol{=\!=}}\phi\left(\boldsymbol{\upsilon}\right)\mathrm ds^2_1 \tag{03}\label{eq03} \end{equation}

Also let a fourth system $\;\mathrm S_4\;$ at rest with respect to $\;\mathrm S_1\;$ with common axis $\;y_4\equiv y_1\;$ and anti-parallel axes $\;x_4\parallel \boldsymbol{-} x_1\;$ and $\;z_4\parallel \boldsymbol{-} z_1\;$. Then the system $\;\mathrm S_4\;$ is moving with respect to $\;\mathrm S_3\;$ with the constant velocity vector $\;\boldsymbol{\upsilon}\;$ along the common axis $\;x_3\equiv x_4$. The configuration of systems $\;\mathrm S_3,\mathrm S_4\;$ is in all respects identical to that of the $\;\mathrm S_1,\mathrm S_2\;$ (this is more clear viewing the Figure from its back). So \begin{equation} \mathrm ds^2_4 \boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)\mathrm ds^2_3\stackrel{\eqref{eq03}}{\boldsymbol{=\!=}}\phi^2\left(\boldsymbol{\upsilon}\right)\mathrm ds^2_1 \tag{04}\label{eq04} \end{equation}

Now the system $\;\mathrm S_1\;$ is at rest with respect to $\;\mathrm S_4\;$ so \begin{equation} \mathrm ds^2_1\boldsymbol{=}\mathrm ds^2_4\stackrel{\eqref{eq04}}{\boldsymbol{=\!=}}\phi^2\left(\boldsymbol{\upsilon}\right)\mathrm ds^2_1 \tag{05}\label{eq05} \end{equation} that is we must have \begin{equation} \phi^2\left(\boldsymbol{\upsilon}\right)\boldsymbol{=}1 \tag{06}\label{eq06} \end{equation} or \begin{equation} \phi\left(\boldsymbol{\upsilon}\right)\boldsymbol{=\pm}1 \tag{07}\label{eq07} \end{equation} But for $\;\boldsymbol{\upsilon=0}\;$ the systems $\;\mathrm S_1,\mathrm S_2\;$ are identical, so in \eqref{eq02} we must have $\;\mathrm ds^2_2 \boldsymbol{=}\mathrm ds^2_1\;$, that is $\;\phi\left(\boldsymbol{0}\right)\boldsymbol{=+}1\;$ and \eqref{eq07} yields \begin{equation} \phi\left(\boldsymbol{\upsilon}\right)\boldsymbol{=+}1\,, \quad \text{for any } \boldsymbol{\upsilon} \tag{08}\label{eq08} \end{equation}

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