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A $\textbf{clock}$ carried by a specific observer $(\gamma, e)$ will measure a $\textbf{time}$ $$ time := \frac{1}{c} \int_{\tau_0}^{\tau_1} d\tau \sqrt{ \eta(v_{\lambda}, v_{\lambda}) } \tag1$$ between the two $''\underline{events}''$ $$ \lambda(\tau_0) \quad \quad \quad \, \text{ "start the clock'' } $$ and $$ \lambda(\tau_1) \quad \quad \quad \, \text{ "stop the clock'' } $$ Here $v_{\lambda}$ is the tangent vector of the the curve $\lambda$ at the point $\lambda(\tau)$

Lets choose coordinates $(x_0=ct,x_1=x,x_2=y,x_3=z)$ such that the trajectory of observer $A$ in spacetime is given by $$\lambda_A=(t,0,0,0) \tag 2$$ and the trajectory of observer $B$ in spacetime is given by $$\lambda_B=(t,d-vt,0,0) \tag 3$$ From these equation

we see that $A$ and $B$ meet at $t= \frac{d}{v}$

In these coordinates the metric is given by $$\eta=c^2dt\otimes dt-dx\otimes dx-dy\otimes dy-dz\otimes dz $$

and the tangent vector of $B$ is given by

$$v^B_{\lambda}=\frac{\partial}{\partial t}-v\frac{\partial}{\partial x} $$

so we have $$n(v^B_{\lambda},v^B_{\lambda})=c^2-v^2 $$ than the time elapsed until their meeting for observer $B$ is $$t_B=\frac{1}{c}\int_0^\frac{d}{v}\sqrt{c^2-v^2}=\frac{1}{\gamma}\frac{d}{v}$$

From the perspective of $B$ ,$(2)$ and $(3)$ are given by $$\lambda_A=(t',-d+vt',0,0) \tag 4$$ $$\lambda_B=(t',0,0,0) \tag 5$$
from this we have that $$v^B_{\lambda}=\frac{\partial}{\partial t'}$$

so we have $$n(v^B_{\lambda},v^B_{\lambda})=c^2$$ Performing the calculus as above we obtain $$t_B=\frac{d}{v}$$

Why I am obtaining different results for $t_B$?

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  • $\begingroup$ why the down vote for this question? $\endgroup$ Apr 3, 2021 at 18:27
  • $\begingroup$ You are considering two different starting points for the integrals. $\endgroup$
    – Javier
    Apr 3, 2021 at 21:39
  • $\begingroup$ @Javier what do you mean by that? $\endgroup$ Apr 3, 2021 at 21:41

2 Answers 2

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Observer-A calculated the elapsed time along an inertial segment between two events on B's worldine.
Did observer-B's calculation for the elapsed time use the same two events?

In the diagram below (for simplicity, using $v_B=-3/5$ so that $\gamma=5/4$ and $k=1/2$),
Obs-A used events $T$ and $R$ (simultaneous according to Obs-A) and thus defined the distance $d=TR$. Then Obs-A calculated $TZ$ and $RZ$.

In B's calculation (for obs-B to arrive at event $Z$),

  • did you start from event $R$ (as obs-A did),
    which is when Obs-B says that Obs-A is a distance $d'=RR_{sim}$ away (using events $R$ and $R_{sim}$ simultaneous according to Obs-B)?
  • Or did you start from event $Q$,
    which is when Obs-B says that Obs-A is a distance $d=QU$ away (using events $Q$ and $U$ simultaneous according to Obs-B)?

robphy-RRGP-properTime-simultaneity

(You can calculate $d'$ by similar triangles.
$UQZ$ and $R_{sim}RZ$ are similar.
In Minkowski spacetime, $RTZ$ and $UQZ$ are congruent.
$v_{B-wrtA}=RT/TZ$ and $v_{A-wrtB}=UQ/QZ$ and $\gamma=TZ/RZ$ and $\gamma=QZ/UZ$.)

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The error was assuming that the coordinate transformation between the systems would leave the metric in the same form.

The transformation between the two frames that express this situation is given by $$t'=\gamma^2\left(t+\frac{vx}{c^2}-\frac{vd}{c^2}\right)$$ $$x'=\gamma^2\left(x+vt-d\right)$$

the chain rule give us that $$v^B_{\lambda}=\frac{\partial}{\partial t}-v\frac{\partial}{\partial x}=\frac{\partial}{\partial t'}$$ So it is true that this is a coordinate system where $B$ is stationary.

Now the curve $A$ in his coordinate system is given by $(t,0,0,0)$ by transformation above in the coordinate system $B$ the curve associate to $A$ is given by $$\left(\gamma^2\left(t -\frac{vd}{c^2}\right), \gamma^2\left(vt-d\right) \right)=\left(t', vt'-d \right)$$ So this is the desired transformation.

The inverse transformation is given by

$$t=\left(t'-\frac{vx'}{c^2}+ k\right)$$ $$x=\left(x'-vt'+c\right)$$

where $k$ and $c$ are constant. From this we have that $$dt=dt'-\frac{vdx'}{c^2}$$ $$dx=dx'-vdt'$$

Using the formulas above we obtain that $$\eta \left (\frac{\partial}{\partial t'}, \frac{\partial}{\partial t'}\right)=c^2-v^2,$$ Finally performing the integral we get the same result

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