Relative Velocities Equal in Std Config? Lorentz Transform

[genxium]

In the wikipedia page of Lorentz Transformation (http://en.wikipedia.org/wiki/Lorent...ormation_for_frames_in_standard_configuration) it's said that

Consider two observers O and O′, each using their own Cartesian coordinate system to measure space and time intervals. O uses (t, x, y, z) and O′ uses (t′, x′, y′, z′). Assume further that the coordinate systems are oriented so that, in 3 dimensions, the x-axis and the x′-axis are collinear, the y-axis is parallel to the y′-axis, and the z-axis parallel to the z′-axis. The relative velocity between the two observers is v along the common x-axis; O measures O′ to move at velocity v along the coincident xx′ axes, while O′ measures O to move at velocity −v along the coincident xx′ axes. Also assume that the origins of both coordinate systems are the same, that is, coincident times and positions. If all these hold, then the coordinate systems are said to be in standard configuration.

The problem I'm having here is that while taking Lorentz Transformation as true, I failed to verify the statement above, i.e. I can't get "O′ measures O to move at velocity −v along the coincident xx′ axes". Here's my calculation:

Suppose that 2 frames $O$ (with observer $A$ at origin) and $O'$ (with observer $A'$ at origin) are put in standard configuration. At time $t$ in frame $O$, $A$ measures that itself is at $P = (x_P, 0, 0, t)$ and $A'$ is at $Q = (x_Q, 0, 0, t)$ where $x_P = 0$. Here $P, Q$ are introduced just as measurement event notation.

Now if denoted $v = \frac{dx_Q}{dt} = \frac{x_Q}{t}$ (second "=" holds because of standard configuration) and $t' = \gamma(v) (t - \frac{v \, x_Q}{c^2}) = \gamma(v) \, t \, (1 - \frac{v^2}{t^2}) = \frac{t}{\gamma(v)}$, then $A'$ measures that $A$ is at $P' = (x_{P'}, 0, 0, t')$ and itself is at $Q' = (x_{Q'}, 0, 0, t')$ where $\gamma(v) = \frac{1}{\sqrt{1 - v^2/c^2}}$, $x_{P'} = \gamma(v) (x_P - v \, t) = - \gamma(v) \, v \, t$ and $x_{Q'} = 0$.

Thus the velocity of $A$ measured by $A'$ is $v' = \frac{dx_{P'}}{dt'} = \frac{x_{P'}}{t'} = - \gamma(v)^2 \, v \neq -v$

Is there something wrong with the calculation? Or did I just have a misunderstanding of the statement in wikipedia?

Any help will be appreciated!

 

[Orodruin]

Is there something wrong with the calculation?

Yes, you assumed that the t' you got from the world line of Q was the same as that you should use for the world line of P. This is not the case, you need to take relativity of simultaneity into account.

Edit: I suggest you instead attempt to compute the world line of P solely by Lorentz transforming how it looks in the unprimed system. You will get an expression for t' as well as xP', which you can relate to the velocity.

 

[genxium]

Thank @Orodruin for the quick reply :)

Did you mean that

In frame O, if A measures 2 events $P = (x_P, 0, 0, t_P)$ and $Q=(x_Q, 0, 0, t_Q)$ "at the same time", i.e. $t_P = t_Q = t$, then the corresponding events $P'$ and $Q'$ are "not simultaneous", i.e. $t_{P'} = \gamma(v) (t - \frac{v \, x_P}{c^2})$ and $t_{Q'} = \gamma(v) (t - \frac{v \, x_Q}{c^2})$ ?

if so I have

$x_{P'} = \gamma(v) (-v t)$ and $t_{P'} = \gamma(v) t$ taking $x_P = 0$, thus $v_{P'} = \frac{x_{P'}}{t_{P'}} = -v$

Hope I get the numbers right this time.

 

[Orodruin]

Correct, in order to see how P moves in the primed systems, you need to take dxP'/dtP' as you did now. There really is not much more to it.