# Lorentz Transformation confusion

In my notes it says, when explaining an immediate consequence of the Lorentz Transformation: The significance of the parameter $v$ is seen as follows. Consider the origin of spatial coordinates $(x', y', z')' = (0, 0, 0)'$ in frame $F'$. From the Lorentz transformation This corresponds to $$\gamma(v)(x - vt) = 0, y = 0, z = 0$$ or $\mathbf x = \mathbf v t$, which is the worldline of an object moving uniformly with respect to $F$ with velocity $\mathbf v = (v, 0, 0)$.

Do this mean that they have said where $P$ is w.r.t. $F'$ (origin of $F'$) and they are now trying to find $P$ w.r.t. $F$? It looks as if they are, and if that is the case, are they doing it this way: Do the reverse Lorentz transformation i.e. from frame $F'$ to frame $F$. Or is it: Find out what the worldline of the origin of $F'$ would have to be in frame $F$ in order to give the Lorentz transformation $(x', y', z')' = (0, 0, 0)'$. Does it matter which way we do it? I know it doesn't, but at this stage, it is hard to explain/see why.

Moreover, how can this result be true, when the origin of frame $F'$ is moving with velocity $v$, and the distance between $F$ and $F'$ must therefore be $\gamma(v)$ Also, how comes we are allowed to define such a $v$ existing. $v$, after all, is distance over time, and the "coordinates" or (loosely) distances $|x-x'|$ depend on $v$ itself???

I am guessing that this is a common trap, and I am also guessing that the answer is because we are doing a passive transformation. But that is just a guess, and I'm not sure I even fully understand it...

Also, note that at this stage in my notes, they have only just stated what the definition of the Lorentz transformation is.

As a deeper guess, I am guessing that you can think of the passive transformations as moving from one space-time coordinate to another one, or, a "ghost" in $\mathbb R^3$ Note space-time here still refers to Newton's version.

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What does it mean to "have said where $P$ is w.r.t. $F$'" or to "find $P$ w.r.t. $F$"? A point $P$ is only in one place, namely at $P$; there's no such thing as where it is w.r.t. some frame. You can describe $P$ using different coordinate systems. The result is different sets of numbers, not different positions.
@Adam: Yes, I think that's what it's concluding. This passage is also not entirely free from abuse of language, e.g. $\mathbf x=\mathbf vt$ isn't the worldline of an object but an equation for the coordinates representing (with respect to frame F) the worldline of an object. That would be nit-picking in most other circumstances, but when you're trying to wrap your head around active and passive transformations, this sort of precision can be crucial. – joriki Apr 18 '12 at 6:27