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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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1 Answer 1

I think there are few areas in which confused language is more widespread and has penetrated deeper into mainstream education than relativity. I can't claim to clear up all your confusions because I have trouble following many of them, but I'm pretty sure that they will all be resolved if you start using precise language and concepts and see through the illusion that any of the handwaving stuff that relativity is often surrounded with actually means anything.

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.

I'm aware that this doesn't come close to answering the myriad of questions you've posed all in one go; but I think if you start clearing up concepts in this way, the fog will start to lift and perhaps you can then express some of the questions in a clear enough form that they admit a clear answer.

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As far as I am aware, we have a point P, a frame F and a frame F'. The coordinates of P in Frame F is different to the coordinates of frame P in frame F'. When I wrote "P w.r.t. F" I really mean "coordinates of P in frame F". –  Adam Rubinson Apr 17 '12 at 20:57
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@Adam: That's exactly what I mean, you wrote something other than what you meant. That's OK when you know what you're doing, but when you're confused about what you're doing it's bound to contribute to your confusion. I'd suggest that you rewrite your question, taking care to always write exactly what you mean. I suspect that a) you'll clear up some of your confusions on your own in this process and b) it will then be a lot easier to help you resolve whatever confusions remain than it is on the basis of the current formulation of the question. –  joriki Apr 17 '12 at 21:07
    
Okay, thanks. I will have another think tomorrow. –  Adam Rubinson Apr 17 '12 at 21:20
    
Okay. The first paragraph of my OP is from my lecture notes. I believe it is concluding that the origin of frame F' has x-coordinate vt in Frame F. Is this what it is concluding? –  Adam Rubinson Apr 17 '12 at 22:39
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@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

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