How to Determine Frame Size with Lorentz Transformation

[genxium]

First time posting in this section. I understand that this question could possibly be an old and common question about Lorentz Transformation, however I failed to find useful discussions or instructions online.

Assuming that there're 2 frames $S, S'$ where $S'$ moves along the $x_{+}$ axis of $S$ at constant speed $v$. The frames coincide at $<0,0,0,0>$ (as well as their rectilinear coordinate axes) for a starting event $P$ and then measure $<x, y, z, t>$ and $<x', y', z', t'>$ respectively for event $Q$.

According to Lorentz Transform I shall have:

$x'=\gamma \cdot (x-vt)$
$t'=\gamma \cdot (t-\frac{vx}{c^2})$

where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Now that $\frac{dx'}{dx} = \gamma$ which indicates that $\frac{dx'}{dx}$ is independent of the direction of $v$ along the $x$ axis. However it doesn't make sense to me here. If I introduce a 3rd frame $S''$ which moves along the $x_{-}$ direction of $S$ at a constant speed $v$, then should I get $\frac{dx''}{dx}=\gamma$ as well and further $dx'' = dx'$ (which should NOT hold bcz $S'$ and $S''$ are dynamic to each other)? Did I make a mistake in the calculation?

I'm quite confused for how "some degree of symmetry" (for relation of $dx, dx', dx''$ above) could be achieved if Lorentz Transformation is true. Any help will be appreciated :)

 

[DrGreg]

When there is more than one variable involved, it's not true that<br /> dx&#039; = \frac{dx&#039;}{dx}\,dx<br />The correct equation is <br /> dx&#039; = \frac{\partial x&#039;}{\partial x}\,dx + \frac{\partial x&#039;}{\partial y}\,dy + \frac{\partial x&#039;}{\partial z}\,dz + \frac{\partial x&#039;}{\partial t}\,dt<br />

 

[genxium]

Hi @DrGreg, I'm not sure how I should interpret your answer. Of course you're right about that $dx' = \frac{\partial x'}{\partial x} dx + \frac{\partial x'}{\partial y} dy + \frac{\partial x'}{\partial z} dz + \frac{\partial x'}{\partial t} dt$, however if $dx, dy, dz, dt$ are mutually independent then $\frac{dx'}{dx} = \gamma$ should still hold. Do you imply that $dx, dy, dz, dt$ can NEVER be mutually independent for any events $P$ and $Q$?

I tried to put the classical thought experiment where $P$ is "emission of light along x+ axis of $S$(as well as $S'$)" and $Q$ is "detection of light somewhere in space-time of $S$(as well as $S'$)" into calculation. Now I have $x=ct$ of $S$ as variable dependency but I'm still lost in the maths :(

Maybe I shall re-describe the question this way: 2 observers Alice and Bob who remain still in frame $S$ and $S'$ respectively where $S$ and $S'$ are the same as stated in my original question. Alice learns Special Relativity and he figures out that currently for 2 specific events $P, Q$ Bob measures larger(or smaller maybe, haven't figured this out) time-elapsed $dt'$ than his own measurement $dt$. Can Alice just reverse the direction of his $x$-axis(i.e. rotate around the $z$-axis) and say that "from now on Bob measures smaller $dt'$ than my $dt$"? In short is measurement dependent upon "choice of coordinate axes" or "alignment of coordinate axes"?

 

[DrGreg]

Of course you're right about that $dx' = \frac{\partial x'}{\partial x} dx + \frac{\partial x'}{\partial y} dy + \frac{\partial x'}{\partial z} dz + \frac{\partial x'}{\partial t} dt$, however if $dx, dy, dz, dt$ are mutually independent then $\frac{dx'}{dx} = \gamma$ should still hold.

To put it more precisely, $dx' = \gamma dx$ is true provided $dt = 0$. (In your question $dy = dz = 0$ always, so we can ignore those.)

Similarly, $dx'' = \gamma dx$ is true provided $dt = 0$. When you come to compare S' with S'', however, you can only conclude that $dx'' = dx'$ whenever $dt = 0$, when really what you are interested is when $dt' = 0$ (or $dt'' = 0$ if you are comparing S' with S'').

 

[genxium]

@DrGreg, that makes sense. Do you minding taking a look at this as well(quoted from my previous reply)?

2 observers Alice and Bob who remain still in frame $S$ and $S′$ respectively where $S$ and $S′$ are the same as stated in my original question. Alice learns Special Relativity and he figures out that currently for 2 specific events $P,Q$ Bob measures larger(or smaller maybe, haven't figured this out) time-elapsed $dt′$ than his own measurement $dt$. Can Alice just reverse the direction of his $x$-axis(i.e. rotate around the $z$-axis) and say that "from now on Bob measures smaller $dt′$ than my $dt$"? In short is measurement dependent upon "choice of coordinate axes" or "alignment of coordinate axes"?