1
$\begingroup$

I am learning Length Contraction from the book Special Relativity and Classical Field Theory by Leonard Susskind.
There are two frames with relative speed $v$ along the x-axis. If my frame is at rest, then the other frame is moving with velocity $v$ towards right, i.e., towards the positive x-axis. The following picture shows the world line of the moving frame w.r.t. my frame, $\overline{OQ}$ is the length of a one-metre ruler in my frame. Length Contraction
The problem is to find out the length of this ruler as measured by an observer in the moving frame with his own scale. The author says that this length is $\overline{OP}$.
My question is, why $\overline{OP}$ is that length? In other words, why would I say that the intersection of the world lines of the 'moving frame' and the 'far end of the ruler' in my frame is the point corresponding to the far end of the ruler in the moving frame? The observer in the moving frame sees the ruler from any point on the axis $x'=0$ ; so why does the point P need to be the perpendicular distance from point Q? More generally, how do I know which point in the moving frame corresponds to a point in my frame?
Please help me with this confusion.

$\endgroup$
0
1
$\begingroup$

First, we need to describe what length of an object is?

The length of a ruler is the distance between the two coordinates of endpoints of the ruler which are measured simultaneously. Now, 2 simultaneous events happening in one frame might not be simultaneous in other and that's where length contraction comes in.

Lorentz transformation is given by: \begin{equation} \begin{aligned} t^{\prime} &=\gamma\left(t-v x / c^{2}\right) \\ x^{\prime} &=\gamma(x-v t) \end{aligned} \end{equation}

So, in the frame $(t,x)$ co-ordinates of O and Q are $(0,0)$ and $(0,1)$ respectively but in $(t',x')$ frame $t'$ co-ordinate would be different. O has co-ordinate $(0,0)$ while Q has (-$\gamma \frac{vx}{c^2}$,$\gamma$).

For measuring length we need $t'$ co-ordinates of the end-points to be the same which we do by setting $t'=0$. So, now O has still the same co-ordinate $(0,0)$ in both frames. But, co-ordinate of other end is given by $(0,x')$ where $x'$ is given by the intersection of the line $t'=0 \Rightarrow t= \gamma \frac{vx}{c^2}$ and $x=1$ which is point P. Thus, the length of ruler in moving frame is given by OP.

$\endgroup$
0
$\begingroup$

The perpendicular line $x=1$ ($x=0$) is the world-line of the end (start) of the meter-stick. You have to follow that line from $Q$ up to $P$ because that is the point that is simultaneous with $O$ in the primed frame, that is, both $O$ and $P$ are at $t'=0$.

So $\bar{OP}$ defines the object in the primed frame; $\bar{OP} < \bar{OQ}$, but that is not obvious from the diagram because the tic marks on both $t'$ and $x'$ are dilated by a scale factor:

$$ F=\sqrt{\frac{1+\tan^2{\alpha}}{1-\tan^2{\alpha}}}$$

where the angle alpha is:

$$ \alpha \equiv \angle POQ$$

so that:

$$ \tan{\alpha} = \frac v c\equiv\beta $$

Hence the length of the moving meter ($L=1\,$m) stick is:

$$ L' = \frac 1 F \times L\sqrt{1+\beta^2}$$ $$ L' = L\sqrt{1+\beta^2}\sqrt{\frac{1-\beta^2}{1+\beta^2}}$$ $$ L' = L\sqrt{1-\beta^2} = \frac L {\gamma} < L$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.