Length observed by a moving frame of reference

Question

Hello im studying special relativity and i was wondering something about length contraction. Let's say someone who is not moving is seeing a rod which has one edge at $x=0$ and the other at $x=a$ so its length is $a$. An observer moving at constant speed $u$ along the $x$-axis uses lorentz transformation to determine the coordinates of these two points in his frame of refrence.

$x'_a=\gamma(a-ut)$, $x'_0=\gamma(0-ut)$ so to find the length of the rod we take the difference of these two points and it gives us $x'_a-x'_0=\gamma a$

I know that $\gamma>1$ so shouldn't the moving reference frame observe the rod to be bigger? Why are we talking about length contraction? I know something is wrong the way i derived this so can anyone explain to me where my mistake lies?

Answer 1

The rod is moving in the observer's reference frame and he has to measure each end at the same time in his frame. You are using the same time, t, in the frame of the first observer who is at rest with respect to the rod. What's simultaneous (same t) in the first observer's frame is not simultaneous in the moving observer's frame.

So the difference in 𝑥′ coordinates that you calculate is not the length in the moving observer's frame.

Answer 2

You want to measure the location of both ends at the same time, let us say when $t'=0$. Given the Lorentz transformations, the point $(x,t)=(0,0)$ maps to $(x',t')=(0,0)$, and $(L,0)$ to $(\gamma L,\gamma v L/c^2)$. But this is the position of the other extreme of the rod at $t'\ne 0$. The position at $t'=0$ is $L'=x'-vt'$, replacing we get $L'=L/\gamma$

Answer 3

The spatial distance between 2 points supposes being taken at the same time. In this case, the moving observer must measure the distance $0a$ at the same time t'.

If we take t = 0 and t'= 0 for x = 0 and x' = 0 things are easier to understand. The moving observer must measure the distance for $t' = 0$.

$0 = t' = \gamma(t - ux_a)$ (for c = 1) => $t = ux_a$

$x'_a = \gamma(x_a - u^2x_a)$ => $x'_a = \gamma x_a(1 - u^2))$

$$x'_a = \frac{x_a\gamma}{\gamma^2} = \frac{x_a}{\gamma}$$

Answer 4

When using Lorentz transformations, you should transform events, that is: specific points in both space and time.

You have done this by transforming (in $S$, the stationary frame):

$$ (t, x) = (0, 0)_S $$ $$ (t, x) = (0, a)_S $$

which is the simultaneous positions of the end of the rod in $S$.

The separation of those events ($c=1$):

$$ \Delta s^2 = \Delta t^2 - \Delta x^2 = (0-0)^2-(a-0)^2 = -a^2$$

is a Lorentz invariant.

They transform to:

$$ (t', x') = (\gamma (t-\beta x), \gamma (x - \beta t))_{S'} = (0, 0)_{S'}$$ $$ (t', x') = (\gamma (0 - \beta a), \gamma(a-\beta0))_{S'} = (-\gamma\beta a, \gamma a)_{S'}$$

with invariant interval:

$$ \Delta s'^2 = \gamma^2(\beta^2)a^2 - \gamma^2a^2 = -\frac{\gamma^2}{\gamma^2}a^2=-a^2$$

so thats good.

But the problem is the $t'$ coordinates. They are not simultaneous in $S'$. In $S'$, the rod is moving, so if you look at the ends at different times, you are going to get the wrong answer.