# Homework - length contraction ($\gamma$ and $u$ are unknown)

I am very sorry for posting these problems in this forum, but i don't know where to post otherwise... I have a weird special relativity problem where i get a relative speed $u$ which is larger that $c$. I dont need an solution but rather some guidance towards my understanding of the special relativity...

Two sticks (both with the proper length $1m$) travel one toward another along their lengths. In the proper system of one stick they measure time $12.5ns$ between two events:

1. right ends are aligned,
2. left ends are aligned.

What is the relative speed $u$ between the sticks?

I first draw the picture:

If i use the Lorentz transformation and try to calculate the relative speed $u$ i get speed that is supposed to be greater than the speed of light:

Here is what i did: \begin{aligned} \Delta x' &= \gamma (\Delta x - u \Delta t) \xleftarrow{\text{Is this ok? I don't think my system is in the standard configuration}}\\ \frac{1}{\gamma}\Delta x &= \gamma (\Delta x - u \Delta t)\xleftarrow{\text{$\Delta x$ is equal to the proper length (in $xy$ the botom stick is standing still)}}\\ \tfrac{1}{\gamma^2}\Delta x &= \Delta x - u \Delta t\\ \left( 1 - \tfrac{u^2}{c^2}\right)\Delta x &= \Delta x - u \Delta t\\ \Delta x - \tfrac{u^2}{c^2}\Delta x &= \Delta x - u \Delta t\\ \tfrac{u^2}{c^2}\Delta x &= u \Delta t\\ \tfrac{u}{c^2}\Delta x &= \Delta t\\ u &= \tfrac{\Delta t}{\Delta x} c^2\\ u &= \tfrac{12.5\cdot 10^{-9}s}{1m} 2.99\cdot 10^{8}\tfrac{m}{s} c\\ u &= 3.74 c \end{aligned}

In the solutions it says i should get $u=0.5 c$, but i get $u=3.74c$. I used the Lorentz transformation found on Wikipedia. These are for the standard configuration, but in my case is it the standard configuration?

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Your problem is in the second to last step where you assumed delta-x is 1m. If the moving meter stick has a contracted length of, say, ~0.8m, then it only need move 0.2m. –  David H Jun 25 '13 at 17:44
But $\Delta x$ is $1 m$. It is stationary in the $xy$ frame and is therefore equal to proper length which is said to be $1m$. Am i wrong? Please explain how am i wrong. –  71GA Jun 25 '13 at 17:59
So when i first write down the Lorentz transformation $\Delta x ' = \gamma(\Delta x -u\Delta t)$ the $\Delta x$ here is not the length of the stick in $xy$? It is some sort of the path right? –  71GA Jun 25 '13 at 18:02
@71GA, see my answer to your 1st comment. –  Alfred Centauri Jun 25 '13 at 18:32
You don't have to be sorry if you show that you put work into the problem :) –  Justin L. Jun 26 '13 at 6:49

There's a more physically intuitive way to do this in my opinion. Note that in your picture, the longer stick has length $L_0 = 1\,\mathrm{m}$ and the shorter stick has length $L=L_0/\gamma_u$ because of length contraction. Therefore, the time $T = 12.5\,\mathrm{ns}$ given in the problem corresponds to the shorter stick traveling a distance $L_0-L_0/\gamma_u$. This gives the equation $$L_0 - \frac{L_0}{\gamma_u} = u T$$ Now simply solve for $u$. I checked this numerically by the way and it gives $u=c/2$.

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I know that $\gamma_u = \tfrac{1}{\sqrt{1-u^2/c^2}}$ but if i put this into your equation i dont know how to get $u$ out of it. How do you usualy do this? –  71GA Jun 25 '13 at 17:56
If i try to construct this a bit diferently (looking at the distances), i get different eq. than you: $\Delta x = \Delta x' - u\Delta t \longrightarrow \boxed{\Delta x - \frac{\Delta x}{\gamma} = u \Delta t}$. Is this due to the direction of speed? Where should i pay attention to the direction of the speed and where not? –  71GA Jun 25 '13 at 18:14
@71GA Yes you need to be careful about the sign in the equation. In particular, note that $\Delta x$ is greater than $\Delta x'$, so $\Delta x - \Delta x'$ must be positive. This means that your equation should be used when the symbol $u$ represents the negative of the speed, and mine should be used when it represents the speed. –  joshphysics Jun 25 '13 at 18:26
could you try and help me with my 1st comment :) –  71GA Jun 25 '13 at 18:29
@71GA Sorry for the delayed response; I see you've already found the answer in math.SE though. Cheers! –  joshphysics Jun 25 '13 at 20:02

But Δx is 1m.

The $\Delta x$ and $\Delta t$ you seek are the change of the coordinates of the right end of the moving stick.

In the unprimed frame, when the left ends are aligned at $t = 0$, the right end of the moving stick is at $x = 1m/\gamma_u$ (the moving stick is contracted in the unprimed frame).

When the right ends are aligned at $t = 12.5ns$, the right end of the moving stick is at $x = 1m$.

Thus:

$\Delta x = 1m - 1m/\gamma_u = (1 - 1/\gamma_u)m$

$\Delta t = 12.5ns$

and

$\dfrac{\Delta x}{\Delta t} = u$

gives the equation in joshphysics' answer

I think your difficulty here is not realizing that, to solve for $u$, you need to pick one end of the moving stick or the other and then find the change in coordinates of that end.

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I understand the theory now, but the biggest problem i have is that i don't know how to calculate anything from the equation Joshphysics provided. I mean $u$ is also hidden in $\gamma$ and when i insert $1/\sqrt{1-u^2/c^2}$ i can't seem to get $u$ out of it... –  71GA Jun 25 '13 at 18:36
@71GA, the algebra is a bit messy. –  Alfred Centauri Jun 25 '13 at 18:41
I know. I posted this in the math forum here: math.stackexchange.com/q/429318 –  71GA Jun 25 '13 at 18:41

Yes, you're in standard configuration. That's not the problem. I don't think you're properly distinguishing between the rod lengths and the distance between events.

Let's go back to the drawing board. You have one stationary rod whose length is $\ell$. The other rod has its ends traveling along a pair of inertial worldlines:

$$s_r(\tau) = \tau \gamma c(e_t - \beta e_x) + \ell e_x, \quad s_\ell(\tau) = \tau \gamma c (e_t - \beta e_x) + \ell e_x - \ell \gamma (e_x - \beta e_t)$$

This has been constructed so that the time component of the right-end worldline is zero when that end is coincident with the stationary rod's right end. Thus, you actually don't even need to find the length of the moving rod. We can just take our equation for $s_\ell$ and see when the x-component is zero.

$$-\tau_\ell \gamma c \beta + \ell - \ell \gamma = 0$$

We find that $\tau_\ell = \ell (1- \gamma)/(\gamma c \beta)$. Put this into the time component equation, and you get

$$s_\ell(\tau_\ell) \cdot e^t = \frac{\ell(1-\gamma)}{\beta} - \ell \gamma \beta$$

Set this equal to $c t$, where $t = 12.5 \, \text{ns}$, and you should eventually get a quadratic in $\beta$, which you can solve.

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