so I'm working on length contraction in relativity theory. I feel pretty confident time dilation and have not really gone over Lorentz Transformations that much.

The question itself lies at the bottom if you want to skip to there. The following just shows my process and conclusion

So, I'm working on a thought experiment where a snake is moving relative to Observer 2. I'm trying to determine the relation between the lengths of each reference frame: snake and Observer 2. Here is a (very messy) illustration: illustration of length contraction

Okay so the variables for each reference frame would be: $$ S_1(snake): \Delta t_1 , L_1 $$ $$ S_2: \Delta t_2 , L_2 $$


Event 1 = the measure at the point at the rear side of the snake. Event 2 = the measure at the point at the front side of the snake.

I am just going to assume that the following equation is true for now after having previously proved it in a time dilation experiment:

$$ \Delta t = \gamma \Delta t_o $$

Since in both reference frames, $S_1 , S_2$, Event 1 and Event 2 do not occur at the same point, we can say the proper time $\Delta t_o$ cannot be found in either of the frames. Likewise, it will be found in some other frame. Therefore, from the equation found in the time dilation experiment, we can reasonably say: $$\Delta t_1 = \gamma \Delta t_o$$ $$\Delta t_2 = \gamma \Delta t_o $$ Where the values of $\gamma$ in each equation are equal because V would be equal in each equation. Then we conclude: $$\Delta t_1 = \Delta t_2 $$ From there, since $L = V \Delta t$, we can say:

$$L_2 = V \Delta t_2 $$

Since both times values are equivalent: $$L_2 = V \Delta t_1 $$ Then, we can replace $\Delta t_1$ with $L_1 \over V$. Plugging that back into the equation: $$L_2 = V {L_1 \over V} $$ and $$L_2 = L_1 $$ This seems to be rather uneventful, and I'm assuming I did something wrong. It would make sense that if time can be different in different frames then length would be too. I'm pretty sure the reason this lame result occurred was because I assumed that the $\Delta t$ in each frame of reference was the same. The text I'm reading states that the following equation: $$L_2 = {L_1 \over \gamma}$$ I know I'm forgetting something, but I can't figure out what it is. Can anyone see what I'm doing wrong?



closed as unclear what you're asking by DavePhD, Brandon Enright, Kyle Kanos, BebopButUnsteady, Prahar May 1 '14 at 16:46

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  • 1
    $\begingroup$ What is $\Delta t_o$? There seems to be only one period of time in the moving reference frame ($\Delta t_o$) which gets converted into two time periods in stationary reference frame ($\Delta t_1$ and $\Delta t_2$). Why? And velocity is distance divided by time, and not length divided by time. $\endgroup$ – bright magus Apr 29 '14 at 20:35
  • $\begingroup$ $$\Delta t_1 = \gamma \Delta t_o$$ $$\Delta t_2 = \gamma \Delta t_o $$ There should be a unique $\gamma_1$ and $\gamma_2$ for $S_1, S_2$ since they're moving at different velocities to one another, and therefore also to frame $S_0$. I feel this question is too localized. $\endgroup$ – John McVirgo Apr 29 '14 at 20:42
  • $\begingroup$ @brightmagus $\Delta t_o$ is the proper time, which can be found in a frame of reference where the two events (i.e. note when back of snake passes observer and when front of snake passes of observer) occur at the same position. Length is analogous to distance in the velocity equation because both can be determined by finding the difference between two points $x_2 - x_1$. $\endgroup$ – Colin Michael Flaherty Apr 29 '14 at 20:53
  • $\begingroup$ "Proper time" of what? Still, why one time period gets converted into two? Does your snake of length L travels a distance of L? $\endgroup$ – bright magus Apr 29 '14 at 21:07
  • $\begingroup$ @brightmagus Proper time when measuring the snake, which could also be seen as measuring the distance between two points, one at the rear of the snake and one at the front of the snake, making this distance equivalent to the length of the snake. $\endgroup$ – Colin Michael Flaherty Apr 30 '14 at 2:10

Normally I recommend beginners to SR always use the Lorentz transformations to work through problems, but actually in this case using the Lorentz transformations is algebraically a bit fiddly, and there's an easier way to do it. We can use the fact that the proper time is Lorentz invarient. For any pair of events $(t_1, x_1)$ and $(t_2, x_2)$ we calculate the change in proper time using:

$$ c^2\Delta \tau^2 = c^2\Delta t^2 - \Delta x^2 $$

where $\Delta t = t_2 - t_1$ and $\Delta x = x_2 - x_1$. The invariance of the proper time means that all observers must agree on the value of $\Delta\tau$. So we'll calculate $\Delta\tau$ in both $S_1$ and $S_2$ and equate them. This diagram shows how we do it:


Start in $S_1$. In this frame we see observer 2 moving past at velocity $v$. We'll choose our origin so observer 2 passes one end of the snake at $(0, 0)$, which means they pass the other end of the snake at $(d/v, d)$ - we get the time $t = d/v$ simply by dividing the length of the snake $d$ by the velocity $v$. The proper time is:

$$ c^2\Delta \tau^2 = c^2 \frac{d^2}{v^2} - d^2 \tag{1} $$

Now look at things in frame $S_2$. In this frame observer 2 is stationary and sees the snake flashing past him at velocity $v$. The time the snake takes to pass is $d'/v$, where $d'$ is the contracted length that we're trying to calculate. Again we choose our origin so the head of the snake passes at $(0, 0)$, and that means the tail of the snake passes at $(d'/v, 0)$. This time the proper time is:

$$ c^2\Delta \tau^2 = c^2 \frac{d'^2}{v^2} \tag{2} $$

Since the value of $c^2\Delta\tau^2$ must be the same in both equations (1) and (2) we can set the right hand side of these two equations equal:

$$ c^2 \frac{d^2}{v^2} - d^2 = c^2 \frac{d'^2}{v^2} $$

The rest is just algebra. We tidy the expression up by rearranging it as:

$$ d^2 \left(\frac{c^2}{v^2} - 1\right) = d'^2 \frac{c^2}{v^2} $$

and dividing both sides by $c^2/v^2$ gives:

$$ d^2 \left(1 - \frac{v^2}{c^2}\right) = d'^2 $$


$$\begin{align} d' &= d \sqrt{1 - \frac{v^2}{c^2}} \\ &= \frac{d}{\gamma} \end{align}$$


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