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Ok so, in the context of special relativity alone, If you were to have a train moving at a speed near the speed of light (say 0.7c), and containing two people, who, with their watches synchronised in the time frame of the train, drop two balls out the window from either end of this train, separated by a distance of 100m in their frame of reference. For an observer in the reference frame of the ground, (at which the speed 0.7c for the train was measured), I believe the two balls should be seen to be separated by a length ~71.4m (by lorentz contraction) while in the air. However once they hit the ground, their velocity should change to that of the ground. Assuming this change occurs instantaneously upon impact, will the balls maintain the same distance separation (71m) as calculated while moving, or would the separation distance return to that of the rest frame of the train (100m), due to the ground becoming the new rest frame of the balls? (Numbers are just for example)

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  • $\begingroup$ Welcome to Stack Exchange. Interesting question: my guess is that, while the balls are contracted, and the train they're dropped from is contracted, the space between the balls isn't contracted. So, once they come to a halt (ignoring any resulting damage), they'll be about 71 meters apart. $\endgroup$ – Daniel Griscom Jan 7 '16 at 23:25
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    $\begingroup$ Short answer: the balls are not released at the same time in the ground reference frame. $\endgroup$ – Alfred Centauri Jan 7 '16 at 23:57
  • $\begingroup$ Thanks for the answers so far, (and the amusing link Daniel), I tend to believe that they will, as you suggest, be left 71m apart, but I cannot quantify why due to not knowing anything about how to describe the path between their moving and inertial frames (with respect to the ground). Alfred, this I am aware of, their time difference is in the order of 10^-7, a fact which I used to calculate their contraction using a lorentz transform method (can be done also by simple length contraction). However my question is more directed toward what happens when the balls slow down $\endgroup$ – Tim Ryan Jan 8 '16 at 0:22
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    $\begingroup$ @urdv has given you the correct answer, but you'd have found it yourself if you'd drawn the spacetime diagram. $\endgroup$ – WillO Jan 8 '16 at 1:27
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Actually the distance that separates the balls as seen from the ground frame depends on when they are released relative to the train's time.

If the balls appear to be released simultaneously in the ground frame, then by all means, they remain separated by the contracted length of the train, $\sim 71$m. But: the fact that their release is simultaneous in the ground frame automatically means that it was not simultaneous in the train frame.

For instance, say the ground frame sees both balls released at time $t_0$ from the ends of the train, which are a distance $L/\gamma$ apart, at $x_0 = vt_0$ and $x_0 + L/\gamma$, where $L$ is the rest length of the train and $\gamma = (1-v^2/c^2)^{-1/2}$. In the train's frame the release of the rear ball occurred at $$ \begin{eqnarray} x'_{rear} &=& \gamma(x_0 - v t_0) = 0\\ t'_{rear} &=& \gamma(t_0 - vx_0/c^2) = t_0/\gamma \end{eqnarray} $$ while the front ball was released at $$ \begin{eqnarray} x'_{front} &=& \gamma((x_0 + L/\gamma) - v t_0) = L \\ t'_{front} &=& \gamma(t_0 - v(x_0 + L/\gamma)/c^2) = t'_{rear} - vL/c^2 \end{eqnarray} $$

If instead the balls are released simultaneously in the train's frame, then their individual releases will not appear simultaneous in the ground frame. The balls will appear to be released at different times and will actually be separated by a distance $\gamma L$, which in your case would be $\sim 140$m.

So again, say the balls are released simultaneously in the train's frame at time $t'_0$, from both the rear end at $x'_0 = 0$ and the front end at $x'_1 = L$. The ground frame will see the rear ball released at $$ \begin{eqnarray} x_{rear} &=& \gamma(x'_0 + vt'_0) = \gamma vt'_0\\ t_{rear} &=& \gamma(t'_0 + vx'_0/c^2) = \gamma t'_0 \end{eqnarray} $$ but the front ball is seen released at $$ \begin{eqnarray} x_{front} &=& \gamma(x'_1 + vt'_0) = \gamma L + x_{rear}\\ t_{front} &=& \gamma(t'_0 + vx'_1/c^2) = t_{rear} + \gamma vL/c^2 \end{eqnarray} $$ So indeed, the distance between the balls is now $x_{front} - x_{rear} = \gamma L \sim 140$m.

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  • $\begingroup$ Thank you very much for the answer, and the explanation, I had confused the concept of distance with length it seems, and this clears it up significantly, Thank you $\endgroup$ – Tim Ryan Jan 8 '16 at 11:34
  • $\begingroup$ @TimRyan Welcome. $\endgroup$ – udrv Jan 8 '16 at 12:16
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Assume the balls are infinitesimally close to the ground when they are dropped so that we don't have to account for any horizontal motion while they are dropping. In the frame of the train the balls are dropped at the space time points $(x_1,ct_1)=(0,0)$ and $(x_2,ct_2)=(100,0)$. The ground frame observer sees the coordinates of these points as: $$ x'_1=cosh(\lambda)x_1+sinh(\lambda)ct_1=0 $$ $$ ct'_1=sinh(\lambda)x_1+cosh(\lambda)ct_1=0 $$ $$ x'_2=cosh(\lambda)x_2+sinh(\lambda)ct_2=cosh(\lambda)100 $$ $$ ct'_2=sinh(\lambda)x_2+cosh(\lambda)ct_2=sinh(\lambda)100 $$ where $\lambda$ is the Lorentz Boost parameter: $$ tanh(\lambda)=\beta=\frac{v}{c}=.7 $$ $$ cosh(\lambda)=\gamma=1.4003 $$ $$ sinh(\lambda)=\beta\gamma=.9802 $$ Therefore, the ground observer sees the balls dropped at different times. The first ball is dropped (and hits the infinitesimally close ground) at $t'_1=0$. The second ball is dropped later at $t'_2=\frac{98.02}{c}$ seconds. The first ball hits the ground at $x'_1=0$. The second ball hits the ground at $x'_2=140.03$ meters. This is not the Lorentz contracted length of $\frac{100}{\gamma}=71.41$ meters which is what the ground observer sees as the difference between the train ends at the same time in the ground frame.

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As soon as the balls join the ground they are in the ground's reference frame, so the Lorentz factor with respect to the ground frame is 1. Therefore, after taking into account the difference in drop timing viewed by the ground frame, the apparent distance must equal the proper distance, so the balls will appear $(100+98)=198\text{m}$ apart (I got about $3.3\cdot10^{-7}\text{s}$ as my time difference).

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udrv's answer is exactly right. Here's how to see it without calculation:

enter image description here

The red and blue broken lines are the worldlines of the two balls. Assuming they're dropped at $A$ and $B$ (simultaneously and 71 meters apart according to the observer on the ground), the second ball follows the solid blue line. The balls hit the ground at $A'$ and $B'$ --- still 71 meters apart according to the observer on the ground.

If the balls are dropped at $A$ and $C$ (simultaneously according to the observer on the train and over 100 meters apart according to the observer on the ground), the second ball follows the dashed blue line. The balls hit the ground at $A'$ and $C$' --- considerably more than 71 meters apart according to the observer on the ground.

The key observation is that the proper time between being released and hitting the ground has to be the same for both balls. That's why the segment connecting their final positions has to be parallel to the segment connecting the events where they're dropped.

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