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The problem I am dealing with is the following. We have two spaceships A and B s.t in $S_E$ their respective speeds are $v_A^E >0$ and $v_B^E<0$ along the x axis.

At $t_0$ in $S_E$ their positions along the $x$-axis are $x_A^E=a$ and $x_B^E=b$ whereas $b-a=l>0$

It is quite easy to calculate the time it takes for the two spaceships to collide in $S_E$ simply with $$\Delta t_E = \frac{l}{v_B^E - v_B^E}$$

We can then calculate the time it takes between the moment the spaceship A arrives at $a$ and the collision in $S_A$ (so $S_A$ being the system of the spaceship A) with $$\Delta t_A = \frac{\Delta t_E}{\gamma_A} $$

Where gamma is the Lorentz factor between $S_A$ and $S_E$

This is the right answer to my problem but then I tried to calculate this quantity in another manner and I got a different answer. The distance $l$ in $S_A$ should be (and this is where I think I'm mistaken but I don't know why) $$l_A = \frac{l}{\gamma_A}$$ and the relative speed of B in A $$v_B^A = \frac{v_A^E - v_B^E}{1 - \frac{v_A^E \cdot v_B^E}{c^2}}$$

Therefore $$ \Delta t_A = \frac{l_A}{v_B^A} = \frac{l}{v_B^A \cdot \gamma_A } = \frac{\Delta t_E}{\gamma_A} \frac{v_A^E}{v_B^A} \neq \frac{\Delta t_E}{\gamma_A} = \Delta t_A $$

I think the mistake lies in the usage of $l_A = \frac{l}{\gamma_A}$ because when thinking about it I wasn't totally sure that $l$ was an proper length. But then again it seems right because $S_E$ is not moving in relation to $l$ so it should be a proper length.

Does any of you know where the error is ? Can you help me out please ?

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  • $\begingroup$ I am wondering whether $l_A=l/\gamma_A$ is the distance spaceshift B has to travel to make collision. I think even in classical newtonian mechanics that would not be correct as the distance B has to travel is only a fraction of the distance $l$ hence the ratio $v_A/v_B$. $\endgroup$ – Ronan Tarik Drevon Aug 14 '17 at 13:46
  • $\begingroup$ Hmm from what I understand (which is arguably not a lot) this is supposed to be taken into account by the change speed between $v_B^A$ and $v_B^E$ because in $S_A$ the vessel is supposed to be static. $\endgroup$ – Paultje Aug 14 '17 at 13:52
  • $\begingroup$ You are correct about where you are mistaken. As for why this is wrong, the answer is that there is no reason it should be right. If you think there is some reason it should be right, try writing out a proof of it from first principles, and the mistake will jump out at you. (Along the way, you''ll be forced to stop and ask yourself what you mean by "the distance $l_A$ in $S_A$." What does this mean exactly? Is it the spatial distance between two events? Or between the two spaceships at some time? If so, what time?.....etc.) $\endgroup$ – WillO Aug 26 '17 at 22:26
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I re thought about your problem and I think that the problem is actually the way you envision solving it. Maybe the problem comes from clock synchronisation :

The distance $L_E = L$ as measured in E separating A and B at $t_E=0$ can not be $L_A=L/\gamma$ as measured in A at $t_A=0$ where I am considering $t_A=0$ corresponding to the event "A and B are separated by $L_E$ in E".

Basically what I am saying is that :

The events "A and B are separated by $L_E$ as measured in E" and "A and B are separated by $L_A = L/\gamma$ as measured in A" are not simultaneous events .

Try to represent your problem on a Minkowski diagram and take the origin of time of A and E synchronised for the event "A and B are separated by L as measured in E". You will see that while the distance between A and B is $L_E$ in the frame of E at $t_E=0$, the spaceshift B is already closer than $L_A=L/\gamma$ at $t_A=0$

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  • $\begingroup$ That's a bit hard to explain without showing the Minkowski diagram but that takes a bit of time to actually draw the diagram. $\endgroup$ – Ronan Tarik Drevon Aug 26 '17 at 20:42

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