# Time to collision for relativistic objects

I was going through some old college materials, came upon a physics final that I had evidently saved, and thought I'd take a look at the questions again for old times' sake. One of them in particular has been troubling me, because I seem to get inconsistent answers depending on approach.

The setup, for our purposes here, are as follows: Two spaceships are on a collision course, directly approaching one another at velocities of 0.80c and -0.60c as measured in an Earth frame of reference. (Hence, gammas of 5/3 and 5/4 respectively)

Their initial separation, also in Earth frame, is $2.52\times10^{12}$ m, or 8400 light-seconds (hereafter just using light-seconds to measure position/distance).

We're told that each ship can evacuate within 90 minutes by their own clock, and want to know if there will be any casualties - that is, what is the time to collision from each ship's perspective?

My first thought, today, was just to rely on the Lorentz transformation. The starting coordinates $(x,t)$ of each ship in the Earth frame are plainly $(0,0)$ and $(8400,0)$.

Simple math says that the the eventual collision should take place 6000 seconds later and 8/14ths of the way through, so the collision coordinates would be $(4800,6000)$.

The time coordinate of the first ship's starting position in its own reference frame is trivially 0; the time coordinate of the collision in the first ship's reference frame is 5/3 $(6000-4800\times0.8)=3600$ seconds, so their time to collision would be an hour.

The time coordinate of the second ship's starting position in its own reference frame is $5/4 (0-8400\times(-0.6))=6300$, and of the collision is $5/4 (6000-4800(-0.6))=11100$, so their time to collision would be $11100-6300=4800$ seconds, an hour and twenty minutes. So far, so good (although apparently everybody dies).

However, I then considered an alternate method - what if I simply look at the distance between the ships in each reference frame, divided by the speed each would see the other as traveling? That combined velocity works out to approximately 0.946c, and using length contraction, the first ship sees the distance to be traversed as 5040 light-seconds, while the second sees it as 6720 light-seconds.

But this means that the first ship would have a time to collision of $5040/.946=5327.7$ seconds, almost an hour and a half, while the second ship would have a time of $6720/.946=7103.6$ seconds, nearly two hours and substantially longer than the collision time in the Earth frame, which is not at all what I'd intuitively expect.

I was pretty well convinced that the first approach was correct, and simply wondering why the second one didn't work. But as it happens, I was able to locate the final's answers on the web, and it actually showed the results from my second approach as being correct. Which is surprising, but I'm not terribly inclined to argue with it.

So I'm wondering, why does using the Lorentz transformation not yield the correct time in this case? Or, on the off chance my old professor actually was wrong, why does the other method fail? Why the disagreement?

Your first method is the correct one, as long as the starting time is measured from the simultaneous moment at $t=0$ in the Earth's frame. As you probably know, it's not very natural to talk about simultaneous moments for spatially separated events, and this is where part of the subtlety in this question lies.

By the way, your first method can be found more simply by just dividing the Earth frame time $6000s$ by the time dilation factor gamma. $$\frac{3}{5}6000s=3600s\qquad \frac{4}{5}6000s=4800s$$

Now let's go to one of the ships' frames, say the one moving at .8c. The time $t'=0$ is when that ship starts its 90 minute timer, but the other ship started its timer at some earlier $t'$!

The event at which the other ship starts its clock is at $$t'=\gamma(t-vx)=\frac{5}{3}(0-.8\times 8400 s)=-11200 s$$ $$x'=\frac{5}{3}(8400s-.8\times 0)=14000s$$ So at $t'=0$ the other ship is at $14000s-.946\times11200s\approx3405s$, which does not agree with your Lorentz contracted value. Continuing on from this distance, you will find it is $3600s$ until collision again, as in your first method.

Why length contraction failed

In order to use the length contraction idea, you need to be transforming a 'proper length.' What that means is that the endpoints of the length are both at rest in the original frame. In your case the endpoints of the distance were the two ships, which were not only not at rest, they were at different velocities.

But just to show where length contraction would come in, let's consider the proper length with endpoints given by the worldlines $x=0, \,x=8400s$. We found the transformed coordinates of the point $x=8400s$ at $t=0$ above. Now in the transformed frame the $x=0$ worldline is moving with velocity $-.8c$, so at $t'=-11200s$ it is at $x'=8960s$, and $14000s-8960s=5040s$ which was the contracted length you found in the second method. You can see it has nothing to do with the distance between the two ships.

• Fascinating. I'll have to think about that some more, but thank you very much! It's good to know that my intuitions were correct after all, about the answer. – Strudel May 9 '18 at 8:59