# The relativistic solution of the ‘bug on band’ problem

• B
The problem goes:
‘One end of a rubber band is attached to a wall. The free end is stretched away from the wall at a rate v. At time zero the band is length L0 and a bug starts crawling along, from the wall, at rate u. How long until the bug reaches the free end?’
(Typically u << v for dramatic effect.)

The classical solution is:
## t = (L_0/v)(e^{v/u}-1) ##

But if we find the time in the wall’s frame of reference with relativistic accuaracy, it gives:
##t = (L_0/v)(\frac{(1+v/c)^{0.5(c/u-1)}}{(1-v/c)^{0.5(c/u+1)}}-1) ##

The exact form is strange but interesting. I’m just posting this in case anyone has any comments on the form of it. It seems vaguely familiar.

## Answers and Replies

pervect
Staff Emeritus
Science Advisor
Is there a question here? I haven't seen this problem before, which is interesting, but it also means I haven't worked out the details or seen them worked out. The approach that comes to mind is to first find the velocity of a point on the band as a function of time, which should be ##v_{band} = v \, \frac{x}{L+v\,t}## , assuming I haven't made an error. This comes from saying that the velocity at the end of the band as a function of time is v, the velocity in the middle of the band is v/2, and in general the velocity of some point on the band is proportional to the distance. As I think about this, this is sort of an idealized rubber band rather than a physical one.

Then in the non-relativistic case we have non-relativistic velocity addition, so the velocity of the bug as a function of time is the velocity of the band at some time plus u, ##v_{band}+u##. In the relativistic case we need to use relativistic velocity addition. Letting ##u1=v_{band}## and u2 = v, we have ##\frac{u1 + u2}{1-u1\,u2 / c^2}##

I'd have to integrate to find the position of the bug as a function of time, then solve for the position of the bug = position of the end of the band. But I haven't done this.

Since the problem has a name, I'd assume there is something written about it, but a quick search didn't turn up anything useful.

It'd also be interesting to know what the time elapsed for the bug would be.

Since the problem has a name, I'd assume there is something written about it, but a quick search didn't turn up anything useful.

It'd also be interesting to know what the time elapsed for the bug would be.

https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

@pervect no question, I just wanted to share how the classical and relativistic answers compare.

Everything you explained is precisely how I solved it. The last thing to mention is that if we work with the variable r = x/(L0 + vt) (the fraction the bug is along the band) instead of x, then both (relativistic and classical) equations become separable, and integration (from r=0 at t=0 up to r=1 at t=tfinal) gives the results in my OP.

I find it curious how the exponential approximates that complicated term.

I also was interested in the time in the bug’s frame, but that integral was a bit more intimidating.
If I’m not mistaken, we would just integrate ∫√(1-(w/c)^2)dt where w is the bugs speed as a function of time (as seen by the wall), which could be found explicitly from the original solution, but I think its messy.

I haven’t tried it, so maybe this integral simplifies. I would be surprised though.