# Solve Relativistic "Bug on Band" Problem

• B
• Hiero
In summary, the problem involves a rubber band attached to a wall with one end and the other end being stretched away from the wall at a rate v. A bug starts crawling along the band at a rate u and the question is how long it will take for the bug to reach the free end. The classical solution involves finding the velocity of a point on the band as a function of time and using non-relativistic velocity addition to find the bug's velocity. In the relativistic case, relativistic velocity addition is used. The time elapsed for the bug can be found by integrating and solving for the position of the bug. There is a curious approximation in the classical solution and the time in the bug's frame can be found by integrating a

#### Hiero

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.

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.

pervect said:
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.

## 1. What is the "Bug on Band" problem and why is it considered a relativistic problem?

The "Bug on Band" problem is a physics thought experiment that involves a bug crawling along a stretched band. The band is stretched at a speed close to the speed of light, making it a relativistic problem because it involves speeds that are a significant fraction of the speed of light.

## 2. How do you solve the "Bug on Band" problem?

The "Bug on Band" problem can be solved using the equations of special relativity, specifically the Lorentz transformation equations. These equations account for the effects of time dilation and length contraction at high speeds.

## 3. What are the assumptions made when solving the "Bug on Band" problem?

Some of the assumptions made when solving the "Bug on Band" problem include: the band is perfectly straight and flat, the bug moves in a straight line along the band, and there is no friction or external forces acting on the bug.

## 4. How does the speed of the band affect the motion of the bug in the "Bug on Band" problem?

The speed of the band has a significant effect on the motion of the bug in the "Bug on Band" problem. As the band approaches the speed of light, the time dilation and length contraction effects become more pronounced, making it more difficult for the bug to reach the end of the band.

## 5. Can the "Bug on Band" problem be applied to real-world situations?

While the "Bug on Band" problem is a theoretical thought experiment, it can be applied to real-world situations, such as understanding the behavior of particles in particle accelerators or the effects of high-speed travel on space-time. It also provides a useful example for understanding the principles of special relativity.