# What is the resolution of the The Bug-Rivet Paradox paradox in special relativity?

1. Dec 17, 2004

### yxgao

Does anyone know a satisfactory resolution to the The Bug-Rivet Paradox?

It is an interesting paradox in special relativity, that cannot be resolved by resorting to timing and reference frames as in the barn-pole paradox. However, I am stumped as to what the resolution is.

Thanks for any responses!

2. Dec 17, 2004

### Ich

The rivet cannot be rigid. So when the back side stops, the front will continue to move until it notices and squash the bug.

http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html

3. Dec 17, 2004

### Staff: Mentor

If I recall correctly, Taylor and Wheeler have a version of this that they call the "Detonator Paradox" (in Spacetime Physics)--just replace the bug with a bomb detonator. Does it explode or not?

Here's a discussion of the bug-rivet paradox, but for some reason it doesn't seem to give the answer: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/bugrivet.html#c1

Bottom line: The bug dies; the bomb explodes. The "paradox" comes from thinking that the rivet is a rigid body: that stopping the head, instantly stops the tip of the rivet. You can show that, from the bug's frame, the tip hits the bug before any "signal to stop" could possibly reach it.

4. Dec 17, 2004

### Andrew Mason

That is certainly one way to look at it. But would that not apply even if the rivet wasn't long enough to reach the bug in the bug's frame?

Ignoring the physics of a rigid body, one could simply observe that when the rivet head strikes the top or face of the bugs 'well' it is no longer moving relative to the bug's frame so length contraction suddenly ends and so does the bug.

AM

5. Dec 18, 2004

### Janus

Staff Emeritus
The problem with this is that it does not cover the following instance:

Assume the well is 1cm deep as measured in its own frame. and that the rivet is 1/2 cm long as measured in its own frame. Thus while the rivet is at rest with respect to the well, the end of the rivet does not reach the bottom of the well.

Now assume that the rivet is traveling at .866c with respect to the well. According to the bug, the rivet contracts to a length of 1/4 cm, well short of the bottom of the well. Thus in the bug's frame, if the length contraction ends the instant the head of the rivet hits, the end of the rivet stops well short of squashing the bug.

From the rivet's frame it is the depth of the well that contracts to a depth of 1cm, Allowing the rivet to squash the bug. So even if you consider that the length contractions become nor longer in effect the instant the rivet stops, the bug is still squashed by the rivet in the rivet's frame and not squashed in the bug frame.

The correct solution works like this:

In the rivet frame, the end of the rivet reaches the bottom of the well and crushes the bug because of the shortened depth of the well. ( The end hits the bottom of the well before the head reaches the top of the well.)

In the bug frame, the head of the rivet hits the top of the well while the end of the rivet is still 3/4cm from the bottom of the well. But because of the speed of light restrictions, the fastest the information that the riviet head has stopped can reach the end of the rivet is at c. Since we are dealing with the bug's frame, this means c with respect to the bug. The end of the rivet continues to move (at .866c) until it gets this information. The time it takes for this information to catch up with the moving end is equal to:

$$t=\frac{0.25cm}{(c-0.866c)}$$

And the distance the end of the rivet travels in this time is

$$t=0.866c\frac{0.25cm}{(c-0.866c)}$$

This works out to about 1.6 cm.

Thus the end of the rivet reaches the bottom of the well before it ever finds out that the head of the rivet has stopped.