Seeking a Resolution to the Bug-Rivet Paradox

yxgao
Messages
122
Reaction score
0
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!
 
Physics news on Phys.org
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.
 
Doc Al said:
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.
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
 
Andrew Mason said:
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

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.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top