Special relativity problems intended to be paradoxes usually involve some careful thinking.   First of all, for special relativity to apply, the two postulates proposed by Einstein have to be true:

 

1)     The bodies involved must have uniform velocity.   No acceleration is permitted.

2)     The speed of light is the same for all observers in all frames.

 

The first postulate is violated in all twin paradox problems because there has to be acceleration. 

There are several great answers to relativity paradoxes in our archives

(http://www.madsci.org/MS_search.html keywords: relativity paradox):

• Twin Paradox (explained by Ken Wharton) http://www.madsci.org/posts/archives/jan2001/980822743.Ph.r.html
• Pole – Barn Paradox (also Ken Wharton), which is similar to your question http://www.madsci.org/posts/archives/jun2001/991868698.Ph.r.html

 

The first postulate is also violated if the system is perfectly rigid and the u-shaped receiver is immoveable.  If the system is rigid then the detonating arm will instantaneously stop moving (decelerate), at which point it will appear to grow in length and hit the switch provided x greater than or equal to y in length.  If we allow a collision and some motion, then we can use relativistic momentum to solve the post collision events, but that is another problem.  In other words, let’s not discuss the collision case because at the instant of collision it violates the first postulate.   Relativistic collisions are fun problems, but one assumes instantaneous accelerations.

 

Also remember that chronology does not have to agree between the two reference frames.  One popular pedagogical example is sending signals from earth to a rocket ship making a 10-year trip (5 out and 5 back).   If the earth sends a signal every year, the rocket ship receives the signals evenly spaced in time but less a year apart (frequency shift), and receives more signals on the return flight than the outbound flight.

 

 Let’s look at the chronology from the two frames and in two cases. We will use the “moving arm reference frame” and the “at rest u-shaped receiver reference frame”, and the case of non-rigid objects and rigid objects.  The only really trouble cases are when the moving arm length is less then the length from the open end (“cap”) to the switch of the u-shaped receiver.

 

 

From the moving arm frame the u-shaped receiver is shorter:

 

y^’ = y_rest/SQRT(1 – (v/c) ²)

 

where y_rest is the depth of the u- shaped receiver measured from the rest frame of the u-shaped receiver, v is the speed of the moving arm, c is the speed of light, and y^’ is apparent length measured from the moving arm frame.

 

Depending on where you are observing this from on the moving arm the information that the switch has been struck will take time to propagate to you. 

 

The fastest this can propagate is the speed of light, c which means

 

time’_know-hit = x/c

 

where x is the length of the moving arm.

 

If things are not rigid (the arm just plows through or collapses) and everything keeps moving forward, the cap end of the moving arm doesn’t know about hitting the switch until time’_know-hit later, and during that time the moving has kept moving and is closer by:

 

D^’_{moved-until-received-signal} = v * time^’_know-hit = x * v /c. 

 

And depending on the lengths of x and y this occurs either before or after the cap of the moving rod should hit the top of u-shaped receiver.

Solve this by comparing D^’_{moved- until-received-signal} to x - y^’.

 

So if x > y’ and D’ > x - y^’ and the order of events is:

 

1)     Hit the switch with moving arm.

2)     Know of the explosion detonation at the back end of the moving arm.

3)     Hit the open end of the u-shaped receiver.

 

If x > y’ and D’ < x - y^’ in moving frame the order of events is:

1)     Hit the switch with the moving arm.

2)     Slam into the end of the u-shaped receiver

3)     Know of the explosion detonation at the back end of the moving arm.

 

In the case that D’ = x – y’ events 2 and 3 are simultaneous in the moving frame.

 

If x < y’ and then in the moving frame the order of events is:

1)     Slam into the end of the u-shaped receiver

2a) Perfectly rigid: slam!! nothing more happens.   The moving arm is now at rest with the u-shaped receiver and y > y’ so x < y.  No contact with the switch occurs.  END.

 

- or -

 

      2b) Non-rigid: The arm plows into the u-shaped receiver, something has to give.

      3)  Hit the switch with the moving arm

4)     Know of the explosion at the back end of the moving arm.

 

In this case we don’t care about the propagation time because it doesn’t influence the order of these events.

 

 

From the u-shaped receiver point of view, the observed length of the moving arm, x’ is shorter:

 

x^’ = x_rest/SQRT(1 – (v/c) ²).

 

Again the location of observation is important.   The events could have a different chronology from the switch end or the open end of the u-shaped receiver, but typically we would think of this frame of reference as centered on a spot out of the plane of motion, centered on the u-shaped receiver.  In this way, the observer is equidistance from both ends of the u-shaped receiver.  Let’s assume this is the case.

 

 

If x’ > y then the order of event is

1)     Moving Arm hits switch.

2)     Explosion occurs.

3)     End of moving arm hits end of u-shaped receiver.

 

If x’ < y then something has to give and it depends on if the system is rigid.

1)     The cap of the moving arm hits the cap of the u-shaped receiver.

2a) Perfectly rigid – the arm instantly grows longer and if x >= y then the explosion occurs or if x < y no explosion happens. END.

 

- or –

 

      2b) Non-rigid – the arm continues to move slamming into the u-shaped receiver, something gives.

      3) The moving arm plows into the u-shaped receiver until it hits the switch.

4) Explosion occurs.

 

 

 

 

I hope this helps with you understand this problem.

 

Sincerely,

 

Tom “Deadbeat Relative” Cull