Discussion in 'Physics & Math' started by DaleSpam, Mar 4, 2006.

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## According to SR, what is the final result of the bug-rivet scenario?

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1. ### DaleSpamTANSTAAFLRegistered Senior Member

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The bug-rivet paradox is an interesting SR problem I found on the HyperPhysics site. At least it is something to discuss besides the twin paradox (you're welcome CANGAS).

Anyway, I thought this might be an interesting discussion. Is this a logical inconsistency in SR? If not then what did HyperPhysics do wrong in their analysis? Does the bug get squished or not? Do the frames agree?

-Dale

3. ### ZephyrHumans are ONERegistered Senior Member

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I vote that the bug dies. A rivet travelling at near light speed has so much kinetic energy that whether the bug is squashed or not, it surely won't survive the heat caused by impact

Or maybe ... (hidden in case of spoiling other people's fun) the rivet isn't the perfectly rigid object they assume?

5. ### DaleSpamTANSTAAFLRegistered Senior Member

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Well, the hyperphysics site wasn't specific, but we can specify a really tough little bug. No heat nor shock wave will hurt the bug. The only way he can die in this scenario is by actually getting hit by the rivet.

-Dale

7. ### przyksquishyValued Senior Member

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Seem to be getting, in both frames, that the rivet squishes the bug then sort of springs back out of the hole when its finished.

Think the next paradox could have a happier ending?

8. ### DaleSpamTANSTAAFLRegistered Senior Member

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How do you figure that?

My wife would emphatically assure you that any ending where the bug gets squished is a happy ending

-Dale

9. ### przyksquishyValued Senior Member

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Not actually sure what happens to the rivet after it kills the bug (I haven't really thought about it carefully yet - besides that's not the paradox), but the bug's definitely a goner. Its easy to see why in the rivet's frame. In the bug's frame, though, not all parts of the rivet stop when the rivet head hits the wall (relativity of simultaneity), and in the time difference the rivet end easily travels far enough to squish the bug (I worked out that if there were no bug an the hole were deep enough, the rivet would stretch to about 1.8cm, both the 'long' way and using the space-time interval formula).

Schrodinger's Cat, DaleSpam's Bug, what next?

Last edited: Mar 5, 2006
10. ### PeteIt's not rocket surgeryRegistered Senior Member

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Is the rivet stretched or compressed?
Say the rivet is very strong in compression, but quite weak in tension... does the rivet break?

11. ### catoless hate, more scienceRegistered Senior Member

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przyk, nice job, could you show that math? I am lazy =]

12. ### przyksquishyValued Senior Member

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I removed the bug and the end of the hole temporarily, and considered what would happen if the only thing that could stop the rivet was the wall coming into contact with the rivet head (The rivet head and the rivet end both come into contact with the wall and the end of the hole respectively, but the wall and rivet will disagree on which happened first).

Assuming that the entire rivet stops at the same time in the rivet frame, we have 2 events:
2: The rivet end stops.
(By stop, I mean that the rivet starts travelling at -0.9c in the frame originally attached to the rivet)

In the rivet frame, the two events occur a distance 8 mm and time 0 s apart.

Method #1:

- The length of the rivet contracts to 8 mm * sqr(1 - 0.9^2) = 3.49 mm in the wall's frame.
- The front of the rivet lags 0.00349 m * 0.9 * c / (c^2 - (0.9c)^2) = 5.51 * 10^-11 s back in time relative to the rivet head.
- During this time it will travel an additional 5.51*10^-11s * 0.9c = 0.0149 m = 14.9 mm
- You add this to the original contracted length of the rivet to get a total length of 18.4 mm.

Method #2:

- The space-time interval S squared is -(0.008 m)^2 = -0.000064 m^2 (since T=0 in the rivet frame)
- L'^2 = c^2 * T'^2 - S^2 (rearrangement of the formula S^2 = T'^2 - c^2 * T'^2, with L' and T' being the space and time intervals in the wall's reference frame)
- L'^2 = c^2 * (5.51 * 10^-11 s)^2 + 0.000064 m^2 (Using the T' calculated in Method 1)
- L' = 0.0184 m = 18.4 mm.

In both cases you get 1.8 cm, which is longer than the 1 cm depth of the hole. Bug is squashed.

That's one way of solving the paradox without consuming too much aspirine, though I'm gonna have a go at solving it more rigourously some time (I think I'm beginning to see a better way of approaching the whole thing).

By the way, is there an equation editor on sciforums I can use?

13. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Just to add some fancy language to the discussion, the fact that the end of the rivet reaches the bug before a signal from the outer wall can tell it it to stop means that the events "rivet hits outer wall" and "rivet hits bug" are spacelike separated i.e. not causally connected. Only when events are spacelike separated can their order in time be relative.

Let me call d the depth of the hole in its rest frame and L the relevant length of the rivet in its rest frame. In the frame where the rivet is moving toward the bug with speed v, the end of the rivet can't possibly be influenced by the rivet head hitting the outer wall until the end has traveled a certain distance x into the hole. This distance x is determined by the following equations. The first equation is x - L/gamma = v t since the end of the rivet, which begins at L/gamma at t = 0, must reach x in time t. The second equation is x = c t since a light signal from the from the outer wall must also reach x in time t. The solution to this equation is x (1 - v/c) = L/gamma or x = sqrt{(1+v/c)/(1-v/c)} L. With L = .8 cm and v = .9 c you get x ~ 3.5 cm, but this is much bigger than d = 1 cm so the end of the rivet couldn't possibly know to stop until after it hits the poor bug. You can also use this equation to figure out what v and L have to be so that the bug does have a chance of surviving (say by attaching light activated rockets to the end of the rivet or something equally silly).

In practice the rivet would be torn apart when its hits the outer as its end continues to hurtle towards the bug. The paradox is resolved if you just wait a bit after the rivet head hits the outer wall in the rest frame of the bug, the bug will shortly be crushed.

14. ### DaleSpamTANSTAAFLRegistered Senior Member

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Unfortunately, no. It is a real pain for this forum.

-Dale

15. ### DaleSpamTANSTAAFLRegistered Senior Member

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It is true that the events are not causally connected, however doesn't the order of their occurence have other physical effects? If the bug gets hit first then the rivet experiences a wave of compression that travels back to the head. If the outer wall is hit first then the rivet gets stretched out, a wave of de-compression traveling forward to the tip.

-Dale

16. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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Assuming the rivet is able to stay intact then all observers agree that it experiences both disturbances: there will be both a compression originating from the end of the rivet and a de-compression originating from the rivet head. However, these events are spacelike separated so observers will disagree about which mechanical disturbance begins first.

17. ### CANGASRegistered Senior Member

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My best ole friend Schroedinger would, at first glance, give the bug a fifty-fifty chance. I will, as soon as practical, carefully study this thing and also consult as many of my roommates as are not stomped yet and try to prove Revatitly is wrong somehow.

You are sure there is not a bug on the rivet?

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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Ah!
This is what I was getting at in my last post - "Say the rivet is very strong in compression, but quite weak in tension... does the rivet break?"

In the bug frame, the rivet head catches and stops while the rivet end keep going. The rivet stretches and is torn apart by its own momentum, as PhysicsMonkey said.

But in the rivet frame, the rivet end hits the bottom of the hole and stops while the rivet head keeps coming. The rivet is compressed, not stretched, and the rivet doesn't break!

What's going on?

19. ### Hurricane AngelI am the MetatronRegistered Senior Member

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Sorry I'm quite new to relativity, and I have related questions;

1) Why is it not possible to look at it from a third perspective? For example see the cross section of the bug, and incoming rivet?

2) If you CAN look at it from that neutral perspective (which I did, and came to the conclusion that the bug will be crushed) you would see the rivet as being super-extended, and therefore be able to crush the bug.. right?

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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You can look at it from any perspective you like... the point is that you have to get the same result from all perspectives, otherwise the theory you're using clearly does not reflect reality.

Last edited: Mar 5, 2006
21. ### PeteIt's not rocket surgeryRegistered Senior Member

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Yep, that was my first take... but I got stuck thinking about the details in the rivet frame. Now unstuck again.

22. ### DaleSpamTANSTAAFLRegistered Senior Member

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How embarassing

, I actually forgot that both would occur and was somehow thinking that only the one that started first would occur.

-Dale

23. ### CANGASRegistered Senior Member

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What is the tensile strength of the rivet?

And, we must keep aware that since the rivet has been moving rapidly, its clock has been slow, according to Revatidly, so it has not been weakened as much by corrosion, cosmic ray strikes, metal fatigue from sub-space turbulance ( see Star Trek for proof ), and other factors which have weakened its stationary twin piece of material, ( the bug cave ).