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Relativistic Car Paradox

  1. Jan 30, 2004 #1
    Here is one that's been bugging me a lot:

    Observer A is zooming along in a very fast car at 0.995c. The "rest length" of the car is about 6 meters. In front of the car is a deep ditch with a "rest length" of 2 meters across. The car has big tires and could drive over the ditch under normal highway speeds.

    Observer B is standing beside the ditch watching the car as it zooms by.

    Due to relativistic length contraction, observer A observes the ditch in front of him to be much less than 1 meter - he should barely even feel the bump.

    But to observer B, the ditch is still 2m, but he observes that the car is much shorter than 2 meters, and will inevitably fall into the ditch.

    What happens to observer A? Does he make it across? Does observer B see him magically float across thin air violating Einstein's postulate: "The laws of physics shall be the same no matter what the frame of reference"?

    Good luck ...
     
  2. jcsd
  3. Jan 30, 2004 #2

    NateTG

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    As typical with these paradoxes, the answer is that synchronicity fails so that the car can actually make it across the ditch.
     
  4. Feb 2, 2004 #3

    turin

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    Not to mention/ask:
    why would you think that a car travelling at 0.995c would fall into a 2 m wide crevace, regardless of how small the tires were. After 2 m, gravity would pull the car downwards by an amount (1/2)(9.8 m/s2)[(2 m)/(0.995c)]2 = 0.0002 pico-meters (which is orders of magnitude smaller than the radius of the H-atom).

    This is NOT a relativistic calculation, and treats gravity as a real force. The bottom line is that the gap would fly by so fast there wouldn't be any time for the car to get pulled into it.
     
    Last edited: Feb 2, 2004
  5. Feb 2, 2004 #4
    Apparently, Quark, you've never seen the Blues Brothers in their '56 Olds (or was it a Chevy) flying across the Hudson River at 120 mph while the draw bridge was open! Nothing magical about it; its called MOMENTUM.

    And a car doing .995c means LOTS of momentum to carry it to the other side of the ditch.

    Creator

    Now if you want a brain bender try figuring out how a car doing .995c rolls on flat wheels.
    Since the 4 foot high wheels shrink only in the horizontal direction, the stationary observer sees the car 'rolling' on vertically flat wheels!!? How do you explain that???
     
  6. Feb 3, 2004 #5

    turin

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    Re: Re: Relativistic Car Paradox

    That is an interesting point, but it isn't quite that simple. If you assume that the wheels are acting normally in the car driver's frame (that is, if you assume that the car driver observes the wheels to have a 0.995c exterior tangential velocity), then, to an observer in the rest frame of the ditch, the bottom of the tire (touching the pavement) is stationary, while the top of the tire is traveling at ~1.000c, and the sides of the tire are traveling somewhere between 0.995c and 1.000c. This causes different parts of the wheel to contract differently. It would look more squished at the top, and wider at the bottom.
     
  7. Feb 5, 2004 #6
    Ok I messed up in the details.

    Forget about the wheels of the car and replace the car with a block on a frictionless table and replace the ditch with a hole in the table.
     
  8. Feb 5, 2004 #7
    Re: Re: Re: Relativistic Car Paradox

    I think that's a good point...I wasn't considering the tangential wheel velocity; the wheel 'flattens' somewhat triangularly; still a paradox how a stationary frame sees it 'rolling'.

    Creator
     
  9. Feb 5, 2004 #8

    turin

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    Re: Re: Re: Re: Relativistic Car Paradox

    I imagine it would look like a really fast bulldozer/tank/snowmobile tread in the shape of a teardrop. It would definitely be strange. Probably never will be observed.
     
  10. Feb 5, 2004 #9

    turin

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    I think you might have missed the point of mine and creator's first posts. It's the same deal.
     
  11. Feb 5, 2004 #10
    I guess I did miss the point. You didn't convince me. Sure it may only drop a small fraction of a pm, but scale the dimensions and it eventually becomes significant. But even that doesn't matter.

    It doesn't matter how little it drops (we're talking about simple ideal objects here - no tires, burms, dirt ramps or draw-bridges, if it drops at all some portion is still going to hit the inside of the ditch/hole. I don't get Creator's first post either:

    Doesn't matter how much momentum you have, gravity is always acting on you in this model and without a resisting force (the table) your going down AS SOON AS the resisting force is gone. It can't fly (I forgot to say that I guess). :smile:
     
  12. Feb 5, 2004 #11
    Re: Re: Relativistic Car Paradox

    Momentum has nothing to do with it. How does momentum "carry it". I hope you guys are joking.

    p=mv normally

    and in this case

    p=ymv (where y is relativity gamma)

    I gave you v. and that's all that counts as Turin showed. It's going to drop just as fast no matter what m is, assuming that the relativistic mass is still a lot less than the earth - otherwise it may even drop faster at relativistic velocities - need to think about that one ...
     
  13. Feb 6, 2004 #12
    This discussion prompts me to ask the related questions
    1) Is the concept of "center of mass" well defined for a uniformly moving object and
    2) If so how would a stationary observer calculate the center of mass of such a uniformly moving object (particularly when aforesaid object is moving close to the speed of light).

    If I had to guess the answer based on this thought experiment (and since I don't have the time this evening to try and formulate this in a nice mathematical way-- nor the relativistic expertise), I would assume the center of mass 'lags behind' the object in motion and can even lag far enough behind so as to be literally behind the object.

    Here's my reasoning (Relativists... be gentle with me):
    If the block is moving fast enough, the hole looks very small... too small in fact for the block's edge and center of mass (as calculated by the block) to both be over the hole at the same time.
    Hence the block will not tip, nor fall into the hole.

    To the outside observer the center of mass of the block is lagging so far behind the block that there is a Wiley Coyote moment when the contracted block is entirely over the hole... but since the center of gravity and the edge are not both over the hole at the same time-- it doesn't fall

    What's more interesting is if the hole is a bit wider, or the block is a bit slower and the block DOES start to tip.... but I leave that analysis to the experts.

    EDIT:
    That also makes me wonder... does the direction of the pull of gravity look different between the block's frame and the bystander's frame? Just to be a bit more precise... if an observer on the block (I guess we should call such an observer a blockhead) erects a pole to show the direction that they feel the pull of gravity, will a pole erected by the stationary observer point in the same direction as the blocks? Will the two poles agree when the block is brought back to the stationary observers inertial frame?
     
    Last edited: Feb 6, 2004
  14. Feb 6, 2004 #13

    turin

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    OK, you're right; this brings up a different issue.

    One issue I think we should confront at this point, though, is whether or not we really want to be talking about gravity. It would probably be better to say that the block is uniformly charged, and that there is an electric field pointing downwards. No, damit, that won't be good either. Hmm, I'll have to think about this (though I probably won't think about it).




    I hadn't even considered this issue, but I'm pretty sure it is "well defined." It may have a surprising definition, though. I'd have to think about that (and I may actually do so).




    I am leaning towards disagreement with this, but it raises an interesting issue. I'm off to think about it.




    You would think so, wouldn't you. But, you are making an assumption that you might not be aware of, but you mention the issue at the end of your post. You are assuming here that gravity is pulling the block striaght down. I don't think that it is.




    I'm pretty sure it does. But this is a GR thing, not an SR thing. There is a lot of confusion/oversimplification about the distinction between SR and GR. GR, to me, seems a gazillion times more bizarre than SR. Don't let people fool you into thinking that GR is just the inclusion of acceleration into SR, because it isn't. SR already allows for acceleration. Even in SR, though, the direction of force generally changes from frame to frame. If we're only talking SR, then I suppose we would have to treat gravitation almost exactly like electromagnetism. In that case, there would be, what some people have called, a gravetomagnetic force (as well as this so called γ factor issue). The gravitational field would be a second rank tensor (that's kind of like a 4x4 matrix). The problem is that mass is not part of a 4-vector. I suppose you could play around with the momentum 4-vector as the source of gravitation. I don't know. I don't think that it is worth playing around with it just to try to resolve this paradox.
     
    Last edited: Feb 6, 2004
  15. Feb 6, 2004 #14
    I hope this won't be too bad... hopefully we can just describe the geodesic in one frame of reference and then pull it back along the Lorentz transformation to describe it in terms of the other (this is NOT my area of expertise so please, anyone correct my errors).

    Last night before dropping off to sleep I think I realized how to calculate the center of mass. How does this sound:
    We let f(x,t) represent the mass distribution of the block in space and time as seen from the observor:
    We call the endpts of the block x0(t) and x1(t)
    Then we use the appropriate integral to calculate the center of mass... something like integral from x0(t) to x1(t) of x*f(x,t) dx divided by integral from x0(t) to x1(t) of f(x,t) dx... or if we've defined our distribution function nicely enough... it could just be integral from -infinity to infinity of x*f(x,t) dx over the integral from -infinity to infinity of f(x) dx.

    Then we use the formulas relating x, x' and t and t' , and a change of variables to recast the integral in terms of x' and t'...
    What disturbs me is the possibility of a dt arising during the change of variables.... but class starts in 5 minutes so I'd better stop musing and run.
     
  16. Feb 6, 2004 #15

    turin

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    It should be more disturbing if no dt arose. This is just subtle evidence that what is simultaneous in one frame is not in another. I still haven't (dove/dived/diven?) into the problem yet. I'll probably get back to it on Monday.
     
  17. Feb 6, 2004 #16
    Great hypothesis curiousbystander. I'm no relativity expert either but you may be on to something. If the object is relativistic I don't think you can simultaneously measure the mass of the object at all points along its length (which you need to do calculate a classical center of mass). Certainly worth some analysis - wish I had more time to try and do it. If I was only a bit smarter, then it wouldn't take so long and I might have the time!
     
  18. Feb 6, 2004 #17
    Observer A is zooming along in a very fast car at 0.995c. The "rest length" of the ca

    Say the occupant of the car/block does notice a small bump. This causes him to accidentally fire his M-16 aimed toward the front of the car. It's a hot round and travels at .010C. The occupant knows he is moving at .995C. He "sees" the bullet now moving toward the front of the car/block at .010. The observer on the ground also makes this observation. Is the bullet traveling at 1.005C?
     
  19. Feb 6, 2004 #18
    Re: Observer A is zooming along in a very fast car at 0.995c. The "rest length" of the ca

    Actally the bullet is going

    w = (u + v)/(1 + uv/c2)

    =0.996c relative to guy on the ground.

    I hope he was wearing safety goggles and hearing protection with that hot a load!
     
  20. Feb 6, 2004 #19
    Good response. Kinda like to ask that one every now and again.

    Thanks.
     
  21. Feb 6, 2004 #20
    I talked to one of my professor's today about the problem. His resolution was the opposite of mine (I think it makes more sense too-- though I didn't quite follow all his reasoning). The center of mass will lead the moving object somewhat... so from the blocks perspective gravity isn't pulling perpendicular to the block anymore... the block will tip and enter the hole at a slant...
    From the bystanders point of view, the block just falls into the hole...

    He also brought up the interesting point that no motion is actually rigid, and information about what's happening at the front of the block can only be transmitted to the back at the speed of light... he also suggested removing gravity and imagining a giant press pushing down so that the block is sandwhiched between the press and the surface... from the bystanders view the press pushes down on the block simultaneously, but from the block point, it starts pushing at the front and works its way back.
     
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