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Car & Garage Paradox

  1. Jan 7, 2009 #1
    A car (original length 20 feet) is traveling towards a garage of length 10 feet at speed of 0.8c . Now at such high speed length contraction occurs & contraction factor is 2 , so from garage's frame of reference the car has shrunk to 10 feet & it should fit in the garage where as from the car's frame of reference the garage is moving towards it at speed of 0.8c & it has shrunk to 5 feet and the car does not fit in the garage . Whats the solution to this paradox ?
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  3. Jan 7, 2009 #2


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    It is another question of "simutaneity". From the garage frame of reference (I just can't bring myself to say "from the garage's point of view"!) the rear of the car is inside the door while the front of the car has not yet hit the rear wall of the garage. From the car frame of reference, the front of the car hits the rear wall of the garage before the rear of the care inside the garage.
  4. Jan 7, 2009 #3
    The man in the car will know that the approaching garage is moving at 0.8c if he has some sort of radar in his car. He can then use the Lorentz transformation to predict the parking configuration (with half of his car sticking outside the garage).

    Similarly, the man in the garage will know the approaching car is moving at 0.8c if he has the same radar device. He can use the Lorentz transformation, too, and predict the same result.

    Both men agree that, when parked, the car is going to be sticking out.
    Both men will agree to disagree about lengths of objects while in different reference frames, because they both took upper level physics courses and had this habit beaten into them by their furious professors.

    EDIT: HallsofIvy is too quick for me, and perhaps grasps the source of the seeming paradox better than I.
  5. Jan 7, 2009 #4


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    This conundrum is usually known as the "barn and pole paradox". Do a Google search on it and you'll find lots of discussion about it!
  6. Jan 7, 2009 #5
    Different observers "see" different things. There are other aspects to this which 'barn and pole paradox" discussions will bring forth. Netwon would not have "allowed" this!!!

    Wikipedia does a decent job on this one at
  7. Jan 7, 2009 #6
    Two people looking at the front and back of a coin see different things, but are looking at the same coin. It's all a matter of perspective. Alas, Newton would be bothered by this.
  8. Jan 7, 2009 #7


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    These two diagrams are accurately plotted. The pole is 88 units long and the barn 80 units when at rest wrt each other. The pole is rushed into the barn and touches a switch at the back which sends a ray of light to the front of the barn to close the door.
    The observer in the barn sees the pole half-in and half-out when the front door closes, but the observer with the pole sees the whole pole through the garage. I trust this diagram, and hope I haven't made a mistake.

    Attached Files:

    Last edited: Jan 7, 2009
  9. Jan 8, 2009 #8
    First of all, length contraction for 0.8c is only 0.6, so not quite the factor two you said. But anyway, let's assume a speed of about 0.87c, so the length contraction is slightly over 2.

    As long as the car is moving, the two observers will disagree on the length of the car and the garage, but this will be solved by the fact that they also disagree on simultaneity.

    The observer in the garage will see a 10 ft car go into the 10 ft garage, so that its front fender will just touch the rear wall simultaneously with the rear fender passing the garage door.

    The observer in the car will see his 20 ft car approach the 5 ft garage, and his front fender will touch the rear wall well before his rear end eventually gets into the garage.

    You have to realise there's a bit missing here: what happens after the car hits the wall?

    Solution 1: the car, given its speed, will go straight through the wall. In that case, no problem, both observers agree. One will say the front breaks the wall as the rear is entering the garage, the other says the front broke through the wall while the rear was still outside, but that's not really a big deal. Time and simultaneity are very subjective things, and there's no real contradiction

    Solution 2: The car comes to a complete stop inside the garage somehow (for example with a series of very strong brakes attached to the garage that grip the sides of the car). Still from the garage point of view, all the brakes act simultaneously and will have to continue to exert a significant amount of force after the car is stopped, since it will want to return to its normal length. The stress may even break the car. Now from the car's point of view, the first break grips the front of the car while the back hasn't entered the garage yet. The rear end keeps moving forward as the next brake grips the car, then the next, etc... until finally the last brake grips the rear end of the car just as it enters the garage. The car was contracted because different parts of it were slowed down at different times. Fortunately the garage expanded during the braking, so that the stress on the car turns out to be the same after all.

    Other solutions are possible, but as long as you define the exact transition consistently in one frame of reference, you will find a perfectly consistent view from the other frame of reference. Make sure you take the speed of light into account, though: if, for example, you would say the car stops because it hit the rear wall, you have to take into account the fact that the rear will not immediately know this because information cannot travel faster than light. It will inevitably contract quite a bit more and end up shorter than the garage from any point of view. That's why I used a system of brakes attached to the garage, which can be activated with any timing as required.
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