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Reltivity Paradoxes

  1. Mar 21, 2003 #1
    Relativity Paradoxes

    Just pondering some paradoxes in Special Relativity.
    The first one deals with length contraction:
    We have a 4-foot long pole and a wall of glass with five feet marked out in front of it on the ground like so:

    <----- 4 ft. ----->

    <- 5 ft. ->

    The pole is moving towards the wall of glass at such a high rate of speed that from the reference frame of the pole the 5 feet in front of the wall of glass is only 2.5 feet long. From the reference frame of the wall of glass, the pole is only 2 feet long. The pole continues moving at a high rate of speed until the end furthest from the wall of glass reaches the end of the 5 foot line, it then stops. From the reference frame of the pole, the pole smashes through the glass, since the pole is 4 feet long and the 5 feet is now 2.5 feet. From the reference frame of the wall of glass, the 4-foot long pole easily fits into the 5 feet since the pole is only 2 feet long. Both reference frames are correct, so we have a paradox: The wall of glass is both smashed and not smashed.
    Special Relativity can't use the relativity of simultaneity to explain this because in the reference frame of the wall of glass, the pole never touches the glass. So how does SR explain this?

    Here's one concerning the speed of light.
    We have the following situation:

    D<-1 light-hr->C<-1 light-hr-> <-1 light-hr-> C

    The observers at the Bs are at rest with respect to each other and with respect to A. A is a light bulb. The distance from A to B is 1 light-hour. The observers at the Cs are moving away from the light bulb A at half the speed of light. D is another light bulb which is at rest with respect to the observer at C on the left. It is one light-hour away from the C on the left. Just when the observers at the Cs reach the Bs, light bulb A emits a flash of light. The light will reach the Cs when they are two light-hours away from A, but from their reference frames it will take only one hour for the light to reach them since the time merely depends on the distance at the time of emission. When the light reaches the Cs, two hours will have elapsed for the Bs from their reference frames. Because time passes at the same rate from the reference frame of the Bs, and the time for the reference frame of the Cs is moving at half that rate, it follows that time is moving at the same rate from the reference frame of the Cs. When we have the light reach observer C on the left, we have light bulb D emit a flash of light. It will take one hour for the light to reach observer C on the left. Whatever the distance is from light bulb D to observer C on the right at the time of emission, it should take that long to reach observer C on the right from his reference frame. In this case it should take 5 hours, but it ends up taking 10 instead. His clock is moving at the same rate as the left C and he is moving away from the left C, therefore in this case because he is moving relative to the left C and his clock isn't running slow, light will travel more slowly relative to him. Let's work this out to make this more clear:

    flash is emitted from lightbulb A:
    clock for left C: 01:00
    clock for right C: 01:00
    clock for left B: 01:00
    clock for right B: 01:00

    light reaches the Bs:
    clock for left C: 01:30
    clock for right C: 01:30
    clock for left B: 02:00
    clock for right B: 02:00

    light reaches the Cs and flash is emitted from D:
    clock for left C: 02:00
    clock for right C: 02:00 (right C is five light-hours from D and should receive the flash in five hours, but he receives it in 10 hours instead.)
    clock for left B: 03:00
    clock for right B: 03:00

    light reaches the left C:
    clock for left C: 03:00
    clock for right C: 03:00

    light reaches where the right C was at the time of emission from D:
    clock for left C: 08:00
    clock for right C: 08:00

    In order for the light to reach the right C in 5 hours his clock would need to move slower than the left C.

    clock reaches the right C:
    clock for left C: 13:00
    clock for right C: 13:00 (The light relative to the right C is moving at .5C)
    Last edited: Mar 21, 2003
  2. jcsd
  3. Mar 21, 2003 #2


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    Yes it can, because your scenario depends implicity on (mis)use of simultaneity!

    In the frame of reference of the wall, you are asserting that the ends of the pole are simultaneously at the mark (5 foot line) and someplace between the mark and the wall.

    In the frame of reference of the pole, you are asserting that the ends of the pole are simultaneously at the mark and someplace past the wall.

    Both observers will agree on the event in space-time when the one end of the pole passes the mark, but since they're in different rest frames, they'll disagree on the simultaneous location of the other end of the pole.

    But you add the twist about stopping the pole; the problem there is that there is a fatal ambiguity! You fairly explicitly declare when the tail end of the pole stops; it stops when it reaches the mark. However, you are implicitly assuming that the front end of the pole stops simultaneously with the tail end. As mentioned before, observers in different reference frames will disagree on that simultaneity, so you have not well-specified where the front end stops.

    Yes, but rememebr the distance at the time of emission appears different to the C's than it does to the B's. Also, B doesn't think that D is exactly 1 light-hour from C. And to make things worse, relativity of simultaneity rears its ugly head; remember you're specifying that the light bulb flashes simultaneously with the events of each C passing their respective B, but that only happens in B's reference frame, each C thinks that all 3 events occurred at different times!

    And as a calculational detail, the Lorentz factor for half the speed of light is sqrt(4/3) = 1.1547. Each B thinks each C clock is running 86.6% of its ordinary rate, not 50%.

  4. Mar 28, 2003 #3
    If I may put my $.02 in. To understand SR first start with this premise:The speed of light in a vacuum is constant for all frames of reference. That is sort or how Einstein started because Michelson and Morely (spelling on the names is probably wrong) tried to measure the speed of the Aether and no matter which way they turned their maching it read the same interference pattern, which basically means that the speed of light was always the same ( this was the super short version).

    Starting from that premise you get time dilation and length contraction because these things have to happen in order for someone to always measure the speed of light as c.
  5. Mar 31, 2003 #4
    I have a relativity paradox!If relativity says that all objects or observers in motion,never catch up to light,that light is a constant,no matter what speed your going or what reference frame your viewing it from.so what if you started 5 objects off accererating at a constant speed toward the speed of light.light in reference to each 5 observers in motion would be at light speed ahead of them.what happenes when all 5 observers suddenly change each of there accereration in reference to each other.light would change its speed in reference to each 5 observers to stay the speed of light ahead,thus the light that started out with each observers at light speed ahead would then have to change in all 5 frames of reference.so then light would be in 5 different places at the same time.also i had a question.if light is a constant because you can never catch up to it,why if its a particle in motion,can't relativity proof that light actually add your motion to its own instead of your to it,and keep light speed ahead of you,keeping you from catching up?
  6. Mar 31, 2003 #5
    This is a tad backwards. You can never "catch up" to light because it is a constant. The reverse is not true.
  7. Apr 1, 2003 #6
    Chosenone-You need to keep the reference frames straight. Any observer no matter how fast relative to anyone else is moving will always measure the speed of light in a vacuum as c. Now another thing is you cannot include people accelerating because these are not inertial reference frames but accelerated reference frames. SR only deals with inertial reference frames, frames that are not accelerated but have and always will move at the same velocity relative to another frame. GR is what you have to invoke in order to take acclerated reference frames into account. That is how you resolve the most famous "relativity paradox" which I will exhaustively explain below.

    Two indentical twin brothers are born on Earth. One gets in a rocked and flys away at close to the speed of light. He returns one year later (as seen by the clock on the spaceship). However he finds his brother is not a year old like him but 75 years old (the age depends upon how close to the speed of light he was going). Now let me clear up a few things. First of all no matter how fast the in the spaceship goes he will always measure c as the speed of light.

    So what you were worried about before with the whole accelerating

    Here comes the paradox: The baby in the spaceship is put next to the window and looks to see his brother then the earth fly away from him really fast. The baby on earth sees his brother fly away from him really fast. So why is it that the one on the ship is younger (time went slower) if they both seem to be in reference frames that went away from each other?

    The answer:Because the one in the rocket ship actually underwent the acceleration whereas the baby on earth never accelerated anywhere (accelerating in his car and planes and even orbital rockets can be neglected).

    So the situation where everyone is accelerating and then they change their acceleration doesn't apply really. However if they are all moving relative to each other (same velocity) then they all accelerate at different rates then stop accelerating at some new different velocity (relative to each other) so now everyone is traveling at a different velocity you will get different clock rates for everyone.

    Also relativity proves that you cannot add its velocity to yours so it always goes C because then an observer on the ground would measure a change in velocity for light. Relativity means that no matter who measures a the speed of a photon it is moving at velocity c. A guy in a spaceship can measure a photon and it is moving at c, a guy on earth measures the same photon and it is moving at c, a guy on mars etc. Always c no matter what.

    We have to be careful because some of this is dipping into why is it like that and that is better left to someone more equipped for that kind of thinking than I.

    Don't take this wrong but you should be a little more careful about how you throw your terms around. Saying that anything accelerates at a constant speed is a bit of a misnomer. Acceleration is the change in velocity in a specific period of time and in very loose terms is actually the change in speed (speed being the magnitude of the velocity) so using it describe a change in acceleration (or no change in your case) is a little sloppy.
  8. Apr 1, 2003 #7
    PLease ignore that part above I started a train of thought then abandoned it but forgot to erase the beginning. Sorry.
  9. Apr 2, 2003 #8
    so what you are saying frackm is that in the reference frame of the ship moving away from the earth,light is c,and the observer on earth is c.the position of the photon moving c is is exactely the same.but what happens when the ship gets near light speed itself.the time dialation is like 60-80 years per second slower than in the earths reference frame.so the earth sees the photon a c,in a certain spot,the ship sees the photon at the same spot as the earth.then as the ship nears light speed.the ships reference frame is moving slower in time,so either the photon starts moving slower in time in reference to the ship,to stay in the earths and the ships reference frame,from where they stared watching the photon.meaning the photon exists in two time frames at the same time or the photon in the ships reference frame is the same,but in the earths the photon disappears to follow the ship.or the photon in the earths stays the same,but the ships,travel faster than light at light to the ship and they loss track of the photon as they start moving slower in time,and the photon is'nt the right one.
  10. Apr 2, 2003 #9
    Actually I kinda didn't really mention length contraction either. That will make a difference as to where each reference frame measures where the photon is. I have a weird feeling about length contraction, I understand it in a general sense but the whole concept just creeps me out so I don't really talk about it. Its like I can't really accept that part of SR. Don't tell my profs.
  11. Apr 4, 2003 #10
    Three clocks paradox

    How about the three clocks paradox:
    You have three clocks A, B, and C which are all synchronised and at rest relative to one another. You then accelerate B. A and C are at rest relative to each other, with B moving away from them. From the reference frame of A and B, B's clock is slower than A's. From the reference frame of B and C, C's clock is slower than B's. Therefore, C's clock is slower than A's. But wait! C and A are at rest relative to each other, so their clocks are running at the same speed. So we can conclude the following:
    C's clock is slower than A's.
    C's clock is the same speed as A's.

    The only way to resolve this is to make one frame the "preferred" frame, which relativity forbids.
    That's why when two observers are in motion relative to one another and you ask the question "Who's clock runs slower?", relativity can't answer that. The common reply is "The observers will never be able to compare their watches." So relativity is protected from it's own internal inconsistencies by preventing any possible tests which might falsify it.
  12. Apr 4, 2003 #11


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    What reason do you have to think that transitivity holds?

  13. Apr 4, 2003 #12


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    Re: Three clocks paradox

    I'm with Hurkyl, How did you reach this conclusion, in particular?
  14. Apr 5, 2003 #13
    B is at rest, C is moving.

    B is at rest and C is moving, therefore C's clock is slower than B's.
    Remember, there is no "preferred" reference frame.
  15. Apr 5, 2003 #14


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    There are two problems.


    "B is at rest and C is moving, therefore C's clock is slower than B's."

    That's only true in the frame where B is at rest. In particular there is a different frame where C is at rest and B is moving, and in that frame C's clock is the faster one.


    You used the logical argument:

    "In frame 1, A's clock is faster than B's"

    "In frame 2, B's clock is faster than C's"


    "In frame 3, A's clock is faster than C's"

    But there is no reason to think that is a valid deduction. If all three observations were done in the same frame then you could take that step, but there's no reason to think it must be true when the three observations are in different frames.

    Consider this flawed argument as a direct analogy:

    The top of a building is further up than sea level.

    If I stand on my head, sea level is further up than the peak of Mount Everest.

    Therefore, the top of a building is further up than Mount Everest.

  16. Apr 12, 2003 #15
    This is a strange position all theoretical physicists are standing in. In the practical of heights measurements the people have taken the see surface as zero point and get the absolute value of the heights.
    It is possible to do ditto in theoretical physics, but physicists prefer to stand in arcane position and are muddling in frames of reference.
    It is understandable if such zero point is absent quite. But it exists. Its particularity is fit for this function because this a constant. This is "c".
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