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Length Contractions

  1. Oct 25, 2009 #1
    I think I finally have a decent understanding of time dilation and simultaneity after researching them and watching countless videos about them. But I'm still confused about why length contractions happen. It seems that they can be explained by simultaneity, but I still don't see how. Can anyone explain this to me?

    Also, lets say that I have a spaceship traveling to the sun at a speed close to the speed of light relative to the earth. As the spaceship passes the earth, the people in the ship see the earth as half its size at rest, and the people on earth see the ship at half its size at rest. How do both frames of reference view the distance left to travel between the ship and the sun? Is it also halved for the people in the ship? Do they get there faster because of it?
     
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  3. Oct 25, 2009 #2

    diazona

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    Yes and yes. The people on the ship see the Earth-Sun distance as 0.5 AU (because that distance is a measurement made in the Earth's reference frame, like the size of the Earth, it gets contracted just like the size of the Earth) and they get there in half the time it would take them to travel 1 AU - according to their clocks, that is. Those of us on Earth would explain the discrepancy in travel time by time dilation.
     
  4. Oct 25, 2009 #3
    Thanks. As for the length contraction explanation, I believe that the answer is that length is defined by the distance between two simultaneous events occurring at the front and back of the object. So if a train moving at relativistic speeds was passing an observer just when a lightning bolt strikes the front and back of the train, that would be the length with respect to the observers frame, which would be smaller than the real length. And if the events were simultaneous to the passenger in the train, the lightning would strike the back first and then the front with respect to the observer, and that would be the length at rest, or the length that the passenger measures? I'm really not sure, so correct me if I'm wrong.
     
  5. Oct 25, 2009 #4

    JesseM

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    Yes, that's right (as long as the strikes were simultaneous in the observer's frame rather than the train's frame, but I assume that's what you meant). However, note that different frames will not only disagree about the length of the train, but they'll also disagree about the spatial distance between those two events of the lightning strikes (and unlike with length, other frames will say the distance between these events is greater than it is in the frame where the two events were simultaneous--in fact it'll be greater by exactly the same factor that appears in the time dilation equation, an equation which tells us that the time between a given pair of events that occur at the same position but different times in one frame will be greater in other frames moving relative to the first).
    Neither of those. The events would happen at different times in the observer's frame, so the distance between them couldn't count as the length of the train in this frame since the train will have moved between the time of each strik. The distance between these two events in the train's frame would be the train's rest length, but as I said above the distance between these two events would be larger in the observer's frame, not smaller like the train's length.
     
  6. Oct 26, 2009 #5
    I'm sorry, but I'm still a little confused. What is the difference between length and spatial difference? And why would it be greater?
     
  7. Oct 28, 2009 #6
    Lol, forget that question, I'd rather not confuse myself at this point...

    Is there a way to explain the length contractions through time dilation? This is what I'm thinking, but I don't know if its right...

    A rocket is passing the earth at a speed close to the speed of light. Measurements of time are taken as soon as both ends of the rocket pass a person on earth. The person on earth should measure the rocket's length as smaller than its proper length.

    d is the proper length, d' is the contracted length, T' is the dilated time which is the rocket in this case, T is the normal time which is the earth's perspective in this case.

    d=vT' is the equation in the rocket's reference frame.
    d'=vT=vT'y where y is the lorentz factor. This is the length in the earth's frame of reference.

    So from the earth's perspective, the rocket should be smaller than d, but vT'y > vT', so obviously I'm doing something wrong...
     
  8. Oct 28, 2009 #7

    JesseM

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    I know you said you wanted to skip this question, but the basic answer is pretty simple: "length" in a given frame is always defined in terms of the distance between the endpoints of an object at a single time (i.e. if the back end is at position x=10 meters at time t=3 seconds in whatever frame you're using, and the front end is at position x=15 meters at the same time of t=3 seconds, that means the length is 15-10=5 meters in this frame), whereas when talking about the distance between an arbitrary pair of events, the events may have happened at different times in whatever frame you're using (for example, if event #1 happened at x=10 meters, t=6 seconds and event #2 happened at x=18 meters, t=7 seconds, then the distance in this frame between the two events is 18 meters, but this is not a length because the two events happened at different times).
     
  9. Oct 28, 2009 #8

    JesseM

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    But in the rocket's frame, the Earth clock will be running slow, so it will elapse less time than the amount of time that elapses in the rocket's frame between the moment the Earth passes the front and the moment it passes the back. This may seem confusing because time "dilation" makes you think it should be more, but the "dilation" refers to the fact that in the rocket's frame the moving Earth clock takes longer to tick forward by a given amount, i.e. if T' is the time in the rocket frame for the Earth clock to tick foward by T, then T' > T. If y is the Lorentz factor (usually written as the greek letter gamma), then T' = yT.
    As I said above, T' = yT which means T = T'/y, so that should be d' = vT = vT'/y. By the way, it's a bit confusing to have the primed d' represent the length of the rocket in the Earth frame, while the primed T' represents the time in the rocket frame...usually the convention is to use primed for measurements in one frame and unprimed for measurements in the other.

    But using your notation (and including my correction about dividing by the Lorentz factor rather than multiplying by it), your derivation does show that the contracted length d' must be equal to vT'/y, and since we know the rest length was d=vT' this shows the contracted length is related to the rest length by d' = d/y. I did notice one unstated assumption in your derivation though: you assume that if the speed of the rocket in the Earth frame is v then the speed of the Earth in the rocket frame is also v (true in relativity and also in classical Newtonian physics, but we could come up with coordinate transformations where it wasn't)
     
  10. Oct 30, 2009 #9
    Alright, let me redo this proof clearly to make sure I got it.

    v = speed of the rocket relative to the earth
    d = length of the rocket measured from the rocket's frame
    T = time it takes for the rocket to completely pass the observer in the rocket's frame
    d' = length of the rocket from the observer's frame of reference
    T' = time it takes for the rocket to completely pass the observer in the observer's frame of reference

    From the rocket's perspective, since the earth's time is dilated, we have T=T'y (I usually make a mistake concerning this so please check that). The rocket measures the distance traveled as d=vT, so this is the length of the rocket in its own reference frame. But the observer on the earth measures d'=vT' from his reference frame. We can rewrite this as d'=vT/y. If we divide d' by d, we get, d'/d=(vT/y)/(vT)=1/y and solving for d', we have d'=d/y. Since y is greater than one d' < d, and therefore, the length of the rocket in the observer's frame is smaller than the length of the rocket in the rocket's frame.

    Is this correct?
     
  11. Oct 31, 2009 #10
    Was that proof correct or not? Someone's been asking me and I don't want to show it to them unless it's right.
     
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