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A question about special relativity, Please Help

  1. Jun 7, 2009 #1
    Hello, everyone

    I am learning special relativity in high school.. I understand the time dilation thing...but i have trouble to understand the length contration..

    I am getting confused why

    Lm=ts*V and Ls=tm*V

    Please help..thank you
     
  2. jcsd
  3. Jun 7, 2009 #2

    diazona

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    I'm not sure what you're referring to by Lm, ts, etc. But in general, length contraction is a very similar effect to time dilation, so if you understand one, it should not be hard to understand the other. The difference is that whereas time intervals appear longer when you are moving relative to the clock, length intervals appear shorter when you are moving relative to the yardstick (or whatever length-measuring device there is).

    Maybe you can explain a little more what you're confused about?
     
  4. Jun 7, 2009 #3
    The problem really is this, how do you measure the length of a moving object, like meter stick?
    After you measure one end, the other end has moved, so its position changed.
    The answer is that you must measure both ends at the "same time". That's the key.
    But at the same time, relative to who? Since two events happen at the same time relative to one frame, do not happen at the same time relative to another frame, if there is relative motion between them
     
  5. Jun 7, 2009 #4
    Thanks to Diazona and Feynmann respond to my question

    Ls is the proper length, and Ts is the proper time

    now, I think I understand the formula..

    however, i still dont really get one concept..

    My textbook says,

    when you look at the length of the spaceship as proper length, then the distance between Planet A and Planet B will shrink.

    If you look at the distance between Planet A and Planet B as proper length, then the length of the spaceship will shrink..

    I understand the first idea, and I dont get the second one

    Thank you for helping
     
  6. Jun 7, 2009 #5
    Since special relativity is really "relative", so its symmetric. You see my length contract and I see your length contracted. Nothing wrong with it. How could it be? after all, how could my length is shorter than yours and your length is shorted than mine? The answer is "it's about "time". Einstein's great insight is "time slow down when moving".
    see this book, It's about "time": https://www.amazon.com/Its-About-Ti...=sr_1_4?ie=UTF8&s=books&qid=1244417503&sr=1-4
     
    Last edited by a moderator: May 4, 2017
  7. Jun 7, 2009 #6

    JesseM

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    When you write Lm=ts*V and Ls=tm*V, what are Lm, tm and V?

    Normally, if Ls is the length of an object in its own rest frame, and Lv is its length in a frame where it's moving at speed v, then the relation between the two lengths is:

    [tex]L_v = L_s * \sqrt{1 - v^2/c^2}[/tex]

    Likewise, if Ts is the proper time between two readings on a clock as measured by the clock itself, and Tv is the time between those same readings in a frame where the clock is moving at speed v (where it is ticking slower, so it takes longer to go from the first reading to the second), the relation is:

    [tex]T_v = \frac{T_s}{\sqrt{1 - v^2/c^2}}[/tex]

    Probably the first means that in the rest frame of the spaceship, the distance from planet A to planet B is less than it is in the planet's own rest frame, and likewise the second would mean that in the rest frame of the planets, the length of the spaceship would be less than it is in the ship's own rest frame.
     
  8. Jun 7, 2009 #7
    Thank you JesseM
    "Probably the first means that in the rest frame of the spaceship, the distance from planet A to planet B is less than it is in the planet's own rest frame, and likewise the second would mean that in the rest frame of the planets, the length of the spaceship would be less than it is in the ship's own rest frame."

    This helps a lot..

    What I confused about the formula is why Ls is calculated by Tv instead of Ts..

    why should we use the moving time to calculate the proper length
     
  9. Jun 7, 2009 #8

    JesseM

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    The proper length Ls isn't related to time in the formulas I posted, it's related to Lv, the length in the frame where the object is moving at speed v. Why would you think length is calculated using time? Does it say that in your textbook?
     
  10. Jun 7, 2009 #9
    Yeah..My textbook use the time dilation relationship to derive the length contraction formula.

    They mutiply both sides by v, which is the speed of the spaceship..

    then V*Ts becomes Lv, and V*Tv becomes Ls
     
  11. Jun 7, 2009 #10

    JesseM

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    But that's just a step in a longer derivation, right? Without seeing the context of what assumptions they're making in the derivation I can't really make sense of that...what's the name of the textbook? Maybe we can find the page on google books.
     
  12. Jun 7, 2009 #11
    haha thank you for helping me

    It is Thomson Nelson Physics 12. By Hirsch, Martindale, Sewart and Barry

    It is in page 574
     
  13. Jun 7, 2009 #12

    JesseM

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    Excellent, turns out I could actually download that section from http://access.mmhs.ca/docs/Science/MMHS%20Web%20folder/physics12/Physics%20CD/physics12/a40.html [Broken]. The key is that Lm = v*ts is not meant to be a general relationship that would hold in any situation, but only in the case of the specific situation they describe on p. 573, where a ship is traveling between two planets, and in the ship's frame the distance between the planets is Lm, while the time between passing each planet as measured by the ship's clock is ts. So naturally in this frame if the second planet is a distance Lm away at the moment the ship is next to the first planet, and the ship remains at rest while the second planet rushes towards it, reaching the stationary ship a time ts later, then since that planet covered a distance of Lm in a time of ts the planet's speed must be v = Lm/ts, just based on the fact that in any frame speed is defined as distance/time (in terms of that frame's coordinates). Likewise, in the frame of the planets it takes the ship a time of tm to get from one planet to the other, and the distance between the planets is Ls in this frame, so the speed of the ship in this frame must be v = Ls/tm. And obviously if you take this equation and multiply both sides by tm you get Ls = v*tm, likewise the first equation can be rearranged to give Lm = v*ts.
     
    Last edited by a moderator: May 4, 2017
  14. Jun 8, 2009 #13
    Thank YOu Jesse M, i think I get it now.

    but..ah..I am not sure how do you know which one is suppose to be the proper length.

    In this scenario, is it true that the captain finds the distance separating the two events to be Lm is because of she thinks the planets are moving, and she is stationary?

    If that is true, so basically the objects that are moving will seem to be shorten to the observers who are stationary?
     
  15. Jun 8, 2009 #14
    If you have to ask this kind of question, that means you know the math equation but you do not understant its physical meaning. I guess you must know the so-called Lorentz transformation by now. He had similar question about time, since there are two "time" in the his equations. One is "time", one is "tau". He did not understand it, so he called one "time" and the other "local time". Lorentz think local time is just a mathematical trick that greatly eased the computational burden of finding new solutions of Maxwell's equations. Einstein cleared things up in one stroke. We now call one the "coordinate time' and the other "proper time", or (Eigen-Zeit) in German, it means its own time.
    No, it is "shorten" if you measure it with ruler, Not just "seem to be shorten" to the observers who are stationary.

    http://www.pitt.edu/~jdnorton/Goodies/rel_of_sim/index.html
     
    Last edited: Jun 8, 2009
  16. Jun 8, 2009 #15

    JesseM

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    Proper length is the distance between the two endpoints of an object (or two separate objects with space between them like the planets) in the frame where the endpoints are at rest.
    Right.
    Well, there's no absolute truth about who's "moving" and who's "stationary" since all speeds are relative, but objects that are moving relative to a given frame of reference will shorten to observers who are stationary relative to that same frame. This explains why if your ship is moving relative to my ship, and both ships have the same proper length, then in my rest frame your ship will shorten relative to mine, while in your rest frame my ship will shorten relative to yours.
     
  17. Jun 8, 2009 #16
    oh yes..that makes perfect sense now..thank you so much
     
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