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B Length Contraction equation derivation

  1. Dec 19, 2016 #1
    I recently saw the derivation of length contraction in Special Relativity . At the end , it said
    x' = (x - vt) γ(gamma)
    x = (x' + vt') γ(gamma)
    Where γ(gamma) is Lorentz transformation . It is = 1/√(1- v²/c²)
    Then derivation continued , with expantion of x' = (x + vt)γ
    As t = 0 in this case
    We end up with , x'(√(1- v²/c²) ) = x
    As √(1- v²/c²) is always between 1 and 0 , x ≤ x'
    Thus , length of an object is contracted of any other observer ( if the speed of object is near the speed of light)
    Now the question is , If we take x = (x' + vt') γ instead of x' = (x + vt)γ
    We end up with x = (x')γ
    and we take x' = (x + vt)γ (as we did above)
    We end up with x' = (x)γ
    How is this possible that
    x = (x')γ and x' = (x)γ

    If i have watched the wrong derevation , please send me the link of correct derevation .
     
  2. jcsd
  3. Dec 19, 2016 #2
    No we do not end up with x = (x')γ because t' isn't 0. You would have to use the Lorentz transformation formula which solves for t' in order to get the t' position of your whichever event that happens at x, t you are trying to map into x', t'


    Check those x-t diagrams

    ssssssssss89abd.png

    The event E1 at x=5ls t=0s would,

    according to the Lorentz transformations, map to..

    x' = γ(x-vt) = 1.1547...(5ls - 0.5c*0s) = 5.7735...ls
    t' = γ(t-(vx/c2)) = 1.1547...(0ls - (0.5c * 5ls) / c2) = -2.88675...s

    ...as you can see in the right diagram.

    You could then think of E1' as the endpoint of a orange ruler which is at rest in this IFR if you wanted to. To measure the length of that ruler, you would need a second endpoint which is simultaneous to the first. If you considered the other endpoint to lie on a worldline which goes through E0'(0,0) then the length of the ruler would equal x' = 5.7735...ls
    ...for this special case as you can see from the diagram.

    But what about if we wanted to know how this ruler would measure from our initial IFR? Again we would need two endpoints which are simultaneous.
    We already have E1(5,0) and since one of the endpoints' worldlines goes through E'(0,0) it also goes through E(0,0) and so we know that this ruler which measures 5.7735..ls in the right diagram/IFR would measure 5ls in the left diagram/IFR
     
  4. Dec 19, 2016 #3

    Ibix

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    @Inderjeet - I'm afraid your presentation is rather sloppy and it's a bit difficult to be certain what you are trying to do. I think you are trying to determine the length of an object when that length is measured in a frame where it is moving. And you are surprised that you get a symmetric result - if you are moving with respect to me, I see you length contracted and you see me length contracted.

    Taking the latter point first, why are you surprised? If I say I am stationary and you are moving, then you say you are stationary and I am moving. So the principle of relativity tells us that any effect I measure in you, you will also measure in me. In any other case, there's an absolute notion of velocity - which we don't see.

    It's not at all clear what you are trying to do with your maths because you haven't explained precisely what you are trying to do. You appear to have a rod that is at rest in the frame using the primed coordinates (##x',t'##), whose length in the unprimed (##x,t##) frame is, indeed, shortened by a factor of ##\gamma##. You also appear to have a rod at rest in the unprimed frame, whose length in the primed frame is shortened by a factor of ##\gamma##. This is a correct result. However, I don't think that this is quite what you intended.

    I think you intended to think about one rod and show that its length was shorter in one frame and longer in the other. To do this, you need to be aware that if the measurement of the positions of the ends in one frame was simultaneous then it won't be simultaneous in the other frame. This means that you can't use both ##t=0## and ##t'=0##. If you do, you are asserting that the rod is stationary in the other frame in both cases - which is why I said you had two rods. As Jeronimus noted, you need to use the full Lorentz transforms on the coordinates of the ends of the rod.

    May I also suggest that you give the rod a length like L, rather than using x as both a coordinate and a length.
     
  5. Dec 19, 2016 #4

    robphy

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    @Jeronimus , in examples when the relative-velocity is not specified,
    rather than use a relative-velocity of 0.5 (which leads to calculations with irrational numbers.. likely requiring a calculator to evaluate to a decimal value),
    it's better to use 0.6 (=3/5) or 0.8 (=4/5) (which leads to calculations with rational numbers [since 3,4,5 is a Pythagorean triple]... which is why many relativity texts use them in examples).
    Your examples will be easier to follow, and your spacetime diagrams will simplify to reveal the graphical method discussed in my Insight.
     
  6. Dec 19, 2016 #5
    It's not possible because it's not possible to have both ##x=(x'+vt')\gamma## and ##x'=(x+vt)\gamma##.
     
  7. Dec 20, 2016 #6
    You mean the signs, correct?
     
  8. Dec 20, 2016 #7
    @Jeronimus Thank you Jeronimus , I figured my mistake . I was presuming t' = t , which was wrong . And you example really helped me .
    @Ibix Will try to be more clear next time . Thanks .
     
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