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Length contraction and time dilation from Lorentz S

  1. Sep 18, 2016 #1
    I'm trying to understand special relativity well enough to explain it to others, ANY others, including myself. I am trying to use Robert Resnick's Introduction to Special Relativity to inform my thinking. In introducing length contraction, he introduces L' as the length measured by an observer in the moving frame and then inserts this result in the Lorentz transformation equation to find L, the length measured by an observer in the stationary frame. (section 2.3 p. 62) Why does he not use the inverse transformation Directly? (Yes, I know: doing so would give length dilation) For time dilation, he does use the inverse transformation directly and gets time dilation, but if he followed the same procedure as he did for length contraction, he would get time contraction.
    My instinct is that if the observer in S' measures L' and (delta x)' while the observer in S measures L and (delta x). To go from S' measurements to S measurements, the most direct rout in the Inverse Transformation while to to from S measurements to S' measurements, the most direct route is the Transformation.
    I realize that I am missing a major conceptual point here and am grateful in advance for any help you can give me. Thank you
     
  2. jcsd
  3. Sep 18, 2016 #2

    vanhees71

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    The point is that length is defined for each of the two observers to bring into coincidence the ends of the rod with his ruler simultaneously. Now say observer A (unprimed coordinates) is in the reference frame, where the rod is at rest. The world lines of the ends of the rod in this frame of reference are given by ##X=(x^0,x^1)## and ##Y=(y^0,y^1)## (I set ##c=1## for convenience and consider only one spatial direction defined by the rod). Now let B (primed coordinates) be an observer, which moves with velocity ##v## against A. The Lorentz transform tells you that for him the world lines look like
    $$X'=\gamma (x^0-v x^1,x^1-v x^0), \quad Y'=\gamma (y^0-v y^1,y^1-v y^0).$$
    Now measuring the length of the rod, he brings the ends of the rod into coincidence with his spatial axis simultaneously, but this doesn't happen at the same time for A but you must have
    $$y^{\prime 0}-x^{\prime 0}=0 \; \Rightarrow \gamma [(y^0-x^0)-v(y^1-x^1)] =0 \; \Rightarrow y^0-x^0=v(y^1-x^1).$$
    Then you get the length of the rod for B as
    $$L'=y^1-x^1 =\gamma [(y^1-x^1)-v(y^0-x^0)]=\gamma (y^1-x^1)(1-v^2)=\frac{1}{\gamma} (y^1-x^1)=\frac{1}{\gamma} L,$$
    i.e., for him the rod is shorter by an inverse Lorentz factor ##1/\gamma=\sqrt{1-v^2}## than what A measured in the restframe of the rod.
     
  4. Sep 18, 2016 #3

    Nugatory

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    The length of an object is defined as the distance between where its two endpoints are at the same time in the frame in which you are measuring the length. Now when one end of the rod is at the origin in the primed frame the other end is at the point (x'=L',t'=0) because that's what it means to say that the length of the rod is L' in the primed frame. However, when you apply the inverse Lorentz transformation to the point (x'=L',t'=0) you do not get something of the form (x=?,t=0) in the unprimed frame; instead you get something with a non-zero value of t. Thus, this point is not what we need to calculate the length in the unprimed frame, namely the location of the far end of the rod in the primed frame at the same time that the near end is at the origin, and the spatial distance between this point and the origin is not the length of the rod in the unprimed frame.

    Instead, we are looking for three points here:
    1) The origin, with coordinates (x=0,t=0) and (x'=0,t'=0)
    2) The point (x=L,t=0) which determines the length of the rod in the unprimed frame
    3) The point (x'=L',t'=0) which determines the length of the rod in the primed frame.

    #2 and #3 are different points, so you cannot use the Lorentz transformations directly to go from one to the other.
     
  5. Sep 24, 2016 #4
    Thank you both for helping me. I would further appreciate each of you letting me know that you have received this note as I don't yet know how to use this site. I don't yet know the difference between a tag and a thread. You get the idea. Thanks again. I will be back for help, honest.

    Mike

    ps. I also need to learn the difference between a reply and an upload
     
  6. Sep 24, 2016 #5
    Thank you both for helping me. I would further appreciate each of you letting me know that you have received this note as I don't yet know how to use this site. I don't yet know the difference between a tag and a thread. You get the idea. Thanks again. I will be back for help, honest.

    Mike

    ps. I also need to learn the difference between a reply and an upload
     
  7. Sep 24, 2016 #6

    Thank you both for helping me. I would further appreciate each of you letting me know that you have received this note as I don't yet know how to use this site. I don't yet know the difference between a tag and a thread. You get the idea. Thanks again. I will be back for help, honest.

    Mike

    ps. I also need to learn the difference between a reply and an upload
     
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