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Lorentz Transformation

  1. Mar 24, 2005 #1
    I am now studying the Lorentz transformation which shares some commonality with the Galiliean transformation. What I'm confused about is how they only seem to transform the x axis. It will help if I write it out. The Galilean transformation looks like:
    [tex] x' = x-vt [/tex]
    [tex] y' = y [/tex]
    [tex] z' = z [/tex]
    [tex] t' = t [/tex]
    Although most of the books I have read do not bother, I will put this in matrix form because I will eventually do that with the Lorentz transformation.
    [tex]
    \left(\begin{array}{cccc}
    1 & 0 & 0 & -v &
    0 & 1 & 0 & 0 &
    0 & 0 & 1 & 0 &
    0 & 0 & 0 & 1
    \end{array} \right)
    \letf(\begin{array}{c}
    x &
    y &
    z &
    t
    \end{array} \right) =
    \letf(\begin{array}{c}
    x' &
    y' &
    z' &
    t'
    \end{array} \right)
    [/tex]
    Pay no attention to the stange layout of the 1 by 4 arrays. I could not figure out how to make them. It is clear that veolocity effects only the x component of the resulting vector. To me that makes no sense. Does anyone know why it is set up this way. I have had no classes on matrices but I work with computer graphics programming and have taught myself much over the summer. I am thinking the book is assuming I also transform the coordinates of the object so that it is traveling only with respect to x and then transform the coordinate space back to the orginal for the final transformation. In that case giving the Lorentz transformation as first listed is not adaquate. Am I right?
     
  2. jcsd
  3. Mar 24, 2005 #2
  4. Mar 24, 2005 #3
    Thanks, I figured out what I needed to do based what was said before. I wanted to know how I simulate a view of a rocket moving at near the speed of light. This would all depend on the way you were looking and how much forshortning the rocket had in its direction of velocity. Although I haven't figured out how to keep track of the rockets displacement using matrices at the moment, I'm sure I'll get there.
     
  5. Mar 24, 2005 #4

    jtbell

    User Avatar

    Staff: Mentor

    Note that if you want to simulate what the rocket actually looks like to the eye of an observer located at a particular point, you need to take into account the fact that light from different parts of the rocket takes different amounts of time to reach the observer's eye. For a start, see

    http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

    then do a Google search on "Terrell rotation" and "Terrell effect" and you'll turn up some more information.
     
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