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Basic derivation of Lorentz Transformation

  1. Sep 26, 2011 #1
    Hey, I'm just learning the ins and outs of special relativity and I'm having a little trouble with the derivation of the Lorentz Transformation. I bought Einstein's book 'Relativity: the special and general theories' and this is described in an appendix. For background, the appendix is linked to at the bottom of this post. (It was posted on marxists.org, I found that pretty funny, even though I knew Einstein was a socialist.) I get lost between
    (x'+ct')=μ(x+ct) and
    a=(λ+μ)/2

    Also, looking ahead, I've gotten lost before (this isn't my first time looking over the appendix, I lost my work from last time, though) between
    dx=1/a
    and x'=a(1-(v2)/c^2))x <-- this one I really can't get (how does he get there?)

    If anyone could be of any help, thanks!
    http://www.marxists.org/reference/archive/einstein/works/1910s/relative/ap01.htm
     
  2. jcsd
  3. Sep 26, 2011 #2

    atyy

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    In defining the mathematical forms of Eq 3 & 4, he sets up the unknown constants λ and μ. The problem then is to determine them.

    Instead of determining λ and μ directly, he will determine them indirectly by determining a=(λ+μ)/2 and b=(λ-μ)/2 first.
     
    Last edited: Sep 26, 2011
  4. Sep 26, 2011 #3
    How did he plug in constants a and b to obtain formulas (5)?
     
  5. Sep 26, 2011 #4

    atyy

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    From a=(λ+μ)/2 and b=(λ-μ)/2, we get a+b=λ and a-b=μ.
     
  6. Sep 26, 2011 #5
    When plugging in a and b, I keep getting x'=ax-act and ct'=bx+bct and getting stuck there. I'm a beginner at this, could you walk me through it (thanks for being patient)? I feel like I'm close but I don't know where to go next.
     
  7. Sep 26, 2011 #6

    atyy

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    By definition:
    a=(λ+μ)/2
    b=(λ-μ)/2

    Adding the two equations gives: a+b=λ, ie. λ=a+b
    Subtracting the two equations gives: a-b=μ, ie. μ=a-b
    Now we know λ and μ in terms of a and b.

    So x'-ct'=λ(x-ct) becomes:
    x'-ct'=(a+b)(x-ct)
    x'-ct'=ax+bx-act-bct (Eq I)

    And x'+ct'=μ(x+ct) becomes:
    x'+ct'=(a-b)(x+ct)
    x'+ct'=ax-bx+act-bct (Eq II)

    If I add Eq I and Eq II, the left hand side (LHS) is x'-ct' + x'+ct' = 2x'.
    If I add Eq I and Eq II, the right hand side (RHS) is ax+bx-act-bct + ax-bx+act-bct = 2ax-2bct.
    Equating LHS and RHS gives 2x'=2ax-2bct.

    So x'=ax-bct, which is the first of Einstein's Eq 5.

    Is that ok so far?
     
    Last edited: Sep 26, 2011
  8. Sep 26, 2011 #7
    That was great! You are a physics god among men.

    Now on to the tougher stuff (for me at least).

    Where does he get dx=1/a from?
     
  9. Sep 27, 2011 #8

    atyy

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    I'm a biologist.

    I don't follow his reasoning either! Is it correct?
     
  10. Sep 27, 2011 #9
    I'm not really sure... it seems to come up in the next term (lost on that one as well), but I don't know what hat he pulled it out of.
     
  11. Sep 28, 2011 #10

    atyy

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    OK, it's correct. Will post an explanation later.
     
  12. Sep 29, 2011 #11
    Good, I couldn't get anywhere with it.
     
  13. Oct 10, 2011 #12
    I don't mean to sound like a jerk here, but could you please post your progress? Thanks.
     
  14. Oct 10, 2011 #13

    atyy

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    I haven't been able to simplify Einstein's explanation, am still working on it.

    But there are many others here who presumably can.
     
  15. Oct 12, 2011 #14

    atyy

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    Let's say we have a rod of length L' moving with constant velocity u with respect to [x',t'] spacetime coordinates. Its left and right ends will trace out parallel lines of spacetime points described by equations of the form x'=xo'+ut' and x'=xo'+L'+ut'. In the particular case that the rod is stationary with respect to [x',t'] spacetime coordinates, u=0, and the parallel lines are x'=xo' and x'=xo'+L'.

    Substituting Einstein's Eq 5 into parallel lines x'=xo' and x'=xo'+L', we get xo'=ax-bct and xo'+L'=ax-bct, which we can rearrange to x=(xo'+bct)/a and x=(xo'+L'+bct)/a to see more clearly that they are parallel lines in [x,t] coordinates. These parallel lines have the form x=xo+vt and x=xo+L+vt, as expected for the left and right ends a rod of length L moving with velocity v with respect to [x,t] coordinates, with xo=xo'/a, L=L'/a and v=bc/a. The equation L=L'/a is Einstein's Eq 7.

    Let us note that since the rod was stationary in [x',t'] coordinates, and moving with velocity v=bc/a in [x,t] coordinates, v is also the velocity of [x,'t'] coordinates with respect to [x,t] coordinates. Hence our equation v=bc/a is Einstein's Eq 6.
     
    Last edited: Oct 13, 2011
  16. Oct 13, 2011 #15
    Where did you get xo=xo'/a and L=L'/a? I follow you completely minus those two.
     
  17. Oct 13, 2011 #16

    atyy

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    Compare these two equations from post #14.

    1) The equation for the spacetime line traced out by the right end of the rod obtained by getting the equation in [x',t'] coordinates then transforming to [x,t] coordinates:
    x=(xo'+L'+bct)/a

    2) The equation for the spacetime line traced out by the right end of the rod in [x,t] coordinates, obtained by the definition that it is a rod moving with velocity v with respect to [x,t] coordinates:
    x=xo+L+vt

    These two equations describe the same thing, so they must be the same equation, which will be the case only if xo=xo'/a, L=L'/a and v=bc/a
     
  18. Oct 13, 2011 #17
    From x' = ax → Δx' = a Δx

    But I = Δx', therefore I = a Δx, or Δx = I/a

    Hope it helps.
     
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