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Deriving Lorentz transformations using perturbation theory

  1. Dec 31, 2015 #1
    1. The problem statement, all variables and given/known data

    Derive the transformations ##x \rightarrow \frac{x+vt}{\sqrt{1-v^{2}}}## and ##t \rightarrow \frac{t+vx}{\sqrt{1-v^{2}}}## in perturbation theory. Start with the Galilean transformation ##x \rightarrow x+vt##. Add a transformation ##t \rightarrow t + \delta t## and solve for ##\delta t## assuming it is linear in ##x## and ##t## and preserves ##t^{2}-x^{2}## to ##\mathcal{O}(v^{2})##.Repeat for ##\delta t## and ##\delta x## to second order in ##v## and show that the result agrees with the second-order expansion of the full transformations.

    2. Relevant equations

    3. The attempt at a solution

    We need to find the third term from ##x' = \frac{x+vt}{\sqrt{1-v^{2}}} = (x+vt)(1+\frac{v^{2}}{2}+\cdots)=x+vt+\frac{xv^{2}}{2}+\cdots##, and

    we need to find the second and third terms from ##t' = \frac{t+vx}{\sqrt{1-v^{2}}} = (t+vx)(1+\frac{v^{2}}{2}+\cdots)=t+vx+\frac{tv^{2}}{2}+\cdots##.

    Using ##x'=x+vt## and ##t'=t+\delta t, \delta t = rx+st+p##,

    we have ##t'^{2}-x'^{2}=(t+rx+st+p)^{2}-(x+vt)^{2}##

    ##t^{2}-x^{2}=((s+1)t+rx+p)^{2}-(x+vt)^{2}##

    ##t^{2}-x^{2}=(s+1)^{2}t^{2}+2(s+1)(t)(rx+p)+(rx+p)^{2}-x^{2}-2xvt-(vt)^{2}##

    ##0=(s^{2}+2s-v^{2})t^{2}+r^{2}x^{2}+(2r(s+1)-2v)tx+2p(s+1)t+2rpx+p^{2}##

    For the constant term, ##p=0##, so the terms in ##x## and ##t##.

    For the term in ##x^2##, we have ##r=0##, so we get a non-zero term in ##xt##.

    Where have I made the mistake?
     
    Last edited: Dec 31, 2015
  2. jcsd
  3. Dec 31, 2015 #2

    jambaugh

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    your variation of t should depend on v. [itex]\delta t = v(ax+bt)[/itex]. Then be sure you are ignoring terms of order [itex]O(v^2)[/itex].
     
  4. Dec 31, 2015 #3
    Using ##x'=x+vt## and ##t'=t+\delta t, \delta t = v(ax+bt)##,

    we have ##t'^{2}-x'^{2}=(t+v(ax+bt))^{2}-(x+vt)^{2}##

    ##t^{2}-x^{2}=t^{2}+(2av)xt+(2bv)t^{2}-x^{2}-(2v)xt##, from which we deduce that ##a=1,b=0##.

    Using ##x'=x+vt +\delta x## and ##t'=t+vx, \delta x = v^{2}(ax+bt)##,

    we have ##t'^{2}-x'^{2}=(t+vx)^{2}-(x+vt+v^{2}(ax+bt))^{2}##

    ##t^{2}-x^{2}=t^{2}+(2v)xt+v^{2}x^{2}-(x+vt)^{2}-2(x)(v^{2}(ax+bt))##

    ##0=(2v)xt+v^{2}x^{2}-(2v)xt-v^{2}t^{2}-2ax^{2}v^{2}-2bv^{2}xt##, from which we deduce that ##a=\frac{1}{2},b=0##.

    Is my approach correct?
     
  5. Dec 31, 2015 #4

    jambaugh

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    First stage, yes. That's what I got. For the second order expansion, you didn't consider additional variation of t. I'd also suggest using new constants (not a and b again).
    In fact it would be good practice to keep the same definitions of [itex]\delta x[/itex] and [itex]\delta t[/itex] but just extend to higher order in v.

    First stage: [itex] \delta x = vt + O(v^2) [/itex] and [itex]\delta t = v(ax+bt)+O(v^2)[/itex] .
    You solved and got a=1, b=0.

    Second stage: [itex] \delta x = vt + v^2(c x + d t+O(v^3), \delta t = vx + v^2 (ex + f t)+ O(v^3)[/itex] substitute in, solve for terms and compare with power expansion as instructed.

    This format is a suggestion for clarity. You might also use subscripted constants at this stage to keep the alphabet soup in hand. [itex] c_1, c_2, c_3, c_4[/itex] instead of c,d,e,f.
     
  6. Dec 31, 2015 #5
    Alright, I've reworked my solution:

    ##x'=x+\delta x, t'=t+\delta t##

    1st order in ##v##:

    ##\delta x =vt+\mathcal{O}(v^{2}), \delta t = v(a_{1}x+a_{2}t)+\mathcal{O}(v^{2})##

    So, ##t'^{2}-x'^{2}=(t+v(a_{1}x+a_{2}t))^{2}-(x+vt)^{2}##

    Using ##t'^{2}-x'^{2}=t^{2}-x^{2}## and expanding the right hand side only to first order in ##v##,

    ##t^{2}-x^{2}= t^{2}+2tv(a_{1}x+a_{2}t)-x^{2}-2xvt##

    ##0=(2va_{2})t^{2}+(2va_{1}-2v)xt##

    Taking the ##t^2## term, ##2va_{2}=0 \implies a_{2}=0##.

    Taking the ##xt## term, ##2va_{1}-2v=0 \implies a_{1}=1##.


    2nd order in ##v##:


    ##\delta x =vt+v^{2}(b_{1}x+b_{2}t)+\mathcal{O}(v^{3}), \delta t = vx++v^{2}(b_{3}x+b_{4}t)+\mathcal{O}(v^{3})##

    So, ##t'^{2}-x'^{2}=(t+vx++v^{2}(b_{3}x+b_{4}t))^{2}-(x+vt+v^{2}(b_{1}x+b_{2}t))^{2}##

    Using ##t'^{2}-x'^{2}=t^{2}-x^{2}## and expanding the right hand side only to second order in ##v##,

    ##t^{2}-x^{2}= (t+vx)^{2}+2(t+vx)(v^{2})(b_{3}x+b_{4}t)-(x+vt)^{2}-2(x+vt)v^{2}(b_{1}x+b_{2}t)##

    ##t^{2}-x^{2} = t^{2}+2tvx+v^{2}x^{2}+2tv^{2}b_{3}x+2tv^{2}b_{4}t-x^{2}-2xvt-v^{2}t^{2}-2xv^{2}b_{1}x-2xv^{2}b_{2}t##

    ##0=(v^{2}-2v^{2}b_{1})x^{2}+(2v^{2}b_{4}-v^{2})t^{2}+(2v^{2}b_{3}-2v^{2}b_{2})tx##

    Taking the ##t^2## term, ##v^{2}-2v^{2}b_{1}=0 \implies b_{1}=\frac{1}{2}##.

    Taking the ##x^2## term, ##2v^{2}b_{4}-v^{2}=0 \implies b_{4}=\frac{1}{2}##.

    Taking the ##xt## term, ##2v^{2}b_{3}-2v^{2}b_{2}=0 \implies b_{3}=b_{2}##.

    Therefore, ##b_{3}## and ##b_{2}## remain undeteremined.

    Are my solutions correct?
     
  7. Jan 6, 2016 #6

    jambaugh

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    I think b3 and b2 should be zero. One moment while I trace your work..., yep your expansion and term cancellation is correct. I suppose we must examine the third order terms to solve for these variables. So, with your b1 and b4 values determined expand further, ignoring terms of O(v^4) and higher. (This kinda makes sense, its to second order twice since we have 2 vars each perturbed to within 2nd order.)

    [Edit] PS Isn't perturbation expansion fun!!!!
     
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