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Problem with signs in the Lorentz Transformation equations

  1. Apr 27, 2015 #1
    1. The problem statement, all variables and given/known data

    The following is Exercise 2.1 from from Ray d'Inverno's 'Introducing Einstein's Relativity.'

    (a) Write down the Galilean transformation from observer ##S## to observer ##S'##, where ##S'## has velocity ##v_1## relative to ##S##.
    (b) Find the transformation from ##S'## to ##S## and state in simple terms how the transformations are related.
    (c) Write down the Galilean transformation from ##S'## to ##S''##, where ##S''## has velocity ##v_2## relative to ##S'##.
    (d) Find the transformation from ##S## to ##S''##.
    (e) Prove that the Gailiean transformations form an Abelian (commutative) group.

    2. Relevant equations

    3. The attempt at a solution


    (a) My answer is $$x' = x - v_1 t,\ y'=y,\ z'=z,\ t'=t,$$ which matches with the textbook answer. So, no problem here.
    (b) My answer is $$x = x' + v_1 t,\ y=y',\ z=z',\ t=t',$$ which again matches with the textbook answer. So, no problem here.
    (c) My answer is $$x''=x'- v_2 t,\ y''=y',\ z''=z',\ t''=t',$$ but the textbook answer is $$x''=x'+ v_2 t,\ y''=y',\ z''=z',\ t''=t'.$$
    (d) My answer is $$x''=x- ( v_ 1 + v_2 ) t,\ y''=y,\ z''=z,\ t''=t,$$ but the textbook answer is $$x''=x+ ( v_ 1 + v_2 ) t,\ y''=y,\ z''=z,\ t''=t.$$

    Can someone please tell me if my answers are wrong, or if the textbook answers are wrong?
     
  2. jcsd
  3. Apr 27, 2015 #2

    BvU

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    They way you put it the book is wrong and you are right.

    It's a small miracle that you agree in a) and b) :wink: since both you and the book assume ##v_1## is along the x-axis of ##S## and ##S'##, and that both clocks and coordinate systems coincide at t = 0.
     
  4. Apr 28, 2015 #3
    Alright, then, here's my attempt to part (e) of the question.


    The group elements of the Galilean transformations are the velocities ##v## of ##S'## relative to ##S##.

    Consider the boost from ##S## to ##S'##, and then the boost from ##S'## to ##S''##, so that ##x''=x- ( v_ 1 + v_2 ) t,## so that ##v = v_1 + v_2##. So, the group of Galilean transformations is closed.

    Consider the boost from ##S## to ##S''##, and then the boost from ##S''## to ##S'''##, so that ##x''' = x'' - v_3 t = x - (v_1 + v_2 + v_3) t##. Also, consider the boost from ##S## to ##S'##, and then the boost from ##S'## to ##S'''##, so that ##x''' = x' - (v_2 + v_3) t = x - (v_1 + v_2 + v_3) t##. So, the group multiplication operation is associative.

    ##v = 0## is the identity element, because it transforms ##(x, y, z, t)## into ##(x, y, z, t)##.

    ##-v## is the inverse of ##v##, because the consecutive application of ##v## and ##-v## transforms ##(x, y, z, t)## into ##(x, y, z, t)##.

    The above four properties prove that the Galilean transformations form a group.

    Furthermore, consider a boost by velocity ##v_1##, and then a boost by velocity ##v_2##, so that ##x''' = x'' - v_2 t = (x' - v_1 t) - v_2 t = x' - (v_1 + v_2) t##. Also, consider a boost by velocity ##v_2##, and then a boost by velocity ##v_1##, so that ##x''' = x'' - v_1 t = (x' - v_2 t) - v_1 t = x' - (v_1 + v_2) t##. So, the group of Galilean transformations is Abelian (commutative).


    Do you think my proof is correct?
     
  5. Apr 28, 2015 #4

    BvU

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    Does it help if I say yes ? You really want to convince yourself, and it shouldn't make a difference if I say "well, it looks good to me !" (which I am perfectly happy to do, even if it probably is against PF rules ("we look at them more as guidelines :wink: ")
     
  6. Apr 28, 2015 #5
    Thank you!
     
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