Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: The lorentz transformations usage

  1. Aug 30, 2010 #1
    There's something about the lorentz transformations which is somewhat confusing to me, and that is how to treat the "x" coordinate. Supposing I have some spaceship which is moving from earth to some other planet located at a distance "D" (from earth) with a velocity v. Now, the spacetime coordinates of the events "1. leaving earth" and "2. reaching the planet" are (the spaceship frame is {S'} and that of earth is {S} ) :

    Leaving earth:

    [tex] (x_{1},t_{1})=(x'_{1},t'_{1})=(0,0) [/tex]

    Reaching the planet:

    [tex] (x_{2},t_{2})=(D, \frac{D}{v} ) [/tex]

    [tex] (x'_{2},t'_{2})=(0 , \gamma (t_{2} - (v/c^{2})x_{2})=(0 , \gamma (t_{2} - (v/c^{2})D) [/tex]

    Now comes the confusing point which is how to treat [tex] x_{3} [/tex] which corresponds to the event of returning back to earth in the earth's frame. (in the spaceship frame it is [tex] x'_{3} = 0 [/tex] )

    The Lorentz transformations relates coordinates and not distances so [tex] x_{3} = 0 [/tex] because the spaceship returns to the origin of earth and [tex] t_{3} = \frac{2D}{v} [/tex]. However, as I have seen in my notes:

    [tex] x_{3} = 2D [/tex]

    , that is, the distance that this spaceship travels is what is accounted for and not its coordinate.

    Can anyone clear this point for me?
     
    Last edited: Aug 30, 2010
  2. jcsd
  3. Aug 30, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I think that must be a mistake in your notes. Assuming the turn-around is instantaneous, then in "earth coordinates" you will have x = 0, t = 2D/v for the spaceship returning on earth.

    [Be aware, by the way, that you are basically working out the twin paradox.]
     
  4. Aug 30, 2010 #3
    But if [tex] (x_{3} , t_{3} ) = ( 0 , 2D/v ) [/tex] then for [tex] t'_{3} [/tex] we'll have:

    [tex] t'_{3} = \gamma (t_{3} - (v/c^{2}) x_{3} ) = \frac{2D/v}{\sqrt{1-(v/c)^{2}}} [/tex]

    while the answer should be:

    [tex] t'_{3} = \frac{(2D/v) - (2Dv/c^{2})}{\sqrt{1-(v/c)^{2}}} [/tex]

    that is, [tex] x_{3} = 2D [/tex] and not [tex] x_{3} = 0 [/tex]

    Why?
     
    Last edited: Aug 30, 2010
  5. Aug 30, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Did you account for different Lorentz transformations because the ship is now traveling in the opposite direction?
     
  6. Aug 30, 2010 #5
    Last edited by a moderator: May 4, 2017
  7. Aug 30, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Because there aren't only two frames S and S'. When the ship turns around, there's now a third frame, S''. There's a different set of Lorentz transformations that relate the coordinates in S to the coordinates in S''.
     
  8. Aug 30, 2010 #7
    I haven't got that ... why have you introduced a third frame? and what is it?
     
  9. Aug 30, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

  10. Aug 30, 2010 #9
    wikipedia uses some terminology that I haven't heard of :D

    anyway, I noticed that if I use that same table with the invariant then I get the right answer:

    Invarinat: [tex] l^{2}= (\Delta x)^{2} - c^2 (\Delta t)^{2} [/tex]

    [tex] 1 \to 2 [/tex]

    [tex] 24^{2} c^{2} - c^{2} 25^2 = 0^{2} - c^{2} (\Delta t')^{2} , t'_{1}=0[/tex]

    [tex] t'_{2}=7 [yr] [/tex]

    =============================

    [tex] 2 \to 3 [/tex]

    [tex] (0-24)^{2}c^{2}-(50-25)^{2}c^{2}=0^{2}-c^{2}(\Delta t')^{2} , t'_{2}=7 [/tex]

    [tex] t'_{3}=14 [yr] [/tex]

    which is strange because this invariant is obtaind from the Lorentz tansformation (isn't it?).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook