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The lorentz transformations usage

  • Thread starter PhMichael
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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?
 
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Answers and Replies

CompuChip
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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.]
 
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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?
 
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vela
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Did you account for different Lorentz transformations because the ship is now traveling in the opposite direction?
 
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vela
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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''.
 
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I haven't got that ... why have you introduced a third frame? and what is it?
 
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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?).
 

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