How to Treat the x Coordinate in Lorentz Transformations?

[PhMichael]

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:

(x_{1},t_{1})=(x'_{1},t'_{1})=(0,0)

Reaching the planet:

(x_{2},t_{2})=(D, \frac{D}{v} )

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

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

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

x_{3} = 2D

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

Can anyone clear this point for me?

 

[CompuChip]

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.]

 

[PhMichael]

But if (x_{3} , t_{3} ) = ( 0 , 2D/v ) then for t'_{3} we'll have:

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

while the answer should be:

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

that is, x_{3} = 2D and not x_{3} = 0

Why?

 

[vela]

Did you account for different Lorentz transformations because the ship is now traveling in the opposite direction?

 

[PhMichael]

If I pick the following:

D=25 c \cdot year , v=0.96c

Then:

http://img153.imageshack.us/img153/9742/relx.jpg

Uploaded with ImageShack.usWhy is the distance traveled used and not the coordinate itself?

 

[vela]

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''.

 

[PhMichael]

I haven't got that ... why have you introduced a third frame? and what is it?

 

[vela]

The Wikipedia has a good explanation of the twin paradox from the traveling twin's point of view.

http://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity

 

[PhMichael]

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: l^{2}= (\Delta x)^{2} - c^2 (\Delta t)^{2}

1 \to 2

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

t'_{2}=7 [yr]

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

2 \to 3

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

t'_{3}=14 [yr]

which is strange because this invariant is obtaind from the Lorentz tansformation (isn't it?).