# Homework Help: The lorentz transformations usage

1. Aug 30, 2010

### 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?

Last edited: Aug 30, 2010
2. Aug 30, 2010

### 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.]

3. Aug 30, 2010

### PhMichael

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

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

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

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

Why?

Last edited: Aug 30, 2010
4. Aug 30, 2010

### vela

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

5. Aug 30, 2010

### PhMichael

Last edited by a moderator: May 4, 2017
6. Aug 30, 2010

### vela

Staff Emeritus
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''.

7. Aug 30, 2010

### PhMichael

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

8. Aug 30, 2010

### vela

Staff Emeritus
9. Aug 30, 2010

### 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?).