Question on the mathematics of the Lorentz Transformation

roosh

This is an extremely basic question on the mathematics of the Lorentz transformation, but I just wanted to see what the answer is.

If we take the formulae for transforming the time co-ordinates:

t' = γ(t - vx/c²)
t = γ(t' + vx'/c²)

where:
γ = 1 / √(1 - v²/c²)


Presumably we can switch the labels above where t is relabeled t' and vice versa, as long as we consistently carry this into the calculations; this isn't really an issue, just a side question.

I'm just wondering why in one equation it is -vx/c, but in the other it is +vx/c?


I have it in my head, but not sure where it has come from, that it has to do with the fact that the reference frames are moving in opposite directions; is that wide of the mark?

dlouth15

roosh wrote: »

I have it in my head, but not sure where it has come from, that it has to do with the fact that the reference frames are moving in opposite directions; is that wide of the mark?

That is basically it as far as I can see. Since we are dealing with vector quantities, If S' is moving at velocity say [latex]v=v_x[/latex] relative to S then S is moving at velocity [latex]v=-v_x[/latex] relative to S'.

roosh

dlouth15 wrote: »

That is basically it as far as I can see. Since we are dealing with vector quantities, If S' is moving at velocity say [latex]v=v_x[/latex] relative to S then S is moving at velocity [latex]v=-v_x[/latex] relative to S'.

Is there a difference then if you've got a third reference frame moving at an angle to the other two?

Morbert

roosh wrote: »

Is there a difference then if you've got a third reference frame moving at an angle to the other two?

You can freely transform from one reference frame to the another regardless of their velocities relative to each other. Moving at an angle isn't a problem. You just rotate your coordinates.

[edit] - A recommendation I would make when learning the lorentz transformations is to let c = 1. This makes the lorentz transformations much clearer, and also better exhibits the symmetry between space and time.

[latex]t' = \gamma ( t - v x)[/latex]
[latex]x' = \gamma ( x - v t) [/latex]

roosh

Morbert wrote: »

You can freely transform from one reference frame to the another regardless of their velocities relative to each other. Moving at an angle isn't a problem. You just rotate your coordinates.

[edit] - A recommendation I would make when learning the lorentz transformations is to let c = 1. This makes the lorentz transformations much clearer, and also better exhibits the symmetry between space and time.

[latex]t' = \gamma ( t - v x)[/latex]
[latex]x' = \gamma ( x - v t) [/latex]

In the equations here and above though, is there a reason why the positive value of the measured velocity is taken in one equation, while the negative is taken in the other?

citrus burst

roosh wrote: »

In the equations here and above though, is there a reason why the positive value of the measured velocity is taken in one equation, while the negative is taken in the other?

Direction. One is written in term of t the other in terms of t'

roosh

citrus burst wrote: »

Direction. One is written in term of t the other in terms of t'

It's the direction bit I'm wondering about. If we have two relatively moving reference frames then according to the co-ordinates of both, they will both say that the other is moving in the same direction; that is, if, according to reference frame t, a relatively moving reference frame, t', is moving to the right, in the positive X direction, according to t', t will also be moving to the right in the positive X direction.

They could be distinguished in terms of the physical objects they are moving towards, such that the directions are different, but why would one direction require the negative value of the measured velocity, while the other require the positive. If we introduce a third relatively moving reference frame, such that each one is moving towards a different physical object, what is the justification for using the negative value of the measured velocity?

Morbert

roosh wrote: »

It's the direction bit I'm wondering about. If we have two relatively moving reference frames then according to the co-ordinates of both, they will both say that the other is moving in the same direction; that is, if, according to reference frame t, a relatively moving reference frame, t', is moving to the right, in the positive X direction, according to t', t will also be moving to the right in the positive X direction.

They could be distinguished in terms of the physical objects they are moving towards, such that the directions are different, but why would one direction require the negative value of the measured velocity, while the other require the positive. If we introduce a third relatively moving reference frame, such that each one is moving towards a different physical object, what is the justification for using the negative value of the measured velocity?

You're introducing a rotation by postulating that the two observers are facing each other. That's fine, but it adds an extra layer to the transformations that only complicates the maths. The transformations, as tendered by Wikipedia, pertain to reference frames oriented the same way, but moving at velocities relative to each other.

roosh

Morbert wrote: »

You're introducing a rotation by postulating that the two observers are facing each other. That's fine, but it adds an extra layer to the transformations that only complicates the maths. The transformations, as tendered by Wikipedia, pertain to reference frames oriented the same way, but moving at velocities relative to each other.

ah OK, sorry, I was working off the representation from that video which outlines the derivation of the Lorentz transformation with Albert and Henry; it shows both facing each other, although it does occasionally show them facing the same way.

Is the negative value taken then because according to one of them the other is moving in the negative X direction (or whichever axis it may be)?

Morbert

roosh wrote: »

ah OK, sorry, I was working off the representation from that video which outlines the derivation of the Lorentz transformation with Albert and Henry; it shows both facing each other, although it does occasionally show them facing the same way.

Is the negative value taken then because according to one of them the other is moving in the negative X direction (or whichever axis it may be)?

Yes, though it's more general than that. Think of it this way: When you shift or "transform" your perspective to that of another, everything will have the opposite velocity added. If you are standing on a road, and you start to walk "forward", everything that was once stationary will now appear to be moving "backwards". If you start to walk backwards, everything will appear to move forward. If you turn left, the scenery will all appear to turn right, Etc.

roosh

Morbert wrote: »

Yes, though it's more general than that. Think of it this way: When you shift or "transform" your perspective to that of another, everything will have the opposite velocity added. If you are standing on a road, and you start to walk "forward", everything that was once stationary will now appear to be moving "backwards". If you start to walk backwards, everything will appear to move forward. If you turn left, the scenery will all appear to turn right, Etc.

ah, I gotcha.

thanks for that.