Lorentz Transformations Confusion

[laser]

TL;DR

Two frames S and S'. Does it matter if S' is moving to the right or to the left of S?

Two simple equations, yet I can't find any sources that explains it very well. If you have any suggestions on where I could find this then please link them! $$ \Delta{t'}=\gamma(\Delta{t}-\frac{v}{c^2}\Delta{x}) $$ Does it matter if S' is moving away to the right, or to the left, of S? I logically think not, because why would that matter? Surely that wouldn't change the value of $\Delta{t'}$. But then if you switch S' and S, S' being stationary and S moving away from S', you get a different equation: $$ \Delta{t}=\gamma(\Delta{t'}+\frac{v}{c^2}\Delta{x'}) $$

 

[Ibix]

Given your notation, S' is moving in the $+x$ direction if $v$ is positive and in the $-x$ direction if it's negative. Your second expression reverses that, which is a notation I would recommend against - increasing $x$ with negative $v$ is a confusing definition.Edit: no longer correct following an edit to the original post.

 

[laser]

Given your notation, S' is moving in the $+x$ direction if $v$ is positive and in the $-x$ direction if it's negative. Your second expression reverses that, which is a notation I would recommend against - increasing $x$ with negative $v$ is a confusing definition.

edited: latex wasn't showing so had me confused on the primes

 

[PeroK]

TL;DR Summary: Two frames S and S'. Does it matter if S' is moving to the right or to the left of S?

Two simple equations, yet I can't find any sources that explains it very well. If you have any suggestions on where I could find this then please link them! $$ \Delta{t'}=\Gamma(\Delta{t}-\frac{v}{c^2}\Delta{x}) $$ Does it matter if S' is moving away to the right, or to the left, of S? I logically think not, because why would that matter? Surely that wouldn't change the value of $\Delta{t'}$. But then if you switch S' and S, S' being stationary and S moving away from S', you get a different equation: $$ \Delta{t'}=\gamma(\Delta{t}+\frac{v}{c^2}\Delta{x}) $$

This transformation is based on the following:

1) S and S' share a common direction along the x/x'-axes.

2) The origin of S' coincides with the origin of S (i.e. $x = 0, t = 0$ corresponds to $x'=0, t' = 0$.)

3) The origin of S' moves at velocity $v$ along the x-axis. By convention, we tend to draw the positive x-axis to the right.

4) The velocity $v$ can be positive or negative. When $v$ is negative, that corresponds to motion to the left.

 

[Ibix]

edited: latex wasn't showing so had me confused on the primes

It's a known bug if there isn't LaTeX already on the page. Going into preview mode and refreshing the page kicks the parser into action.

 

[PeroK]

TL;DR Summary: Two frames S and S'. Does it matter if S' is moving to the right or to the left of S?

Two simple equations, yet I can't find any sources that explains it very well. If you have any suggestions on where I could find this then please link them! $$ \Delta{t'}=\gamma(\Delta{t}-\frac{v}{c^2}\Delta{x}) $$ Does it matter if S' is moving away to the right, or to the left, of S? I logically think not, because why would that matter? Surely that wouldn't change the value of $\Delta{t'}$. But then if you switch S' and S, S' being stationary and S moving away from S', you get a different equation: $$ \Delta{t}=\gamma(\Delta{t'}+\frac{v}{c^2}\Delta{x'}) $$

The second equation should be the inverse transformation.

If $$ \Delta{t'}=\gamma(\Delta{t}-\frac{v}{c^2}\Delta{x}) $$then$$ \Delta{t}=\gamma(\Delta{t'}+\frac{v}{c^2}\Delta{x'}) $$

 

[Orodruin]

What do you mean by ”matter”? S’ can be defined as either moving to the left or to the right relative to S. In the standard setup it is defined as moving to the right. The two choices are different as in the frame S’ is different between the two choices. For example, the two choices of S’ do not have the same simultaneity convention etc.

 

[Ibix]

Following your edit, then, you have the forward and inverse transforms. In the standard configuration, $v$ is defined to be the velocity of $S'$ with respect to $S$. Naturally, $S$ has the velocity $-v$ with respect to $S'$, so the inverse transforms are the same as the forward ones (because of the principle of relativity) but with the sign on $v$ flipped (because that's how we've defined $v$).

 

[laser]

This transformation is based on the following:

1) S and S' share a common direction along the x/x'-axes.

2) The origin of S' coincides with the origin of S (i.e. $x = 0, t = 0$ corresponds to $x'=0, t' = 0$.)

3) The origin of S' moves at velocity $v$ along the x-axis. By convention, we tend to draw the positive x-axis to the right.

4) The velocity $v$ can be positive or negative. When $v$ is negative, that corresponds to motion to the left.

Ah, I think I see something consistent here?

If event 2 is on the left of the origin, then $\Delta{x}$ is negative. $v$ is also negative. If event 2 is on the left of the origin, both $\Delta{x}$ and $v$ are positive. So the computation works out to be the same when S' is moving towards event 2! (although I am not sure it is accurate to say this, as S' could have already passed the coordinate of x2 before it happened, so let's just say that when S' passes x2?)

And if S' doesn't pass x2, then the opposite happens.

 

[laser]

What do you mean by ”matter”?

As in, do we get the same answer following computation?

 

[Orodruin]

An inertial system is never moving towards an event. All events are in all inertial frames. An inertial frame cannot ”pass” an event.

As in, do we get the same answer following computation?

following what computation? It is not clear what you are attempting to do.

 

[laser]

An inertial system is never moving towards an event. All events are in all inertial frames. An inertial frame cannot ”pass” an event.

following what computation? It is not clear what you are attempting to do.

Take for example, a bomb on an island and a plane flying over the island. The plane can be considered an inertial frame. The bomb is the event. So the inertial frame is passing the event?

The lorentz transformation computation.

 

[PeroK]

Ah, I think I see something consistent here?

If event 2 is on the left of the origin, then $\Delta{x}$ is negative. $v$ is also negative. If event 2 is on the left of the origin, both $\Delta{x}$ and $v$ are positive. So the computation works out to be the same when S' is moving towards event 2! (although I am not sure it is accurate to say this, as S' could have already passed the coordinate of x2 before it happened, so let's just say that when S' passes x2?)

And if S' doesn't pass x2, then the opposite happens.

You are overthinking this. These transformations transform coordinates from one frame to another. $v$ is some arbitrary velocity and $x, t$ are arbitrary coordinates. This is not a specific physical scenario.

Imagine instead in geometry you have the x-y axes and a set of x'-y' axes that are rotated at an angle $\theta$. Then, you have the coordiante transformation:
$$x' = x\cos \theta + y\sin \theta$$$$y' = -x\sin \theta + y\cos \theta$$You should look at the Lorentz Transformations as the spacetime equivalent of a spatial rotation (spatial coordinate transformation).

 

[Orodruin]

The plane can be considered an inertial frame.

Common confusion. No, the plane is not an inertial frame. It has a rest frame, but the plane exists in all frames. The event occurs also in the rest frame of the plane, it is just a single event. The bomb may be moving towards the plane in that frame, but the event itself is a position at a particular time regardless of frame - it does not have any motion.

 

[Sagittarius A-Star]

The bomb is the event.

As others said, that's not logical. The explosion is the event.