Lorentz Transformations and their Inverse

[NanakiXIII]

I have these past few weeks been steadily studying the different aspects of the theory of Special Relativity. I started with the Lorentz transformations and, thinking I understood them, went along and studied other parts of the theory. However, along the line it has become apparent to me that my understanding of the transformations is a bit blurry, especially when I got around to using the inverse Lorentz transformations, and I'm hoping someone could help me straighten things out.

As I see it, there are four "different" sets of co-ordinates when dealing with two reference frames S and S' moving relative to each other:

- S co-ordinates as seen from S
- S co-ordinates as seen from S'
- S' co-ordinates as seen from S'
- S' co-ordinates as seen from S

Now, I'd like to clear up for myself once and for all, between which of these the Lorentz transformations transform. And also, what purpose do the inverse transformations serve when the situation is symmetrical and reversible? If anyone could provide me with a clear definition, I'd much appreciate it.

 

[robphy]

You might wish to first study the Euclidean analogue of your problem... a simple rotation from one set of orthogonal axes to another set (sharing the same origin [and orientation]).

 

[Jorrie]

Two, not four!

... As I see it, there are four "different" sets of co-ordinates when dealing with two reference frames S and S' moving relative to each other:...

Think two, not four sets of coordinates! It's normally only the x,t set and the x', t' set that matter in your stated problem.

As robphy said, try your hand at transforming backwards and forwards from one set of Euclidean coordinates to another one that is rotated...

Jorrie

 

[NanakiXIII]

In Euclidean co-ordinates, you simply mean the transformation x'=x-vt? Well, there, you simply deal with co-ordinates in two reference frames. However, since in Special Relativity you have to deal with time dilation and length contraction, the way an observer in S views co-ordinates in S' isn't the same as how an observer in S' views co-ordinates in S', is it?

 

[nakurusil]

In Euclidean co-ordinates, you simply mean the transformation x'=x-vt? Well, there, you simply deal with co-ordinates in two reference frames. However, since in Special Relativity you have to deal with time dilation and length contraction, the way an observer in S views co-ordinates in S' isn't the same as how an observer in S' views co-ordinates in S', is it?

You are getting all mixed up:

1. Time dilation and length contraction are CONSEQUENCES of the LT
2. LT give you the correct and complete transforms between spacetime coordinates, you don't need to add time dilation/length contraction (see point 1).

 

[MeJennifer]

However, since in Special Relativity you have to deal with time dilation and length contraction, the way an observer in S views co-ordinates in S' isn't the same as how an observer in S' views co-ordinates in S', is it?

That is correct, and that is exactly what you use the Lorentz tranformations for.

 

[NanakiXIII]

Well, now one person is saying I'm all wrong, and one is saying I'm right. That's rather confusing.

 

[robphy]

To get at the main issue of your initial question, do what I said: Study the Euclidean case of a rotation [neither a Galilean transformation (suggested in #4) nor a Lorentz Transformation]. Euclidean rotations involve cos(theta) and sin(theta), protractors,...

Take two sheets of graph paper and stick a thumbtack through the origins. Turn one sheet so that there is a nonzero angle between the corresponding graph-paper axes. How you do describe the coordinates of another thumbtack you place on the sheets? How does each sheet describe the x-axes and the y-axes? Given one set of coordinates of the thumbtack on one sheet, determine [predict] the coordinates on the other sheet.

 

[JesseM]

I have these past few weeks been steadily studying the different aspects of the theory of Special Relativity. I started with the Lorentz transformations and, thinking I understood them, went along and studied other parts of the theory. However, along the line it has become apparent to me that my understanding of the transformations is a bit blurry, especially when I got around to using the inverse Lorentz transformations, and I'm hoping someone could help me straighten things out.

As I see it, there are four "different" sets of co-ordinates when dealing with two reference frames S and S' moving relative to each other:

- S co-ordinates as seen from S
- S co-ordinates as seen from S'
- S' co-ordinates as seen from S'
- S' co-ordinates as seen from S

Now, I'd like to clear up for myself once and for all, between which of these the Lorentz transformations transform. And also, what purpose do the inverse transformations serve when the situation is symmetrical and reversible? If anyone could provide me with a clear definition, I'd much appreciate it.

The easiest form of the transformations is when you set them up so that the origin of each system is moving along the x-axis of the other, with the origin of the S' system moving along the x-axis at velocity v as seen in the S system (v is positive if the origin of S' moves in the +x direction, negative if the origin of S' moves in the -x direction), and the origin of the S' system moving along the x'-axis at -v as seen in the S' system. Also have it so the y-axis of S is parallel to the y'-axis of S', and the z-axis of S is parallel to the z' axis of S', and arrange things so that at t = t' = 0 in both coordinate systems, their two origins coincide at a single point in space.

In this case, if you have the coordinates of an event x,y,z,t in the S system and you want to know that event's coordinates in the S' system, the transformation is:

$$x' = \gamma (x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma (t - vx/c^2)$$
where $$\gamma = 1/\sqrt{1 - v^2/c^2}$$

Likewise, if you have the coordinate of an event x',y',z',t' in the S' system and you want to know that event's coordinates in the S system, the transformation is:

$$x = \gamma (x' + vt')$$
$$y = y'$$
$$z = z'$$
$$t = \gamma (t' + vx'/c^2)$$

with $$\gamma$$ defined in the same way (v is still the velocity of S' along the x-axis of the S system, which of course has the same magnitude but the opposite sign as the velocity of S along the x'-axis of the S' system).

Transforming S to S or S' to S' is just the identity transform, of course--if you know the coordinates x,y,z,t of an event in the S system, then it's just a tautology that that event's coordinates in the S system are x,y,z,t.

 

[bernhard.rothenstein]

direct and inverse Lorentz transformations

The easiest form of the transformations is when you set them up so that the origin of each system is moving along the x-axis of the other, with the origin of the S' system moving along the x-axis at velocity v as seen in the S system (v is positive if the origin of S' moves in the +x direction, negative if the origin of S' moves in the -x direction), and the origin of the S' system moving along the x'-axis at -v as seen in the S' system. Also have it so the y-axis of S is parallel to the y'-axis of S', and the z-axis of S is parallel to the z' axis of S', and arrange things so that at t = t' = 0 in both coordinate systems, their two origins coincide at a single point in space.

In this case, if you have the coordinates of an event x,y,z,t in the S system and you want to know that event's coordinates in the S' system, the transformation is:

$$x' = \gamma (x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma (t - vx/c^2)$$
where $$\gamma = 1/\sqrt{1 - v^2/c^2}$$

Likewise, if you have the coordinate of an event x',y',z',t' in the S' system and you want to know that event's coordinates in the S system, the transformation is:

$$x = \gamma (x' + vt')$$
$$y = y'$$
$$z = z'$$
$$t = \gamma (t' + vx'/c^2)$$

with $$\gamma$$ defined in the same way (v is still the velocity of S' along the x-axis of the S system, which of course has the same magnitude but the opposite sign as the velocity of S along the x'-axis of the S' system).

Transforming S to S or S' to S' is just the identity transform, of course--if you know the coordinates x,y,z,t of an event in the S system, then it's just a tautology that that event's coordinates in the S system are x,y,z,t.

Numbering the equation above then (1) and (4) (direct Lorentz transformations)solved for x' and t' lead directly to (6) and (9) (inverse transformation and vice versa). What is the name of that property.

 

[NanakiXIII]

So, what I gather is that one way of looking at inverse Lorentz transformation is that there's just a plus sign because the velocity is negative. Am I in any way correct here? (I'm aware that the inverse transformations can be derived from the normal transformations, I'm just looking to know whether the thought described above is correct.)

Numbering the equation above then (1) and (4) (direct Lorentz transformations)solved for x' and t' lead directly to (6) and (9) (inverse transformation and vice versa). What is the name of that property.

Is that a Socratic question? I haven't a clue.

 

[Hootenanny]

So, what I gather is that one way of looking at inverse Lorentz transformation is that there's just a plus sign because the velocity is negative. Am I in any way correct here? (I'm aware that the inverse transformations can be derived from the normal transformations, I'm just looking to know whether the thought described above is correct.)

Yes, this is correct. You can think of it as the S frame is traveling along the x axis, with a velocity v in the negative direction.

Numbering the equation above then (1) and (4) (direct Lorentz transformations)solved for x' and t' lead directly to (6) and (9) (inverse transformation and vice versa). What is the name of that property.

A transformation of this type is said to be invertible; i.e. if a transformation has an inverse it is said to be invertible. I am guessing this is what Bernard means by his comment.

 

[NanakiXIII]

Then I would be interested in an explanation of the use of the inverse transformation used in this post:

https://www.physicsforums.com/showpost.php?p=1178108&postcount=3

If one would scroll down to the section starting with:

Now, how does your clock behave in his frame, S'?

In this section the inverse Lorentz transformation is applied, introducing the plus sign, but the velocity is also negative:

$$t = 1.6667 [3 + (0.8)(-2.4)] = 1.8$$

Both the Lorentz transformation and the velocity are inverted. Why shouldn't it be just one of these two?

 

[Hootenanny]

Note that the velocity is positive; but the displacement is negative.

 

[NanakiXIII]

Could you elaborate on that? Aren't velocity and displacement linked by definition?

 

[Hootenanny]

Okay, let's examine the inverse lorentz transformation;

$$t = \gamma (t^\prime + v x^\prime / c^2)$$

Now, v is the velocity of the S' frame in your frame of reference. This is positive, agree?

x' is your position in the S' frame. Now, if v is positive it follows that x' must be negative.

Think of it this way; you and your friend start by standing on the same spot. Now, your friend starts to walk away from you with a velocity v at t = 0, in your frame of reference. Now, after some time t>0, will your position be positive of negative from your friends frame of reference?

Do you follow?

Aren't velocity and displacement linked by definition?

Indeed they are. However, if one is positive it is not necessary for the other to be positive.

 

[NanakiXIII]

Ah, of course. I was thinking that since we're now transforming from S' to S, v would be the velocity of S in S' and thus inverted. But of course, if you were to look at it that way, you'd have to use the normal Lorentz transformation and insert that negative velocity. Thanks for clearing that up, Hootenanny.

 

[Hootenanny]

Ah, of course. I was thinking that since we're now transforming from S' to S, v would be the velocity of S in S' and thus inverted. But of course, if you were to look at it that way, you'd have to use the normal Lorentz transformation and insert that negative velocity. Thanks for clearing that up, Hootenanny.

No problem, my pleasure. I often find that in elementary SR, properly defining the reference frames is the most difficult part of the problem. If you can properly define you frames and time intervals at the outset your 60% there.

 

[NanakiXIII]

Yes, I believe that was and still is my problem is this matter. I still do not fully understand between what things the Lorentz transformations are converting. The aforementioned equation, for example. It determines the time elapsed on Earth in S', yet it calculates t, which is a co-ordinate of S.

It's just all rather confusing and I lose track of what it is I'm doing.

 

[Hootenanny]

The aforementioned equation, for example. It determines the time elapsed on Earth in S', yet it calculates t, which is a co-ordinate of S.

You are correct in what you are saying; however, I would add something to your comment;
It determines the time elapsed on Earth in S', yet it calculates t, which is a co-ordinate of S as seen from the S' frame. I.e. this is the time that the twin sees your clock (on earth) reading. Does that make sense?

 

[NanakiXIII]

Well it only brings me back to my initial assumption that there are in fact four sets of co-ordinates.

- S co-ordinates as seen from S
- S co-ordinates as seen from S'
- S' co-ordinates as seen from S'
- S' co-ordinates as seen from S

This, I was told, is false.

Think two, not four sets of coordinates! It's normally only the x,t set and the x', t' set that matter in your stated problem.

And, if it is not false, that still leaves a problem. Earlier in the explanation in that post I linked to, a Lorentz transformation is used to determine that t'=3 in S'. So here, the Lorentz transformation converts from t seen in S to t' seen in S'. Yet the equation we discussed transforms from t' seen in S' to t seen in S'. That doesn't seem very consistent.

On a side note, I think one of my problems is also that I do not myself fully understand what it is I don't understand and thus cannot describe my problems adequately.

 

[bernhard.rothenstein]

Yes, I believe that was and still is my problem is this matter. I still do not fully understand between what things the Lorentz transformations are converting. The aforementioned equation, for example. It determines the time elapsed on Earth in S', yet it calculates t, which is a co-ordinate of S.

It's just all rather confusing and I lose track of what it is I'm doing.

I think that you would get more understanding expressing the LT as a function of dt, dt',dx,dx' and asking if dx and dx' represent proper lengths or non-proper ones and id dt,dt' represent proper time intervals or coordinate time intervals?

 

[Jorrie]

Your'e almost there!

... I still do not fully understand between what things the Lorentz transformations are converting. ...

Hootenanny has done an excellent job to get you this far! If he would pardon me for throwing a few loose remarks that just might help to answer the question above (maybe it has already been said…)

Two inertial observers that move relative to each other (the x,t frame and the x',t' frame) can both observe (or properly measure) the time interval and the space interval between events A and B. They will not agree on the time interval or the space interval.

The Lorentz transformation simply tells us how to convert (or transform) the space and time interval values that the one observer measured to what the other observer would have measured. Depending on from who's measurement you transform to who's, you just have to get the sign of your relative velocity right. Right?

As Hootenanny said, decide who is the reference frame and who is moving and in which direction (generally a velocity vector). Then plug the reference frame's time and space interval measurements into the transformation equation and you know what the other frame would have measured.

Regards, Jorrie

 

[JesseM]

Yes, I believe that was and still is my problem is this matter. I still do not fully understand between what things the Lorentz transformations are converting. The aforementioned equation, for example. It determines the time elapsed on Earth in S', yet it calculates t, which is a co-ordinate of S.

It's just all rather confusing and I lose track of what it is I'm doing.

You're talking about this equation from this post:

$$t = 1.6667 [3 + (0.8)(-2.4)] = 1.8$$

Looking at the post, he did it this way because of simultaneity issues--he wanted to know the time on Earth at the "same moment" that the twin had been traveling 3 years in the S' frame (using the S' frame's definition of simultaneity, of course), so he found the Earth's position in the S' frame when the time in the S' frame was 3 years. Then he had the event x'=-2.4, t'=3 years, and he wanted to figure out the corresponding t coordinate for this event in the S frame (which would be the time on the Earth's clock at this event), so he used the inverse Lorentz transformation on x' and t'.

 

[NanakiXIII]

he wanted to know the time on Earth at the "same moment" that the twin had been traveling 3 years in the S' frame

Wasn't it already established that this time was 5 years?

I think that you would get more understanding expressing the LT as a function of dt, dt',dx,dx' and asking if dx and dx' represent proper lengths or non-proper ones and id dt,dt' represent proper time intervals or coordinate time intervals?

Here's another one of my problems. My knowledge of physics and mathematics isn't very elaborate (think hasn't-finished-high school-yet level) and this post looks like complete technobabble to me.

Two inertial observers that move relative to each other (the x,t frame and the x',t' frame) can both observe (or properly measure) the time interval and the space interval between events A and B. They will not agree on the time interval or the space interval.

The Lorentz transformation simply tells us how to convert (or transform) the space and time interval values that the one observer measured to what the other observer would have measured.

This sounds plausible to me, and is how I initially approached this, but then how come the elapsed time in S is 5 years, then calculated that the elapsed time in S' is 3 years, then calculated that the elapsed time in S is 1.8 years. Either those 5 and 1.8 years mean different things, in which case we're not simply dealing with t in S and t' in S', or these calculations are incorrect. I'm assuming the former. But then what does each signify and why can we use the Lorentz transformations to both get from 5 to 3 years and from 3 to 1.8 years?

 

[JesseM]

Wasn't it already established that this time was 5 years?

No, that was in the S frame. Are you aware that the Lorentz transformation indicates that different frames define simultaneity differently, so two events that happen at the "same time" in one frame happen at different times in another? In the S frame, the event of the traveling twin reaching the turnaround point, with his clock reading t'=3 years, happens at the "same time" the Earth's clock reads t=5 years. But in the S' frame, the event of the traveling twin reaching the turnaround point at t'=3 years happens at the "same time" as the Earth-twin's clock reading 1.8 years. Each clock has been at rest in their respective frames up until the moment of the turnaround, so each twin measures the other twin's clock to have been slowed down by a factor of 0.6 due to time dilation.

 

[NanakiXIII]

I suppose simultaneity might just be the key to my problem. However, more questions arise.

Since the velocity is constant, you could also say that the Earth is moving away from the spaceship, and thus turn this entire business around. It's supposed to be symmetrical. In this case I'd assume the spaceship measures 5 years for the trip. So why doesn't it? Why can we take the Earth's frame and say the spaceship measures 3 years?

Secondly, we're going from 5 years to 3 to 1.8... couldn't we be doing this until the end of time? Do everything done for calculating that 1.8 years, but apply it to those 1.8 years. We'll measure that for the spaceship, 1.8*.6 years will have elapsed, won't we? What's the significance of this then, if it can be done perpetually?

 

[JesseM]

I suppose simultaneity might just be the key to my problem. However, more questions arise.

Since the velocity is constant, you could also say that the Earth is moving away from the spaceship, and thus turn this entire business around. It's supposed to be symmetrical. In this case I'd assume the spaceship measures 5 years for the trip. So why doesn't it? Why can we take the Earth's frame and say the spaceship measures 3 years?

It's symmetrical in the sense that each one measures the other one's clock to be ticking slower. But if you pick a particular event on the traveling twin's worldline, then all frames must agree on what the traveling twin's clock reads at that particular event, namely 3 years. If you had instead picked the event of the Earth-twin's clock reading 3 years, and the traveling twin hadn't turned around, then the situation would also be symmetrical in the sense that in the Earth's frame, this event would be simultaneous with the event of the traveling twin's clock reading 1.8 years, while in the traveling twin's frame this event would be simultaneous with the event of the traveling twin's clock reading 5 years.

Secondly, we're going from 5 years to 3 to 1.8... couldn't we be doing this until the end of time? Do everything done for calculating that 1.8 years, but apply it to those 1.8 years. We'll measure that for the spaceship, 1.8*.6 years will have elapsed, won't we? What's the significance of this then, if it can be done perpetually?

Sure, in the Earth twin's frame, the event of the Earth twin's clock reading 1.8 years is simultaneous with the event of the traveling twin's clock reading 1.8*0.6 = 1.08 years; in the traveling twin's frame, the event of the traveling twin's clock reading 1.08 years is simultaneous with the event of the Earth twin's clock reading 1.08*0.6 = 0.648 years; and so forth. But I don't really see that this has any significance, aside from showing you that each measures the other one's clock to be running slower than their own.

 

[NanakiXIII]

It's symmetrical in the sense that each one measures the other one's clock to be ticking slower. But if you pick a particular event on the traveling twin's worldline, then all frames must agree on what the traveling twin's clock reads at that particular event, namely 3 years.

I was going to say that if both agree that 3 years have elapsed, then doesn't that already mean that the traveling twin is younger (3 years older versus 5 for the twin on Earth), but I suppose this is countered by the fact that they do not both agree on how much time has elapsed on Earth, correct? However, if all frames must agree on what a clock reads at an event, then shouldn't they also agree on what the clock on Earth reads?

 

[JesseM]

I was going to say that if both agree that 3 years have elapsed, then doesn't that already mean that the traveling twin is younger (3 years older versus 5 for the twin on Earth), but I suppose this is countered by the fact that they do not both agree on how much time has elapsed on Earth, correct? However, if all frames must agree on what a clock reads at an event, then shouldn't they also agree on what the clock on Earth reads?

No, because "the traveling twin turns around" specifies a unique event on the traveling twin's worldline, but "the time on Earth at the same moment the traveling twin turns around" does not specify a particular event on the Earth's worldline, it could refer to different events depending on which frame's definition of simultaneity you use.

 

[NanakiXIII]

Ah, I see. Well, though it hasn't come through fully just yet, I do believe my mistaken thoughts lie with the relativity of simultaneity, as you pointed out in post #26. Some pondering and going over what new knowledge I acquired is probably in place for now. Thanks to everyone who replied.

 

[NanakiXIII]

Well, this next question is about a somewhat different topic, but still about the Lorentz transformations. A little bit down the page in the post I linked to earlier, there are also used these "generalized" Lorentz transformations. They look to me like two Lorentz transformations simply subtracted, but why would you do that? How does that generalize the transformation?

 

[Hootenanny]

Well, this next question is about a somewhat different topic, but still about the Lorentz transformations. A little bit down the page in the post I linked to earlier, there are also used these "generalized" Lorentz transformations. They look to me like two Lorentz transformations simply subtracted, but why would you do that? How does that generalize the transformation?

In the Lorentz Transformations (LT) you have met so far, we have assumed that at t=t'=0 then x=x'=0 , in other words the origins of both reference frames (spatial and temporal coordinates) coincide at the start. The more general version jtbell quoted allows us to consider situations where the origins of the two reference frames do not coincide at t = 0. It may be more intuitive if you consider time and position intervals or changes in time and displacement (as Berhard mentioned previously). If we take jtbell's general LT for position;

$$x_1^{\prime \prime} - x_0^{\prime \prime} = \gamma [(x_1 - x_0) - v (t_1 - t_0)]$$

$$\text{Define:}\hspace{1cm}\Delta x^{\prime \prime}:= x_1^{\prime \prime} - x_0^{\prime \prime}\hspace{1cm}\Delta x := x_1 - x_0 \hspace{1cm}\Delta t := t_1 - t_0$$

Then we have;

$$\Delta x^{\prime \prime} = \gamma [\Delta x - v \Delta t]$$

Hence, we obtain an expression for a change in coordinates, i.e. if we travel some distance $\Delta x$ in the S frame, which takes $\Delta t$ seconds. Then in the S'' frame traveling at constant velocity v in the positive x direction, the measured distance will be $\Delta x^{\prime\prime}$. The expression is just a subtraction of two LT's at different coordinates, hence we obtain intervals. Does that make sense?

 

[JesseM]

Well, this next question is about a somewhat different topic, but still about the Lorentz transformations. A little bit down the page in the post I linked to earlier, there are also used these "generalized" Lorentz transformations. They look to me like two Lorentz transformations simply subtracted, but why would you do that? How does that generalize the transformation?

Those equations just more general in the sense that you're no longer assuming the spatial origins of the two coordinate systems (in one dimension, the spatial origins are just x=0 and x''=0) coincide at the time t=t''=0 (which was one of the conditions I mentioned in my first post on this thread), but instead just assuming you know the coordinates of some "reference event" in both coordinate systems (with the reference event having some coordinates $$x_0$$, $$t_0$$ in one system and some other coordinates $$x_0^{\prime \prime}$$ and $$t_0^{\prime \prime}$$ in the other...they could be anything, like $$x_0$$ = 3 light years and $$t_0$$ = 27 years).

 

[NanakiXIII]

Thanks, Hootenanny and JesseM, I believe I understand these generalized transformations now. I'll be testing this understanding soon, but in the mean time I have another question. When substituting the values into the generalized transformation, jtbell comes to the conclusion that x" = -2.4. I figured this was the position of Earth in S''. However, a few lines down:

just after the turnaround, you are 8.2 light years in front of him

Where did this 8.2 come from? Also, if the Earth is now 8.2 ly away, how can the traveling twin bridge this gap in 1.8 years?EDIT: No, I do have another question about those generalized transformations. If we can't use the Lorentz transformations between co-ordinates of S and S'', then why can we use them between intervals on S and S''? What's the difference?