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Lorentz Transformations and their Inverse

by NanakiXIII
Tags: inverse, lorentz, transformations
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NanakiXIII
#1
Jan22-07, 11:22 AM
P: 392
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.
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robphy
#2
Jan22-07, 11:51 AM
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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
#3
Jan22-07, 12:06 PM
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Quote Quote by NanakiXIII View Post
... 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
#4
Jan22-07, 12:28 PM
P: 392
Lorentz Transformations and their Inverse

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
#5
Jan22-07, 12:46 PM
P: 329
Quote Quote by NanakiXIII View Post
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
#6
Jan22-07, 01:03 PM
P: 2,043
Quote Quote by NanakiXIII View Post
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
#7
Jan22-07, 01:09 PM
P: 392
Well, now one person is saying I'm all wrong, and one is saying I'm right. That's rather confusing.
robphy
#8
Jan22-07, 04:45 PM
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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
#9
Jan22-07, 05:07 PM
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Quote Quote by NanakiXIII View Post
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:

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

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:

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

with [tex]\gamma[/tex] 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
#10
Jan22-07, 10:16 PM
P: 997
Quote Quote by JesseM View Post
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:

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

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:

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

with [tex]\gamma[/tex] 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
#11
Jan23-07, 08:00 AM
P: 392
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
#12
Jan23-07, 08:29 AM
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Quote Quote by NanakiXIII View Post
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
#13
Jan23-07, 08:45 AM
P: 392
Then I would be interested in an explanation of the use of the inverse transformation used in this post:

http://www.physicsforums.com/showpos...08&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:

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

Both the Lorentz transformation and the velocity are inverted. Why shouldn't it be just one of these two?
Hootenanny
#14
Jan23-07, 08:52 AM
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Note that the velocity is positive; but the displacement is negative.
NanakiXIII
#15
Jan23-07, 08:57 AM
P: 392
Could you elaborate on that? Aren't velocity and displacement linked by definition?
Hootenanny
#16
Jan23-07, 09:08 AM
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Okay, lets examine the inverse lorentz transformation;

[tex]t = \gamma (t^\prime + v x^\prime / c^2)[/tex]

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
#17
Jan23-07, 09:19 AM
P: 392
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
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Jan23-07, 09:22 AM
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Quote Quote by NanakiXIII View Post
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.


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