
#1
Jan2207, 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 coordinates when dealing with two reference frames S and S' moving relative to each other:  S coordinates as seen from S  S coordinates as seen from S'  S' coordinates as seen from S'  S' coordinates 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. 



#2
Jan2207, 11:51 AM

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PF Gold
P: 4,108

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]).




#3
Jan2207, 12:06 PM

PF Gold
P: 717

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



#4
Jan2207, 12:28 PM

P: 392

Lorentz Transformations and their Inverse
In Euclidean coordinates, you simply mean the transformation x'=xvt? Well, there, you simply deal with coordinates 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 coordinates in S' isn't the same as how an observer in S' views coordinates in S', is it?




#5
Jan2207, 12:46 PM

P: 329

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). 



#6
Jan2207, 01:03 PM

P: 2,043





#7
Jan2207, 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.




#8
Jan2207, 04:45 PM

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PF Gold
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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 graphpaper axes. How you do describe the coordinates of another thumbtack you place on the sheets? How does each sheet describe the xaxes and the yaxes? Given one set of coordinates of the thumbtack on one sheet, determine [predict] the coordinates on the other sheet. 



#9
Jan2207, 05:07 PM

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P: 8,470

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 xaxis 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 courseif 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. 



#10
Jan2207, 10:16 PM

P: 997





#11
Jan2307, 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.)




#12
Jan2307, 08:29 AM

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PF Gold
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#13
Jan2307, 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: [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? 



#14
Jan2307, 08:52 AM

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




#15
Jan2307, 08:57 AM

P: 392

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




#16
Jan2307, 09:08 AM

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PF Gold
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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? 



#17
Jan2307, 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.



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