Talk:Lorentz transformation

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Contents

[edit] Error in Section "Details"

Quote:

Then, we can write

s^2 = \begin{bmatrix} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}\ .

Endquote.

This is clearly not true. Because if you multiply it out, you'll get another vector (which is easy to see when you know a bit of matrix algebra: A n x n matrix multiplied with a n x 1 matrix gives a n x 1 matrix). Which is probably not what the author meant to say. The square of the norm of a vector xμ

x^{\mu} = \begin{bmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}

with respect to a metric ημν is:

s2 = ημνxμxν

where Einstein's Summation Convention is used. More explicit:

s^2 = \sum_{\mu=1}^4 \sum_{\nu=1}^4 \eta_{\mu\nu} x^{\mu} x^{\nu}

So in this case with

\eta_{\mu\nu} = \begin{bmatrix} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}

We get:

s2 = ημνxμxν = − (cΔt)2 + (Δx)2 + (Δy)2 + (Δz)2 = − c2(Δt)2 + (Δx)2 + (Δy)2 + (Δz)2
OK, I think its fixed. Please check it. PAR 14:30, 17 September 2006 (UTC)
Ok, there's some style details. I find it unfortunate how the spacetime events A and B are notated by upper case letters. The norm with respect to the minkowski metric is most naturally written in tensor co/contravariant notation as with the s2 = ημνxμxν, where xμ denotes a vector x and where its components are denoted by x1,x2 or generally xν . There's another clash with the use of x as in x,y and z of the spatial components. This might confuse readers. I'm not sure how to go about this. FlorianPaulSchmidt 19:16, 17 September 2006 (UTC)

[edit] Confusing

The below, added to the History section, is confusing:
c is then identified to the propagation speed of electromagnetic radiation in vacuum. (Comment from before 2003)

[edit] Merge

We must merge this with Lorentz transformation equations


Brute-force merged: now the equations need thorough checking, because they are subtly different in each sub-version... -- The Anome 00:04, 13 Aug 2003 (UTC)

[edit] Proper foundation

"Proper foundation for its application" ... I'd love to watch a debate between the author and this bunch who seem to be staking their careers on Lorentz being right after all: "The Einstein assumptions leading to the Special and General Theory of Relativity are shown to be falsified by the extensive experimental data. Contrary to the Einstein assumptions absolute motion is consistent with relativistic effects, which are caused by actual dynamical effects of absolute motion through the quantum foam, so that it is Lorentzian relativity that is seen to be essentially correct." I suggest "proper foundation" is not NPOV; but WP has been sufficiently abusive to me I'm doing no more than to leave this note. -- fmr Kwantus (March 2004)

[edit] Lorentz Transformation is Flawed

The Lorentz transformation is actually nothing but a 'cheated' Galilei Transformation: it uses a vectorial velocity addition for light signals (which contradicts the invariance of the speed of light) and then this error is being 'rectified' by changing the original length and time units (see my page http://www.physicsmyths.org.uk/lightspeed.htm for more).

In order to understand the interpretation of the invariance of the speed of light, behold the eigenvectors of the Lorentz transformation.
217.81.157.237 20:29, 10 Jan 2005 (UTC)

In your website page illustrated example your hypothesis and conclusion seem to be one and the same: the invariance of the speed of light. Then any "demonstration" would be redundant (i.e. already in the hypothesis). The whole discussion about the Lorentz transformation is the physical validation or invalidation of his mathematical expression of t' in relation to t. If possible please post relevant material of how he arrived at the proposed equation.--Lucian 20:00, 27 May 2005 (UTC)

Yeah, I fail to understand how that website proves anything. Perhaps the original poster should read up on the Lorentz group to learn how the Lorentz transformation isn't a 'cheated' Galilean transformation, but rather an extension of rotations. The most intuitive way to understand the derivation of Lorentz transformations is to see it as finding a transformation that preserves c (rotations) and extending that in a really simple way, rather than pulling a new transformation out of thin air, which is usually how it's presented to freshmen undergrads). --Laura Scudder | Talk 20:34, 27 May 2005 (UTC)

The error made in the website is the statement:

"Since obviously the same amount of rope has reeled off at the car as well, the marking at the latter does at this moment also read x1 and since the rope has reeled off with the same speed, the car-clock does also shows T1'=T1=x1/v."

The same amount has not reeled off. In the GBO frame, the GBO-rope is moving at velocity v  while the car-rope is at rest. This means that the GBO-rope is Lorentz-contracted, while the car-rope is not. At any given time in the GBO frame, the GBO-rope will measure a greater distance to the car than the car-rope. This is not a paradox, because the spatial distance between two events is not absolute, but depends on intertial frame in which it is measured. 69.143.43.101 15:47, 28 August 2005 (UTC)

Yes, see the ladder paradox for a more in depth explanation. Fresheneesz 19:02, 17 April 2006 (UTC)

[edit] Derivation of Lorentz Transformation

I am missing a derivation of the Lorentz transformation on Wikipedia. I think this is so vital for the development of Special Relativity (both historically and didactically) that one should not have to refer to external links for this.

done... —Preceding unsigned comment added by 131.114.192.226 (talk) 13:53, 11 September 2007 (UTC)

[edit] Merging?

Shouldn't this be merged with Lorentz-Fitzgerald contraction hypothesis? This article is about relativity and that one is about luminiferous aether, but it's basically the same contraction used for two different purposes and it seems like it should be one article. Ken Arromdee 18:05, 5 October 2005 (UTC)

Maybe - but the contraction hypothesis should be included in the Lorentz transformation article, not vice versa. PAR 23:49, 5 October 2005 (UTC)

[edit] Triangle of Velocities proves that invariant speed is mathematical error

The article at http://www.masstheory.org/triangle_of_velocities.pdf presents detailed analysis of Lorentz transformation, it's derivation procedure and formal mathematical proof that introducing "invariant" speed is indeed nothing but mathematical error.

I'd be interesting to know if the person who posted this understands what the author of "triangle of velocities" wrote. I read it, and I think he made a mistake in assumptions about the variables - ie what variable corresponds to what - which i admit is very confusing (which is why I wrote the variable definitions out on this page for the SC transforms). Fresheneesz 21:16, 17 April 2006 (UTC)

The article is funny but not serious, and reflects serious mistakes in physics knowledge. The assert x=v+c is wrong. We can speak about (x,t) and (x',t') but when introducing third velocities, we need to use velocity transforms.The Lorentz transforms are valid for spacetime point transforms, not for third velocity transforms. For these cases, Lorentz transforms helps us to get the accurate 'velocity transform'. The article is not serious nor matematicamente rigorous. On the other hand, the speed of the light is not already an observation, is an international definition: 2.99798458E8 m/s.Mel Viso —Preceding unsigned comment added by 80.25.164.213 (talk) 09:47, 29 January 2008 (UTC)

Was it necessary to say, in the definitions of x1 and x2, that the "point" was stationary in one frame and moving relative to the other? Couldn't these be the coordinates of a point moving relative to both frames (measured relative to each frame) in which case dx1 / dt1 is the speed of the point in one frame and dx2 / dt2 its speed in the other? Is there any reason why one of these speeds must be zero? E4mmacro 06:42, 20 April 2006 (UTC)

It depends what you're measuring. If one dx/dy is 0 in one frame, then it will neccessarily not be 0 in another frame (a frame at a different velocity). Fresheneesz 09:29, 20 April 2006 (UTC)
I think that the definitions as written were confusing. Making the point stationary in one frame may appear to be a simplification, but gives a (false?) impression that this was necessary, that the transformations do not apply unless the point is stationary in one of the frames. E4mmacro 21:23, 20 April 2006 (UTC)
I agree, I did a terrible job on that. I was trying to get it all straight in my head. I've tried to think about it more, but my laziness is getting the better of me. I think it works if you consider (t,x,y,z) and the other one to be just two different coordinate systems, but for me - that doesn't help me think about it. If I could get off my ass (metephorically) and rework it, I think I would go back to saying that they're simply points in a coordinate system, but it needs to be well explained - because otherwise they're just confusing equations. Fresheneesz 11:14, 23 April 2006 (UTC)

[edit] Remove prime notation and put transforms at top

I know everyone likes to use prime notation, but noone likes reading it. We should use subscripts here. Perhaps like this with front and back subscripts: 0Xs . Also, the transforms should go at the top for reference, not burried under the "details" section. User:fresheneesz 128.111.95.147 22:30, 10 April 2006 (UTC)

I did it. PLEASE, someone look over it and tell me if I botched anything - because I don't know if I got it all right. Fresheneesz 21:14, 17 April 2006 (UTC)

[edit] Variable definitions

I finnally put all the variable definitions down for the SC lorentz transforms - but I would appreciate someone *checking* them to make sure I did it right. It was a very confusing process. Fresheneesz 21:17, 17 April 2006 (UTC)

[edit] time dilation and simultaneity

What exactly is this sentence from the article supposed to mean?

Time dilation was also used to prove that simultaneity varies between reference frames.

How was time dilation used to prove relativity of simultaneity? Is it just trying to say the Lorentz transformations incoporate both time dilation and reletivity of simultaneity? There is some related discussion in the relativity of simultaneity page. E4mmacro 06:28, 20 April 2006 (UTC)

[edit] Not cartesian but...

The first equation under "details" only holds for certain coordinate systems, but I forgot the name of them. The equivalent of Cartesian coordinate systems in 3-space. PAR 00:52, 21 April 2006 (UTC)

I'll bet the word is either "Lorentzian" or "Minkowskian" coordinates. -lethe talk + 01:27, 21 April 2006 (UTC)

[edit] Simpler

I have been studying special relativity in college for the past semester, and I've learned these equations. That said, they looked a great deal foreign to me when I saw them on this page. In class, I worked with x, x', t, t', v, v', and B (beta). I never saw the hyperbolic transformation, and it seems to me like--although we should keep the original equations for history's sake--we ought to also provide the easier equations. After all, wouldn't it benefit the Wikicommunity at large to add a small section at the bottom with the simpler algebraic equations? -Jess V

They're there, but lower on the page. You would have noticed them if the "easier" equations were higher. I think they *should* be higher on the page - I had the same problem as Jess V when I first came to this page. Fresheneesz 05:28, 8 May 2006 (UTC)

[edit] Lorentz transormations vs Length contraction

I have been very confused with how the lorentz transformations translate into length contraction. I know about how the time component goes away since length contraction involves changes in time and distance, and the change in time is 0 since both events are measured at the same time (in one person's reference frame). However, it seems to me that if we used lorentz transformations we could get *both* L = \frac{L_0}{\gamma} and L = \gamma L_0 \ . I don't understand why its one and not the other, can anybody help? Fresheneesz 05:32, 8 May 2006 (UTC)

I only see this now. This is a common confusion, and it would be good to explain in the article space that two distant clocks (with (t2 - t1 = 0) are read in the stationary frame for establishing two readings of position x2 and x1;
and that the readings of corresponding clocks in the moving system are offset due to relativity of simultaneity. The boundary condition t2 - t1 = 0 is essential. Harald88 15:07, 17 September 2006 (UTC)

[edit] Animated lorentz transformation

Image:Animated_Lorentz_Transformation.gif

I added this animation to the Links section. It's a big file so I wasn't sure it should be on the main page. Thanks, Jonathan (JDoolin, 13 May 2006)

Nice picture, but it is indeed too big. Maybe you could make a non-animated version with a link to the full version or change the image license to allow someone else to do so? Han-Kwang 20:51, 12 July 2006 (UTC)

[edit] Order changed to make it easier for non-experts to follow

The article started out with the section "Lorentz boost in 2 dimensions". This text would be unintelligible to a reader who didn't already understand the formulation, in that it used undefined terms and undefined mathematics (i.e., if the reader didn't already know what "the form xy" is, the article is incomprehensible.) I moved this to come second, and added "in matrix form" to clarify (possibly I should have said "tensor" or "linear algebraic form," but since tensors aren't brought up yet either...)

The word "form" should be clarified with a link to the wikipedia article bilinear form, I think.

[edit] History section

Anyone else think History should be at the top as an easy lead in to the subject?--Light current 07:23, 7 October 2006 (UTC)

[edit] Causality implies the Lorentz group

Quote:

If space is homogeneous, then the Lorentz transformation must be a linear transformation. Also, since relativity postulates that the speed of light is the same for all observers, it must preserve the spacetime interval between any two events in Minkowski space

However it is very common to require linearity from Lorentz transformation, really we don't have to make such strong assumptions. In an 1964 paper, E.C. Zeeman proved that the causality preserving group is equal to the Lorentz group. That is, instead of requiring linearity and isometry from the coordinate transformation between inertial observers, it is enough to require causality preservation. Linearity and isometry are mathematical consequences!

The paper: E. C. Zeeman, Journal of Mathematical Physics -- April 1964 -- Volume 5, Issue 4, pp. 490-493

This result is also found in Naber, Gregory L., The Geometry of Minkowski Spacetime, Springer-Verlag, New York, 1992. ISBN 0-387-97848-8 (hardcover), ISBN 0-486-43235-1

XCelam 08:11, 25 December 2006 (UTC)

[edit] Two errors

There is a serious misinterpretation of a paper by Zeeman, E. C. in the section titled, Derivation. The error of that section is this claim:

"[Pal's] derivation invokes the natural, but unnecessarily strong assumption of homogeneity of spacetime. A deeper analysis [3] shows that the Lorentz-transformation is the consequence of the only requirement of keeping causality relations on spacetime."

The abstract of Zeeman's paper simply states:

"Causality is represented by a partial ordering on Minkowski space, and the group of all automorphisms that preserve this partial ordering is shown to be generated by the inhomogeneous Lorentz group and dilatations." ©1964 The American Institute of Physics.

In other words, Zeeman began with Minkowski space, which was already known to be a homogeneous spacetime.

The second error is not as serious. The statement, "If space is homogeneous, then the Lorentz transformation must be a linear transformation" is only misleading. It all depends on whatever definition of the Lorentz transformation you begin with. For instance, we could adopt the opening sentence of the article as a definition. In that case, "the Lorentz transformation is a set of equations that converts back and forth between two different observers' measurements of space and time." That would be an error. The truth is, observers are free to use nonlinear clock synchronization schemes and therefore nonlinear coordinate transformations if they like. That choice wouldn't be practical but there is no law of physics against it. See exercise 1 and 2 of the reference Generalized Lorentz Transformations for a valid counterexample. --e.Shubee 17:10, 29 January 2007 (UTC)

Zeeman's theorem isn't a tautology. This is quite clear if we think on the fact that Zeeman's statement is valid only if the dimension of spacetime is greather then 2. In 1 space + 1 time dimension there are automorphisms of spacetime which keep the causal structure but aren't linear, consequently they aren't element of the Lorentz-group. XCelam 16:41, 30 January 2007 (UTC)
Can you tell me Zeeman's definition of spacetime for which his theorem is true? --e.Shubee 20:34, 31 January 2007 (UTC)
The "homogeneity of spacetime" in the sense of Palash B.Pal's paper means that the translation of a rigid rod does not change its length. Zemman does not make such assumptions, and what is more, in the 1+1 dimensional case (in which case Pal derives the Lorentz-transformation) isn't valid at all. It can be that the rod is translated but it's length changes in a reference frame, in spite that the coordinate transformation between the frames preserves causality relation. This means that Zeeman's condition is definitely weaker than Pal's one (beause Zeeman's condition alone does not results in the Lorentz-transformation, only together with the dimension>2 condition, while Pal's condition alone results in the Lorentz-transformation in any dimensions).
But you have asked me about the definition of Zeeman's spacetime. Zeeman defines the Minkowskian spacetime in terms of coordinates. This can be done whatever is the transformation rule between the observers' coordinates. Zeeman does not suppose an absolute Minkowskian spacetime with an observer-independent Minkowski-metric. Every observer defines his affine structure and his metric by means of his coordinates and there isn't an assumption that the Minkowskian or Euclidean distance is preserved during a transformation between the observers or during moving a rod. And there is no assumtion that a straight line transforms to a straight line or a uniform motion transforms to a uniform motion. The only criterium is that the trasformation preserves the causality relation. So in this case the "homogeneity of the Minkowski-space" means here nothing because every observer defines the Minkowski-form on his own coordinates and there is no predefined connection between these coordinates. In other words, this can be done independently what realy the spacetime is and whatever the transformation rules are between the observers. XCelam 07:40, 1 February 2007 (UTC)

I believe you're saying that Zeeman assumed a spacetime structure weaker than Minkowski space. Please define Zeeman spacetime. I asked if you can tell me Zeeman's definition of spacetime for which his theorem is true. That was my question. Please realize that you haven't stated a precise definition yet. --e.Shubee 12:07, 1 February 2007 (UTC)

No, I say that Zeeman doesn't define general spacetime structure, only each observer defines his own Minkowski-structure on his coordinates. The definition itself is the standard definition: R^4 and an nondegenerate, symmetric, indefinite bilinear form on it with index 1. XCelam 13:45, 1 February 2007 (UTC)
I have no clue or idea what an observer-dependent Minkowski structure means. What does it mean physically? --e.Shubee 13:54, 1 February 2007 (UTC)
Perhaps the same as the observer-dependent Euclidean structure :-) XCelam 14:03, 1 February 2007 (UTC)

I can't imagine what an observer-dependent Euclidean space would be if it isn't Euclidean space. If the meaning of a Zeeman-Euclidean space refers to an uncountable number of Euclidean spaces, then that's fine but what does that mean physically? I don't see the point of editors editing Wikipedia articles if they don't understand the subject. --e.Shubee 14:27, 1 February 2007 (UTC)

I don't understand yor problem. You can define an Euclidean metric on x,y,z,t coordinates, but this will be observer-dependent, isn't it? Can you imagine this? XCelam 14:37, 1 February 2007 (UTC)
Euclidean spaces have nothing to do with observers. Euclidean spaces come predefined and they all have the same metric. I can see the physical meaning of an infinite collection of Euclidean spaces and each copy representing one inertial frame of reference. That is the essential starting point of The Axiomatization of Physics - Step 1. But what defines clock time throughout all the Euclidean spaces in Zeeman's universe? There has to be some mathematical link or physical connection between the definition of time in one inertial frame and some another frame. Disjointed realities have no meaning. --e.Shubee 16:00, 1 February 2007 (UTC)

According to JSTOR, Zeeman's theorem states that the group of self-transformations of Minkowski space, preserving causality, is the orthochronous inhomogeneous Lorentz group with dilations. That's an interesting result but no one is calling it a derivation. I therefore believe that your interpretation of Zeeman's theorem is incorrect. --e.Shubee 18:39, 1 February 2007 (UTC)

[edit] Update

The paper by H. J. Borchers and G. C. Hegerfeldt titled The structure of space-time transformations generalizes Zeeman's theorem. The physics is clearly stated to be that the constancy of light alone gives us the Lorentz transformation. If I remember correctly, Vladimir Fock proved that in Appendex A of his book, Theory of Space, Time and Gravitation. The 2nd edition of that book is dated June 1964. According to one review at JSTOR, the second edition of The Theory of Space Time and Gravitation by V. Fock is very little different from the first English edition published in 1959. I'm certain that we're talking about a very old result. --e.Shubee 19:27, 1 February 2007 (UTC)

[edit] An example for a observer-dependent Minkowski-structure and for a causal automorphism which isn't linear

Quote:

I have no clue or idea what an observer-dependent Minkowski structure means. What does it mean physically? --e.Shubee 13:54, 1 February 2007 (UTC)

Dear Shubee, I show you an example of a possible world having 1 time and 1 space dimension, in which there isn't absolute Minkowskian, nor Euclidean structure, still each observer can define his Minkowski-form and define a "causality relation" based on it. In this fancy world, the coordinate transformation rule preserves the observer's causality relations (in spite that there isn't an invariant Minkowski structure), but isn't linear, hence it isn't a Lorentz-transformation.

This world is the following.

Each observer measures his x and t coordinates by his resting measuring rods and by his resting clocks. Each observer defines his Minkowski-form as Q(x,t) = x^2 - t^2 \, and a causality relation (x_1,t_1)<\cdot (x_2,t_2) \, defined by t_1 \,<t_2 and Q(x_2-x_1,t_2-t_1) = 0 \, (we don't know what is the physical meaning of this form and this relation but we can't prohibit them to define anything they want.) The physicists of this world realize that the transformation rule between their (x,t) \, coordinates and the (x',t') \, coordinates of an observer movig with velocity v \, with respect them is

x' = \frac{1}{2}(1+v^2)(t+x)(1+v^2(t+x)^2) - \frac{1}{2}(1+v^4)(t-x)(1+v^4(t-x)^4) \,,
t' = \frac{1}{2}(1+v^2)(t+x)(1+v^2(t+x)^2) + \frac{1}{2}(1+v^4)(t-x)(1+v^4(t-x)^4) \,.

This transformation preserves their causality relation, but has nothing to do with the Lorentz-transformation, and what is more, it isn't linear at all.

Zeeman's theorem states that such a strange world isn't possible in higher dimensions; if dim > 2 then all causality-preserving transformation is composed from a Lorentz-transformation, a translation and a dilatation (multiplication with a scalar).XCelam 12:03, 3 February 2007 (UTC)

You have explained that clearly. Each observer uses the quadratic form x^2 - t^2 \, to mimic the definition of simultaneous events in Minkowski space. Causality is preserved for all observers, not because x^2 - t^2 \, is invariant but because x'^2 - t'^2 = f(x,t,v)(x^2 - t^2)\,. The proof then shows that all causality preserving automorphisms must be linear in higher dimensions. I still don't think that such a gimmick should be called a derivation of the Lorentz transformation unless it is called that in the literature. And you have shown why it is incorrect to say "the Lorentz-transformation is the consequence of the only requirement of keeping causality relations on spacetime." The proof doesn't presuppose a general definition of causality as that sentence implies. The proof presupposes a Minkowskian definition of causality. Why is that remarkable?
I also believe it's unfair to say that "[Pal's] derivation invokes the natural, but unnecessarily strong assumption of homogeneity of spacetime" but that Zeeman's theorem provides "A deeper analysis." We're talking about two completely different ways of coming up with the Lorentz transformation. Zeeman's theorem might be more difficult than Pal's derivation but this is not an instance where it is mathematically correct to say that one assumption is stronger than the other. --e.Shubee 19:14, 3 February 2007 (UTC)
Hmmm... So, it seems unfair. OK, I will try to reformulate this sentence. However, that is an indubitable fact that causality-preserving condition is weaker in mathematical sense that homogeneity, because the former works only together with the dim>2 condition, while the latter works in dim=2 case too. But I will try to refine my sentence. XCelam 21:30, 3 February 2007 (UTC)

[edit] Error in the article cited in section "Derivation"

The derivation in the cited Nothing but Relativity article contains a serious error.

On Page 2 equation (9) is wrong.

There stands:

Suppose there is a rod placed along the x-axis such that its ends are at points x1 and x2 in the frame S, with x2 > x1. In the frame S′, the ends will be at the points X(x1, t, v) and X(x2, t, v), so that the length would be l′ = X(x2, t, v) − X(x1, t, v) .

This relation would be valid only if the simultaneity relation was independent of the frame. But it isn't in the theory of special relativity.

XCelam 02:33, 10 January 2007 (UTC)


I don't think it is an error. In the notation of the NbR article, the uppercase X represents a function of lowercase x, t, and v. The uppercase function performs the Lorentz transformation from the base frame of reference to the frame of reference that is moving at relative velocity v. Thus, by definition, this function X, which is what the article attempts to derive, provides the correct spatial location of the event in the moving frame of reference given the event's position and time in the base frame of reference. There is no violation of simultaneity whatsoever. First Harmonic 02:42, 10 January 2007 (UTC)
Please think over again. Length is defined by the distance of simultaneous events.
The whole transformation is:
x′ = X(x, t, v)
t′ = T(x, t, v)
So, the (x_1,t) \, coordinates transforms to (X(x_1,t,v),T(x_1,t,v)) \,, while the (x_2,t) \, coordinates to (X(x_2,t,v),T(x_2,t,v))\,.
If T(x_1,t,v)\neq T(x_2,t,v) then these two events aren't simultaneous in S'. Therefore you don't get any length if you subtract the x coordinates of them. XCelam 07:52, 10 January 2007 (UTC)

[edit] Removed Eugene Shubert's original, unpublished research.

I removed the original research by Eugene Shubert because it is massively inappropriate for Wikipedia. I am making a new section on this because I anticipate much arguing between myself and Eugene over the matter.

Jowr 05:21, 9 June 2007 (UTC)

Good spot - I hadn't noticed its presence on this article. It is not only massively inappropriate - in the first place it is massive nonsense. He might have put this reference in other articles as well. DVdm 06:26, 9 June 2007 (UTC)

[edit] Untitled image?

The image "Animated Lorentz Transformation Frame.png" at the top of the page should have at least some description shouldn't it? --Beast of traal 21:25, 9 June 2007 (UTC)Beast of traal

: I agree; also, I think a better image could be found to grab the readers attention in a more informative way. MP (talk•contribs) 21:25, 6 November 2007 (UTC)

[edit] Suggestion for derivation section

I have included a derivation show/hide box for the derivation section for primarily 2 reasons:

  • Better aesthetics.
  • There should clearly be more derivations in the article (such as some of the other standard ones), and the show/hide option is ideal for this as it won't make the article inordinately long.

As an aside, the derivation intro. should be improved (I'll try to do this myself, sometime) as there currently appears to be a discussion that goes off into a tangent. MP (talk•contribs) 21:36, 6 November 2007 (UTC)

[edit] Error in caption of GIF animation

The caption of Diagram 1 ("Views of spacetime along the world line of a rapidly accelerating observer") says: The lower quarter of the diagram shows the events visible to the observer. Upper quarter shows the light cone- those that will be able to see the observer. I think this is wrong: one can not see oneself in the past. The light cone in this 1+1 dimensional Minkwoski space are the two diagonal lines. The only events that are visible to an observer in the origin are represented by the dots that pass through the diagonal lines in the lower part of the diagram. A beautiful animation, by the way. JocK (talk) 18:14, 31 December 2007 (UTC)

[edit] Comment by 66.152.149.8

In the matrix Beta must equal v/c² and gamma must equal (1-(v/c)²)^½ else the whole transformation is lost —Preceding unsigned comment added by 66.152.149.8 (talk • contribs)

Please note the presence of the factor c in ct and ct'. I have reverted your edit. DVdm (talk) 14:51, 19 February 2008 (UTC)
Even with ct it makes it impossible to get back the transformation. if t'= gamma(t-Beta*x). in the first matrix t' will then only be =gamma(t-v*x/c) when what we need is =gamma(t-(v*x)/(c)²). And as for ct'=gamma(t-Beta*x) as stated currently on the site give t'=gamma((t/c)-(v*x)/(c)²) witch is also incorrect. Do you not agree? —Preceding unsigned comment added by 66.152.149.8 (talk) Tjd195 (talk) 22:05, 19 February 2008 (UTC)
never mind I have noted my mistake —Preceding unsigned comment added by Tjd195 (talk • contribs) 22:30, 19 February 2008 (UTC)

[edit] First equation of Lorentz Transformation

Lorentz modified t’ = t – (vt/c) to get his first equation. He made t as a function of variant x by t(x) = x/c. That is fine, according to mathematical regulation of function he could change the function t’(t) to t’(t(x)) = t(x) – (v t(x)/c) or simply t’(x) = ((c-v)/c^2)x. But then that is not what he wanted. That new function t’(x) simply means to the observer at the position x in the stationed system, what will be the arrival time of the ray which was emitted from the origin of the moving system at the time t = t’ = 0 when two origins met.

Since that is not what he wanted, he substituted only one of the variant, the second variant, and leave the first variant as an additional variant in the original function. That means, he actually changed the function t’(t) to a new function t’(t, x) because of that mathematical error. Logically both “If A&B then C” and “IF A then C” could be true if B is not related to the reasoning. For example, both “If it is an apple and it is red then it is a fruit.” and “If it is an apple then it is a fruit.” are valid. But “If t’(t) = ((c-v)/c)t and t(x) = x/c then t’(t, x) = t – (vx/c^2).” is valid while “IF t’(t) = ((c-v)/c)t then t’(t, x) = t – (vx/c^2).” is not true.

Actually, when a man measures the actual time period t of an event happened in an inertial system which is moving toward him at constant speed v, and the measured result is t’, then t’ = ((c-v)/c)t. If the inertial system moves away from him at constant speed v, then t’ = ((c+v)/c)t. Time relationship is just so simple.

John C. Huang (talk) 18:12, 16 March 2008 (UTC)

My mistake. Here is the correction:

Lorentz derived his first equation from his two second equations of his transformation and inverse transformation. However, my reasoning provided another possible time formula.

John C. Huang (talk) 04:00, 4 April 2008 (UTC)

[edit] Vandalism in 'Matrix Form' section.

In the first matrix shown in the section in question, a fifth line states "lolololol". I'm pretty sure there shouldn't be any "lol" in a matrix... I'd change it, but I don't know how the math formatting works. mj_sklar (talk) 22:23, 7 May 2008 (UTC)

[edit] Transformation in hiperbolic form

In the section of the Lorentz tranformation in hiperbolic form there's only the matrix of transformation in the x axis. There shoud be the tranformation for a movement towards any direction. I now this matrix but I'm not familierized with mathematical scripture in wikipedia so if anyone could write it would be useful and would complete the article. I wanted also to say that the hiperbolic form (using rapidity) should be the first one in the article, beacouse it's the most powerful and the most useful one for those who work on general relativity.It should also say that the Lorentz transformation is a Tensor and put a link to the page explaining them.

--Eudald (talk) 10:13, 17 May 2008 (UTC)

[edit] On proper Lorentz transformations

In my opinion, there is a mistake in the second following sentence : Lorentz transformations with \det ({\Lambda^\mu}_\nu)=+1 are called proper Lorentz transformations. They consist of spatial rotations and boosts and form a subgroup of the Lorentz group.

Because the PT symmetry, which reverse space's and time's directions, is a element of the proper Lorentz transformations group.

Heartily. LyricV (talk) 12:01, 1 June 2008 (UTC)