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Talk:Lorentz transformation

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Contents

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y 1 Error in Section "Details"y 2 Confusingy 3 Mergey 4 Proper foundationy 5 Lorentz Transformation is Flawedy 6 Derivation of Lorentz Transformationy 7 Merging?y 8 Triangle of Velocities proves that invariant speed is mathematical error y 9 Remove prime notation and put transforms at topy 10 Variable definitionsy 11 time dilation and simultaneityy 12 Not cartesian but...y 13 Simplery

14 Lorentz transormations vs Length contractiony 15 Animated lorentz transformationy 16 Order changed to make it easier for non-experts to followy 17 History sectiony 18 Causality implies the Lorentz group

o 18.1 Two errorso 18.2 Updateo 18.3 An example for a observer-dependent Minkowski-structure and for a

causal automorphism which isn't linear


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y 19 Error in the article cited in section "Derivation"y 20 Removed Eugene Shubert's original, unpublished research.y 21 Untitled image?y 22 Suggestion for derivation sectiony 23 Error in caption of GIF animationy 24 Comment by 66.152.149.8y 25First equation of Lorentz Transformationy 26 Vandalism in 'Matrix Form' section.y 27 Transformation in hiperbolic formy 28 On proper Lorentz transformationsy 29 Group Postulate Derivation

o 29.1 There is nothing wrong in dividing by zeroo 29.2 Inline math renderingo 29.3 We need more citations and less toying around with derivationso 29.4 Do we really need instead of 1 / c2?

[edit] Error in Section "Details"

Quote:

Then, we can write

Endquote.

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

with respect to a metric is:

s2 = x

x

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


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So in this case with

We get:

s2 = xx = (ct)2 + (x)2 + (y)2 + (z)2 = c2(t)2 + (x)2 + (y)2 + (z)2

OK, I think its fixed. Please check it. PAR14:30, 17 September 2006 (UTC)Ok, there's some style details. I find it unfortunate how the spacetime events A and Bare notated by upper case letters. The norm with respect to the minkowski metric ismost naturally written in tensor co/contravariant notation as with thes2 = x

x,wherex denotes a vector x and where its components are denoted byx1,x2 orgenerallyx . There's another clash with the use of x as in x,y and z of the spatialcomponents. This might confuse readers. I'm not sure how to go about this.FlorianPaulSchmidt19:16, 17 September 2006 (UTC)

[edit] Confusing

The below, addedto the Historysection, isconfusing:c is then identified to the propagation speed ofelectromagnetic 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 subtlydifferent 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 andthis bunch who seem to be staking their careers on Lorentz being right after all: "The Einsteinassumptions 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 isconsistent with relativistic effects, which are caused by actual dynamical effects of absolutemotion through the quantum foam, so that it is Lorentzian relativity that is seen to beessentially correct." I suggest "proper foundation" is not NPOV; but WP has been sufficientlyabusive to me I'm doing no more than to leave this note. -- fmr Kwantus (March 2004)


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[edi L entz T nsformation is Flawed

The Lorent transformation is act all nothing but a 'cheated' Galilei Transformation:it usesa vectorial velocit addition forlight signals (which contradicts the invariance ofthe speed oflight and then this erroris being 'rectified' by changing the originallength and time units (see

my page http://www.physicsmyths.org.uk/lightspeed.htm for more).In orderto understand the interpretation ofthe invariance ofthe speed oflight beholdthe eigenvectors ofthe Lorent 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 andthe same:the invariance ofthe speed oflight. Then any "demonstration" would be redundant(i.e. already in the hypothesis). The whole discussion aboutthe Lorent transformation is the

physical validation orinvalidation of his mathematical expression oft' in relation to t. Ifpossible please post relevant material of how he arrived atthe proposed equation.--Lucian20:00, 27 May 2005 (UTC)

Yeah, I failto understand how that website proves anything. Perhaps the originalposter should read up on the Lorent groupto learn how the Lorent transformationisn't a 'cheated' Galilean transformation, but rather an extension of rotations. The mostintuitive way to understand the derivation of Lorent transformations is to see it asfinding a transformation that preserves c (rotations) and extending thatin a reallysimple way, ratherthan pulling a new transformation out ofthin air, which is usuallyhow it's presented to freshmen undergrads). --Laura Scudder| Talk20:34, 27 May2005 (UTC)

The error made in the website is the statement:

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

The same amount has not reeled off. In the GB frame, the GB -rope is moving at velocityv while the car-rope is at rest. This means thatthe GB -rope is Lorent -contracted, while thecar-rope is not. At any given time in the GB frame, the GB -rope will measure a greaterdistance to the carthan 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 itis 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, 17April 2006 (UTC)

[edit] Derivation ofLorentz Transformation

I am missing a derivation ofthe Lorentz transformation on Wikipedia. I thinkthis is so vitalforthe development of SpecialRelativity (both historically and didactically) that one shouldnot have to referto externallinks forthis.


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

[edit] Merging?

Shouldn'tthis be merged with Lorentz-Fitzgerald contraction hypothesis? This article is aboutrelativity and that one is aboutluminiferous aether, butit's basically the same contractionused fortwo different purposes and it seems like it should be one article. Ken Arromdee18:05, 5 October 2005 (UTC)

Maybe - butthe contraction hypothesis should be included in the Lorentztransformation article, not vice versa. PAR23:49, 5 October 2005 (UTC)

[edit] Triangle of Velocities proves that invariant speed is

mathematical error

The article athttp://www.masstheory.org/triangle_of_velocities.pdfpresents detailed analysisof Lorentz transformation, it's derivation procedure and formal mathematical proofthatintroducing "invariant" speed is indeed nothing but mathematical error.

I'd be interesting to know ifthe person who posted this understands whatthe author of"triangle of velocities" wrote. I read it, and I think he made a mistake in assumptionsaboutthe variables - ie what variable corresponds to what - which i admitis veryconfusing (which is why I wrote the variable definitions out on this page forthe SCtransforms). Fresheneesz 21:16, 17 April 2006 (UTC)

The article is funny but not serious, and reflects serious mistakes in physics knowledge. Theassert 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 spacetimepointtransforms, not forthird velocity transforms. Forthese cases, Lorentz transforms helpsus to getthe accurate 'velocity transform'. The article is not serious nor matematicamenterigorous. On the other hand, the speed ofthe lightis not already an observation, is aninternational definition: 2.99798458E8 m/s.Mel VisoPreceding unsigned comment added by80.25.164.213 (talk) 09:47, 29 January 2008 (UTC)

Was it necessary to say, in the definitions ofx1 and x2, thatthe "point" was stationary in oneframe and moving relative to the other? Couldn'tthese be the coordinates of a point movingrelative to both frames (measured relative to each frame) in which casedx1/dt1is the speedofthe pointin one frame and dx2/dt2its speed in the other? Is there any reason why one ofthese 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 willnecce arily not be 0 in another frame (a frame at a different velocity). Fresheneesz09:29, 20 April 2006 (UTC)I thinkthatthe definitions as written were confusing. Making the point stationary inone frame may appearto be a simplification, but gives a (false?) impression thatthiswas necessary, thatthe transformations do not apply unless the pointis stationary inone ofthe frames. E4mmacro 21:23, 20 April 2006 (UTC)


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I agree, I did a terriblejob on that. I was trying to getit all straightin my head. I'vetried to think aboutit more, but my laziness is getting the better of me. I thinkit worksif you consider (t,x,y,z) and the other one to bejusttwo different coordinate systems,

but for me - that doesn't help me think aboutit. If I could get off my ass(metephorically) and reworkit, I think I would go backto saying thatthey're simply

points in a coordinate system, butit needs to be wellexplained- because otherwise

they'rejust 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 usesubscripts here. Perhaps like this with front and back subscripts:0Xs . Also, the transformsshould go atthe top for reference, not burried underthe "details" section. User:fresheneesz128.111.95.147 22:30, 10 April 2006 (UTC)

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

[edit] Variable definitions

I finnally put allthe variable definitions down forthe SClorentz transforms - but I wouldappreciate 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 referenceframes.

How was time dilation used to prove relativity of simultaneity? Is itjusttrying to say theLorentz transformations incoporate both time dilation and reletivity of simultaneity? There issome related discussion in the relativity of simultaneity page. E4mmacro 06:28, 20 April2006 (UTC)

[edit] Not cartesian but...

The first equation under"details" only holds for certain coordinate systems, but I forgotthe

name ofthem. The equivalent ofCartesian coordinate systems in 3-space. PAR00:52, 21April 2006 (UTC)

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

[edit] Simpler


7/21

I have been studying special relativity in college for the past semester, and I've learned theseequations. That said, they looked a great deal foreign to me when I saw them onthis page. Inclass, 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 theWikicommunity 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 s Length contraction

I have been very confused with how the lorentz transformations translate into lengthcontraction. 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 aremeasured 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* and . I don'tunderstand 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 inthe article space that two distantclocks (with (t2 - t1 = 0) are read in the stationaryframe 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. Harald8815:07, 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 themain 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 doso? Han-Kwang 20:51, 12 July 2006 (UTC)

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

follow


8/21

The article started out with the section "Lorentz boostin 2 dimensions". This text would beunintelligible to a reader who didn't already understand the formulation, in thatit usedundefined terms and undefined mathematics (i.e., ifthe reader didn't already know what"theform xy"is, the article is incomprehensible.) I moved this to come second, and added "inmatrix form"to clarify (possibly I should have said "tensor" or"linear algebraic form," butsince tensors aren't brought up yet either...)

The word "form" should be clarified with a linkto the wikipedia articlebilinear form, I think.

[edit] History section

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

[edit] Causality implies the Lorentz group

Quote:

Ifspace is homogeneous, then the Lorent transformation must be a linear transfor mation.

Also,since relativitypostulates that the speedoflightis thesamefor allobservers,it must

preserve the spacetime intervalbetween any two eventsin Minkowskispace

Howeveritis very common to require linearity from Lorentz transformation, really we don'thave to make such strong assumptions. In an 1964 paper, E.C. Zeeman proved thatthecausality preserving group is equalto the Lorentz group. Thatis, instead of requiring linearityand isometry from the coordinate transformation between inertial observers, itis enough torequire causality preservation. Linearity and isometry are mathematical consequences!

The paper:E. C. Zeeman, Journal ofMathematical Physics -- April 1964 -- Volume 5, Issue4, pp. 490-493

This resultis also found inNaber, Gregory L., The Geometry ofMinkowski Spacetime,Springer-Verlag, New York,

99

. ISBN 0-387-97848-8 (hardcover),ISBN 0-486-43

35-

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 ofthat section is this claim:

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

The abstract of Zeeman's paper simply states:


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"Causality is represented by a partial ordering on Minkowski space, and the group ofall automorphisms that preserve this partial ordering is shown to be generated by theinhomogeneous Lorentz group and dilatations." 1964 The American Institute ofPhysics.

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

homogeneous spacetime.

The resultlinking the light-cone preserving transformations with the Lorentztransformatins is (a) not something that"just Zeeman came up with", (b) runs muchdeeperthat some "misconception of one person" as you're painting it as, (c) has beenaround along time (since A. A. Robb 1915) in the literature, and (d) is a more-or-less

part ofthe standard folklore in mathematical physics and is widely accepted, and (e)is the basis ofthe name causal group. The correct statementis this:in any affinespace (of dimension 3 or greater), the transformations preserving a light conestructure comprise the causal group. This consists ofthe inhomogeneous Lorentztransformations, along with global scale change (dilations). For 4 dimensional affinespaces, this comprises a 11-parameter group. This can be generalized to projectivespaces (in which case, one gets the 15-parameter conformal group).The underlying geometric basis forthe result was laid out backin 1915 by A. A.Robb, who showed that one can systematically reconstructthe spatial part ofthespace-time geometry (e.g. definition of congruence, angles, orthogonality, etc.) fromthe light cone structure, alone. Underlying this are such constructions as the Robbrectangle, which is used to define orthogonality between spacelike and timelikeintervals (and, indirectly, spatial orthogonality).Of course you're beginning in a Minkowski space (more precisely:in an affine space

possessing an invariantlight cone structure). That's the whole setting ofthe discussionin this article! Calling Zeeman misleading on this countis, itself, misleading.

Nobody's talking about deriving Minkowski space (whateverthat means) here;justthe Lorentz transformation; and allthe objections you're raising are thereforeirrelevant. -- Mark July 2, 2008Preceding unsigned comment added by 64.24.187.116 (talk)06:28, 2 July 2008 (UTC)

The second erroris not as serious. The statement, "Ifspace is homogeneous, then the Lorent transformation must be a linear transformation"is only misleading. It all depends onwhatever definition ofthe Lorentz transformation you begin with. Forinstance, we couldadoptthe opening sentence ofthe article as a definition. In that case, "the Lorent transformation is a set ofequations thatconverts back andforth between two different

observers'measurements ofspace andtime."That would be an error. The truth is, observersare free to use nonlinear clock synchronization schemes and therefore nonlinear coordinatetransformations ifthey like. That choice wouldn't be practical butthere is no law of physics

againstit. See exercise 1 and 2 ofthe reference Generalized Lorentz Transformations for avalid counterexample. --e.Shubee 17:10, 29 January 2007 (UTC)

Zeeman's theorem isn't a tautology. This is quite clearif we think on the factthatZeeman's statementis valid only ifthe dimension of spacetime is greatherthen 2. In 1space + 1 time dimension there are automorphisms of spacetime which keep thecausal structure but aren'tlinear, consequently they aren't element ofthe Lorentz-group. XCelam 16:41, 30 January 2007 (UTC)


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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 thatthetranslation of a rigid rod does not change its length. Zemman does not make suchassumptions, and whatis more, in the 1+1 dimensional case (in which case Palderives the Lorentz-transformation) isn't valid at all. It can be thatthe rod is translated

butit's length changes in a reference frame, in spite thatthe coordinate transformationbetween the frames preserves causality relation. This means that Zeeman's conditionis definitely weakerthan Pal's one (beause Zeeman's condition alone does not resultsin the Lorentz-transformation, only together with the dimension>2 condition, whilePal's condition alone results in the Lorentz-transformation in any dimensions).But you have asked me aboutthe definition of Zeeman's spacetime. Zeeman definesthe Minkowskian spacetime in terms of coordinates. This can be done whateveris thetransformation rule between the observers' coordinates. Zeeman does not suppose anabsolute Minkowskian spacetime with an observer-independentMinkowski-metric.Every observer defines his affine structure and his metric by means of his coordinatesand there isn't an assumption thatthe Minkowskian or Euclidean distance is preservedduring a transformation between the observers or during moving a rod. And there isno assumtion that a straightline transforms to a straightline or a uniform motiontransforms to a uniform motion. The only criterium is thatthe trasformation preservesthe causality relation. So in this case the "homogeneity ofthe Minkowski-space"means here nothing because every observer defines the Minkowski-form on his owncoordinates and there is no predefined connection between these coordinates. In otherwords, this can be done independently what realy the spacetime is and whateverthetransformation rules are between the observers. XCelam 07:40, 1 February 2007(UTC)

I believe you're saying that Zeeman assumed a spacetime structure weakerthan Minkowskispace. Please define Zeeman spacetime. I asked if you can tell me Zeeman's definition ofspacetime for which his theorem is true. That was my question. Please realize that youhaven'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 observerdefines his own Minkowski-structure on his coordinates. The definition itselfis thestandard definition:R^4 and an nondegenerate, symmetric, indefinite bilinear form onit with index 1. XCelam 13:45, 1 February 2007 (UTC)I have no clue oridea what an observer-dependentMinkowski structure means. Whatdoes it mean physically? --e.Shubee 13:54, 1 February 2007 (UTC)Perhaps the same as the observer-dependent Euclidean structure :-) XCelam 14:03, 1February 2007 (UTC)

I can'timagine what an observer-dependent Euclidean space would be ifitisn't Euclideanspace. Ifthe meaning of a Zeeman-Euclidean space refers to an uncountable number ofEuclidean spaces, then that's fine but what does that mean physically? I don't see the point ofeditors editing Wikipedia articles ifthey don't understand the subject. --e.Shubee 14:27, 1February 2007 (UTC)

I don't understand yor problem. You can define an Euclidean metric on x,y,z,tcoordinates, butthis will be observer-dependent, isn'tit? Can you imagine this?XCelam 14:37, 1 February 2007 (UTC)


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Euclidean spaces have nothing to do with observers. Euclidean spaces comepredefined and they all have the same metric. I can see the physical meaning of aninfinite collection of Euclidean spaces and each copy representing one inertial frameof reference. That is the essential starting point ofThe Axiomatization of Physics -Step 1. But what defines clock time throughout all the Euclidean spaces in Zeeman'suniverse? There has to be some mathematical link or physical connection between the

definition of time in one inertial frame and some another frame. Disjointed realitieshave no meaning. --e.Shubee16:00, 1February 2007 (UTC)

According to JSTOR, Zeeman's theorem states that the group of self-transformations ofMinkowski space, preserving causality, is the orthochronous inhomogeneous Lorentz groupwith 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.Shubee18:39, 1February 2007 (UTC)

[edit] Update

The paper by H. J. Borchers and G. C. Hegerfeldt titled The structure of space-timetransformations generalizes Zeeman's theorem. The physics is clearly stated to be that theconstancy 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 andGravitation. The 2nd edition of that book is dated June 1964. According to one review atJSTOR, the second edition ofThe Theory of Space Time and Gravitationby V. Fock is verylittle different from the first English edition published in 1959. I'm certain that we're talkingabout a very old result. --e.Shubee19:27, 1February 2007 (UTC)

[edit] An example for a obser er-dependent Minkowski-structure and for a

causal automorphism which isn't linear

Quote:

Ihave no clue or idea whatan observer-dependentMinkowskistructure means Whatdoes itmean physically? --e.Shubee1

:54, 1 February2007 (UTC)

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

This world is the following.

Each observer measures hisx and tcoordinates by his resting measuring rods and by his

resting clocks. Each observer defines his Minkowski-form as and a

causality relation defined by and (we don't knowwhat is the physical meaning of this form and this relation but we can't prohibit them todefine anything they want.) The physicists of this world realize that the transformation rule


12/21

between their coordinates and the coordinates of an observer movig withvelocity with respect them is

,

.

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).XCelam12:03, 3 February 2007(UTC)

You have explained that clearly. Each observer uses the quadratic form tomimic the definition of simultaneous events in Minkowski space. Causality is

preserved for all observers, not because is invariant but because

. The proof then shows that all causalitypreserving automorphisms must be linear in higher dimensions. I still don't think thatsuch a gimmick should be called a derivation of theLorentz transformation unless it iscalled that in the literature. And you have shown why it is incorrect to say "theLorentz-transformation is the consequence of the only requirement of keepingcausality relations on spacetime." The proof doesn't presuppose a general definition ofcausality as that sentence implies. The proof presupposes a Minkowskian definition ofcausality. Why is that remarkable?I also believe it's unfair to say that "[Pal's] derivation invokes the natural, butunnecessarily strong assumption of homogeneity of spacetime" but that Zeeman'stheorem provides "A deeper analysis." We're talking about two completely differentways of coming up with the Lorentz transformation. Zeeman's theorem might be moredifficult than Pal's derivation but this is not an instance where it is mathematicallycorrect to say that one assumption is stronger than the other. --e.Shubee19:14, 3February 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 inmathematical sense that homogeneity, because the former works only together withthe dim>2 condition, while the latter works in dim=2 case too. But I will try to refinemy sentence. XCelam 21:30, 3 February 2007 (UTC)

[edit] Error in the article cited in section "Deri ation"

The derivation in the citedNothing but Relativity article contains a serious error.

On Page 2 equation (9) is wrong.


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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 thelength 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 uppercaseXrepresents a function of lowercasex, t, and v. The uppercase function performs theLorentz transformation from the base frame of reference to the frame of reference thatis moving at relative velocity v. Thus, by definition, this functionX, which is what thearticle attempts to derive, provides the correct spatial location of the event in the

moving frame of reference given the event's positionand time in the base frame ofreference. There is no violation of simultaneity whatsoever.First Harmonic 02:42, 10January 2007 (UTC)Please think over again. Length is defined by the distance ofsimultaneous events.The whole transformation is:x = X(x, t, v)t = T(x, t, v)

So, the coordinates transforms to , while the

coordinates to .

If then these two events aren't simultaneous in S'.Therefore you don't get any length if you subtract the x coordinates of them.XCelam07:52, 10 January 2007 (UTC)

[edit] Remo ed Eugene Shubert's original, unpublished

research.

I removed the original research by Eugene Shubert because it is massively inappropriate forWikipedia. I am making a new section on this because I anticipate much arguing betweenmyself 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 massivelyinappropriate - in the first place it is massive nonsense. He might have put thisreference in other articles as well. DVdm 06:26, 9 June 2007 (UTC)

[edit] Untitled image?

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


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: I agree; als

, I think a better image c

uld be f

und t

grab the readers

attenti

n in a m

re inf

rmative way. MP(talkc

ntribs)21:25, 6 N

vember 2007

(UTC)

[edit] Suggestion for derivation section

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

y Better aesthetics.y There should clearly be more derivations in the article (such as some ofthe other

standard ones), and the show/hide option is ideal forthis as it won't make the articleinordinately long.

As an aside, the derivation intro. should be improved (I'lltry to do this myself, sometime) asthere currently appears to be a discussion that goes offinto a tangent. MP(talkcontribs) 21:36, 6

November 2007 (UTC)

[edit] Error in caption of GIF animationThe caption of Diagram 1 ("Views ofspacetime along the worldline ofa rapidly acceleratingobserver") says:The lower quarter ofthe diagram shows the eventsvisible to the observer.Upper quartershows the lightcone- those thatwillbe able to see the observer. I thinkthis iswrong: one can not see oneselfin the past. The light cone in this 1+1 dimensionalMinkwoskispace are the two diagonallines. The only events that are visible to an observerin the originare represented by the dots that pass through the diagonallines in the lower part ofthediagram. 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/c and gamma must equal (1-(v/c) )^ else the wholetransformation is lostPreceding unsigned comment added by 66.152.149.8 (talk contribs)

Please note the presence ofthe factor c in ctand ct'. I have reverted your edit. DVdm(talk) 14:51, 19 February 2008 (UTC)Even with ctit makes itimpossible to get backthe transformation. ift'= gamma(t-Beta*x). in the first matrix t' willthen only be =gamma(t-v*x/c) when what we needis =gamma(t-(v*x)/(c) ). And as for ct'=gamma(t-Beta*x) as stated currently on thesite give t'=gamma((t/c)-(v*x)/(c) ) 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 February2008 (UTC)

[edit] First equation ofLorentz Transformation

Lorentz modified t = t (vt/c) to get his first equation. He made t as a function of variant xby t(x) = x/c. Thatis fine, according to mathematical regulation of function he could changethe function t(t) to t(t(x)) = t(x) (v t(x)/c) or simply t(x) = ((c-v)/c^2)x. Butthen thatisnot what he wanted. That new function t(x) simply means to the observer atthe position x in


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the stationed system, what will be the arrivaltime ofthe ray which was emitted from theorigin ofthe moving system atthe time t = t = 0 when two origins met.

Since thatis not what he wanted, he substituted only one ofthe variant, the secondvariant, and leave the first variant as an additional variantin 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&Bthen C and IF A then C could betrue ifBis not related to the reasoning. For example, both Ifitis an apple and itisred then itis a fruit. and Ifitis an apple then itis a fruit. are valid. But Ift(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)tthen t(t, x) = t (vx/c^2). is nottrue.

Actually, when a man measures the actualtime period t of an event happened in an inertialsystem which is moving toward him at constant speed v, and the measured resultis t, then t= ((c-v)/c)t. Ifthe inertial system moves away from him at constant speed v, then t =((c+v)/c)t. Time relationship isjust 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 andinverse transformation. However, my reasoning provided another possible time formula.

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

[edit] Vandalism in 'Matri 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 mathformatting works. mj_sklar(talk) 22:23, 7 May 2008 (UTC)

[edit] Transformation in hiperbolic form

In the section ofthe Lorentz tranformation in hiperbolic form there's only the matrix oftransformation in the x axis. There shoud be the tranformation for a movementtowards anydirection. I now this matrix but I'm not familierized with mathematical scripture in wikipediaso if anyone could write it would be useful and would complete the article. I wanted also tosay thatthe hiperbolic form (using rapidity) should be the first one in the article, beacouse it'sthe most powerful and the most useful one forthose who work on general relativity.It should

also say thatthe Lorentz transformation is a Tensor and put a linkto the page explainingthem.

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

[edit] On proper Lorentz transformations


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In my opinion, there is a mistake in the second following sentence: Lorentztransformations

with are calledproper Lorentz transformations. Theyconsistofspatial

rotations and boosts and form asubgroup ofthe Lorentz group.

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

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

[edit] Group Postulate Deri ation

Firstly, calling this Group Postulate Derivation seems to be just a fancy way of saying thatthe transform must be linear. Is this right ? If so, then why not say that instead ?

Thank you Bakken for adding the Group axioms.Delaszk (talk) 07:45, 4 November2008 (UTC)

Secondly the following statement seems to be incorrect: "Theoretically it can be eitherinfinitely large, which gives Galilean transformation and Euclidean world with absolute time,or it can be finite, which gives Lorentz transformation and Minkowski world of specialrelativity."

Given two transforms with finite v and v' then the entries of the transform matrix must befinite and so the entries of the inverse matrix must be finite therefore the expression:

must be finte therefore must be finte and hence c isfinite.

The Galilean transforms may be a limit of the Lorentz as c tends to infinity, but an actualvalue of inifinity is inconsistent with this derivation of the Lorentz transformation.Delaszk(talk) 13:20, 2 November 2008 (UTC)

Those matrices are the inverses of each other, and have finite entries. I'm tweaking the

derivation to make the point more rigorous. (If (v) is zero and v(v) isn't, isn'tfinite. OTOH, (v) cannot be zero, as 2 + v = 1, and so v(v) is nonzero when v is

nonzero.)--Army1987 (t c)14:48, 2 November 2008 (UTC)The changes you have made to the article have inverted the numerator anddenominator. v is now on the denominator which would force c to equal infinity.Delaszk (talk) 15:25, 2 November 2008 (UTC)That was exactly what I intended. And they don't "force" anything as(v) is nonzero.(In the case v = 0 you have a zero-over-zero situation, but in that case the matrixshould obviously be the identity.) --Army1987 (t c)15:52, 2 November 2008(UTC)


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Ok, but is a constant for any value of v, so in particularitis aconstant for any value of v for which delta is nonzero, so ifthere exist any v for whichdelta is nonzero then c is finite. So c could only be infinite if delta was zero for every

possible lineartransform. Delaszk(talk) 16:17, 2 November 2008 (UTC)

In Galilean relativity, (the upper right entry in the matrix) is zero for allv, so there isno inconsistency. --Army1987 (t c) 16:24, 2 November 2008 (UTC)So Galilean relativity may be self-consistent but only by disallowing certaintransforms. Galileo was unaware that he was implicity ignoring such transforms. Ifyou recognise the existence oftransforms with nonzero delta then you get specialrelativity with a finite max-speed built-in. Delaszk(talk) 16:34, 2 November 2008(UTC)

Indeed. (v) equals Kv/1 Kv2;Kis 0 in Galilean relativity and 1/c2in specialrelativity. (BTW, I've fixed the derivation so thatit never divides by terms whichcould ever possibly be zero.) --Army1987 (t c) 17:16, 2 November 2008(UTC)I'm going to rewrite the section on Group Postulate Derivation. As the abovediscussion shows, if you start with a generallineartransformation then you force theexistence of a finite max-speed. There is no choice in the matter. A finite max-speedis a necessary consequence. The Galilean transformations are something elsealtogether which you get by restricting the types of motion that you are willing toaccept. The factthat putting c=infinity into Lorentz transform gives you the Galileantransform is irrelevant because we've already shown that c cannot equalinfinity.Changing the article to avoid talking about division, obscures this fact. So I'm goingto rewrite the article based on a previous edit.Delaszk(talk) 18:04, 2 November 2008(UTC)

I can't understand that. By the principle of relativity you getthe transformations I've

shown. IfKis 0 you get Galilean relativity and ifitis positive you get special relativity with c= 1/K. And there is no a priori reason why Kshould be nonzero that'sjust anexperimental fact. --Army1987 (t c) 18:08, 2 November 2008 (UTC)

If you recognise the existence of nonzero delta in the above discussion (and thederivation does assume this atthe beginning by using a generallineartransform inwhich delta can be anything) then c must be finite.This is part of a comment by Delaszk(of 18:19, 2 November 2008 (UTC)), which was interrupted bythe following:Indeed, "anything". Which includes the zero function. --Army1987 (t c)13:19, 4 November 2008 (UTC)Itis finite if any nonzero delta exist. You cannotlet K=0 unless you also go backto

the beginning ofthe derivation and disallow nonzero delta. The form ofthe Galileanand Lorentz transforms may be similar butthe derivations of each have differentstarting points. c is a constant and has only one value - it cannot vary between finiteand infinitejust by taking different values of delta. The only way to change its valueis to change the starting point ofthe whole derivation, but given the starting pointthatwe have, then c is finite. Delaszk(talk) 18:19, 2 November 2008 (UTC)We must however avoid dividing by zero which is an invalid operation, therefore wedo indeed put delta in the numerator and gamma in the denominator. Gamma is never


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zero, so the expression is always finite. Delaszk(talk) 20:50, 2 November 2008(UTC)In fact all arguments based on the possibility of having zero on the denominator areall complete nonsense. You cannot derive any valid conclusions by having zero on thedenominator. I have now rewritten the article in a way which makes it obvious that chas to be finite. Delaszk(talk) 21:24, 2 November 2008 (UTC)

But c doesn'thaveto be finite. c happensto be finite. THAT is the difference, and itisnot a trivial one.Headbomb { WP Physics} 03:06, 3 November 2008(UTC)If you allow frames of reference to be rotated with respectto one another (and whyshouldn't you ?) then you must have nonzero delta in the upper-right element ofthetransform matrix whenever you have such a rotation. If you let K=0 then delta mustalways be zero therefore K=0 is inconsistent with allowing rotations, therefore Kisnot equalto zero. I suppose the consideration of rotated frames is the crux ofthematter from which the finite-speed emerges. You only get K=0 and hence absolute-time if you disallow frames to be rotated. Delaszk(talk) 08:23, 3 November 2008(UTC)This is not correct, Delaszk: rotations do not affectthe time coordinate. The rotationmatrix has zero components with indexes [0,1],[0,2],[0,3] and [1,0],[2,0],[3,0] and aunit element [0,0]. Rotational part ofthe transformation is identical for Lorentz andGalilean transformation.Ifthatis not correctthen there is stillthe matterthat according to the New Scientistarticle about Feigenbaum's paper, special relativity is a consequence of allowingcertain rotations which arise in a natural way. The references are atTemporarilyreverted material pending evaluation. Delaszk(talk) 07:59, 4 November 2008 (UTC)

[edit] There is nothing wrong in dividing by zero

Delaszk, writing 1 /f(z) instead off(z) does not change a thing. The physical meaning iscompletely the same. The singularities, if genuine, reveal certain physicalmeaning. You can'tavoid a genuine singularity in a physicaltheory by mathematicaltransformations.

For example:the elements ofthe Lorentz matrix diverge at . This is a genuinesingularity which prevents transformations with relative velocity larger, than the speed oflight. It makes the Lorentz group non-compact with allthe consequences. Bakken (talk)10:13, 3 November 2008 (UTC)

That argumentis a red herring. Yes is indeed a genuine singularity butthathas nothing to do with dividing by zero. Dividing by zero is wrong, especially since it

only occurred here by mistake, and has nothing to do with any genuine singularity.Whatis really going on here is this:Calculating the product oftwo transformationmatrices, one with vthe other with v' and comparing the diagonal elements andcancelling like terms gives: (v)v'(v') = (v')v(v). The only question is where do yougo afterthis equation. Dividing through by zero is certainly notthe answer as this isinvalid and produces contradictions. Instead you divide through by the gamma termswhich are nonzero and therefore valid to divide by.Delaszk(talk) 11:36, 3 November2008 (UTC)


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you seem to believe thata = b and 1 /a = 1 /b are very differentthings. They areactually the same. See, if, say, a = 0 then the left-hand term ofthe second equation isinfinitely large. The equation can then only be satisfied when the right-hand side isalso infinitely large. That means the b = 0. See, the second equation has the samemeaning as the first equation even in the case where a = b = 0. Trust me, there isnothing wrong in dividing by zero. In physics we do it quite often. Bakken (talk)

12:33, 3 November 2008 (UTC).I can't believe I'm reading this. Itis a factthata = b and 1/a = 1/b are very differentthings. The first equation says something abouttwo numbers. The second equationsays something abouttwo non-zero numbers. If, say, a = 0, then the second equationis invalid. You can't even say something aboutits left-hand term to begin with. See,the second equation has NO meaning in the case where a = b = 0. Trust me,everything is wrong in dividing by zero. You are nottalking about physics. At bestyou are talking aboutlaziness and carelessness. DVdm (talk) 12:47, 3 November 2008(UTC):) think aboutthe second equation in this way: 1 / (a + i) = 1 / (b + i) where is a

small number. Then it has a perfect meaning fora = b = 0, does it not? See, very oftenthe integral of a function converges, despite the factthatthe function itself diverges.

Many Green's function diverge, the Coulomb cross-section diverges, the classicalpotential energy ofthe electron diverges, the S-matrix diverges at eigenstates oftheHamiltonian. I am sorry, but division by zero simply means that a function has a poleatthis point. Many functions have poles. Bakken (talk) 13:03, 3 November 2008(UTC)Sure, you can thinkaboutit as you like, and I'm sure that every mathematician does it,

butthat does not mean we can write itthat way. Yes, many functions have poles.These are cases where a function has no value, since division by zero is simply notallowed. And yes, there's a lot of divergence around, and it's all properly covered withthe clean and decent mathematics oflimits. I notice that your most recent change says"When v --> 0, ...". Thatis just about barely acceptable. Had you written "When v =0, ...", it would have been totally unacceptable. Anyway, I have made anotherlittlechange to remove the predicate "barely". Cheers, DVdm (talk) 13:24, 3 November2008 (UTC)That's all right with me, to write ... :) See, division by zero is simply "notdefined". There is no such thing as "not allowed"in physics, I don'tthink so. But Iagree that division by zero has to be treated carefully no matter how obvious it mightseem to somebody... Cheers, Bakken (talk) 14:00, 3 November 2008 (UTC)Considertwo analytical functions,f(z) andg(z), of a complex variablez, possibly withzeros and poles. Can you tell me, whatis the difference betweenf(z) =g(z) and 1 /f(z)= 1 /g(z)? Bakken (talk) 14:04, 3 November 2008 (UTC)For z for which f(z)=0 or g(z)=0, the string "1/f(z)=1/g(z)" represents nonsense and is,in properlingo, "not even wrong":-) DVdm (talk) 17:55, 3 November 2008 (UTC)

:) well, I canjust as well say that forz for whichf(z) has a pole, the stringf(z) =g(z)"represents nonsense". What's the principal difference? :) Iff(z) is an analyticalfunction,f(z) and 1 /f(z) contain the same information. You can't say one is any betterthan the other. They are equivalent. Bakken (talk) 18:19, 3 November 2008 (UTC)For z for which f has a pole, there is no such thing as f(z). Your phrase "forzforwhichf(z) has a pole"is nonsense before you can even take breath to finish yoursentence. Also note that"The reciprocal of an analytic function that is nowhere zerois analytic". DVdm (talk) 18:34, 3 November 2008 (UTC)


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I am afraid the discussion descends into tautology. I customarily use phrases like "theanalytical function 1 / (1 +z2) has poles at " and everybody so farunderstood me without problems. If you insist on a more precise language, be myguest,just correct whatever sentences you deem ambiguous. Bakken (talk) 18:47, 3

November 2008 (UTC)Yes, I agree. You customarily use "lazy and careless"language. I try to avoid it. But

as I already said, I can live with the current still somewhat sloppy but sufficientlyacceptable wording. Cheers, DVdm (talk) 18:56, 3 November 2008 (UTC)

If v=0 then the equation : cannot be derived from the equation (v)v'(v')= (v')v(v)

It makes no sense to say thatthe first equation is valid for all v. Itis only valid for nonzero v.Saying itis valid for still doesn'tlet you say the equation is valid for v=0. Delaszk(talk) 14:58, 3 November 2008 (UTC)

Dear Delaszk, let's say we have different opinions on that: you cannot divide by zero,while I can. I suggest we make both opinions visible, by saying thatuser:Delaszk

believes the equation is not valid in the limit . Be my guest, write it down, butdon't delete context you disagree with or do not understand -- it's vandalism. Bakken(talk) 16:46, 3 November 2008 (UTC)I did not say it was invalid in the limit. I said itis not valid for v=0 and you cannotargue thatitis. You cannot everin any context divide by zero. You might be using the

phrase "divide by zero"to mean something else, but I mean the phrase "divide byzero"to mean divide by zero. You also accuse me of vandalism for deleting content Idisagree with when you deleted the content about nonzero v because you disagreedwith it. So you are criticising me for doing exactly what you did. Delaszk(talk) 17:40,3 November 2008 (UTC)I did not delete it, I have rewritten itto make it correct. You can see it on the bytecount ofthe article. If you insist on claiming thatthe equation is not valid fortheidenticaltransformation, please, write it down. Bakken (talk) 18:14, 3 November2008 (UTC)You are playing with words to avoid being seen as a hypocrite. I wrote stuff aboutnonzero v. You changed it. I changed it back. You accused me of vandalism for doingthis. You need to explain why itis okto divide by zero, otherwise I will revertitagain. Delaszk(talk) 18:25, 3 November 2008 (UTC)I don't understand: your phrase "This is valid for nonzero v since (v) is alwaysnonzero, as 2 + v = 1"is not deleted, itis there in the article. What are you talkingabout? Bakken (talk) 21:20, 3 November 2008 (UTC).First of all, I admitthatthere can be contexts in which division by zerois possible, butI'm not sure thatthis article is such a situation. I don'tthinkthat using limits can bedescribed as being the same thing as dividing by zero.

Now in answerto your question: The equation : is derived abouthalf way down the section where it says "Thus, ( v) = (v) and comparing the twomatrices, we get".

Putting gamma=0 into : gives 0=1 which is absurd therefore gammacannot be zero. Now since I was concerned about not allowing zero on the


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denominator of: I don't need to worry about gamma because I'vejust shown thatitis nonzero. The only other factorto worry aboutthat could make thedenominator zero is if v=0. So I don't want v=0 because then the equation doesn'texist, but for all values of v for which the equation exists we don't need to worry

about gamma becoming zero as it can't, therefore for allthese nonzero values of v theexpression is a well-defined finite constant and not some undefined thing like 0/0.Delaszk(talk) 20:40, 3 November 2008 (UTC)well, in my language the equation exists all right forv = 0, as in my language thefunctionsf(z) and 1 /f(z) carry exactly the same information. But I have realized nowthatitis rather offensive for you. And you know what, ifit really is so important for

you,just do it -- consider separately and v = 0 and I won't object. I am sorry, Ididn't realise division by zero can be so a touchy issue. I always divided by zero ifneeded and never gotinto problems so far. I am sorry for my insensitive stubbornness.Bakken (talk) 21:18,

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