Talk:Special relativity/Archive 19

History of this talk page
This is just a heads up that I've undeleted the history of this talk page, formerly at Talk:Special relativity/Archive2 and moved it to Talk:Special relativity/Page history. Neither the history page nor the the redirect should be deleted because they contain backlinks. I could not do a straight history merge to this page because the histories overlap. Graham 87 01:25, 3 June 2008 (UTC)

Notation Issues
In order of importance:


 * replace 4D with 4-space or spacetime: The use of the term 4D implies that the time dimension has the same properties as spatial dimensions
 * Consider replacing Minkowski imaginary notation with covariant and contravariant notation. As far as I can tell, without being an expert.  Using imaginary numbers for the time axis is outdated.
 * Consider using the metric with spatial components having the - sign. I am not an expert on the subject but I think this is the most popular metric and it is pedagogically useful to work toward a common notation.  Personally I prefer this metric because it leads to positive intervals for time-like events.

TStein (talk) 16:02, 18 June 2008 (UTC)


 * I am not aware of any reason why anyone would think that "4D" means anything different than "4-space" or "spacetime". I used "4D" because it makes the titles shorter.
 * As far as I know, there is no use of imaginary numbers in the article now.
 * I am using -+++ because the spatial components of most tensors are more important and it is better to make them match the non-relativistic values rather than have minus signs or whatever. This signature is used in many references also. JRSpriggs (talk) 01:34, 19 June 2008 (UTC)

I didn't mean to be belligerent with my suggestions, I apologize if I came across that way with my quick and blunt suggestions.

I will start with the last comment and work up.

I was only making a suggestion with the choice of +--- over -+++. It is mostly six dozen of one and half dozen of the other. It does bother me that there is no consistency. On wikipedia, for instance this article and Lorentz transformation use -+++, while spacetime and four-vector uses +---. Changing from 1 to the other involves minus signs popping up in weird places.

I will confess to a little drive by reviewing, especially with the imaginary number. When skimming through the article my eye easily caught the one one use of imaginary number (which is after the second equation in the section entitled the geometry of space time) and missed your better discussion earlier. It is the only time that the word 'imaginary' is used on the page so it should be easy to find if you want to kill it.

As far as the use of the term 4D, it is ambiguous. You are right that it is used sometimes to signify 3 dimensions of space + 1 of time. I strongly oppose that because the term is also used for four spatial dimensions such as for tesseracts. See spacetime as well where it discusses 3+1 dimensions. None of the textbooks I have which discuss relativity, which I admit is small number including Griffiths, Jackson, and Goldstein, use the term 4D and for that reason, I believe. It is a small correction, but it makes the article just a little bit stronger.

Looking further I notice as well that this articles uses (t,x,y,z) for the contravariant vector while Lorentz transformation as well as the other articles cited use, in my experience at least, the more common and dimensionally consistent (ct,x,y,z).

On a lighter note, I wish someone would decide whether or not to use 4-vector or four-vector, 4-force of four-force (not just for this article but everywhere). I am ambivalent, but it is kind of irritating to see both being used. TStein (talk) 03:51, 19 June 2008 (UTC)
 * This metric signature standardization can only be done by agreement of lots of people. Doing it here, just within the Wikpedia context, might be possible for us.  The place to start would be to propose a standard for consistency for the Category:physics, perhaps. Wwheaton (talk) 04:06, 19 June 2008 (UTC)
 * I thought the -+++ standard had been adopted for general relativity articles, taking their cue from Misner, Wheeler and Thorne. --Michael C. Price talk 05:26, 19 June 2008 (UTC)


 * To TStein: If you want to raise the issue of a standard for the signature, I suggest that you do so at either Wikipedia talk:WikiProject Relativity or Wikipedia talk:WikiProject Physics.
 * I see that you are correct that ict is mentioned. However, after using it to analogize boosts to rotations, it is dismissed as an inferior method.
 * Personally, I prefer to use SI units consistently. This requires that we use (t,x,y,z) rather than (ct,x,y,z) for the position vector. In turn, that requires that we use different units for time components of tensors than for their spatial components. There is a simple rule for doing this as I explained on another talk page:
 * Actually, the components of a tensor should not all have the same units. In SI units, time is measured in seconds and distance is measured in meters. Thus the position 4-vector has units of [seconds, meters, meters, meters]. More generally, a factor of (seconds/meter)^(# of contravariant time indices &minus; # of covariant time indices) must be multiplied by the units of the purely spatial components to get the units of components involving time.


 * See the reference "Post, E.J., Formal Structure of Electromagnetics: General Covariance and Electromagnetics, Dover Publications Inc. Mineola NY, 1962 reprinted 1997." for more details.
 * If you want to replace "4D" by "spacetime", feel free to do so. JRSpriggs (talk) 08:03, 19 June 2008 (UTC)


 * As far as I can tell, it is the modern convention to have all four-vectors having the same dimension for both the temporal and their spatial components. The displacement 4-vector should not be an exception.  See almost any other article on wikipedia that covers special relativity here for example four-vector for instance.  Of the all the four-stuff on the four-vector page only four-momentum followed your convention.  A quick search of google showed maybe 20% of the pages using your convention and some of those I think are using natural units where the factor of c is absorbed into the unit for time in which case time and space have the same units (inverse MeV if you like).


 * The problem may be of different fields using different conventions. All I have to go on are well established books in electricity and magnetism and classical mechanics.  (Griffiths: Introduction to Electrodynamics, third edition and Jackson: Classical Electrodynamics, 2nd edition.  My Goldstein text is at home.)


 * I am not an expert on relativity, although I do use four-vectors. This is why I will probably not edit the article even though it is much faster then making the argument on this page.  I am just a physicist who is frustrated because a field cannot seem to gets its act together to speak in one language.  Having to deal with different factors of negative signs and c floating around randomly is a pain. (It is a small one, but real nonetheless.)  It must be even worse to someone without experience with this sort of stuff.  TStein (talk) 15:12, 19 June 2008 (UTC)


 * I prefer +--- over -+++, the same as TStein, but for different, more mundane, reason. It is because Landau & Lifshitz and their school use this metric in the LL book and papers, and I read mainly their works and make my calculations with their metric. I realize that Anglo-Americans use -+++ (Misner, J. Synge (Irish)), and whenever I use their results, I have to revert the metric, but this is a rare event for me. For people who use mainly Anglo-American literature it is the opposite, I suppose. A question of practicality, as is Latin vs Greek indices (LL use Greek indices for 3D, Latin for 4D). --Lantonov (talk) 16:00, 19 June 2008 (UTC)
 * BTW Landau & Lifshitz weren't consistent in their usage. Their Classical Theory of Fields 1962 uses -+++, but the 1971 edition uses the +--- convention. --Michael C. Price talk 23:38, 19 June 2008 (UTC)
 * I do not have the 1962 edition so I don't know about it. I have the 1966, 1971, and 1988 editions, and there it is +---. Also I have almost all the papers of the group after 1960 and there it is +--- too. I guess this is because they use mainly the time variable to differentiate when decomposing the metric, and in some cases, as in deriving the Kasner metric, time is the only argument. Also, their extensive use of synchronous frame may be a factor. Inconvenience is that the metric determinant is negative so they have to write $$\scriptstyle{\sqrt{-g}}$$ in tensor densities. I am talking about GR, but maybe there can be some more compelling reasons from SR to use +---. I haven't worked enough in SR to say. I think it's more of a preference, as here, and it depends on the specific problem at hand. One tends to avoid minus signs as far as possible. For example, LL introduce "spatial metric tensor" γαβ = − gαβ to obviate the sign problem. --Lantonov (talk) 05:49, 20 June 2008 (UTC)
 * I prefer -+++ since it makes it easier to work with the spatial components, before extending to the time component. I wasn't aware that it was an Anglo-American standard -- I only wish it was.--Michael C. Price talk 16:14, 19 June 2008 (UTC)

(unindent) As a matter of philosophical principle, the criterion which should guide us in choosing one notational convention over another is &mdash; does it make the job of the practical (experimental) physicist easier or harder? Imagine Alice setting up her laboratory and performing an experiment. She makes notes of the results and publishes a report describing the set-up and results. Bob reads Alice's report and tries to duplicate her work in his laboratory. Our goal in choosing a notation should be to simplify the work of Alice and Bob as much as possible (assuming they follow our notational convention). Since they probably already are used to using SI units and have access to equipment calibrated to them, we should also use SI units. The components of our tensors should (as far as possible) agree with the usual non-relativistic quantities with which they normally work. If discarding the myth that the components of a tensor must have the same units helps us do this, as it does, then we should discard that myth. Also, introducing extra minus signs should be kept to a minimum which implies (as I see it) that the spatial components of the metric be positive. JRSpriggs (talk) 08:20, 20 June 2008 (UTC)


 * The signature +--- has historic precedence in hyperbolic quaternions introduced in 1891 by Alexander Macfarlane (mathematician). Refer to split-complex number for insight.Rgdboer (talk) 21:14, 18 July 2008 (UTC)


 * I don't necessarily think that the discussion about +--- vs -+++ should go here, even though I was the one stupid enough to open that can of worms here. That discussion belongs on the working group page I believe since it affects all relativity articles and not just this one.  I will respond to JRSpriggs response about using SI units since this is one of maybe 2 articles that I have seen that use that notation.
 * I agree with you, JRSpriggs, that the SI units are better for experimental work for the most part. It may even turn out that you convention will dominate in the future the same way that SI units took over E&M despite Jackson's and others best efforts to keep more theorist friendly units.  There are a number of good reasons to not use SI units here, though.  The first and foremost reason is that this is not about philosophy but about speaking the same language.  The standard is to define an x_0 = ct, at least in my experience.  You make an interesting case, but is this article really the place to make a stand on philosophical principles?  (I could be wrong about  x_0 being a defacto standard. A quick google search showed more x_0 = t than I thought would be there, but it wasn't near 50%.)
 * My second point is that while your notation is more experimentalist friendly, the standard notation is more theorist friendly. It has less constants to accidentally drop for instance.  Correct me if I am wrong but most of the use cases of SR in science are in theory, not experiment.
 * TStein (talk) 03:11, 19 July 2008 (UTC)

I also prefer the +--- metric, but I also think that it is not important. What is important is that it is clearly mentioned what conventions we use. I also don't agree with using SI units, that's very unnatural and leads to messy equations. We should simply put c = 1 and use consistent units for space and time. But this isn't really a big deal because you can always pretend that c = 1 even if it isn't the intended interpretation of that equation.

We should not per se define units in a consistent way on wikipedia. That would be an unnecessary burden. People don't benefit from that at all, because if you are unable to make the trivial switch from one set of units to another, you can forget about understanding special relativity and in particular the more technical parts of the theory in which the units actually matter. Count Iblis (talk) 22:17, 18 July 2008 (UTC)


 * It is not just about being able to make the trivial switch from one set of units to another vs being able to understand special relativity. It is about putting an extra burden (small as it is) on someone who is (for now) struggling to understand the content.  A toddler learning to walk might trivially be able to chew gum but forcing him to chew gum while taking his first steps might be more than he can handle.


 * I don't think that working toward a common notation is an unnecessary burden. Certainly being bludgeoned from on high about this issue when so many other more important problems exist is an unnecessary burden.  It is also not worth degenerating into a fight over.  But working toward consensus is how wikipedia works best and a consistent set of units and notation adds a level of professionalism and reduces the burden on new readers.  (And quite frankly it reduces the irritation of certain more experienced but cranky readers who can easily make the switch but don't want to be forced to.)  I have already plugged up this talk page too much with this so I will shut up now.
 * TStein (talk) 03:11, 19 July 2008 (UTC)


 * Yikes, I just realized I completely misinterpreted what Count Iblis was saying and refuted what I said earlier. I thought that the unnecessary burden referred to working toward a common notation for all SR articles.  My response favored working toward everyone using the same units.  I stand by that statement even though it contradicts my belief that we should not use SI units in SR, here.  The more important goal in my opinion is unifying the SR articles.  I will shut up for real, now.
 * TStein (talk) 03:23, 19 July 2008 (UTC)

Poetic, Pop Science
I removed the following from the introduction: Special relativity overthrows Newtonian notions of absolute space and time by stating that time and space are perceived differently by observers in different states of motion:

This ranges from poetic, to misleading to just plain wrong.
 * D'know how poetic it is, but how is it wrong? Brews ohare (talk) 16:18, 16 September 2008 (UTC)


 * It is true, but the tone is too negative. It should be balanced by pointing out that there is an invariant separation between events. JRSpriggs (talk) 16:58, 16 September 2008 (UTC)


 * I've added a "balancing" quotation and citation following your suggestion. Brews ohare (talk) 20:52, 16 September 2008 (UTC)


 * it's not correct, and encyclopedias are not for views - don't balance it, take it out!

The lorentz transformation transforms spatial and temporal coordinates - they are very well defined. Blablablob (talk) 21:36, 16 September 2008 (UTC)

The "As opposed to Newtonian spacetime" at the beginning of point 2 suggests that the remainder of the sentence is false in classical physics. This is wrong. If I'm on board of a moving ship, and I jump up, and I land back on the same part of the floor, the spatial separation between the event "I take off" and the event "I land" is zero in the ship's reference frame, but nonzero in the sea's reference frame; and that would be true even if c were infinite. See Galilean transformation: t doesn't change (so "Unlike Newtonian space-time" in point 1 is correct), but x does. A r m y 1 9 8 7 ! ! ! 23:12, 16 September 2008 (UTC)


 * I suspect that "nonsimultaneous" should be "simultaneous". For simultaneous events the spatial distance between the events is the same for all observers according to Newtonian mechanics, but not in special relativity. But, of course, in a different frame the events are no longer simultaneous in special relativity... Count Iblis (talk) 23:37, 16 September 2008 (UTC)

This is the original text

So, Sklar invokes a Neo-Newtonian space-time separate from the Newtonian Space-Time. The Newtonian one is presumably defined relative to an absolute rest frame, so the counterargument by Army1987 is then not allowed in the Newtonian Space-Time, but but it is allowed in the Neo-Newtonian Space-Time. Count Iblis (talk) 23:59, 16 September 2008 (UTC)

External link
Can someone take a look at this external link, included in the "External Links"? It seems to promote some (fringe) theory. It has also been included in Introduction to special relativity. -- Crowsnest (talk) 21:35, 15 September 2008 (UTC)


 * I think we can immediately remove all of them (1, 2, 3) on the basis of Original Research, WP:Source and WP:spam.
 * I'll leave a message on the talk page. DVdm (talk) 10:28, 16 September 2008 (UTC)

Hyperbole
There's a cliche out there to the effect of "Relativity overthrew Newton". Almost universally in commercial, pop science material, its salient feature is that it is sensationalist.

Consider the correspondence principle in quantum mechanics and, in special relativity, the guiding principle that the laws of physics are the same for all inertial observers. We have not abandoned F=ma, nor conservation of momentum, nor equal and opposite, and space is still isotropic and homogeneous. We've have extended our knowledge. `overthrow` is too much. Blablablob (talk) 21:32, 16 September 2008 (UTC)


 * There is no doubt that the viewpoint of relativity is radically different, even if the predictions are the same as Newton in an appropriate range. E = m c2  is pretty radical. Trading time for space and making space-time interval the invariant is pretty different. What is wrong with saying so? Brews ohare (talk) 04:59, 17 September 2008 (UTC)

Inconsistent conventions on placement of c
Recent edits by appear to be aimed at making section Special relativity and subsection Special relativity consistent with some other articles (especially Four-vector) which follow the convention he prefers concerning the placement of c. This is contrary to the decision which was made in the section "c in the section Special relativity#Physics in spacetime" of Talk:Special relativity/Archive 13. More to the point, he has (so far) done a incomplete job of changing those sections, so that they are now internally inconsistent. See also the discussions at Wikipedia talk:WikiProject Relativity and Talk:Stress-energy tensor. I do not intend to help him convert from the convention I prefer to the convention he prefers. So, if he does not make the section internally consistent soon, I will have no alternative but to revert his changes. JRSpriggs (talk) 08:15, 21 September 2008 (UTC)
 * As I said earlier, there are plenty of textbooks that set the Minkoswki metric to +/- x (-+++), none that I can find that set it to -1/c^2,+++, as JRSpriggs prefers. The one reference that JRSpriggs supplies involves OR, is outdated AND it admits it is a minority POV. --Michael C. Price talk 07:08, 22 September 2008 (UTC)
 * Personally, I prefer to use orthonormal bases (AKA "using the same units for all components of a vector"), but let's avoid opening that can of worms. Just use whatever convention you like most, as long as you are consistent. A r m y 1 9 8 7 ! ! ! 09:41, 21 September 2008 (UTC)
 * Just use natural units and put c = 1 everywhere. Seriously, If I had to use this page as a quick reference, I would put c = 1 in the equations, and then do whatever computations I needed to do in Natural units and then put c back (if necessary) in the final result. Count Iblis (talk) 14:29, 21 September 2008 (UTC)
 * Except that lots of untrained people find natural units confusing. Are we aiming for wider access or not? --Michael C. Price talk 07:10, 22 September 2008 (UTC)
 * Of course, if we used c = 1, we would clearly state that before... --A r m y 1 9 8 7 ! ! ! 11:55, 22 September 2008 (UTC)
 * We can put c = 1 when we start to discuss the technical formalism involving tensors. It is extremely confusing to have dimensionally incompatible quantities for the components of a tensor and then have to compensate for that by including the conversion factors in the transformation matrices. People who can understand these sections will have no difficulties understanding natural units, so it is a small price to pay to eliminante the c's everywhere and clean up the mess.


 * I would guess that there are more people who are able to understand natural units but who now cannot make their way through the sections involving tensors, simply because things look more complicated with all the factors of c's scattered everywhere. Count Iblis (talk) 12:56, 22 September 2008 (UTC)


 * I understand Count Iblis' point, but I feel that the total elimination of c shuts out more people. I have made the article consistent which should make it slightly easier. --Michael C. Price talk 18:36, 22 September 2008 (UTC)


 * To MichaelCPrice: Please stop misrepresenting my position. (1) I use
 * $$\eta_{\alpha\beta} = \begin{pmatrix}

-c^2 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$
 * for the Minkowski metric &mdash; notice that there is no reciprocal applied to the -c2, contrary to what you keep saying.
 * (2) As I explained at Wikipedia talk:WikiProject Relativity, what I am doing is not original. It comes from a text by E.J.Post where it is also not original since he cites two earlier works by H.Dorgelo and J.A.Schouten. As for it being a minority view, these are the authors who have seriously investigated the issue of units. Most other authors either ignore it or carelessly give into to the myth (or prejudice) of the public concerning units rather than try to get it correct.
 * You still have not made the sections of this article in question internally consistent. JRSpriggs (talk) 12:20, 22 September 2008 (UTC)
 * Note the term metric can refer to either $$\eta_{\alpha\beta} $$ or $$\eta^{\alpha\beta}$$. I hesitate to point out this blindingly obvious fact, except that JRSpriggs seems to be making a big deal out of it.  I also note that JRSpriggs has not addressed the issue that the reference he gives admits that his preference is a minority view, and has been for the last 40+ years.  No one uses it today -- except JRSpriggs. --Michael C. Price talk 14:44, 22 September 2008 (UTC)
 * The convention in the article is now referenced, internally consistent and consistent with four vector. The Minkowski metric and lambda are now dimensionless, as they are in the literature. --Michael C. Price talk 18:31, 22 September 2008 (UTC)
 * There still are inconsistencies in the "Electromagnetism in 4D" subsection. Try looking at the equations related to the Lorentz force (and others also). JRSpriggs (talk) 20:27, 22 September 2008 (UTC)
 * Will do. --Michael C. Price talk 20:35, 22 September 2008 (UTC)

The sklar quotes are wrong
There are two problems with the sklar quote (at least):


 * 'well defined': we do know how to make all calculations in space and time for inertial observers; this part of the quote only clouds the issue.


 * There is simply no such thing as "the distance" between events that are nonsimultaneous: this is completely, and utterly wrong - the Lorentz Transform is the means by which you can calculate and compare distances between arbitrary events, and for arbitrary inertial observers.
 * The term "the distance" is in quotes for a reason - it stands for an antiquated notion (i.e. Galilean invariance and absolute space and time). Brews ohare (talk) 04:54, 17 September 2008 (UTC)

Further, even if you go inside the logic of this mystical, pseudo-science, it implies that there is no such thing as distance what so ever! Simultaneity is relative, so two events can not be simultaneous for all observers.
 * I don't follow your implication. Can you spell it out? Brews ohare (talk) 04:54, 17 September 2008 (UTC)

Hey, no distance, no physics - to say it is wrong is an understatement.
 * Huh? Brews ohare (talk) 04:54, 17 September 2008 (UTC)

I've already reverted these twice, so it's time to look to the community. If you see Sklar quotes in the introduction, please remove them. They don't help, and the introduction without them is actually pretty good.
 * I don't see any problem with these quotes. They point out significant differences introduced by relativity. They are perfectly clear and accurate. And they are supported by citations to published work by (at least in one case) UC Press - not so shabby. Brews ohare (talk) 04:54, 17 September 2008 (UTC)

Regarding future arguments on this, the best that can be said of the Sklar quotes is that something else is meant - even then, they are done in a sensationalist, book selling style that is inappropriate for an encyclopedia, but especially for Special Relativity. Blablablob (talk) 00:00, 17 September 2008 (UTC)


 * What is sensationalistic about these quotes? Please spell out your problems instead of providing a hysterical reaction. You can check the quotes (and their context) yourself simply by clicking on the links provided. Brews ohare (talk) 04:54, 17 September 2008 (UTC)

The Sklar quotes are useful and correct
Here are the quotes with links where more extensive context can be found on Google Books:

Please register your thoughts and position on including these quotations.

For inclusion

 * 1) Personally, I find these remarks helpful, well expressed and accurate. The first book is endorsed by reviews in Physics Today and in Nature. It is described by Prof Butterfield in his course reading recommendations at Cambridge as "The best overall introduction avoiding all formalism". The author himself is a UC Irvine Distinguished Fellow. Both books are on reading lists from all over the world (try Googling them). A laudatory review for both books can be found at Mind. I vote that they be included in the article. Brews ohare (talk) 05:08, 17 September 2008 (UTC)

For exclusion

 * 1) I partly agree with what you say about differences between Newtonian and Minkowski spacetime, but I have already pointed out that the "as opposed to Newtonian spacetime" part is wrong. (Unless that refers to Isaac Newton's own philosophical views; but "Newtonian", especially in scientific context, is extremely likely to be taken to mean "of classical physics", and in Galilean relativity there is no notion of absolute distance between nonsimultaneous events.) "We can say of two events neither that they simultaneous nor that they are spatially coincident" − We cannot say that they're spatially coincident in Galilean relativity, too. I think that quotes are confusing at best. A r m y 1 9 8 7 ! ! ! 09:49, 17 September 2008 (UTC)
 * As opposed to Newtonian spacetime, the notion of the spatial separation of nonsimultaneous event locations is not well defined To me this means that invariance in SR of only $$ds^2 = d\mathbf{r}^2 -c^2 (dt)^2 $$ means a variety of spatial separations exists depending upon the observer's time interval dt. That is not disputable, so what do you think the quotation means? Brews ohare (talk)
 * I object to the "as opposed to Newtonian spacetime" phrase. If someone jumps on a ship and lands back, the spatial separation between the event "he jumps" and the event "he lands" is different for an observer on the ship than for an observer on land. That's all right. But that would be all right even if c were infinite, so there is no difference with classical physics. --A r m y 1 9 8 7 ! ! ! 15:38, 17 September 2008 (UTC)

Comments
I think you (Brews ohare and Blablablob) are disagreeing more on terminology than on content. For example, as for "well defined space and time": space and time are well-defined for each inertial frame (except in black holes and before the Big Bang, but we're not talking about GR...), and there are well-defined rules for converting them for different frames. But I think Brews by "well-defined" means "defined indipendently of any arbitrary choice", and this is wrong. (But the same thing happens to spatial coordinates in classical physics. The y-component of a vector is only "well-defined" when we choose in which direction the y-axis points. The main difference between SR and Galilean relativity is relativity of simultaneity. So, while I think its useful to say "SR overthrew absolute time", adding "and space" at the end is not terribly useful, and very confusing. A r m y 1 9 8 7 ! ! ! 09:49, 17 September 2008 (UTC) In both cases, I'm using standard configuration (parallel axes, coincident origins at t = t&prime; = 0, x-axis oriented in the direction of the relative motion). There are complications if we introduce rotations and translations, but they are the same in both cases. The formulae about intervals stay the same, anyway.

The only differences are: the &gamma; factor (length contraction and time dilation, relativistic mass, and indirectly E = mc2, the magnetic force, and other stuff); the first entry in the bottow row of the matrix (relativity of simultaneity); and the form of invariant intervals. All other things (isotropy of space, principle of relativity — inverting those matrices just changes the sign of v, etc.) are unchanged. A r m y 1 9 8 7 ! ! ! 10:15, 17 September 2008 (UTC)
 * Thanks for the thoughts. I am still confused though. Here are three questions:
 * <<"SR overthrew absolute time", adding "and space" at the end is not terribly useful, and very confusing.>> This wording is not part of the quotations, It is text inherited from an earlier time, and I don't know where it came from. I added the quotes to put some context around this textual prologue. Perhaps this prologue to the quotes should be changed. Are the quotations themselves OK?
 * In my mind what the first quotation says is exactly what you say: there are separate equations for &Delta;t and &Delta;r prior to relativity and only one equation afterward. What part of the quote suggests something else?
 * —This is part of a comment by Brews ohare which was interrupted by the following:
 * I shouldn't have written that one about space. See the example about the person jumping on a ship, above. A r m y 1 9 8 7 ! ! ! 15:57, 17 September 2008 (UTC)
 * How about the second quotation: is it less controversial than the first one? Brews ohare (talk) 14:23, 17 September 2008 (UTC)
 * Nothing of what it says is wrong, but I think adding it to the lead section is overkill. We could just summarize it in a sentence and put the quote itself in the References section. A r m y 1 9 8 7 ! ! ! 15:57,

17 September 2008 (UTC) BTW, Sklar is not a flake: his books are well received and he is a UC Irvine Chancellor's Distinguished Fellow. Sklar is the author of several other books and a long string of professional pubs in journals. Brews ohare (talk) 14:53, 17 September 2008 (UTC)
 * This does not make him infallible. Sir Arthur Stanley Eddington wasn't a flake, either, but this doesn't mean that the Eddington number argument about the fine-structure constant is not bullshit. And many statements in popular science books (even ones by Stephen Hawkings) are semplified to the point of trespassing the borders of falsehood. As Albert Einstein said, "everything should be made as simple as possible, but not simpler". A r m y 1 9 8 7 ! ! ! 15:57, 17 September 2008 (UTC)
 * For me, your recent edits have fixed matters. Brews ohare (talk) 16:26, 17 September 2008 (UTC)


 * Regarding Galilean relativity, if $$\Delta t \, = \, 0 \,,$$ then $$\Delta x^2 + \Delta y^2 + \Delta z^2 \, = \, \Delta x'^2 + \Delta y'^2 + \Delta z'^2$$ does hold after all. JRSpriggs (talk) 19:34, 21 October 2008 (UTC)

Reversal on the derivation of the relativistic force formula
I do not know why the paragraph I wrote has been labelled "unhelpful" and has been reverted. As far as I know:

-the derivation I inserted is correct

-the current formula (the one JRSpriggs reverted to) is incorrect DS1000 (talk) 23:45, 18 September 2008 (UTC)
 * It is identical to yours, as dv/dt is the tangential acceleration at, and v &middot; a = vat. Also, your derivation is very confusing, using two different meanings for m, and first stating that f = ma and then confuting it. A r m y 1 9 8 7 ! ! ! 00:16, 19 September 2008 (UTC)


 * OK, I cleaned it up. DS1000 (talk) 00:59, 19 September 2008 (UTC)
 * It was still wrong, but I really cleaned it up. JRSpriggs (talk) 07:42, 19 September 2008 (UTC)
 * It was correct, it was just in a form unfamiliar to you. DS1000 (talk) 14:06, 1 October 2008 (UTC)

BTW, I think there is lots of redundancy in sections "Mass, momentum, and energy", "Relativistic mass" and "Force". They could be unified into one or two sections without losing much information. I'm going to take a stab at it. A r m y 1 9 8 7 ! ! ! 10:05, 19 September 2008 (UTC)


 * To : I see now that you were correct about that formula. I do not know how I failed to see that, except perhaps that the sloppiness of your formatting may have discouraged me from seriously examining the content. I am sorry. JRSpriggs (talk) 17:18, 1 October 2008 (UTC)


 * FWIW, I made a few more changes for consistency, like moving the numeric factor (v.a) to the left of the vector. DVdm (talk) 18:01, 1 October 2008 (UTC)

Question on explaining conservation of mass in the context of E = mc^2
In the text a "confusing" tag was placed by someone wondering if we could explain why it is, if energy, mass, and momentum are all separately conserved (in closed systems), how it can be that mass is said to be converted to energy. I have attempted to explain. Einstein's famous m = E/c^2 equation applies to systems only when p = 0 for the system, and thus any system mass decrease is actually invariant mass decrease. We "explain" this by noting that such systems must NOT be closed, if the mass is to decrease! Even in an atomic bomb, the mass decrease only happens because you let invariant mass OUT, as light and heat, and THAT carries away the mass. If you didn't let the bomb cool, the mass would stay the same. If you do let it cool (which of course you always do), the heat transfers the "missing" mass to things it heats up, but THEY gain the mass. Invariant mass is always conserved, period (or otherwise it wouldn't be invariant). Atomically, the "mass defect" in binding energy also is carried away by kinetic energy and photons, which ultimately carry away the invariant mass. But they have it-- the mass never goes away unless you let it out of your system and lose track of it. It's always somewhere. S B Harris 05:04, 12 October 2008 (UTC)


 * Yes, in a closed system, all components of the four-momentum (E/c2, p) = (∑i γimi, ∑i γimivi) are conserved, and so is invariant mass $\sqrt{E^{2} &minus; (cp)^{2}}$/c2. But nuclear plants aren't closed systems. -- Army1987 (t — c) 10:42, 12 October 2008 (UTC)


 * Precisely. A gigawatt-day is 8.64e13 J, which is close to the 9e13 J represented by a gram of mass. So in the inertial frame of a gigawatt powerplant (nuclear or not-- it's the same for a coalfired plant that large!), the plant sends a gram of electricity out into the world every day. This isn't a gram of electrons! It's a gram of heat, light, and work that comes from the electric field made available by the plant's generators. If you draw your system boundaries around the plant, it loses a gram a day, but the town it powers, gains that gram (not counting escaping light). If you draw your system boundary around the Earth and don't let the streetlight out, no mass is lost or gained, but is (as usual) strictly conserved, even for a powerplant. The same for a bomb. A big thermonuke can make pounds of heat, but if you count objects that absorb the heat, no mass is lost in the system as a whole. Similarly, also not "closed" is the fusion process whereby nucleons get crammed together to make nuclei in nucleogenesis. And yet that missing mass (packing fraction) is held popularly to have "disappeared" and to represent the binding energy (deficit). Well, it didn't disappear, it's just been removed before you got there, and now is someplace else (back in the remnants of the supernova where the nucleas was created, perhaps) Another popular thing that backs this misconception is that mass-conservation, like all conservation, not only requires a closed system, but also a single observer. If you have a U-235 nucleus split and the parts fly off, it's sometimes said that the two fission products have less mass than the original. Well, as a system, they don't have less mass while they're in flight, as seen by the original observer. That system of fission products in flight has the same invariant mass it always did. But, you can try to game it by stopping the products (in which case their kinetic energy escapes, and you lose mass that way, because the system isn't closed), OR you can game it by looking at the products in flight one by one, in the rest frame of each of them. But that's summing rest masses, and the reason this is illegal even before they are stopped, is because you have to switch to various inertial frames (those of each of the fission fragments) to get these numbers. So you've gone from one observer at the beginning, to at least 2 different ones at the end. Obviously if mass isn't conserved in THAT process, it's not the fault of Einstein. S  B Harris 17:48, 12 October 2008 (UTC)
 * ==In Asimov's study of "large numbers" he developed the concept of a "scene" having a value in relation to the importance of it's occurrence in the "spacetime continuum". And in Galalean spacetime and Euclidean Physics this scene value is made up of three independent values of space plus one of time and has lots of possibuilities which he then explored. Please note that our our concepts of motion, and momentum, and kinetic energy and angular momentum all argue that that the increments of dimension and time are independent of each other. Then Lorentz and Einstein came along with the proposition that time is not an independent variable with respect to any dimension, but rather a dependent variable, such that a "scene" of the motion of a ray of light to its ultimate destination is all in the same scene. This is a difficult concept to swallow WFPMWFPM (talk) 15:33, 21 October 2008 (UTC)
 * ==Particularly in view that when we set up an experiment to find out how much of our time it takes for light to go to the moon and back, we get approx. 2.5 seconds of our time. which argues against the zero time increment concept. But that's what an extension of mathematical concepts into physics will do for you.WFPMWFPM (talk) 17:53, 21 October 2008 (UTC)

Temporarily reverted material pending evaluation
I have temporarily reverted the the following material (from User:Delaszk) simply because it is prima facie contrary to my own preconceptions, and therefore strongly suspect in my opinion. I apologize if this is simply my own ignorance, but I would appreciate some validation by other editors before it goes into the article. Better a little delay than nonsense, I guess. Wwheaton (talk) 09:24, 30 October 2008 (UTC)


 * Maximum speed limit as a consequence of Gallileo's relativity


 * Mitchell Feigenbaum and others have revisted Gallileo's writings on relativity and have found that a maximum speed limit and the acceleration of the expansion of the universe can be derived from first principles from the idea of relativity of motion and are thus mathematically inevitable. Therefore Einstein's postulate that light travels at the same speed relative to all observers is superfluous. Light is merely the first and most obvious phenomenon that achieves the maximum speed that Gallileo's relativity implies, and the expansion of the universe and the accelerating expansion do not require invoking any special mechanism or have anything to do with the distribution of matter.

I had a look at Feigenbaum's paper. I doesn't add anything really new. (We all already know that the parameter c in the Lorentz transformations has, a priori, nothing to do with light.) It is no wonder that the paper has not ever been published, and that it resides in the "physics" (aka nonsense) subarchive of the arXiv. At the very least this article in non notable. I see o reason why wikipedia should make any reference of it. (TimothyRias (talk) 10:05, 30 October 2008 (UTC))
 * Correct, see Lorentz transformation – but there is no a priori reason why c2 couldn't be infinitely large, giving Galilean relativity. -- Army1987 (t — c) 10:11, 30 October 2008 (UTC)

Some points I'd like to make about this:
 * This work is based on Gallileo's principle of relativity: that the laws of physics look the same to all observers. It is NOT based on Newton's notion of absolute time or the Gallilean transformations.


 * Feigenbaum is not the first to suggest that Gallileo could have derived special relativity, but he is the main author that the New Scientist article refers to. It is still true that special relativity can be derived without Einstein's second postulate whether or not you like Feigenbaum's paper. Perhaps other references should be used instead of or as well as Feigenbaum's but the paragraph as a whole should still be included in wikipedia. The section should be renamed Maximum speed limit as a consequence of Gallileo's principle of relativity to clarify matters.

"However, following Dyson [12], we are led to observe here that «Minkowski in his 1908 lecture failed to carry his argument to its logical conclusions». Indeed, the argument that guided Minkowski to drop the commutativity of boosts thereby allowing him to interconnect boosts and space rotations into the Lorentz group has left spacetime translations out of the picture. On the other hand, it is only logical that the same argument should be extended to translations, by dropping their commutativity as well. Indeed, it appears that, even at the beginning of the 20th century, astronomers would not have had any a priori compelling reason to believe that the commutativity of translations would not eventually break down provided one would look sufficiently far away in spacetime. The reason why such a possibility did not occur to scientists then is very likely due to the fact that nobody had yet any clear knowledge at that time of the existence of other galaxies, in the universe, beyond ours. In fact, it was only through his observations with the 100 inch Hooker telescope at Mount Wilson Observatory, carried out during the years 1923 and 1924, that Hubble was able to establish beyond doubt that most of the nebulae, earlier observed with less powerful telescopes and thought to belong to the Milky Way, were not part of our galaxy, being instead themselves galaxies in their own right, beyond our own. Therefore, since the picture one had of the universe at the time of Minkowski was that of an island galaxy immersed in an otherwise infinite void, the idea of non commuting space and time translations would have indeed appeared quite unnatural to everybody, Minkowski himself included. Nevertheless, it is perhaps a bit surprising that no «act of audacity on the part of higher mathematics » arose suggesting that translations were not, after all, commutative. In the words of Dyson [12], an opportunity was missed by Minkowski and by his fellow mathematicians. Had someone made the further daring leap of renouncing the commutativity of translations, a quite logical step after the commutativity of boosts had been abandoned, and acknowledged that Lorentz transformations and spacetime translations should not «lead their lives entirely apart» but be instead interconnected, one would have realized that the true kinematical group of nature is not the Poincaré group P(c) but the de Sitter group dS(c; �), � being an unspecified “cosmological constant” to be determined by observation [13]." Delaszk (talk) 11:41, 30 October 2008 (UTC)
 * Regarding the derivation of the expansion of the universe from first principles I'd like to quote this paragraph from the paper of S. Cacciatori, V. Gorini, and A. Kamenshchik:


 * Just a few notes.
 * I don't really mind (nor oppose to) the material as it is clearly covered elsewhere and not much new indeed. But if the material is to be permanently removed, it should be removed from Delaszk's other contributions as well: Gravitation, Galilean invariance, Principle of relativity, Fundamental Speed and perhaps more to come.
 * I'm pretty confident that there is a policy against copying material word by word in multiple articles.
 * I do mind the link to the New Scientist article though. It requires registration, and that is an external links no-no.
 * DVdm (talk) 11:43, 30 October 2008 (UTC)


 * If the external link is a no-no then that is easily replaced with a simple reference to the offline version of the magazine.
 * The exact same paragraph word for word in four different articles may be repetitive but wikipedia is not paper and the paragraph is clearly relevant and pertinent to each of the articles.Delaszk (talk) 11:54, 30 October 2008 (UTC)


 * Suggestion: Remove the material from all the artcles, except from Fundamental Speed where you elaborate on it with what you have written above. If it survives, then simply point to it from the other articles. Good luck - DVdm (talk) 12:03, 30 October 2008 (UTC)

The material is now only in the article Galilean invariance and is pointed to from other articles with:

Maximum speed limit and other consequences of Gallileo's principle of relativity

It has been shown that special relativity and the accelerating expansion of the universe are just consequences of Gallileo's principle of relativity, requiring no further postulates or data. See Galilean invariance. Delaszk (talk) 15:43, 30 October 2008 (UTC)


 * This is better, although I don't think this deserves an entire section with header and all in these articles. I.m.o. it should be much less conspicuous, by either integrating in the prose, or by merely adding a pointer in the  sections of these aeticles. As it is now, i.m.o. we have multiple cases of WP:Undue weight. DVdm (talk) 16:05, 30 October 2008 (UTC)


 * I think it should be more than just a see also. It appears as the main topic on the cover page of the latest New Scientist and it seems very significant. The magazine article says that a consensus amongst physicists has emerged that Einstein's second postulate is unnecessary even if this hasn't yet filtered into the textbooks. A see also or a throwaway comment amongst the prose is likely to be overlooked by the reader and therefore not give enough weight. Delaszk (talk) 16:21, 30 October 2008 (UTC)


 * Hm... as it is now, it looks like "HEY PEOPLE! LISTEN UP!! I'VE GOT SOMETHING EXTREMELY IMPORTANT TO TELL YOU!!!". I'm pretty sure it will not survive in this form. I'll leave it to other editors to make suggestions. DVdm (talk) 16:29, 30 October 2008 (UTC)
 * "hasn't yet filtered into the textbooks"? As far as I remember, I first learned about the derivations that take the principle of relativity to derive the Lorentz transformations up to the parameter that turns out to be 1/c&sup2; in the Sexl/Urbantke text book (Relativity, groups, particles), which was published 16 years ago. And the basics of these derivations (and attempts to find a set of independent axioms for special relativity) weren't exactly new then. New Scientist articles should always be taken with a grain of salt.


 * That said, it is true that these statements are not usually included in the more basic text-books, which are mostly concerned in getting the Lorentz transformation derived quickly and then move on to dynamics and other applications. So I'm all for adding something on the axiomatic basis of special relativity. But let's not pretend this is a new discovery. Chapter 3 of http://arxiv.org/abs/math-ph/0602018 has some of the original articles. Markus Poessel (talk) 21:37, 30 October 2008 (UTC)


 * That text is grossly misleading. Galileo's principle of relativity, as expressed at the time of Galileo, assumed no maximal speed. It is true, as discussed in Lorentz transform, that if one makes the further necessary assumption that a finite maximal speed exists then that assumption plus Galileo gives Lorentz transforms, but that is by definition a "further postulate" necessarily motivated by data. Also, all of this is disconnected from the accelerating expansion which is even further removed for Galilean concepts, and dependent on yet other data.  Dragons flight (talk) 16:33, 30 October 2008 (UTC)


 * Who said anything about assuming a maximal speed ????? A maximal speed is a consequence of Galileo's principle of relativity. That is the whole point that the references are making. Ditto for the expansion of the universe.Delaszk (talk) 17:47, 30 October 2008 (UTC)


 * That you say it, doesn't make it so. Galileo's relativity works just fine with assuming, as he did, there is no maximal speed.  Your Feigenbaum reference says special relativity is "an extension of Galileo's thoughts".  It is not that special relativity follows intrinsically from Galileo (it doesn't), but rather that special relativity follows from Galilean transforms plus an additional assumption which can be phrased in a variety of ways but ultimately is equivalent to postulating the existence of maximal speed.


 * Also I am wondering if you are confusing Hubble expansion (which is arguably related to Galilean precepts of isotropy) with dark energy which explains the accelerating universe and requires several additional assumptions. Dragons flight (talk) 20:03, 30 October 2008 (UTC)


 * Yes, a maximal speed is a consequence of Galileo's principle of relativity. But nothing requires it to be finite, a priori. In principle, it could well be infinite, in which case you get Galilean transformations. -- Army1987 (t — c) 14:54, 31 October 2008 (UTC)


 * As far as I can see, "there either is a finite maximal speed or speeds can have any value at all" isn't a consequence of Galileo's principle of relativity – it's just as trivially true as any other statement of the form "A or not-A". Markus Poessel (talk) 17:14, 31 October 2008 (UTC)
 * I meant that from the Galilean principle of relativity you can retrieve the form of Lorentz transformation, but that principle doesn't tell you whether the term 1/c2 that appears is nonzero or not. -- Army1987 (t — c) 01:01, 1 November 2008 (UTC)
 * OK, I just misunderstood you, then. No disagreement there. Markus Poessel (talk) 09:23, 1 November 2008 (UTC)
 * The New Scientist article seems to say that there is a finite speed. Bearing in mind your advice to take the magazine with a pinch of salt, here is a quote from the article:

"if Frank's world is aligned with yours - if the north and east of both you and Frank point in the same direction - and Kate's world is similarly aligned with Frank's, you might think that Kate's is aligned with yours. The problem is, mathematical logic alone does not permit that conclusion ... it in fact allows a distinct possibility that Kate's world could be rotated with respect to yours. ... The possibility of such rotations turns out to have far-reaching consequences. Ignore them, and Galileo's relativity pops out. Allow them, and the algebra works out very differently: the mangled space-time of Einstein's relativity emerges, complete with a definte but unspecified maximum speed that the sum of individual relative speeds cannot exceed." Delaszk (talk) 17:26, 31 October 2008 (UTC)


 * The use of the phrase "the mangled space-time of Einstein's relativity" should have made it clear that this article was not written by a scientist but rather by someone who hates relativity. It is just trying to make Einstein's accomplishments look trivial or absurd. In other words, it has a (poorly) hidden anti-relativity agenda. JRSpriggs (talk) 22:30, 31 October 2008 (UTC)


 * Far from being anti-relativity, the opposite is the case. That you get a maximum speed for free, rather than having to assume it, makes relativity all the more natural. It elevates relativity to an even more unassailable position. It takes nothing away from Einstein's achievements for coming up with special relativity in the first place. The use of the word mangled is not being derogatory. I suppose you would prefer the more fashionable word: warped. The article does admittedly use the title: "Why Einstein was wrong about relativity" but that is perhaps just to sell more magazines with an eye-catching headline. In that respect the only agenda of the magazine is to make a profit. To say this is an anti-relativity agenda is wide of the mark. There may exist people who are anti-relativity but this article is not an example of such a position. Delaszk (talk) 07:59, 1 November 2008 (UTC)


 * About that title and your remark "... perhaps just to sell more magazines with an eye-catching headline.": I think that this title is precisely the reason why the pointer will not survive in any encyclopedia. Perhaps indeed nor the article, nor the author, nor the publisher is/are anti-relativity, but I think the one who decided to use that title, deserves a litle kick somewhere. DVdm (talk) 10:56, 1 November 2008 (UTC)


 * I'm not happy with the "mangled space-time" either, and I don't see that allowing said rotations implies finite maximum speed (demanding non-trivial such rotations does, but that doesn't follow from the relativity principle, as far as I can see). But let's not dwell on the New Scientist bit - where in the Feigenbaum article does it say that Galilean relativity implies a finite maximum speed? Markus Poessel (talk) 09:23, 1 November 2008 (UTC)


 * Delaszk, from the Galilean principle of relativity, you can deduce that the transformations must have the form
 * $$\begin{cases}

t' &= \frac{t + Kvx}{\sqrt{1 + Kv^2}} \\ x' &= \frac{x - vt}{\sqrt{1 + Kv^2}} \\ y' &= y \\ z' &= z \end{cases}$$
 * and if you don't want time to be able to go in opposite directions for different observers, K cannot be positive. Now, K has the dimensions of an inverse squared speed, and, if it is negative, you can show that $\sqrt{&minus;1/K}$ is the maximum possible speed if you don't want to violate causality.
 * But why the hell do you claim that K cannot be zero? Galilean relativity worked just fine for centuries, so, if there was any internal inconsistency in it, someone would have noticed it, don't you think so?-- Army1987 (t — c) 10:46, 1 November 2008 (UTC)

(unindent) To Delaszk: Both "mangled" and "warped" are derogatory terms. The preferred term is "curved", but even that applies only to general relativity where one has gravity. In special relativity, spacetime is flat (see Minkowski space). And Minkowski space is actually simpler in structure than the 3+1 dimensional structure of Galilean-Newtonian space+time. What you fail to understand is that "supporting" relativity by making false claims for it, is actually opposing it by making it seem self-contradictory and incomprehensible by ordinary people. Lies undermine the truth even (and especially) if they are made to seem favorable. JRSpriggs (talk) 13:33, 1 November 2008 (UTC)


 * I am going to back off from this discussion and from making any further edits to the relativity articles since I do not have the knowledge to pursue this, but Feigenbaum's paper says in the opening paragraph "It is my purpose here to show precisely that in these seminal concepts of Galileo lies the special theory of relativity with no further additional physical insight or knowledge.". Delaszk (talk) 15:16, 1 November 2008 (UTC)


 * On second thoughts there is something else I can say about this. The derivation of the Lorentz transformation does seem to imply a finite c. An infinite c does seem to be impossible. For details see here: Talk:Lorentz transformation/Archive 1. Delaszk (talk) 13:24, 2 November 2008 (UTC)


 * I've answered there. In that derivation, a constant was defined of the ratio of two quantities; the numerator could be shown to be nonzero, but the denominator couldn't. I've fixed it using its inverse. -- Army1987 (t — c) 15:05, 2 November 2008 (UTC)


 * You cannot derive any valid conclusions if you divide by zero. The article Lorentz transformation has now been rewritten so that it is obvious that c must be finite. Delaszk (talk) 21:30, 2 November 2008 (UTC)


 * The derivation there had an error in it which created the illusion that Galilean-Newtonian kinematics was excluded. I fixed it. JRSpriggs (talk) 00:11, 3 November 2008 (UTC)


 * I don't agree that it is an illusion. We may be talking at cross-purposes here. The first thing to say is that K is a finite constant. Whether K is zero or not depends on whether delta in the numerator is either always zero or always nonzero (we are excluding the case v=0). So accepting a necessary max-speed depends on showing that delta must be nonzero. So now we come to the central point of all this: if you consider a frame of reference which is rotated with respect to another then you must have nonzero delta in the transform matrix. Since the expression $$\frac{\delta(v)}{v\gamma(v)}$$ is a constant then the existence of any nonzero delta forces every delta to be nonzero. Allowing any rotation implies everything is a rotation. This is not compatible in any with Galilean-Newtonian kinematics which does not consider the possibility of frames of reference being rotated with respect to one another. Delaszk (talk) 08:45, 3 November 2008 (UTC)


 * Galilean-Newtonian kinematics has no problem dealing with frames of reference that are rotated (by constant Euler angles) with respect to each other. That's why we have conservation of angular momentum in classical mechanics. If you want a systematic treatment, take a look at math-ph/0602018. The difference you're talking about here is that, for the Lorentz group, a succession of boosts can be the same as a rotation. But the fact remains that you still need some further input beyond the relativity principle to arrive at the Lorentz transformations instead of the Galilei transformations. Which further input you choose (boosts able to generate a rotation, finite maximal speed, constancy of light speed) doesn't make much of a difference, but there must be some additional input. Markus Poessel (talk) 13:55, 4 November 2008 (UTC)


 * My argument may be in error based on a misinterpretation of the New Scientist article. However my misunderstandings do not reflect on the validity of Feigenbaum's paper or other author's who also claim that the we only need the principle of relativity. I have started a new article to describe the work of all those who claim this. The phrase "solely based on the principle of relativity" does however need some clarification which is what I intend the article to be. (See Special relativity via single postulate). Delaszk (talk) 12:04, 6 November 2008 (UTC)
 * Well, if "a physical law" include Maxwell's equations, the content of that article is true (and that's what Einstein himself did in later papers, IIRC). But, of course, the principle of relativity doesn't give you Maxwell's equations. -- Army1987 (t — c) 12:12, 6 November 2008 (UTC)


 * You don't need to go further than Feigenbaum's abstract, where he clearly states that, from relativity, he can obtain the Lorentz transformations "with a parameter 1/c&sup2;, some undetermined, universal constant of nature" - no mention that his parameter somehow must be non-zero. The main text reaches this question on page 23, where Feigenbaum writes explicitly: "there can be no conceptual way to decide the value of 1/c&sup2;" – and then goes on to explore ways of fixing the parameter by reference to experiment. Which is fine – as I said before, there are different ways of getting that extra information. But the point is that you do need extra information, and Feigenbaum is very clear about that. Markus Poessel (talk) 18:03, 7 November 2008 (UTC)
 * I think the point is that the relevance of the second postulate to the universe we live in has to be confirmed by experiment anyway, so why bother making the postulate in the first place. Just derive the Lorentz/Galilean transformations and by experiment determine the value of c without reference to a second postulate. Delaszk (talk) 19:12, 7 November 2008 (UTC)

(Un-indenting.) To be sure, the principle of relativity should also be subject to experimental tests. In particular the mechanical part of it used to derive special relativity up to the value of 1/c&sup2; (beyond that, the relativity principle is more of a general program than a specific postulate - it's something that's supposed to hold for all physical laws, even the ones you haven't discovered yet). But it's important to state that Galilean relativity isn't sufficient to fix the Lorentz transformations on its own. You need something more, either an additional postulate (to be tested experimentally like all other postulates) or a free parameter (to be fixed experimentally). And that, again, is not a new result. And it's why is rather misleadingly named. What you call that additional ingredient is a rather minor point, but you shouldn't make people think there isn't an additional ingredient. That said, it looks to me that Special relativity via single postulate is pretty much a POV split. You didn't agree with what was written in special relativity, so you started another article devoted to the statements that were deleted from the main article. That's not how it should go. The information that you can get most of the structure of the theory from the relativity principle should really go into the main article. Markus Poessel (talk) 21:09, 7 November 2008 (UTC)


 * Consider two approaches:


 * 1) Formulation of special relativity based on: principle of relativity + universal lightspeed
 * 2) Derivation of lorentz transformations up to a free parameter.


 * The first approach produces a theory with a free parameter c to be determined by experiment.


 * The second approach produces a theory with a free parameter -1/c^2 to be determined by experiment.


 * When the values of the free parameters are fixed by experimental measurements then the two approaches produce the same theory. So special relativity can be developed without extra information.


 * The usefulness of the second postulate is that it makes it explicit from the start what the conclusion is going to be. It gives a shortcut to deriving the Lorentz transformations. Postulating universal lightspeed is a public relations exercise so that educationally it makes relativity easier to understand. Delaszk (talk) 08:50, 8 November 2008 (UTC)


 * Where to start? The usual approach 1) doesn't have c as a free parameter. It explicitly says that c is the same c as in Maxwell's equations, and thus essentially fixed before you even begin to derive the consequences of your postulates.


 * Secondly, what is the experimental input from the measurement (in particular the crucial bit, the one that says 1/c&sup2; isn't infinite) other than "extra information" that goes beyond the principle of relativity?


 * Also, postulating a universal speed of light is not a "public relations exercise". It's the way all this was historically developed. Einstein's starting point (certainly in his original article) was the combination of Maxwell's equations with the principle of relativity. Also, don't forget that you need to define some standard coordinates with the usual definition of simultaneity before you get the Lorentz transformations in their usual form. There are other ways of doing that ("standard cannons" that fire particles - and you need isotropy), but light is certainly a convenient choice. Markus Poessel (talk) 13:37, 9 November 2008 (UTC)


 * To be specific, the value of c depends via Maxwell on the value of the permittivity of free space. Theory does not specify the value of the permittivity. So we have:
 * Original approach results in the theory up to the free parameter pair: c and permittivity.
 * Feigenbaum's approach results in the theory up to free parameter -1/c^2.
 * In both cases you must determine the numerical value of the free parameters by experiment, and when you have done so you end up with the same theory. The only difference is that the Feigenbaum's approach doesn't use the second main postulate, so compared to the original two-postulate formulation it can be described as a single-postulate formulation.


 * I agree that the first approach was the way it was historically developed. The wikipedia article is accurately reflecting the way it was originally formulated. I think that the effect of Einstein's second postulate was that it helped to familiarise people with the theory. Delaszk (talk) 15:44, 9 November 2008 (UTC)


 * Again, where to begin? The original approach doesn't have any free parameters, in the sense of parameters that still need to be determined. It explicitly postulates that a previously known constant (the speed of light) plays a previously unsuspected role for kinematics and mechanics. That's different from deriving the Lorentz transformations from mechanical relativity, without any indication of what the parameter 1/c&sup2; is.
 * If you want to be precise, you would need to call the two varieties the "two-postulate" and the "one-postulate-one-experiment" formulation.
 * I still have no idea what you mean by "helped to familiarise people with the theory". It's not as if, in 1905, there were different versions, and the one with the constancy-of-light-speed was somehow easier understandable, or more popular, and "won out" for that reason.
 * But let's go back to what's Wikipedia-relevant. I still don't think there's any course of action other than getting rid of that POV-fork-like article of yours, and adding a sentence about what can be deduced from the relativity principle alone to the main article. Markus Poessel (talk) 18:26, 9 November 2008 (UTC)


 * In the original formulation you don't know the value of c until you do an experiment. In the other approach you likewise don't know the value of c. When the experiments are done the resulting theories are the same. There is no point in calling it "one-postulate-one-experiment" unless you also call the original the "two-postulate-one-experiment".
 * The remark about familiarisation is that it is a good thing that Einstein came up with the second postulate because if he had derived the Lorentz transformations using some other approach it would have been less immediately clear what the theory said and perhaps taken longer for people to accept it. Anyway there have been various formulations over the years and Einstein's has won out as it is still the one people refer to.
 * The single-postulate point of view is a notable POV and in my view deserves more than a sentence. Delaszk (talk) 19:10, 9 November 2008 (UTC)


 * There are, in fact, more postulates involved. The existence of inertial frames is one. Then there are the different formulations of the usual two postulates. Do you count "Maxwell's theory obeys the relativity principle" as a new postulate? If your "one postulate" is only kinematical/mechanical relativity, do you need an additional postulate to get relativistic electrodynamics? There are people who've written more systematic books that split special relativity into independent axioms, starting with events, postulating the existence of free particles, postulating isotropy and so on (showing that they're independent is quite a feat, btw!). Given the different levels of rigorosity, and the different ways of formulating things, there are many different "numbers of postulates" you can assign to special relativity. Which, I think, shows that the "number of postulates" isn't all that interesting from a physics point of view. The interesting bit is what you can derive from mechanical relativity alone, namely more than people think, but yes, I think that merits no more than one or two sentences in the overall special relativity article. And some sentences under "relativity principle", perhaps. Markus Poessel (talk) 22:22, 9 November 2008 (UTC)


 * The article now includes the phrase "claim may be misleading because actually these formulations rely on various unsaid assumptions such as isotropy and homogeneity of space. Nevertheless the form of the Lorentz transformations can be derived without first postulating the univeral lightspeed."


 * I intend that the article will now discuss general alternative formulations rather than just those formulations which claim single-postulate. Therefore the title has been changed to: Special relativity (alternative formulations). Delaszk (talk) 11:44, 10 November 2008 (UTC)


 * Better, but still a rather strange lemma. There's a continuum of derivations, and no clear-cut division between "standard" and "alternative". Also, some statements in that article are still rather odd. Why call it "Feigenbaum's derivation" when it's been done long before (and I've given you the references)? And the "bizarre rotations", as far as I can see, are a standard and certainly not bizarre part of the Lorentz group, which also has been known for pretty much a century. What would, in my estimation, make sense would be a lemma "Axiomatization of special relativity" as a spin-off of a corresponding section in the main article. But not this hodge-podge. Markus Poessel (talk) 11:52, 11 November 2008 (UTC)


 * The whole section on Feigenbaum has now been deleted since his paper is just one more example of the single postulate claim. Your reference Algebraic and geometric structures of Special Relativity has been added. I agree that there could be a new article Axiomatization of special relativity but the current article is now more of an overview of the variety of theories about special relativity. Delaszk (talk) 12:38, 11 November 2008 (UTC)

Removed unneeded second derivation of Kinetic Energy
I undid DS1000's two attempts (this and this) at adding another derivation of the Kinetic Energy formula. I think the original one was better and largely sufficient. This is an article about Special relativity, not about derivations of relativistic kinetic energy. DVdm (talk) 19:07, 30 October 2008 (UTC)


 * My derivation is much better than the existent one. Since you are so pedantic as to object to two derivations, I have replaced the older one with the newer one. I hope that this is satisfactory and it will not escalate into an editing war. DS1000 (talk) 22:01, 30 October 2008 (UTC)

Both derivations are only heuristic arguments. What is missing in this article is a real derivation of the equations for relativistic energy and momentum. This cannot be "derived" in the current approach; what you have now is a circular reasoning. We should use the well known argument by R.C. Tolman involving an elastic collision to derive this. This is what is done in almost all of the textbooks. There is no shortcut, if there were then the textbooks wouldn't bother setting up Tolman's argument. Count Iblis (talk) 23:38, 30 October 2008 (UTC)


 * (moved DS1000's comment from my talk page - DVdm (talk)):
 * Since you seem to object to having two derivations I left only one : the cleaner one that does not involve acceleration. I hope that you will leave it alone and not escalate this to an editing war. :-) DS1000 (talk) 22:06, 30 October 2008 (UTC)


 * I have reverted again. Please check with other's opinions first. And remember wp:3RR. By the way, there are multiple glaring errors in what you tried to insert. As an exercise, see if you can find them before you try to reinsert this (sorry) mess. You can use this talk page for your trials. Hint: I am not even alluding to the silly typo with the square root.
 * Other than that I agree with Count Iblis that this article doesn't really need a derivation to begin with. This is probably the reason why the word "derivation" wasn't used. It is more of a feasability statement, and as such it is nicely building on the preceding paragraph on Force, expressed in terms of acceleration, which i.m.o. is a very good reason to keep it here. It has survived in this form since quite a while, so I don't see any harm in keeping it the way it was.
 * DVdm (talk) 09:26, 31 October 2008 (UTC)


 * There was just another obvious typo that occured when I transcribed from my notes, you simply can't follow two lines of elementary calculus. You would have easily inferred it if you followed the computations. Too bad, here is the correct formula for your education:

$$\Delta K=W=\int_0^v \mathbf{F}\cdot d\mathbf{x}=m \int_0^v (v^2d \gamma+\gamma \mathbf{v}d\mathbf{v})=m \int_0^v (d(\gamma v^2)-\frac {\gamma dv^2}{2})=m \gamma v^2-m \int_0^v \gamma v dv=$$


 * $$=m \gamma v^2-m \int_0^v \frac { v dv}{\sqrt {1-\frac{v^2}{c^2}}}=m c^2(\gamma-1)$$
 * DS1000 (talk) 14:23, 31 October 2008 (UTC)


 * That leaves one remaining glaring error. See if you can find it. And don't forget to explain how you get from F.dx to the next step. This comes naturally in the original derivation. DVdm (talk) 14:47, 31 October 2008 (UTC)


 * Dirk, can't you fill in the elementary steps? Here, so you can stop whining now:
 * $$v^2d \gamma+\gamma \mathbf{v}d\mathbf{v}= v^2d \gamma+\gamma vdv=d(\gamma v^2)-\gamma d(v^2)+\frac{1}{2}\gamma d(v^2)=d(\gamma v^2)- \frac{1}{2}\gamma d(v^2)=d(\gamma v^2)- \gamma vdv$$ DS1000 (talk) 15:05, 31 October 2008 (UTC)
 * DVdm was referring to the previous step (the third equals sign above, not the fourth). -- Army1987 (t — c) 15:14, 31 October 2008 (UTC)


 * $$ \mathbf{F} = m \frac{d(\gamma \, \mathbf{v})}{dt} = m \left( \frac{d \gamma}{dt} \, \mathbf{v} + \gamma \frac{d\mathbf{v}}{dt} \right).$$
 * $$ d \mathbf {x}=\mathbf {v} dt $$


 * You can do the dot product now all by yourselves. DS1000 (talk) 15:19, 31 October 2008 (UTC)


 * There is no need to say that I am whining. I can find the steps alright. I am having the readers of this article in mind. You have 4 equalities with almost trivial steps, and you have 2 equalities with a much less trivial step. I'm sure that everone else would agree to prefer the original "derivation", as it follows 100% naturally from the preceding subsection, without the need for any further clarification.
 * And you still haven't found the most glaring error in there. You can't find it? DVdm (talk) 15:44, 31 October 2008 (UTC)


 * I wrote the preceding subsection (on force). There isn't any error in the derivation I showed, nor is it "heuristic", it is simple,straightforward calculus but you can continue to confabulate. DS1000 (talk) 16:02, 31 October 2008 (UTC)


 * Well, I'm sorry, there still is an (i.m.o. painful) error in your string of equalities. If you look very carefully, you should find it. DVdm (talk) 16:32, 31 October 2008 (UTC)


 * I don't profess to be perfect, I checked the computations, the result is correct and so appear to be the intermediate steps. Has the thought of "you" being wrong crossed your mind? I am sorry, I don't have time to play games, so, if you think you found an error, let's see it. :-) DS1000 (talk) 17:27, 31 October 2008 (UTC)


 * Look closer. You'll be thankful if you find it without my help. DVdm (talk) 18:16, 31 October 2008 (UTC)


 * So, from "multiple glaring errors" you are down to one.Ever considered that it is "you" that might be wrong? Naaah! I think you are mixing up science with playing games. I suggest you take your games elswhere, like a children playground :-) DS1000 (talk) 18:25, 31 October 2008 (UTC)


 * Closer. DVdm (talk) 19:02, 31 October 2008 (UTC)

← I hope you are not referring to the missing dot in vdv and to the fact that the bounds of integration are always shown as 0 and v regardless of which is the actual variable of integration in each step. Or are you? -- Army1987 (t — c) 01:04, 1 November 2008 (UTC)


 * Ah, you spoiled it. The more or less sloppy integral bounds notation I could live with, but the missing dot is, at least i.m.o., serious. Glad that you spotted it. Perhaps DS1000 didn't spot it because he thought I was just joking or teasing. Or perhaps unlike me he didn't find it worth bothering. No big deal - people differ.
 * Sorry for the fuss, DS1000. I notice you removed that Apeiron thing elsewhere. Good job - but expect a fight. Cheers, DVdm (talk) 10:38, 1 November 2008 (UTC)


 * I missed this post, so you are FINALLY admitting that there is no error in my derivation. Too bad that you kept this travesty going for so many days, did you imagine for a second that I did not know what an inner product was ? That it needed a dot between the two vectors? Jeez DS1000 (talk) 19:32, 1 November 2008 (UTC)
 * Sorry for spoiling it, but I believed (hoped?) that you might be referring to something more subtle that I failed to see. If your point against that derivation was only the missing dot, the only reasonable behaviour would be adding it. (But I agree that many of the steps in DS1000's derivation were way too nontrivial to be implicit. But my point is, adding that bloody dot between the v and the dv wouldn't by itself make that derivation suitable for the article, so repeatedly referring to that "glaring error" was not very useful, other than for "intimidating" (for lack of a better word) DS1000.) -- Army1987 (t — c) 11:01, 1 November 2008 (UTC)
 * No, that missing dot was the least of my points against. I could have inserted it without all the fuss. I agree, I can imagine that my attitude looked a bit intimidating. I'd like to apologise for that, specially to DS1000. DVdm (talk) 11:08, 1 November 2008 (UTC)


 * What are you two talking about? It is obvious that $$\mathbf{v}d\mathbf{v}$$ is a dot product. Adding the dot between the vectors doesn't change anything. You mean that BOTH of you guys don't know that:
 * $$\Delta K=W=\int_0^v \mathbf{F}\cdot d\mathbf{x}=m \int_0^v (v^2d \gamma+\gamma \mathbf{v}. d\mathbf{v})=m \int_0^v (d(\gamma v^2)-\frac {\gamma dv^2}{2})=m \gamma v^2-m \int_0^v \gamma v dv=$$


 * ....this is because


 * $$v^2d \gamma+\gamma \mathbf{v}. d\mathbf{v}= v^2d \gamma+\gamma vdv$$


 * .....which in turn is due to the WELL KNOWN, ELEMENTARY identity taught to freshman students:


 * $$\mathbf{v}. d\mathbf{v} = vdv$$


 * So, the purported "error" has been all along an error indeed, but on YOUR side, Dirk. Congratulations, this is simple vector algebra that is being taught to freshman students!You could have seen that  I already used the same exact formula in the derivation of force, the paragraph just above the derivation of energy.  DS1000 (talk) 15:06, 1 November 2008 (UTC) —Preceding unsigned comment added by DS1000 (talk • contribs) 15:02, 1 November 2008 (UTC)
 * Could you re-read my posts more carefully, please? -- Army1987 (t — c) 16:06, 1 November 2008 (UTC)


 * I read them, there is nothing of any validity. Could any of you two stop playing games and write some math? It looks like both of you are mistaking your ignorance for an error in my derivation. DS1000 (talk) 16:16, 1 November 2008 (UTC)
 * I said in my post that I failed to see any other error other than the missing dot, and hoped DVdm's point was something more serious than that. Of course, I realize that defending you wasn't the right thing to do. And I do not understand what you mean by "my ignorance", since there is was no evidence in my post that I was unaware that v&thinsp;dv = v &middot; dv (which I wasn't). -- Army1987 (t — c) 16:35, 1 November 2008 (UTC)


 * OK, I understand.Dirk keeps PRETENDING that he'a found something more serious, what he's found is his own ignorance. So, you agree that there is no error, it is only Dirk that continues his childish game. So, perhaps you can ask him to point out the "glaring error", eh? 18:54, 1 November 2008 (UTC)  —Preceding unsigned comment added by DS1000 (talk • contribs)
 * Dirk has already stated that the "error" was indeed the missing dot, unlike I believed, and he apologized. He later almost retracted his apologies, and given your attitude he wasn't terribly wrong in doing so. -- Army1987 (t — c) 19:21, 1 November 2008 (UTC)
 * I'm almost tempted to think that apologising wasn't the right thing to do either. Almost. Ah well. DVdm (talk) 16:52, 1 November 2008 (UTC)


 * Dirk, your ignorance is only superseded by your unlimited arrogance. In two dimensions:


 * $$vdv=\sqrt {v_x^2+v_y^2}* (v_x*dv_x+v_y*dv_y)/\sqrt {v_x^2+v_y^2}=v_x*dv_x+v_y*dv_y=\mathbf{v}. d\mathbf{v}$$


 * They teach the above to freshmen, you must have skipped class. So, the whole derivation is correct, the alleged "error" is just in your ignorance of freshman math. Or is it the missing dot between v and dv ? :-) DS1000 (talk) 18:52, 1 November 2008 (UTC)
 * And, for any real inner product space,
 * $$v \stackrel{\scriptstyle\mathrm{def}}{=} \sqrt{\mathbf{v} \cdot \mathbf{v}} \quad \to \quad v^2 = \mathbf{v} \cdot \mathbf{v};$$
 * $$2v\,dv = d(v^2) = d(\mathbf{v} \cdot \mathbf{v}) = d\mathbf{v} \cdot \mathbf{v} + \mathbf{v} \cdot d\mathbf{v} = 2 \mathbf{v} \cdot d\mathbf{v}.$$
 * The problem is not correctness, the problem is that few readers are going to be able to understand the non-obvious steps such as $$\mathbf{F}\cdot d\mathbf{x}=m(v^2d \gamma+\gamma \mathbf{v}\cdot d\mathbf{v})$$. -- Army1987 (t — c) 19:14, 1 November 2008 (UTC)


 * No, if you look the problem is that Dirk has been LYING for a few days already that he's found "glaring errors" in order to justify deleting my derivation. I don't give a damn about the derivation but I care about people being honest. This is freshman physics. Readers are expected to have a certain level, no? Anyways, I have found someplace, tucked in, his note admitting that he was wrong all along and that the derivation is correct. We would have been much better off if he didn't play these childish games for days. :-) DS1000 (talk) 19:20, 1 November 2008 (UTC)

← The fact that "readers are expected to have a certain level" (by whom? see WP:MTAA...) doesn't mean that you can skip so many steps in the derivation, like that. I wouldn't object to something like:

W = \int_{x_0}^{x_1} \mathbf{F} \cdot d\mathbf{x} = \int_{t_0}^{t_1} \frac{d}{dt}(\gamma m\mathbf{v}) \cdot \mathbf{v}dt = m \int_{t_0}^{t_1} \left(\frac{d\gamma}{dt}\mathbf{v}\cdot\mathbf{v}  + \gamma \frac{d\mathbf{v}}{dt}\cdot\mathbf{v} \right)dt = m \int_{t_0}^{t_1} \left(v^2\frac{d\gamma}{dt}  + \gamma v\frac{dv}{dt} \right)dt =$$ $$m \left( \left. v^2\gamma\right|^{t_0}_{t_1} - \int_{t_0}^{t_1} \gamma \frac{2vdv}{dt} + \int_{t_0}^{t_1} \gamma  v\frac{dv}{dt} \right)=$$ $$m \left( \left. v^2\gamma\right|^{t_0}_{t_1} - \int_{t_0}^{t_1} \gamma v \frac{dv}{dt} \right) =$$ $$m \left( \left. v^2\gamma\right|^{t_0}_{t_1} - \left. v^2\gamma\right|^{t_0}_{t_1} + \int_{t_0}^{t_1}v \frac{d(\gamma v)}{dt}  dt \right)=$$ $$m \int_{v_0}^{v_1} v \frac{d}{dv} \left(\frac{v}{\sqrt{1-v^2/c^2}}\right) dv =$$ $$m \int_{v_0}^{v_1} \left( \frac{v}{\sqrt{1-v^2/c^2}} + \frac{v^3}{c^2\sqrt{1-v^2/c^2}^3} \right) dv = \dots = m\gamma_1 c^2 - m \gamma_0 c^2.$$ -- Army1987 (t — c) 20:34, 1 November 2008 (UTC)

You made the computations unnecessarily complicated, you know that dv/dt*dt=dv, right? So, it is not necssary to carry those t_0 and t_1 limits with you around, nor is it necessary to integrate wrt t, you can integrate wrt v. Working in v parameter instead of t would had enabled you to drastically reduce the number of steps. Using total differentials instead of derivatives further reduces the number of steps. Compare your derivation to mine. :-) But the main point is that all this could have been avoided if Dirk didn't play his silly games for so long. DS1000 (talk) 20:59, 1 November 2008 (UTC)
 * What if v(t) is not monotonic? Split the integral into parts for when the particle is speeding up and parts for when the particle is slowing down, swap signs or bounds when needed, and re-add? -- Army1987 (t — c) 21:28, 1 November 2008 (UTC)
 * You end up integrating over v ANYWAYS.See your last two integrals? The exercise was to show the kinetic energy gained when the particle accelerates from 0 to v, no? Just a simple exercise but if you insist on using your derivation, it is all yours, kid. Post it, I am sure it will meet with Dirk's approval :-)DS1000 (talk) 21:46, 1 November 2008 (UTC)

← I was deliberately using as many steps as reasonably possible to make a point, but what about something like:
 * $$W = \int_{x_0}^{x_1} \mathbf{F} \cdot d\mathbf{x} = \int_{t_0}^{t_1} \frac{d}{dt}(\gamma m\mathbf{v})\cdot\mathbf{v}dt$$$$\ = m \int_{t_0}^{t_1} \left(\frac{d\gamma}{dt}\mathbf{v}\cdot\mathbf{v}  + \gamma \frac{d\mathbf{v}}{dt}\cdot\mathbf{v} \right)dt$$$$\ = m \int_{t_0}^{t_1} \left( v^2\frac{d\gamma}{dt} + \gamma v\frac{dv}{dt} \right)dt$$$$\ = m \int_{v_0}^{v_1} \left( v^2\frac{d\gamma}{dv} + \gamma v \right) dv$$$$\ = m \int_{v_0}^{v_1} \left( \frac{v^3/c^2}{\sqrt{(1-v^2/c^2)^3}} + \frac{v}{\sqrt{1-v^2/c^2}} \right) dv = \dots$$$$\ = m\gamma_1 c^2 - m \gamma_0 c^2.\,\!$$ — Army1987 (t — c) 02:06, 2 November 2008 (UTC)


 * You missed the point, you have to integrate over v. There is no way around it. This is what I showed you DS1000 (talk) 06:11, 2 November 2008 (UTC)
 * Oh dear. $$\int_{v_0}^{v_1} \left( \frac{v^3/c^2}{\sqrt{(1-v^2/c^2)^3}} + \frac{v}{\sqrt{1-v^2/c^2}} \right) dv$$ is an integral over what? -- Army1987 (t — c) 09:31, 2 November 2008 (UTC)


 * Over v, don't you see your limits of integration? Don't you see the dv? Jeez DS1000 (talk) 14:45, 2 November 2008 (UTC)
 * So what was your "you have to integrate over v" about given that that was exactly what I did? BTW, I've seen that in the article you wrote the power as Fv. While you might not like the dot for the scalar product, it is standard notation. Fv means something else (see dyadic tensor—it's true that it doesn't make sense in that context, but that's not an excuse for using nonstandard notation). -- Army1987 (t — c) 15:34, 2 November 2008 (UTC)


 * It was in answer to your earlier post "What if v(t) is not monotonic? Split the integral into parts for when the particle is speeding up and parts for when the particle is slowing down, swap signs or bounds when needed, and re-add?". Could you please stop generating irrelevant diversions and try to understand my posts?
 * As to the "Fv" and such, pettiness will not take you very far in physics,kid. First off, adding dots in dot products is not necessary, second off, wasting your life with such pettiness is not very productive. Writing two vectors next to each other is ALSO a standard form in denoting dot products. Lighten up, kid :-) DS1000 (talk) 15:53, 2 November 2008 (UTC)
 * As for the first paragraph, yes, you're right – let's stop this. As for the second paragraph, that standard is so common that I've never seen it used anywhere before. The only times I encountered that notation, it denoted dyadic tensors. -- Army1987 (t — c) 16:22, 2 November 2008 (UTC)

We can also do a partial integration, similar to DS1000's original derivation in the second step.

$$\frac{d}{dt}(\gamma m\mathbf{v})\cdot\mathbf{v} = \frac{d}{dt}(\gamma m\mathbf{v}\cdot\mathbf{v}) - (\gamma m\mathbf{v})\cdot\mathbf{\frac{dv}{dt}}$$

Then when integrating the second term over time you write it as the integral over v of 1/2 m gamma(v) d(v^2), you don't expand 1/2 dv^2 as v dv, because gamma(v) is actually a function of v^2, so it is the same as integrating a term 1/sqrt(1-x/c^2) over x. Count Iblis (talk) 02:29, 2 November 2008 (UTC)
 * Even without writing d(v2) explicitly that way:
 * $$W = \int_{\mathbf{r}_0}^{\mathbf{r}_1} \mathbf{F} \cdot d\mathbf{r}$$ $$= \int_{t_0}^{t_1} \frac{d}{dt}(\gamma m\mathbf{v})\cdot\mathbf{v}dt$$ $$= \left. \gamma m \mathbf{v} \cdot \mathbf{v} \right|^{t_1}_{t_0} - \int_{t_0}^{t_1} \gamma m\mathbf{v} \cdot \frac{d\mathbf{v}}{dt} dt$$ $$= \left. \gamma m v^2 \right|^{t_1}_{t_0} - m\int_{v_0}^{v_1} \gamma v\,dv$$ $$= m \left( \left. \gamma v^2 \right|^{t_1}_{t_0} - c^2\int_{v_0}^{v_1} \frac{2v/c^2}{2\sqrt{1-v^2/c^2}}\,dv \right)$$ $$= \left. m\left(\frac {v^2}{\sqrt{1-v^2/c^2}} + c^2 \sqrt{1-v^2/c^2} \right) \right|^{t_1}_{t_0}$$ $$= \left. \frac {mc^2}{\sqrt{1-v^2/c^2}} \right|^{t_1}_{t_0}$$ $$= \left. {\gamma mc^2}\right|^{t_1}_{t_0}$$ $$\displaystyle= \gamma_1 mc^2 - \gamma_0 mc^2.$$-- Army1987 (t — c) 09:31, 2 November 2008 (UTC)

Propose removal of paragraph about application to cyclotrons in subsection on Kinetic energy
Although I still strongly think that the original "derivation" of kinetic energy, building on the preceding subsection on force, was much more suited in its original context, the current new derivation is healthy. Of course it does remain a showcase of elementary calculus and algebra, but if that is what we need in an article about special relativity, then there's no point in my trying to remove it against what seems to have become some sort of consensus.

On the other hand, the part about the immediate application of the above in cyclotrons, a nice derivation it may be, but I really think this does not belong here. It is unsourced, it looks like original research or at best as a well worked out text book exercise, and i.m.o. most important, it does not provide any added value to the main article. Perhaps, without the reproduction of the entire derivation, it belongs in the cyclotrons article. Remember that Kinetic Energy is just a subsection. Thoughts? DVdm (talk) 10:31, 3 November 2008 (UTC)


 * I didn't expect any different, Dirk. First you fought tooth and nail to remove the current derivation for kinetic energy, PRETENDING to have found "glaring errors", now you are back trying to get the application of the formula out. It is important to show applications and it is nice to show connections with related domains, like the Abraham-Lorentz force. At least this time you just didn't remove things without asking anybody. Leave things alone. Please. DS1000 (talk) 15:04, 3 November 2008 (UTC)


 * The equations in the article (and more) can be found in this non-exhaustive set of references:


 * 1. C. S. Roberts and S. J. Buchsbaum, “Motion of a chaged particle in a constant magnetic field and a trasnverse electromagnetic wave propagating along the field”, Phys. Rev. 135, A381 (1964)
 * 2. V.G. Bagrov, D.M. Gitman and A.V. Jushin, Solutions for the motion of an electron in electromagnetic field, Phys. Rev.D, 12, 3200 (1975)
 * 3. H.R.Jory, A.W.Trivelpiece, J.Appl.Phys. "Charged particle motion in large. amplitude electromagnetic fields",39,3053 (1968)
 * 4. R. Ondarza-Rovira, “Relativistic motion of a charged particle driven by an elliptically polarized electromagnetic wave propagating along a static magnetic field”, IEEE Transactions on Plasma Science, Vol. 29, 6, 903 (2001)
 * 5. J Kruger and M Bovyn, “Relativistic motion of a charged particle in a plane electromagnetic wave with arbitrary amplitude”, J. Phys. A: Math. Gen., Vol.11, 9 1841 (1976)
 * 6. H. Takabe, “ Relativistic motion of charged particles in ultra-intense laser fields”, Journal of Plasma and Fusion Research, Vol.78, 4, 341(2005)
 * 7. H P Zehrfeld, G. Fussmann, B.J. Green, “Electric field effects on relativistic charged particle motion in Tokamaks”, Plasma Phys. 23 473 (1981)
 * 8. R. Giovanelli, “Analytic treatment of the relativistic motion of charged particles in electric and magnetic field”, Il Nuovo Cimento D, Vol. 9, 11,1443 (1987)
 * 9. A. Bourdier, M.Valentini, J.Valat, “Dynamics of a relativistic charged particle in a constant homogeneous magnetic field and a transverse homogeneous rotating electric field”, Phys. Rev. E 54, 5681 (1996)
 * 10. S.W.Kim, D.H.Kwon, H.W. Lee, “Relativistic cyclotron motion in a polarized electric field”, Jour. of Kor. Phys. Soc., Vol. 32, 1, 30 (1998)
 * 11. L.B.Kong, P.K. Liu, “Analytical solution for relativistic charged particle motion in a circularly polarized electromagnetic wave”, Phys. Plasmas 14, 063101 (2007)
 * DS1000 (talk) 15:34, 3 November 2008 (UTC)


 * To DS1000: There is no need to take this personally, or to look at it as if it were some game. If this part is going to stay here, please add the refs to the text? Thanks.
 * ( Thanks DS1000. I have moved the refs into the section - DVdm (talk) )
 * Do we need all eleven of them, considering they very likely state the same things with almost the same words? Wouldn't one or two of them be enough? -- Army1987 (t — c) 12:16, 4 November 2008 (UTC)
 * In my opinion we shouldn't even mention the entire cyclotron thing to begin with. But apparently that's just me of course, so never mind what I think :-) - DVdm (talk) 12:46, 4 November 2008 (UTC)


 * Does anyone else have thoughts about the relevance of this elaborated derivation in an article about Special relativy?
 * DVdm (talk) 17:10, 3 November 2008 (UTC)


 * I think the added text is too large. It doesn't explain any physics, it is just an exercise in applying the formula for force. We should focus on the steps that are missing in this article that deal with the physics, like a derivation of the formula for relativistic momentum and energy.
 * Detailed worked out examples of problems can be useful if they illustrate certain widely used mathematical techiques in special relativity. Instead of the cyclotron example, we could discuss some collision problem and demonstrate how you can efficiently solve that problem using four-momentum algebra. Count Iblis (talk) 17:48, 3 November 2008 (UTC)


 * Not really, it shows how different forces (Lorentz, Abrahms-Lorentz and the "conventional" F=dp/dt) come together with the kinetic energy in order to help solve the practical problems of particle separation in cyclotrons. This doesn't mean that we can't discuss some collision problems as well. DS1000 (talk) 18:41, 3 November 2008 (UTC)

Army1987, I just noticed how you took care of the showcase of the calculus and algebra showcase. Very Neat! DVdm (talk) 13:21, 4 November 2008 (UTC)

Parallel force
Delaszk, you insist that a sentence like "Acceleration requires a component parallel to the existing velocity in addition to a component in the direction of the acceleration."
 * Uniform circular motion has acceleration without a component parallel to the existing velocity.
 * Furthermore you say that "Acceleration requires a component ... in the direction of the acceleration". That's like saying that a vector has a component in its own direction. Duh.

So, you have a combinatin of a completely faulty statement and the most trivial tautology one could possible imagine. Forgive me, don't take this personal, but I really think this deserves to be called nonsense. DVdm (talk) 11:24, 9 November 2008 (UTC)
 * You are right, the statement I orginally made was nonsense. The corrected version of the statement is however true and is a consequence of the Lorentz factor according to the book. Delaszk (talk) 11:33, 9 November 2008 (UTC) Sorry, I changed my posting above at 11:16 local time to correct it with the word force. You replied at 11.24 local time, sorry if you didn't notice the change. We are now talking at cross-purposes. Delaszk (talk) 11:37, 9 November 2008 (UTC)

Originally the statement about parallel component was indeed nonsense. I had omitted the word force as I hadn't read the book properly. Can we now agree that the modified statement is a consequence of the Lorentz factor, and therefore put the statement into the article. The statement should now read: "Acceleration requires a force not only in the direction of the acceleration but a force which also has a component parallel to the existing velocity.". Delaszk (talk) 12:13, 9 November 2008 (UTC)


 * ... I don't think I can help you. Perhaps someone else can, but frankly... I doubt it. Sigh. Sorry. - DVdm (talk) 13:13, 9 November 2008 (UTC)


 * Well then, let me quote the reference. First of all, relativistic momentum P is given by: $$ \mathbf{P} = \frac{m\mathbf{v}}{\sqrt{1 - \frac{v^2}{c^2}}}$$ and Newton's second law is: $$ \mathbf{F} = {\mathrm{d}\mathbf{p} \over \mathrm{d}t}$$. Now, at the top of Page 80 of "Introducing Special Relativity" by W.S.C.Williams, published by Taylor and Francis it says: "However, the factor $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ present in the formula for P, leads to a situation in which an acceleration generally needs a force with a component parallel to v in addition to that in the direction of the acceleration." It then gives an exercise for the reader to complete: "Show that to produce an acceleration of a particle in a particular direction requires a force F not solely in the direction of the acceleration but with a component along the direction of the particle's existing velocity." Delaszk (talk) 15:09, 9 November 2008 (UTC)


 * I didn't ask for a reference. What you are trying to say here, is old hat. Check for instance Special relativity and its edit history. I actually worked on it. What I am trying to tell you is that I cannot help you with your problems (1) expressing yourself properly, and (2) finding the proper places where to put your comments. This is not about parallel forces - this is about you. Please take some time off to think about that... DVdm (talk) 16:16, 9 November 2008 (UTC)


 * I made a mistake in what I wrote and I admit it. However, I think that the Consequences section should mention the parallel force component, but thankyou for pointing out the section Special relativity which gives the details. Unless there is a disagreement then I will put a statement about this in the consequences section. I accept that it is nothing new and that you have worked on this before. Delaszk (talk) 17:17, 9 November 2008 (UTC)

Delaszk's statement
I have edited Delaszk's statement
 * "By using the same units to measure both space and time (e.g. the metre), then speed becomes a dimensionless quantity (e.g. metre per metre). By the principle of relativity, equations with similar form to the Lorentz transformations are derived. Maxwell's Equations then fix the numerical value of the speed of light in these units to be the dimensionless number 1. This can be thought of as saying: a time of 1 metre is the time it takes for light to travel one metre. See Special relativity via single postulate."

into
 * "By using the same units to measure both space and time (e.g. the metre), speed becomes a dimensionless quantity (e.g. metre per metre) and the value of the speed of light in these units becomes the dimensionless number 1."

for the following reasons: Comments? - DVdm (talk) 11:49, 9 November 2008 (UTC)
 * 1) "By using ..., then speed becomes..." is grammatically wrong. The word "then" should not be there.
 * 2) The phrase "By the principle of relativity, equations with similar form to the Lorentz transformations are derived." doesn't make sense. And "with similar form to" is grammatically wrong.
 * 3) The phrase "Maxwell's Equations then fix...": Either the postulate of constant lightspeed, or experiment, combined with the definition of speed and the choise of the units fixes the number 1 for the speed of light.
 * 4) The phrase "This can be thought of as saying that..." is trivially obvious.
 * 5) Delaszk's (private) project Special relativity via single postulate is not related to the units of speeds.


 * Note: I think that the sentence I put in place of the original, doesn't even belong in that section either. It can be removed. DVdm (talk) 12:08, 9 November 2008 (UTC)

The whole paragraph is basically an introduction to the article Special relativity via single postulate which I think deserves a link from the main article. Perhaps you disagree that it deserves a link or that the link should be in the Status section. But as a summary of that article the paragraph is as accurate as I could make it.
 * Special relativity via single postulate is not a private project. Anyone is free to edit any wikipedia article. Anyone who has read any literature which discusses the single postulate issue will be able to usefully edit that page.
 * One of the main references about the single postulate approach is the Hsu&Hsu book and they make the definition of units central to their thesis. So the article is related to units of speed.
 * The phrase "This can be thought of as saying that..." may be trivially obvious to you, but it was not obvious to me until I read it in the book, and I think it usefully explains in words what the situation is.
 * The references in the article Special relativity via single postulate do not postulate the constant lightspeed. The Hsu&Hsu book does however derive transformations from the relativity principle which are equivalent to the Lorentz but with a maximal speed of 1. Combining this with Maxwell's equations then shows that light travels at this maximal speed.
 * There may be some issues with grammar which I happily accept can be improved. Delaszk (talk) 12:48, 9 November 2008 (UTC)

Taiji relativity
To respond in full to the above section, I'd like to clarify a point here about units of time and speed. In Hsu&Hsu's taji relativity formulation, all 4 coordinates of spacetime are measured in the same units. The transformations are then derived using just the principle of relativity and have a maximal speed of 1, which is quite unlike the derivation of the Lorentz transformations in which the parameter may still be zero. So this is not the same as other "single postulate" derivations. However the relationship of taiji time "w" to standard time "t" must still be found, otherwise it would not be clear how an observer would measure taiji time. The taiji transformations are then combined with Maxwell's equations to show that the speed of light is independent of the observer and has the value 1 in taiji speed. Since we can measure the speed of light by experiment in m/s to get the value c, we can use this as a conversion factor. i.e. we have now found an operational definition of taiji time: w=ct. Delaszk (talk) 19:23, 11 November 2008 (UTC)


 * Brilliant! Before someone beats you to it, be the first to create another new article: Taiji time. You can also link to it from your Special relativity (alternative formulations). Hurry! DVdm (talk) 19:46, 11 November 2008 (UTC)


 * I assume that DVdm was being sarcastic.
 * This approach in which "all 4 coordinates of spacetime are measured in the same units" begs the question. How do you measure time in meters in Galilean-Newtonian space+time? You cannot. JRSpriggs (talk) 05:11, 12 November 2008 (UTC)


 * Exactly. In space + time you must add the postulate of maximum speed but in spacetime it is already there. Delaszk (talk) 09:54, 12 November 2008 (UTC)


 * Inverse mass can also be measured in the same units as length by interpreting Planck's constant as a dimensionless conversion factor. And if put G = 1, then mass itself is dimensionally the same as a length, everything has become dimensionless.


 * You can regain dimensions and units even if all quantities are dimensionless. You can redefine physical quantities by introducing arbitrary dimensionless rescaling parameters and rewrite the theory in terms of these rescaled variables only. At this stage everything is still dimensionless. But then you consider the singular limit in which some rescaling parameters become zero or infinite. In this singular limit the connection between some variables gets lost.


 * If in this theory intelligent beings subjectively experience the extremely rescaled physical variables, then the laws of physics they discover will reflect that. So, they will formulate their laws of physics using incompatible physical quantities by assigning incompatible dimensions to the variables. Only much later will they discover the existence of the very small or very large "conversion factors". But by that time, the perception that the different appeantly incompatible physical quantities are actually incompatible is so much part of their culture that they'll not overturn this. They'll assign dimensions to the rescaling constants to make the conversion "dimensionally correct", rather than abolish the dimensions altogether. Count Iblis (talk) 15:05, 12 November 2008 (UTC)


 * w metres = (c m/s)* t seconds
 * That seems perfectly dimensionally correct to me.


 * Let r= distance. Then taiji speed = r metres / w metres = r/w dimensionless. Delaszk (talk) 15:59, 12 November 2008 (UTC)


 * Also it is not just due to the choice of units that there is a maximum speed. It is the principle of relativity, that Hsu&Hsu say, when applied to 4d spacetime, implies the invariance of the 4d-spacetime interval $$s^2=w^2-r^2$$ and this leads to the coordinate transformations involving the factor $$1\over\sqrt{(1-\beta^2)}$$ where beta is the magnitude of the velocity between two inertial frames. The difference between this and the spacetime interval $$s^2=c^2t^2-r^2$$ in Minkowski space seems to be that $$s^2=w^2-r^2$$ is invariant purely by the principle of relativity whereas $$s^2=c^2t^2-r^2$$ requires both postulates.Delaszk (talk) 18:14, 12 November 2008 (UTC)


 * Actually in the above there is nothing wrong with defining time in metres in Newtonian kinematics. Why not say 1 metre of time=the time it takes for a car to travel 1 metre. The reason Hsu&Hsu get transformations with a finite maximum speed must be that Hsu&Hsu are taking the "principle of relativity" in spacetime to mean invariance of laws under 4-dimensional transformations. Delaszk (talk) 11:37, 14 November 2008 (UTC)
 * Perhaps you have not noticed, but some cars travel faster than others. Furthermore, the same car can travel at different speeds at different times, or stop altogether. So your "definition" of how to measure time in Newtonian space+time is nonsensical. JRSpriggs (talk) 06:14, 15 November 2008 (UTC)
 * Before atomic clocks or the recognition of the constancy of light, time used to be measured with things like pendulum clocks or clockwork clocks. Conventionally this was expressed in angular rotation e.g. half and hour, quarter of an hour etc. There is no difference between angular rotation and linear distance. Simply define the unit of time to be the length of the arc that the tip of the second hand travels for some particular angle of rotation. Delaszk (talk) 09:32, 15 November 2008 (UTC)

Lorentz transformation without second postulate
It seems there is a lot of discussion about single-postulate derivations of special relativity ;-). For some historical informations, see: History of special relativity. And look into the section "Lorentz transformation without second postulate" for the approaches of Ignatowski and Frank/Rothe. However, according to Pauli (1921), Resnick (1967), and Miller (1981), those models were insufficient.... --D.H (talk) 09:57, 12 November 2008 (UTC)
 * And, according to Einstein, the constancy of the speed of light is contained in Maxwell's equations. --Michael C. Price talk 10:49, 13 November 2008 (UTC)
 * That section includes the phrase "Ignatowski was forced to recourse to electrodynamics to include the speed of light.". Exactly, the trio of "principle of relativity+Maxwell+numerical values from experiment" gives special relativity and this should be compared with "principle of relativity+second postulate+Maxwell+numerical values from experiment". Since Einstein's 1905 paper is all about electrodynamics he is assuming Maxwell's equations, and the theory isn't practically applicable without numerical values. When compared like with like, from the point of view of asking what is knowable, the second postulate can be deduced. If you restrict your attention to just the standalone theory of relativity then yes you need the postulate. But given all the available knowledge we don't need to postulate it. In other words different domains of knowledge are overlapping and thus taken together have more information than necessary. Delaszk (talk) 10:21, 12 November 2008 (UTC)


 * In your formulation "principle of relativity+Maxwell+numerical values from experiment", how do you determine how the Maxwell fields are supposed to transform? Markus Poessel (talk) 09:04, 13 November 2008 (UTC)


 * My argument is based on the following:
 * Experimental results rule out the validity of the Galiliean transformations.
 * That just leaves the Lorentz transformations with a finite maximal speed V.
 * Given a maximal speed V, the only consistent way of combining PofR with Maxwell's equations is to identify Maxwell's parameter :$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \ . $$ with the aforementioned maximal speed V.
 * We are now at the same starting point as if we had postulated the constancy of light, so we proceed to develop all the usual results of special relativity, including the transformation of the electric and magnetic fields. Delaszk (talk) 16:52, 13 November 2008 (UTC)
 * To Delaszk: You seem to be assuming that the postulates are something independent of or added to the experimental results. On the contrary, the postulates summarize the experimental results. JRSpriggs (talk) 06:20, 15 November 2008 (UTC)
 * This argument is about what it is possible to derive. Einstein derived the Lorentz from the assumption about light. However the finite maximal speed can be deduced without assuming anything about light. From group postulates the only consistent choices are either the Galilean or Lorentz. But measuring the addition of velocities of two particles shows Galilean transformations do not match actual results, then the only other set of transformations consistent with the PoR, isotropy etc are the Lorentz transformations. If it turns out that the photon has mass that would affect a lot of physics but it would not ruin special relativity because the existence of the maximal speed is not dependent on the constancy of light. Delaszk (talk) 10:05, 15 November 2008 (UTC)

There is absolutely no doubt about it that Einstein derived special relativity on the basis of the constancy of the speed of light, and that without such a postulate, the special theory of relativity would fall. There is however some argument surrounding how Einstein stumbled upon the concept that speed of light is a universal constant. He himself claims that it is implicit in the Maxwell-Hertz equations. This was discussed here on an earlier thread, and one editor supplied the source in question. That source was not however Maxwell's original papers. Maxwell himself never said anything about the speed of light being constant. David Tombe (talk) 12:43, 15 November 2008 (UTC)


 * This argument is not about what Einstein did or did not do. It's about what is possible. The Lorentz transformations can be derived before making any prior assumptions about light. For one thing, time dilation shows that the Galilean transformations don't fit experimental results. The theory of special relativity would not fall if light wasn't constant. If the speed of light was very close to the maximal speed this would still give the impression that it was constant. It would mean that Maxwell's equations weren't accurate. I'm not suggesting light isn't constant but if it weren't it would not mean the fall of special relativity. Delaszk (talk) 14:31, 15 November 2008 (UTC)

We'll have to agree to differ on that. The entire derivation of the special theory is based on the universal constancy of the speed of light. It is based on the fact that light doesn't obey the Galilean vector addition of velocities. Once we take that fact away, we can no longer derive the theory. David Tombe (talk) 15:42, 15 November 2008 (UTC)


 * As much as I disagree with much of what Delaszk has written, in this particular instance, he is correct. You can derive the core of special relativity - the Lorentz transformations and all that follows from them - from the principle of relativity as it applies to the way that different inertial observers relate their observations to each other, plus the assumption that there is a maximal speed. You don't need to talk about light. The argument works just as well in a universe where there are no electromagnetic fields. Markus Poessel (talk) 16:09, 15 November 2008 (UTC)


 * Well, at least we agree on something :-)


 * I would further add that "the assumption that there is a maximal speed" is not really an assumption but hinges simply on determining a numerical value for the constant 1/c^2. If any situation is found in which it is nonzero, then it must always be nonzero and then there must be a maximal speed. The maximal speed is a consequence of the form of the equations.


 * The theory is all there, and to practically apply it, all you need to do is experimentally determine the numerical value of the parameter. Delaszk (talk) 17:36, 15 November 2008 (UTC)


 * But as soon as we agree on something, we go in circles again... all assumptions must be tested experimentally. The relativity principle, as far as that is possible, and any assumptions that are made about the parameter 1/c&sup2;. And before you say "it's not an assumption, it's just a measurement": the main point is that it is an additional piece of information. As long as you don't have that information, your framework encompasses both special relativity and Galilean relativity - two very different theories. And if you want to have the minimum set of requirements that define special relativity, this piece of information is crucial. Markus Poessel (talk) 21:06, 16 November 2008 (UTC)


 * If you choose Galilean relativity by measuring a zero value of the parameter, it's not extra information that is needed to choose instead special relativity - it is just the same information measured more accurately.


 * And if you take spacetime to be a single thing then according to various references, Galilean relativity isn't even an option. Delaszk (talk) 16:59, 17 November 2008 (UTC)

Relativity of simultaneity
I have twice reverted changes to the relativity of simultaneity which have added the words 'appear to' to the text. Relativity has nothing to do with appearances, it is to do with measurements. In a single reference frame, or to a single observer, simultaneity is well defined in SR. Martin Hogbin (talk) 20:31, 27 December 2008 (UTC)