Talk:Derivations of the Lorentz transformations

Transfer of content
Apart from the lead, "see also" section, and categories, I didn't write the content. See here. Thanks. M&and;Ŝc2ħεИτlk 11:41, 27 May 2013 (UTC)

Error in From physical principles > Galilean and Einstein's relativity > Principle of relativity > Principle of relativity
After reversing the axes R′ sees R moving in the positive x′ direction. The transformation must yield x = 0 if x' = vt', but it doesn't.

Furthermore the axes need to be reversed back again to get the original orientation.

GHT153 (talk) 22:33, 2 April 2014 (UTC)


 * ✅ - See . There's no need to reverse axes. - DVdm (talk) 08:53, 3 April 2014 (UTC)

From physical principles > Galilean and Einstein's relativity > The speed of light is constant
The speed of light is constant in all directions. The derivation covers only the case of a light beam moving in the positive x-direction. The other direction is missing.

BTW, multiplying equations does not belong to any set of mathematical tools. It's quite simply nonsense.

GHT153 (talk) 23:17, 26 April 2014 (UTC)


 * No need for directions here. If you look at that part of the section Derivations of the Lorentz transformations, you see that it is about the speed being the same in two different reference frames, independent of direction. The transformation is demanded to guarantee that t=x/c (speed c in one frame) and t′=x′/c (speed c in the other frame).
 * BTW, don't you think that the equations A=B and X=Y always imply AX=BY? That's a direct consequence of the most basic mathematical tool we have: substitution. Proof: start from AX. First substitute the value B for A. You get AX=BX. Then substitute the value Y for X. You get BX=BY. So you have two equalities AX=BX and BX=BY. A final subsitution gives the result AX=BY. QED. The last step is also called "the transitivity of the identity relation". See Equality (mathematics). Pretty basic. - DVdm (talk) 09:20, 27 April 2014 (UTC)


 * There are two different results, from which only the first is presented in the derivation, that is to say
 * $$x' = \gamma \left( 1 - v/c \right) x$$
 * for a light beam moving in the positive x-direction (x=+ct), and
 * $$x' = \gamma \left( 1 + v/c \right) x$$
 * for a light beam moving in the negative x-direction (x=-ct).
 * So direction is indeed important and it needs to be discussed, how the different results are compatible with homogeneity and isotropy of space.


 * To solve a set of equations substitution is fact the most basic tool whereas multiplication cannot be used.
 * For example, think of the two equations y=1-x and y=1+x.
 * Substitution gives exactly one solution, i.e. (x,y)=(0,1).
 * Multiplication of the two equations makes you believe that there is a solution set (x,y)=(x,±sqrt(1-x^^2)).


 * BTW, the quadratic equation for gamma must have two distinct roots which is ignored here.
 * GHT153 (talk) 21:44, 7 May 2014 (UTC)


 * Ha, but the original equations are not solved for variables x and x' , and neither is the new multiplied equation. The two transformation equations must both be satisfied for particular values of x and x' . So the multiplied equation must also be satisfied for those same values. It turns out that the multiplied equation has a common factor xx'  on both sides, so any pair (x,x') with a non-zero product can be divided out of it. Doing so results in a new equation independent of x and x' , from which the originally unknown &gamma; can be solved. This is one of the most elementary standard techniques for finding unknown coefficients in equations, and specially in transformations: have the equation(s) satisfy known values of the variables, and solve the resulting equatons for the unknown coefficients. Kindergarten stuff, so to speak.
 * Note that in this case we have demanded that the transformation equations satisfy both x=ct and x'=ct′ , resulting in the value for &gamma;. If you do the same thing for a light ray in the other direction, then you demand that the transformation equations satisfy both x=-ct and x'=-ct′ . Doing so gives the same result for gamma. - DVdm (talk) 22:22, 7 May 2014 (UTC)
 * Additional note #1 — Those who, in spite of the above explanations, are somehow allergic to multiplying the equations $$\quad x'= \gamma\left(1 - v/c\right) x \quad$$ and $$\quad x= \gamma\left(1 + v/c\right) x', \quad$$ can simply substitute the x'  expression of the former equation into the latter, to get $$\quad x = \gamma^2 \left(1 - v^2/c^2\right) x. \quad $$As this must be valid for every x, just take x=1, to get the familiar value of gamma. Elementary maths.
 * Additional note #2 — BTW, only the positive root is chosen in order to have the primed and unprimed axes point in the same direction. Taking the negative root for gamma would result in a perfectly workable transformation, but with the x and x'  axes pointing in opposite directions, and time running forward in one frame while running backward in the other frame, which would be in conflict with the way we usually build our clocks. - DVdm (talk) 08:43, 8 May 2014 (UTC)


 * To summarize the discussion:
 * The kindergarten tool (aka multiplication of equations) sometimes works (your wordy explanations) and sometimes does not work (see my counter-example above). In any case, the reliability of the tool must be verified by one of the other well known standard tools.
 * Anyway, if you can find any reliable sources about solving a set of equations by multiplying the equations, please let me know.


 * The most important result is hidden by the kindergarten tool, it hides the relationship between x' and x, that is to say
 * $$\quad x'/x = \gamma\left(1 - v/c\right) \quad$$ for x=ct, and
 * $$\quad x'/x = \gamma\left(1 + v/c\right) \quad$$ for x=-ct.
 * Again, it is a highly remarkable result that for a light beam, which should travel at constant speed in all directions in a homogeneous space, the distance it travels in a given amount of time is different for different directions. That is a contradiction in itself. GHT153 (talk) 22:44, 21 May 2014 (UTC)


 * The two equations
 * $$\quad x'/x = \gamma\left(1 - v/c\right) \quad$$ for x=ct, and
 * $$\quad x'/x = \gamma\left(1 + v/c\right) \quad$$ for x=-ct.
 * together most definitely do not establish the relationship between x' and x, as you say. Taken together as a system of equations, x=ct and x=-ct, can only be valid together for x=t=0, which is the event where these two light signals intersect, in which case the two equations with x in the denominator aren't even valid.
 * The first equation establishes the relationship between coordinates x' and x for all events taking place on one light signal (with equation x=ct), whereas the second equation establishes another relationship between coordinates x' and x for all events taking place on another light signal (with equation x=-ct). In both cases they produce the same value for &gamma;, as you can verify. The idea of the section is to deduce the value of &gamma;, and there is no need to do it twice.


 * Anyway, we don't need "reliable sources about solving a set of equations by multiplying the equations", because we are not solving a set of equations. We are deducing the value of &gamma;, and I must repeat: those who, in spite of the above explanations, are somehow allergic to multiplying the equations $$\quad x'= \gamma\left(1 - v/c\right) x \quad$$ and $$\quad x= \gamma\left(1 + v/c\right) x', \quad$$ can simply substitute the x'  expression of the former equation into the latter, to get $$\quad x = \gamma^2 \left(1 - v^2/c^2\right) x. \quad $$As this must be valid for every x, just take x=1, to get the familiar value of &gamma;. Elementary maths.


 * But you asked for a source? Exactly the same technique as in the article is used here on pages 12-13 in , Extract of page 12, where they substitute x and x′ of the pair x=ct and x'=ct'  into the equations, giving $$ct'= \gamma (c-v) t$$ and $$ct= \gamma (c+v) t'$$, and then, "multiplying both these equations with each other", produces the value of &gamma;. And, by the way, of course, note that substituting x and x′ of the pair x=-ct and x'=-ct'  will produce the same value of &gamma;.
 * You want another source? Here you go, on pages 236-237 in , Extract of page 237, where Born does exactly the same thing, multiplying the two equations together to obtain the expression for &gamma;, here represented as 1/&alpha;.
 * For your convenience I have added a properly sourced remark to the section in the article.


 * So, as far as I can see, the summary of the discussion is that you seem to have no idea about (1) the physical meanings of the coordinates in these equations, (2) the physical meanings of the equations themselves, (3) coordinate transformations in general, and (4) how the derivation of an unknow transformation coefficient actually works.


 * Also note (per wp:talk page guidelines) that this is not the place where we discuss the subjects of our articles or explain elementary techniques to our readers. If you have a proposal to make a specific change to the article, and you have a reliable source to back it up, then by all means bring it up here, but stop abusing article talk space for challenging well-established and properly sourced article content. I have put another warning on your talk page. - DVdm (talk) 07:42, 22 May 2014 (UTC)


 * The derivation shown here is needlessly complex and difficult to follow. The step of equating the time squared coefficients is particularly likely to lose readers. I found it easier it do the math myself than follow the presentation.  I imagine it is the same for others.  I propose to replace the derivation with a better one if nobody else gets to it first.  Just as we have latitude in selecting words, I think we have some latitude in presenting routine algebraic steps.


 * Aside from realizing that r' does not have to equal r, I think the most significant part of the derivation is realizing the mapping must have a linear form. The article alludes to a proof based on inertia.  That proof should be included.  — Preceding unsigned comment added by 50.4.144.200 (talk) 13:22, 30 October 2014 (UTC)

The first sentence
The first sentence seems to be ungrammatical. — Preceding unsigned comment added by 78.147.57.77 (talk) 14:20, 8 May 2014 (UTC)


 * Please sign your talk page messages with four tildes ( ~ ). Thanks.
 * Yes, thanks. Attempted to fix it. Feel free to hone. - DVdm (talk) 14:49, 8 May 2014 (UTC)

Error in: From physical principles > Spherical wavefronts of light
The following Equation:

\left[ {\gamma^2} - \frac{ \left( 1 - {\gamma^2} \right)^2 c^2}{ {\gamma^2} v^2} \right] x^2 - \left[ 2 {\gamma^2} v - 2 \frac{ \left( 1 - {\gamma^2} \right) c^2}{ v}\right] t x + y^2 + z^2 = \left[ c^2 {\gamma^2} - v^2 {\gamma^2} \right] t^2 $$

has an error in sign inside the second bracket on the lest side.

The above equation is obtained from the equation on the previous line, which is:



\left[ {\gamma^2} - \frac{ \left( 1 - {\gamma^2} \right)^2 c^2}{ {\gamma^2} v^2} \right] x^2 - 2 {\gamma^2} v t x + y^2 + z^2 = \left( c^2 {\gamma^2} - v^2 {\gamma^2} \right) t^2 + 2 \frac{ \left[ 1 - {\gamma^2} \right] t x c^2}{ v} $$

by subtracting the last term: $$2 \frac{ \left[ 1 - {\gamma^2} \right] t x c^2}{ v}$$ from both sides.

Therefore, if you check it carefully, you can see the sign inside the 2nd bracket on the left side should be positive since there is a negative sign in front of this bracket.

Thanks ZvikaFriedman (talk) 13:36, 21 December 2014 (UTC)


 * Ah yes, you're right. Sorry for having reverted your correction. I reverted my revert. Thanks for spotting. - DVdm (talk) 14:09, 21 December 2014 (UTC)

From Experiments - Rather a Disproof of Special Relativity?
The notation used in this chapter differs from the rest of this page. To avoid confusion the translation formulas should read:

$$\begin{align} t' & =a(v)t+\varepsilon(v) x\\ x' & =b(v)(x-vt)\\ y' & =d(v)y\\ z' & =d(v)z \end{align}$$

Substituting the "experimentally determined" constants (functions of v)

$$1/a(v)=b(v)=\gamma$$, $$d(v)=1$$

and the assumed Einstein sychronization

$$\varepsilon(v)=-v/c^{2}$$

into the translation formulas gives the transformation equations:

$$\begin{align} t' & =\frac{1}{\gamma}t-\frac{v}{c^{2}} x\\ x' & =\gamma(x-vt)\\ y' & =y\\ z' & =z \end{align}$$

which are definitely NOT the Lorentz transformations described on the rest of this page.

The Lorentz transformation gives for the speed of light $$c'=c$$, while this one will result in $$c'=\gamma^{2}c$$ which is the result of the Galilei transformation (for the 2-way speed of light).

So what does the article do in this context? Rather re-invent Newtonian physics under an inappropriate headline (others would call it fraud) or rather experimentally disprove special relativity with "great precision"? GHT153 (talk) 22:46, 24 March 2015 (UTC)


 * The section, and the article referred to, explain how test theories can be used to test the "real" transformation. Indeed, the resulting transformation is not the same. That is the point. Also have a look at Test theories of special relativity. - DVdm (talk) 05:38, 25 March 2015 (UTC)


 * Both the section and the main article claim that the above transformation, independent of any test results, is the Lorentz transformation. This is not the case. So the test theory presented here is completely worthless and should be deleted. Any objections? GHT153 (talk) 23:16, 30 March 2015 (UTC)


 * As far as I can see, the section and the main article, both properly sourced, do not claim that the above transformation is the Lorentz transformation. - DVdm (talk) 06:58, 31 March 2015 (UTC)


 * Both say that for $$1/a(v)=b(v)=\gamma$$, $$d(v)=1$$ and $$\varepsilon(v)=-v/c^{2}$$ the given transformation converts into the Lorentz transformation. The problem is that the resulting transformation, as shown above, is not the Lorentz transformation. GHT153 (talk) 21:32, 5 April 2015 (UTC)


 * Ah, but indeed for those values they do result in the Lorentz transformation.
 * The test equations of the article section and in the main article
 * $$\begin{cases}\begin{align}

t &= a(v)T+\varepsilon(v) x\\ x &= b(v)(X-vT)\\ y &= d(v)Y\\ z &= d(v)Z \end{align}\end{cases}$$
 * together with
 * $$1/a(v)=b(v)=\gamma$$, $$d(v)=1$$ and $$\varepsilon(v)=-v/c^{2}$$
 * give
 * $$\begin{cases}\begin{align}

t &= \frac{T}{\gamma} -\frac{v}{c^{2}} x\\ x &= \gamma (X-vT)\\ y &=Y \\ z &= Z. \end{align}\end{cases}$$
 * The $$x, y, z$$ equations are part of the LT. Substituting the expression for $$x$$ in the first equation

t = \frac{T}{\gamma} -\frac{v}{c^{2}} x $$
 * results in
 * $$\begin{align}

t&=\frac{T}{\gamma} -\frac{v}{c^{2}} \ \gamma (X-vT)\\ &=\gamma \left( \frac{T}{\gamma^{2}} + \frac{v^{2}}{c^{2}} T - \frac{v}{c^{2}}X \right)\\ &=\gamma \left( T ( \frac{1}{\gamma^{2}} + \frac{v^{2}}{c^{2}} ) - \frac{v}{c^{2}}X \right)\\ &=\gamma \left( T ( 1 - \frac{v^{2}}{c^{2}} + \frac{v^{2}}{c^{2}} ) - \frac{v}{c^{2}}X \right)\\ &=\gamma \left( T - \frac{v}{c^{2}}X \right) \end{align}$$
 * which is the time part of the LT. So what is the problem? - DVdm (talk) 09:07, 6 April 2015 (UTC)

Misleading illustration
The illustration from reference 2 seems to be misleading in the discussion of derivation of Lorentz Transformation. I have made a Note near the illustration. Though the figure is a good starting point, it leaves out later considerations used in the derivation such as coinciding origins at start, meaning x=x'=0 at t=t'=0. To have the moving observer detect the event later than stationary observer after moving vt distance (required for Time Dilation), the event must have happened at x <= 0 (less than or equal to zero), not at x > 0. Please some expert verify and add appropriate changes. Thanks. - Achandrasekaran99 (talk) 21:26, 30 April 2015 (UTC)


 * I don't see a problem. As can be seen in the equations in the speech bubbles in section From physical principles, the figure assumes the origins coincidence event to have x=x'=0 at t=t'=0. The shown event is just some general event. - DVdm (talk) 19:30, 30 April 2015 (UTC)


 * I understand that this figure is 66 years old and possibly seen by thousands of scientists and millions of students. Still even as a general event it violates Special Relativity. This shown general scenario allows r'>r, r'=r and r't, t'=t and t't which is indisputable. Thus this general event does not conform to Special Relativity and not a suitable model to be placed here. Surely, it is a nice illustration but seems to lack in obvious scientific accuracy which adds lots of difficulty when one tries understand in depth. - Achandrasekaran99 (talk) 13:59, 1 May 2015 (UTC) Clarification: The times I mention here are elapsed times and not clock times. - Achandrasekaran99 (talk) 14:22, 1 May 2015 (UTC)


 * I put a note on Maschen's talk page asking him to join in. Maybe you and he can work out how to modify the figure so as to eliminate the points that you find confusing. Stigmatella aurantiaca (talk) 15:41, 1 May 2015 (UTC)


 * Thank you for referring. Looking forward to discuss. My understanding is that the event took place at t=t'=0, at which point x=x'=0. The motion is along x axis. Then the event must have happened in 0,y,z (and 0,y',z') plane to get the result with t'>t. Random event locations will allow t'>t, t'=t or t' t follows from the Lorentz transformations not casualty. The coordinates can be anything. How does moving the event to 0yz plane fix anything? There is nothing to be gained from this since the event happens for any y, z and the event does not necessarily occur at x = x’ = 0 and t = t’ = 0. What is relevant is that the frames coincide with x = x’ = 0 and t = t’ = 0 and the motion is along the x and x’ axes. Also, where is the 0yz plane (or the 0y’z’ plane)? If you mean the set of coordinates in the unprimed frame where x = 0 (or in the primed frame x’ = 0) then neither plane contains the event. The observer in primed frame will notice it to be at rest, and the unprimed frame to move in the opposite direction (negative x and x’ directions), and the event does not move with either the 0yz or the 0y’z’ planes. M&and;Ŝc2ħεИτlk 10:19, 2 May 2015 (UTC)


 * Looking at their insistence on getting "the result with t'>t", it seems that user Achandrasekaran99 has a misconception about transformations and events. I don't think that this article's talk page is the place to explain--see wp:TPG. Perhaps the user talk page User talk:Achandrasekaran99 is more suitable for this. - DVdm (talk) 10:40, 2 May 2015 (UTC)


 * Looks like I am accused of pushing my agenda here! As I clarified, t' and t are elapsed times and not clock times. "Time dilation" clearly means t' > t. That is not my invention, it is the Relativity. Anyway, I do not want to continue in the face of such accusation. Thanks all. Anyway, I have pointed out what I wanted to point out as much clearly as possible. My job is done. - Achandrasekaran99 (talk) 17:19, 3 May 2015 (UTC)


 * Nobody is accusing you of anythng, certainly not of agenda pushing. I can't see why or how you might possibly be doing that. Just expressing concerns about some misconceptions you seem to have, again expressed in your statement that "time dilation clearly means t' > t". Perhaps someone is prepared to help you by addressing this on your talk page. If not, you might try asking on some public forum, like for instance this one. Cheers. - DVdm (talk) 17:35, 3 May 2015 (UTC)


 * Sorry that I took asking to discuss in my personal talk page as personal agenda accusation. I clarified twice that I was referring to elapsed times not coordinates, but it has been overlooked. So, let me say it better: delta-t' > delta-t is a must (when we had t=t'=0 as start, they become equivalent; I was not referring to coordinates). In this light, please read my messages again and if you feel everything is fine in this page, we can leave peacefully. Thanks for clarifying that there was no accusation. - Achandrasekaran99 (talk) 14:10, 4 May 2015 (UTC)


 * No problem. But yes, everything is fine on this page. Your worries about delta-t' > delta-t being a must really rests on a misconception of your part. Alas, this article talk page is not the place to get that cleared out. - DVdm (talk) 15:56, 4 May 2015 (UTC)


 * Thanks for your honest efforts to improve Wikipedia! It takes a bit of effort to read this diagram correctly, and I can understand how you might have misread it, but it is really OK. Stigmatella aurantiaca (talk) 18:47, 4 May 2015 (UTC)

Landau & Lifshitz
I have seen plenty of "derivations". Almost all are shaky. I have only browsed the ones here, so I might have missed some gold. But there is one that is as solid as you could ever expect in deriving from physical principles mathematics, and that is the one in L&L, The Classical Theory of Fields.

The thing is to first solidly establish invariance of the interval (not just light-like null intervals). After that you can't really fail. Just hand-wavingly making a "linear ansatz" somewhere doesn't do it. Edit: What I mean is that invariance of the interval is vital because otherwise the Lorentz transformation would be the wrong solution.

L&L brings it to the infinitesimal level,
 * $$ds^2 = A(v, ...)ds'^2.$$

From there, it is rock solid based on homogeneity and isotropy of spacetime (eqns 2.4–2.6).

Once invariance of the interval is established, one could conceivable go to group theory right away and just pick the right group (see Classical group)  No, not really helpful here   Yes, it is very helpful once one realizes that a linear solution drawn directly from classical group to a simplified problem actually does solve the general problem.. Or one could continue to follow L&L. They, in fact, do write down the right transformation right away (because they read another reduced invariant interval off and use group theory), but the connection between the "rapidity", or whatever it is called (L&L call it "angle", with the quotation marks), and the real physical quantities is brought out. (eqns 4.2–4.3)

I'd add it myself if I had the time, but if someone else is itching, please go ahead. (And put it at the top!) YohanN7 (talk) 09:30, 1 July 2015 (UTC)

I'm actually at it now. YohanN7 (talk) 16:18, 2 July 2015 (UTC)


 * Done. I provided the invariance of the interval as a consequence of the second postulate of relativity, outlined the general setup for recycling (will edit article to remove redundancy). I did also took the liberty of putting the L&L solution at the top since I think it is superior, except perhaps Einstein's solution (which is even shorter, but requires a fair amount of voodoo that us mere mortals aren't allowed to practice). We should not give the reader the impression that it requires experiments of thought (by that I mean beyond what is outlined in "standard setup" in the article) to derive the LT. It is a relatively simple problem mathematically. YohanN7 (talk) 18:21, 3 July 2015 (UTC)


 * Good work as always but this article should keep the elementary derivations using algebra (and SR postulates). Using infinitesimals is indeed general and rigorous, but most people encounter the algebraic ones when they first learn SR. M&and;Ŝc2ħεИτlk 17:58, 4 July 2015 (UTC)


 * Thanks. I suppose we can change to finite intervals, $d → Δ$. It should matter little. As a whole, the section must to stay because w/o the requirement of invariance, there are non-Lorentz transformations satisfying $c = const$. These are non-linear, but then we must present non-rigorous arguments about tidal forces or whatever that nobody will truly understand. As it stands, we can require that the transformation is linear and solves the "simplified version", otherwise the interval is not invariant, even if the "simplified version" (see article in "setup") is. One advantage is that it relieves the individual sections of motivating linearity, even if the sources "help" in that matter. (There have been a complaint or two above.) YohanN7 (talk) 19:52, 4 July 2015 (UTC)

Spherical wavefronts of light
You write in the article: "Similarly, the equation of a sphere in frame O′ is given by x′²+y′²+z′²= r′², so the spherical wave front satisfies x′²+y′²+z′²= c²t′². "

It is very important to know that the term x′²+y′²+z′²= c²t′² does not represent some unique wave front (as given by x²+y²+z²= c²t²) because of the trivial fact that for any point of the moving O'-frame t' is covariant to x'.

Just look: All points of the static O-frame share the same time (t is the same value for all x-values). Hence the constraint to select the spherical wave is valid for all.

All points of the moving O'-frame have different time values(t' is different for all different x'-values). The constraint to select a spherical wave is assigned a single t'-value which is fulfilled by a single point only. Hence the constraint to select the spherical wave does introduce phantoms: There are as many different spheres as different time-values t' are given in O'. (And LT asserts that these different t'-values do exist simultaneously in O'!)

Hence there is no prove that Lorentz transformation gives a valid picture of two frames in relative motion as there is no valid invariant (or at least some invariant to be proved as anything is relative, even the time).

09:05, 23 August 2016‎ special:contributions/84.59.24.88 (talk)‎


 * In the O&prime; frame, there is only one time t&prime; and only one x&prime; coordinate, and the equation of a spherical wave front of light using the space and time coordinates in this frame is
 * $${x'}^2 + {y'}^2 + {z'}^2 = (ct')^2$$
 * The Lorentz transformations preserve the form of a spherical wave front.
 * And what does "t' is covariant to x'" even mean?? 'M'&and;Ŝc2ħεИτlk 12:40, 23 August 2016 (UTC)

Error in From physical principles > Galilean and Einstein's relativity > The speed of light is constant
This part (consequence):

At any time after t = t′ = 0, xx′ is not zero, so dividing both sides of the equation by xx′ results in

Is different than the preceding part (premise):

Hence the transformation must yield x′ = 0 if x = vt.

The premise states that x' = 0, which means that xx' = 0 resulting in a mathematical division-by-zero fallacy. — Preceding unsigned comment added by 155.4.133.178 (talk) 18:59, 23 November 2016 (UTC)


 * Please sign all your talk page messages with four tildes ( ~ ). Thanks.
 * Those are two entirely different situations. The first situaton looks at events satisfying x'=0, resulting in some restriction to the unknown parameters. The second situation looks at events taking place on a light ray (satisfying t=x/c and t′=x′/c). After the event that satisfies t=t′=0, both t and t' are non-zero, and so are x and x', so for those events we can safely divide by xx'. No problem. - DVdm (talk) 19:29, 23 November 2016 (UTC)

Simpler/simplest derivation and problem with Transformations of Time section
This is my first attempt to contribute to Wikipedia, so please forgive any unknowing breaches in etiquette.

I feel this page would greatly benefit from a simpler derivation, more like the one for Time Dilation at https://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity

I happen to know a very simple derivation requiring just one diagram and a few simple lines of algebra, taking the homogeneity and isotropy of spacetime and time dilation as premises, so if the reader can follow the above Time Dilation derivation available at the above link, they can use it to easily generalise it to the full set of Lorentz transforms.

Having struggled to understand the Derivations page in its current format, my feeling is that this most simple derivation should probably be presented first, in an analogous manner to the Time Dilation page.

This would also resolve the problem with the Transformation of Time section, which is problematic. I am struggling to specify my objection in completely concrete terms; hopefully the following captures it:

The page suggests arriving at a general solution for t' in terms of x and t by "Substituting term by term into the earlier obtained equation for the spatial coordinate", but offers no reason why this particular algebraic procedure should be used, when other substitutions also generate formulae for t' in terms of x and t in the context of the light cone that is this section's special case.

For instance, one could equally well generate a formula for t' in terms of x and t by just substituting ct' for x' in x' = gamma * (x - vt), and leaving the right hand side alone, giving t' = (gamma/c) ( x - vt), which would be true on the light cone centred at the co-origin but quite wrong everywhere else.

So, I don't think the Transformation of Time actually derives the claimed result, as it stands (unless I've missed something critical here?)

Should I proceed by working up the suggested derivation, including the solution for t', and then posting it on this talk page for discussion?

Your kind attention, guidance and patience is appreciated. Kebl0155 (talk) 17:46, 7 January 2017 (UTC)


 * You missed a few things:


 * 1) In that section, the idea is to find the hitherto unknown constants A and B in the earlier mentioned linear equation t'=Ax+Bt. For the first equation x'=gamma x + bt, we already have found the values of the constants gamma and b. Now we must find A and B. We can do that by using the fact (postulate) that for all events that satisfy x'=ct', we also have that x=ct, and insert that into the equation t'=Ax+Bt. So we are algebraicly eliminating all {x,t,x',t'} from the four equations { t'=Ax+Bt, x=ct, x'=ct', x'=gamma(x-vt) }, solving for A and B. Now we have two equations that produce the values of x' and t' for all values of x and t.
 * 2) When you are "substituting ct' for x' in x' = gamma(x-vt), and leaving the right hand side alone, giving t' = (gamma/c)(x-vt)", you take an equation that is valid for all events, and restrict it to events on the light cone only. You can continue and insert x=ct into it, and get t'=sqrt((c-v)/(c+v)) t, which is not the "transformation of time", as it is valid on the light cone only. In other words, you get equations that produce x' and t' for values of x and t that satisfy x=ct only.
 * 3) The section Time_dilation is not a derivation of the "transformation of time" equation. In fact, it presents a way to derive something that would be much easier derived as a consequence of the Lorentz transformation—see, for instance in Special_relativity. That's not the subject of this article. At this point in this article, we don't have that transformation yet—we are deriving it.
 * - DVdm (talk) 11:36, 8 January 2017 (UTC)


 * This second point is indeed the root of my objection to the process described in the live version of the section, and thank you for cutting to the quick so quickly. Because this section proceeds from the specific case of the light cone, the results it derives here are only derived or proven to apply to the light cone.  I think part of the difficulty is that it talks about 'the preceding equations' without specifying which preceding equations - I was using the immediately preceding ones in the Principle of Relativity section that is just above it.  Other substitutions are possible, giving different formulae that give correct results for the light cone, but nowhere else, as you have so rightly interpreted and clarified upon my original somewhat opaque objection.  Thank you for that.  The result that is given in this section is indeed the general result, but that is not proven by the current live version of this section.  Rather, as it currently stands, this section simply demonstrates the more modest fact that the general result works on the light cone.  It is not a derivation or proof of the general result per se.  I think probably that's because it doesn't specify which preceding equations adequately.
 * I have not been able to determine whether your additional words in the first point above constitute proof of the general result. I suspect that it doesn't, because if one eliminates all {x,t,x',t'} from the four equations it is hard to see how one has anything left at all.  Perhaps that's just hard for me to see, and is obvious to minds finer than mine.  However, looking at the initial results this section is trying to reach, and substituting just for {t,t'} in the specific set of equations you mention in your comment above, one gets the four equations x'/c = Ax+Bx/c, x=cx/c, x'=cx'/c, x'=x*gamma*(x-vx/c).  The middle two are not helpful.  The one on the right does give the first desired result, x' = x*gamma*(1 - v/c).  I couldn't use that to get rid of the A and B in the one on the left, because I'm left with one variable and two unknowns, but I could use it with an appeal to symmetry in spatial directions to invoke the standard procedure of swapping primes and the sign of v to get x = x' * gamma(1 + v/c), which is the second desired result.
 * I feel like I should update this section to add clarity. Certainly I was not able to reach this without your further clarification on which specific set of equations should be used.  Solving for A and B certainly seems more difficult than it appears.  Indeed, I'm not convinced it's possible without invoking swapping-primes-and-sign-of-v.  Am I right in my instinct to make some edits? I am new here and defer to your experience. Kebl0155 (talk) 13:09, 11 January 2017 (UTC)
 * I agree with everything you have written there, but I feel I may also have more to say on this matter. Can we please park that until we have reached agreement on the "The speed of light is constant section"?  Respectfully, Kebl0155 (talk) 13:09, 11 January 2017 (UTC)


 * I have reformatted your reply below mine. Please don't weave comments in other comments. See wp:THREAD and wp:INDENT.
 * This is a standard way to fix constants in systems of linear equations, by demanding that the equations satisfy known conditions. In this case some of the known conditions are given by knowledge of how light cone event coordinates transform. If a set of equations must be vaild for all (x,t), then it must certainly be valid for some specific (x,t). That demand puts restraints on the unknown constants. In this case it is the last step that helps fix the constants. It looks like you don't understand this elementary algebraic procedure. If you keep having a problem with this standard practice, then please note that article talk pages are not for explaining (parts of) the subject to others—see wp:talk page guidelines, so we cannot go deeper into this here. You can always ask at our wp:Reference desk. - DVdm (talk) 14:58, 11 January 2017 (UTC)
 * I made some slight changes to clarify the procedure at least a little bit. - DVdm (talk) 15:27, 11 January 2017 (UTC)
 * Oh yes that's much better! The expression of the proof is now formal, at least to my mind. Thank you so much! Thank you also for the link to the guidelines that I have now read; I can see that my initial question was appropriate, as was your response, but perhaps not my subsequent response, as I could also have just fixed it, given your response. Your fix is much better than the one I would have applied, so I'm glad you got there first. Thank you again. Kebl0155 (talk) 16:17, 11 January 2017 (UTC)
 * OK well now that's squared away - and thank you again - I'd like to return to the subject of your Point 3. I completely agree that once one knows the Lorentz transforms, then Time Dilation is pretty much immediately obvious from them.
 * What I am trying to say is that if one doesn't know the Lorentz transforms, as I must assume many of the readers of this 'pedia will not, then that is unfortunately useless.
 * For this reason I very much admire the Simple inference of time dilation section of the Time dilation page. As a lay person, or rusty fool like me, one can read this, and just this, and know all about Time Dilation in a precise, mathematical way, without having to know much of anything else at all, except Pythagorus.  It's just one diagram, which is simple and clear, and a few lines of maths that are easy to follow.  For me, it is a wonderful starting point for anyone who knows nothing about relativity to begin assimilating the knowledge.
 * What I am proposing is the creation of a "Simple inference of Lorentz Transformations due to Time Dilation" on this Derivations page, featuring a back link to the Time Dilation simple inference. I have this written down with one diagram that's even simpler than the one on the Time Dilation page, and a similarly brief amount of basic maths.
 * That way someone who knows nothing about the Lorentz Transforms, or indeed Relativity, can come to understand them completely from reading two very short sections.
 * I do feel this would be of genuine benefit for fools like me, or indeed anyone attempting to derive Special Relativity from first principles.
 * I have looked for authoritative versions of this derivation online. References are few and far between; it appears it is not well known.  The closest I've seen so far is [], page 2, section C and III.  This particular reference is lacking the simple diagram and is more verbose than what I have in mind, but the method is similar; I am hoping to find additional references, which are probably out there somewhere as it's unlikely that I came up with something this simple and elegant myself.
 * That way someone could use Wikipedia to learn and derive Special Relativity to the level of Lorentz Transformations from a standing start in about 20 minutes.
 * I thought I might work this up in my Sandbox and that you might perhaps be persuaded to take a look and share your thoughts. Would that be appropriate? As before I defer to your superior knowledge of etiquette in these matters.  Respectfully, Kebl0155 (talk) 17:45, 11 January 2017 (UTC)
 * That's a very nifty derivation here. Never seen it before. I have no problem having it in the article, in, let's say, a section that simply goes like "the LT can be derived from time dilation and length contraction, by using the vector equation OM = OO' + O'M, ...", closely following the remainder of that section C. As TD and LC are already derived in their articles (resp. here and here), there would be no need to derive them here again. In this encyclopedia we use our wikilinks for that. - DVdm (talk) 18:54, 11 January 2017 (UTC)
 * I thought so too. I'm so pleased you like it.  Am adding it now. Will repost here when done so you can tell me all the stylistic and etiquette ways in which they way I've added it is not the wikipedia way, if you would be so kind - I really want to get things right moving forwards. I'm almost certainly going to put it in the wrong place, for instance. Kebl0155 (talk) 21:51, 11 January 2017 (UTC)
 * That's done. I put it first in the physical list because it's shortest.  Feel free to move it if that's not right. There's a brief definition of terms; it draws some from the Standard Configuration too.  Some of the earlier preamble got shoved to the next one. I couldn't work out how to mark up the lines over the OM etc. in the original source; they are one dimensional vectors so technically they should be there.  Could also be left in Bold I suppose, but then we lose the feel that we are moving from a multi-dimensional proof to a one-dimensional one (two if you include time).  Not sure how important the mark up is though; I don't think the current exposition lacks clarity.  How did I do? Kebl0155 (talk) 00:17, 12 January 2017 (UTC)
 * That looked good to me, but I had to drastically copyedit it to avoid obvious copyright problems—see WP:COPYOTHERS. - DVdm (talk) 08:11, 12 January 2017 (UTC)
 * Huh. Out of all the ways that I could have messed up, that one really didn't occur to me! Thanks so much for the link.  I am learning so much from you - thank you.  I might do a little diagram for this proof later, given that it is geometric in origin. Kebl0155 (talk) 10:21, 12 January 2017 (UTC)

I haven't read all the above. I like the new section, but my first question would be how the heck do we know TD and LC? From Time dilation:
 * Time dilation can be inferred from the observed constancy of the speed of light in all reference frames...

Good! Parsing Length contraction quickly, I found Proof using time dilation. Does this close the circle? If so, it should be pointed out in the new section filling in the gap (gap = how the heck do we know ..). YohanN7 (talk) 10:42, 12 January 2017 (UTC)
 * Yes, the two sections Time_dilation and Length_contraction close the circle (as mentioned above). I have put the wikilinks to the subsections in the article, together with a little tweak: . Looks very sound now. - DVdm (talk) 10:57, 12 January 2017 (UTC)


 * Nice! YohanN7 (talk) 11:01, 12 January 2017 (UTC)
 * And link notifications added:, . - DVdm (talk) 11:06, 12 January 2017 (UTC)
 * Well that's grand then. If you like that one, you will probably like the even simpler one it appears I have devised using just Time Dilation, very suitable for lay readers, that I intend to post on the main Lorentz Transformation page some time in the next couple of days.  It was the reason that I sought the reference that we have now quoted.  We can talk about it afterwards on that page's talk page if you like.  Perhaps you might refrain from posting it if you reach it yourself until I have had a chance to do mark it up. My aim is to make that page as beneficial for lay readers as the current Simple Derivation on the Time Dilation page. If I have perhaps been overly emboldened by the Being Bold section of the etiquette guide, then please do tell me.  Respectfully, Kebl0155 (talk) 13:50, 12 January 2017 (UTC)
 * When you reply to a talk page thread, could you please start the editor by clicking the edit link next to the section header instead of the one on top of the page? That way we can go directly to that section from our watch list. Same thing in article pages. Thanks.
 * Be very careful not to add your simpler one that you have devised using just time dilation into any article, unless you have a proper source to support it, as Wikipedia does not accept our wp:original research". Very important. - DVdm (talk) 15:15, 12 January 2017 (UTC)
 * Thank you for the note about section editing, and your advice on original research, which is most timely. I believe I have the objection covered. I am about to add the proof (which will be my third attempt to add benefit to this wonderful site), and will immediately post some accompanying comments on that page's talk page in a new section thereafter.  I do hope that will turn out to be all in order.  Respectfully, Kebl0155 (talk) 16:22, 12 January 2017 (UTC)

Negative root missing in: Galilean and Einstein's relativity > Determining the constants of the first equation
The equation xx' = γ2(1 - v2/c2)xx' is reduced to:

γ = 1 / sqrt(1 - v2/c2)

That is only the principal (positive) root. The negative root is missing. (155.4.133.178 (talk) 21:20, 11 January 2017 (UTC))


 * The positive root is taken because the x and x' axes are assumed to point in the same direction, and in both systems time is "running forward", so also in the same direction. Further questions can be answered by reading the sources, or by asking at the wp:reference desk. - DVdm (talk) 22:07, 11 January 2017 (UTC)

From physical principles > Time dilation and length contraction
In F O'M is well known as well, O′M = x-vt.

Thus we have in F: OM = x, OO′ = vt and O′M = x-vt = x′/γ, i.e. x' = γ(x-vt),

and in F': OM = x/γ, OO′ = vt/γ = v't′ and O′M = x′, i.e. x' = (x-vt)/γ.

There is something wrong here since x' = γ(x-vt) = (x-vt)/γ leads to γ=1 and contradicts length contraction. Sources? 178.10.105.212 (talk) 12:20, 25 June 2017 (UTC)


 * Note that in F', OO′ = vt′ ≠ vt/γ because the t and t' here are not related to event M. Applying length contraction by dividing by the Lorentz factor γ can only be done in the context of event M, so to O'M in F, and to OM in F'. The source is provided in the article as https://hal.archives-ouvertes.fr/hal-00132616/document - DVdm (talk) 13:57, 25 June 2017 (UTC)


 * The place for the reference to the source is inappropriate. It is hidden behind the links to Wikipedia articles where is not belongs to. A good place would be at the end of the header.


 * Following your argumentation we have now
 * in F: OM = x, OO′ = vt and O′M = x′/γ, and
 * in F': OM = x/γ, OO′ = vt′ and O′M = x′.


 * OO' not beeing affected by the Lorentz factor leads either to OO'=vt'=vt which is equivalent to t'=t, or rather to OO'=v't'=vt which is equivalent to v'≠v. Both results are far from being that what special relativity wants to tell us.


 * How OM = OO′ + O′M can be fulfilled in both F and F' (except for γ=1) is another question that raises from the article. 188.104.116.90 (talk) 12:35, 27 June 2017 (UTC)


 * I think the citation was already properly placed after the first sentence of the section, but I have moved it now to immediately before the derivation:.
 * In F, OO' = vt. In F', OO' = vt'. That is backed by the source, just like how OM = OO′ + O′M can be fulfilled in both F and F'.
 * Per our wp:Talk page guidelines we should discuss the article here, not the subject, and we are not really supposed to explain details from the sources here, but I guess, to clarify the setup of the source, I can say the following :
 * In F, the distance between the origins O and O' is measured by subtracting the x-coordinates of two events Ot and O't, both taken at time t. The x-coordinates are resp. 0 and vt.
 * In F', the distance between the origins O and O' is measured by subtracting the x'-coordinates of two different events Ot' and O't' both taken at time t'. The x'-coordinates are resp. -vt' and 0.
 * This involves 4 different events, whereas applying length contraction is done with 3 events:
 * To find in F that O'M = x'/γ we need the events O't, O't' and M, where O't and M are simultaneous in F with time t. See red in the diagram.
 * To find in F' that OM = x/γ we need the events Ot, Ot' and M, where Ot' and M are simultaneous in F' with time t'. See green in the diagram.
 * A spacetime diagram makes this clear. - DVdm (talk) 13:43, 27 June 2017 (UTC)


 * The diagram is nice but not correct. Please have a look at my corrected version below, it follows the rules outlined in wp:Minkowski diagram. Since there is no correct way to put in gamma there it has been simply left out (i.e. gamma=1):




 * GHT153 (talk) 14:46, 26 July 2017 (UTC)


 * Your x' lies along the (horizonally drawn) x-axis. That is wrong. Likewise, your x lies along the tilded x'-axis. That is wrong. You also have a line, parallel to the t'-axis, labeled x=x'. That is wrong. You have a v' in your drawing. There is no such variable. It looks like you have no idea what any of the variables mean. An article talk page is not the place where we are supposed to explain the theory to those who don't understand it. As the content you are questioning is properly sourced and approved by other editors (—see above—), you are now disrupting this talk page. You already have a final warning about talk page abuse on your user talk page, so you must let this go here. Next time I will have to report you at the Administrators' noticeboard/Incidents. - DVdm (talk) 15:38, 26 July 2017 (UTC)