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∧Ŝc²ħεИ_τlk 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)

 Fixed - See [1]. 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#Galilean and Einstein's relativity, 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)#Some basic logical properties of equality. 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 γ 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 γ. 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 γ, as you can verify. The idea of the section is to deduce the value of γ, 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 γ, 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 γ. Elementary maths.

But you asked for a source? Exactly the same technique as in the article is used here on pages 12-13 in Gupta, S. K. (2010). Engineering Physics: Vol. 1 (18th ed.). Krishna Prakashan Media. p. 12-13. ISBN 81-8283-098-2., 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 γ. 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 γ.

You want another source? Here you go, on pages 236-237 in Born, Max (2012). Einstein's Theory of Relativity (revised ed.). Courier Dover Publications. p. 236-237. ISBN 0-486-14212-4., Extract of page 237, where Born does exactly the same thing, multiplying the two equations together to obtain the expression for γ, here represented as 1/α.

For your convenience I have added a properly sourced remark ([2]) 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:

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

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:

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

by subtracting the last term: $2{\frac {\left[1-{\gamma ^{2}}\right]txc^{2}}{v}}$ 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:

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

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:

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

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

${\begin{cases}{\begin{aligned}t&=a(v)T+\varepsilon (v)x\\x&=b(v)(X-vT)\\y&=d(v)Y\\z&=d(v)Z\end{aligned}}\end{cases}}$

together with

$1/a(v)=b(v)=\gamma$, $d(v)=1$ and $\varepsilon (v)=-v/c^{2}$

give

${\begin{cases}{\begin{aligned}t&={\frac {T}{\gamma }}-{\frac {v}{c^{2}}}x\\x&=\gamma (X-vT)\\y&=Y\\z&=Z.\end{aligned}}\end{cases}}$

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

${\begin{aligned}t&={\frac {T}{\gamma }}-{\frac {v}{c^{2}}}\ \gamma (X-vT)\\&=\gamma \left({\frac {T}{\gamma ^{2}}}+{\frac {v^{2}}{c^{2}}}T-{\frac {v}{c^{2}}}X\right)\\&=\gamma \left(T({\frac {1}{\gamma ^{2}}}+{\frac {v^{2}}{c^{2}}})-{\frac {v}{c^{2}}}X\right)\\&=\gamma \left(T(1-{\frac {v^{2}}{c^{2}}}+{\frac {v^{2}}{c^{2}}})-{\frac {v}{c^{2}}}X\right)\\&=\gamma \left(T-{\frac {v}{c^{2}}}X\right)\end{aligned}}$

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'<r all equally possible which lead to t'>t, t'=t and t'<t. For Special Relativity, it must be 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. So, simply moving the event to 0yz plane will make it completely accurate, consistent with text and result, making it far easier to grasp. - Achandrasekaran99 (talk) 17:32, 1 May 2015 (UTC)

Hi everyone sorry for the delay. The space and time coordinates shown are those measured in the frames by the observers using their clocks and rulers, and 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∧Ŝc²ħεИ_τlk 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 hereYes, 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∧Ŝc²ħεИ_τlk 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′ frame, there is only one time t′ and only one x′ 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∧Ŝc²ħεИ_τlk 12:40, 23 August 2016 (UTC)

Error in From physical principles > Galilean and Einstein's relativity > Principle of relativity > Principle of relativity

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 first part states that x' = 0, which means that xx' = 0 resulting in a mathematical division-by-zero fallacy.