Talk:Lorentz transformation/Archive 4

Blend/shuffle sections...
New changes:
 * collect the transforms of other quantities in one section and add the general formula in Special relativity (should definitley be here, more than SR), allowing a coherent description of spacetime coords before other physical quantities,
 * blend the short section transformation of the electromagnetic field into Special relativity, both include simalar content on relativistic EM, there is no point seperation,
 * move local (!) sections around slightly to achieve these points,
 * in doing so group The correspondence principle, transformation of the electromagnetic field under the section Special relativity, these are all related by qualitative description,
 * fix my eariler unnoticed italic error in The correspondence principle
 * make Special relativity more neutral - "most astounding" may be true but too much hype, "the breakthrough" is more neutral (WP:NPOV).

F = q(E+v×B) ⇄ ∑ici 09:17, 19 June 2012 (UTC)

trouble with general transform
The current version says:



\begin{bmatrix} c\,t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&-\gamma\,\beta_x&-\gamma\,\beta_y&-\gamma\,\beta_z\\ -\gamma\,\beta_x&1+(\gamma-1)\dfrac{\beta_x^2}{\beta^2}&(\gamma-1)\dfrac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\dfrac{\beta_x \beta_z}{\beta^2}\\ -\gamma\,\beta_y&(\gamma-1)\dfrac{\beta_y \beta_x}{\beta^2}&1+(\gamma-1)\dfrac{\beta_y^2}{\beta^2}&(\gamma-1)\dfrac{\beta_y \beta_z}{\beta^2}\\ -\gamma\,\beta_z&(\gamma-1)\dfrac{\beta_z \beta_x}{\beta^2}&(\gamma-1)\dfrac{\beta_z \beta_y}{\beta^2}&1+(\gamma-1)\dfrac{\beta_z^2}{\beta^2}\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}\,. $$

But if we set beta_y and beta_z equal to zero, we expect to recover the transform in the x direction. Instead, we get y' = 0 and z=0. Something is seriously wrong here.

Pervect (talk) 23:56, 24 August 2012 (UTC)


 * Why is that a problem? If $$\beta_y = \beta_z = 0$$ then the particle has no y or z components of velocity, i.e. $$v_y = v_z = 0$$. In that case the matrix reduces to at top-left 2-by-2 matrix which only acts upon the x coordinate and time. If the relative y-component of velocity is zero then there ought not be any relativistic effect in the y-direction. Likewise for the z-component/direction. Perhaps what worries you is that they use ct instead of simply t? The benefit of using ct is that you get a spacial dimension, i.e. measured in metres. (Speed multiplied by time is distance.) — Fly by Night  ( talk )  00:24, 25 August 2012 (UTC)


 * But we don't get y' = 0 and z'=0. With $$\beta_y = \beta_z = 0$$, we get $$\beta_x = \beta$$, and the (3,3) and (4,4) diagonal values immediately reduce to 1:

\begin{bmatrix} c\,t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&-\gamma\,\beta_x&0&0\\ -\gamma\,\beta_x&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}\,, $$
 * recovering the standard form with the velocity in the x-direction, and giving y' = y and z' = z. - DVdm (talk) 10:28, 26 August 2012 (UTC)

Article upgrade to B class
I think C is too little...

This article is worthy of B class since although as stated in this table. Change back if you disagree. M&and;Ŝc2ħεИτlk 12:03, 27 December 2012 (UTC)
 * much of the material is complete and well-covered, more or less coherently,
 * nothing is seriously irrelevant (perhaps heavy on the derivations... in this case it should be ok, since they explain the origin of the transformation in different ways including Einstein's),
 * not likely to satisfy professional researchers/practitioners etc...
 * mainly well-cited, though some bits of maths aren't (presumably WP:CALC?), a few citations towards the end could be added...

New edits summarized here...
because I couldn't fit in the summary:


 * delete red links which point to a previously deleted article Lorentz transformation under symmetric configuration, for good reason: this article has all that article could ever have, and the title is presumably OR...
 * rewrite the "symmetric configuration" paragraph, not well-written:
 * "symmetric presentation between the forward Lorentz Transformation and the inverse Lorentz Transformation can be achieved if coordinate systems are in symmetric configuration. The symmetric form highlights that all physical laws should remain unchanged under a Lorentz transformation."
 * Coordinate systems are not the reason that physical laws are symmetric under Lorentz transformations... Anything which is frame-invariant (physical laws, proper time, speed of light, rest mass, etc) are independent of coordinates! M&and;Ŝc2ħεИτlk 15:20, 27 December 2012 (UTC)

Error in Derivation > From physical principles > Galilean reference frames > Principle of relativity ?
In that section it says

″According to the principle of relativity, there is no privileged Galilean frame of reference. Therefore, the inverse transformation for the position from frame R′ to frame R must be ... with the same value of γ (which must therefore be an even function of v).″

I think this is incorrect reasoning: γ is not a constant — rather it is a function of v and so the Principle of Relativity here would give us γ with v replaced by minus v. As it turns out, γ is an even function of v, so that the substitution has no effect. But ″therefore″ is not warranted: in fact it needs to be established (otherwise) that γ is an even function.

In Einstein′s derivation he appeals to relativity to measure a unit rod in one frame from the other, in both directions, then using relativity to assert those lengths to be equal. The derivation here attempts to finesse that step; but for the reasons above I don′t think it′s sound.

Comments?

Nunibad (talk) 05:42, 28 December 2012 (UTC)


 * ... No-one says it's a constant. The inverse of a Lorentz transform should necessarily have an identical form to the original except for negating the relative velocity v to −v (and obviously exchange primed/unprimed quantities), else there is no symmetry... So this is possible if and only if γ(v) = γ(−v), i.e. an even function of v, so "therefore" does seem justified... That particular point there and then in the article doesn't have a citation, but there is a citation for the same statement higher up in the article (2nd paragraph in this section), which is enough. M&and;Ŝc2ħεИτlk 10:23, 28 December 2012 (UTC)


 * ... Yes, it′s true that the inverse should have the same form; and it′s also true that γ is an expression in which v occurs; thus, if one were to make that explicit, one would get


 * $$x'=\gamma_v\left(x - vt\right)$$
 * $$x=\gamma_{-v}\left(x' + vt'\right)$$


 * for the two transformations, the first from R to R′ and the second back again, from R′ to R. Although it turns out to be true, later, that γ is shown to be even in v, at this point it has not been established and so we can′t use simply γ in both lines with the subscript v assumed.


 * I′d suggest a small modification that ends up in the same place, viz.


 * According to the principle of relativity, there is no privileged Galilean frame of reference: therefore the inverse transformation for the position from frame R′ to frame R should have the same form as the original. To take advantage of this, we arrange by reversing the axes that R′ sees R moving towards positive x′ (i.e. just as R sees R' moving towards positive x), so that we write


 * $$-x=\gamma\left(-x' - vt'\right)$$


 * which, when multiplied through by -1 becomes


 * $$x=\gamma\left(x' + vt'\right) .$$


 * The speed of light is constant
 * Since the speed of light is the same in all frames of reference...


 * Nunibad (talk) 23:20, 1 January 2013 (UTC)


 * ...Except reversing the axes is non-essential. The symmetry is all that's needed.


 * And how do you plan to establish that γ is an even function by continuing your method (which so far only obtains the inverse transformation)? The current derivation (Galilean and Einstein's relativity) relies on γ to be even, and it should be complete and self-contained (i.e. you can't leave it for another derivation to ensure that γ is even)...


 * I suspect you could follow a similar reasoning to what's already there (multiplying transformation equations), but you'd have equations in products like γvγ−v and would have to use physical conditions (SR assumptions) to solve for this product... which would only complicate the algebra... M&and;Ŝc2ħεИτlk 10:06, 2 January 2013 (UTC)


 * Sorry about the crossed-out comment, I overlooked that you were referring to γ and not γv or γ−v...
 * Given that I don't have major objections... M&and;Ŝc2ħεИτlk 13:31, 2 January 2013 (UTC)


 * ... I don't think the fact that γ is even is in fact needed in the derivation: just two lines later, at


 * Substituting for t and t′ in the preceding equations gives:
 * $$x'= \gamma\left(1 - v/c\right) x, $$
 * $$x= \gamma\left(1 + v/c\right) x'. $$
 * Multiplying these two equations together gives,
 * $$xx' = \gamma^2 \left(1 - v^2/c^2\right) xx'. $$
 * At any time after t = t′ = 0, xx′ is not zero, so dividing both sides of the equation by xx′ results in
 * $$\gamma=\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}, $$
 * which is called the "Lorentz factor".


 * the value for γ is derived, and indeed it is even. But we didn't need to assume its even-ness anywhere I can see.
 * Let me know if I′m mistaken about that. Otherwise ok? Nunibad (talk) 08:46, 3 January 2013 (UTC)


 * Yes, that's why I crossed out my mistaken comment. Feel free to add. M&and;Ŝc2ħεИτlk 08:49, 3 January 2013 (UTC)


 * No, in this context symmetry exactly means
 * $$\gamma_{-v}\left(v\right)=\gamma_{v}\left(-v\right)$$.
 * In the current derivation equal-ness is assumed instead of symmetry
 * $$\gamma_{-v}\left(v\right)=\gamma_{v}\left(v\right)$$,
 * which is different from symmetry and results in even-ness coming in at the back-door.
 * GHT153 (talk) 23:55, 12 May 2013 (UTC)

A clarification needed?
The page, when introducing the mathematics of the Lorentz transformation, simply specifies "The relative velocity between the two observers is v along the common x-axis", after defining two observers/reference frames O and O'. So, without piecing it together, a reader is left to wonder whether v is the velocity of a point stationary in O, as measured in O', or vice versa. This seems like something that should be repeated throughout the article!

(The v corresponding to the article is in fact the velocity of O' as observed in O)

(also, 1st ever post on wikipedia. Heya!) Neurofuzzy1 (talk) 10:40, 1 February 2013 (UTC)


 * Hi, there are diagrams to illustrate what you're saying, although I reworded in places anyway. Thanks, M&and;Ŝc2ħεИτlk 23:12, 1 February 2013 (UTC)

Removal of another derivation
The newest derivation now removed didn't seem too helpful as


 * it takes up too much space, way out of proportion with the rest of the article (WP:LENGTH)
 * There are already four derivations, two of which yield a deeper insight to spacetime itself using hyperbolic functions and group theory, plus two "simpler" ones. WP is not a textbook containing every conceivable derivation. (WP:MOSMATH) M&and;Ŝc2ħεИτlk 06:58, 27 May 2013 (UTC)

Here it is for future reference:

From physical derivation

Consider two inertial frames of reference O and O′, assuming O to be at rest while O′ is moving with a velocity v w.r.t O along the positive direction of x-axis. The origins of O and O′ initially coincide each other. Let a light signal is emitted from the common origin and travels as a spherical wave fronts. Consider a point P on a spherical wavefront is at a distance r and r′ from the origin of O and O′ respectively. According to second postulate of special theory of relativity as the speed of light is same in both the frames so r and r′ will be different only if t and t′ are different,


 * $$\begin{align}

r &= ct \\ r' &= ct' \end{align}$$

The equation of the spherical wavefront in frame O will be,


 * $$x^2 + y^2 + z^2 = r^2.$$

or


 * $$x^2 + y^2 + z^2 = c^2t^2.$$

similarly, the equation of the spherical wavefront in frame O′ will be,


 * $$x'^2 + y'^2 + z'^2 = r'^2.$$

or


 * $$x'^2 + y'^2 + z'^2 = c^2t'^2.$$

since O′ is moving along x-axis, therefore,


 * $$\begin{align}

y' &= y \\ z' &= z \end{align}$$

The relation between x and x′ should be in linear form and in such a way that it should reduce to Galilean transformation at v<<c. Therefore, such a relation can be written of the form:
 * $$\begin{align}

x' &= \gamma \left( x - v t \right)\\ \end{align}$$

and inversely:
 * $$\begin{align}

x &= \gamma \left( x' + v t' \right)\\ \end{align}$$

The above two equations gives the the relation between t and t′ as:
 * $$x = \gamma \left[ \gamma \left( x - v t \right) - v t' \right]$$

or


 * $$t' = \gamma t + \frac{ \left( 1 - { \gamma^2} \right)x}{ \gamma v}$$

substituting the expressions of x′, y′, z′ and t′ in terms of x, y, z and t in spherical wavefront equation of O′ frame we get,


 * $$x'^2 + y'^2 + z'^2 = c^2t'^2.$$

or


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

and therefore,


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

which implies,


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

comparing the coefficients of $$t^2$$ from above equation with the spherical wavefront equation of O frame we yield,


 * $$c^2 {\gamma^2} - v^2 {\gamma^2} = c^2$$

or


 * $${\gamma} = \frac{1}{ \sqrt{1 - \frac{v^2}{c^2}}}$$

Thus we yield the Lorentz transformation from the above expression and is given by,


 * $$\begin{align}

x' &= \gamma \left( x - v t \right)\\ t' &= \gamma \left( t - \frac{vx}{c^2} \right) \\ y' &= y \\ z' &= z \end{align}$$

where $$\ \gamma$$ is called Lorentz factor.

If this derivation must be included, we need consensus that it's better than the others.

Another (better?) alternative, since there are so many derivations already in this article, would be to create a separate specialized article (analogous to Derivation of the Navier–Stokes equations) and move all the derivations there. It could make it's own article since there must be tens of ways. This article would link to it saying:


 * "for the derivations see Derivations of the Lorentz transformations"

M&and;Ŝc2ħεИτlk 07:27, 27 May 2013 (UTC)


 * I agree that a new article Derivations of the Lorentz transformations would be fine. --D.H (talk) 07:39, 27 May 2013 (UTC)


 * May as well just do it. Even if there is consensus to move back in the future, it will be easy to move it back in the derivation section of this article. M&and;Ŝc2ħεИτlk 11:36, 27 May 2013 (UTC)


 * Just my penn'orth here but I enjoy this derivation as it derives the transforms by considering only the basic fundamentals: once you have two reference frames and a constant light-speed, you get Lorentz; it's lovely and basic. It'd be a shame for it just to disappear. PanDTV (talk) 01:07, 5 June 2015 (UTC)


 * There are plenty of derivations in "Derivations of the Lorentz transformations" including the one you like. Including them all in this article would make the article far too long. M&and;Ŝc2ħεИτlk 21:54, 13 September 2015 (UTC)

'Reference' to "Minkowski space" (in the intro)
I've been there. Utter goblin-bobblin for academics. → Guys, what if someone could either do something to that or, more relative here, create a subsection about that thing? It could have a "main article" subtitle, anyway... Josh, linguist (talk) 15:06, 27 September 2013 (UTC)