Talk:Lorentz transformation

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Why not GA ?

The Lorentz Transformation article is very math oriented. But you failed to mention the Geometric Algebra (GA) approach , which is much easier than the tensorial method. For example , I could show you , based on many papers by Hestenes, Doran , ... that all the very basic results of Special Relativity , time dilation , length contraction , time and length units , are contained in a single GA relation

e’0 e0= e’0 . e0 + e’0∧e0

But sorry , I will not do so , because that could be qualified original research !

It’s up to you !

--Chessfan (talk) 16:00, 16 October 2018 (UTC)

→== Examples, please!!!!!!!!!!!!!!!!!!!! ==

For whom are you writing, man???????? Only for yourself, I guess. — Preceding unsigned comment added by Koitus~nlwiki (talk • contribs) 19:42, 8 April 2019 (UTC)

OK I will try.

We consider two references frames $(e_{0},e_{1})and(f_{0},f_{1})$ , both pseudo-orthogonal.

$1/\qquad e_{0}^{2}=f_{0}^{2}=1\qquad e_{1}^{2}=f_{1}^{2}=-1\qquad e_{0}\cdot e_{1}=f_{0}\cdot f_{1}=0$

$2/\qquad f_{0}e_{0}=f_{0}\cdot e_{0}+f_{0}\wedge e_{0}$ → $f_{0}=f_{0}\cdot e_{0}e_{0}+f_{0}\wedge e_{0}e_{0}$

$3/\qquad f_{0}\cdot e_{0}=\gamma$

We define $\gamma$ by (3) and we guess that $\gamma$ is indeed the Lorentz factor.

We notice that the vector $f_{0}$ is decomposed into a vector parallel to $e_{0}$ , and a vector orthogonal to it :

$4/\qquad f_{0}\wedge e_{0}e_{0}=f_{0}-\gamma e_{0}$ → $(f_{0}\wedge e_{0}e_{0})\cdot e_{0}=0$

We guess now that $v$ is the euclidian velocity wich represents the movement of the f system :

$5/\qquad v=(f_{0}\wedge e_{0})/(f_{0}\cdot e_{0})$

In fact v is a bivector , and :

$5bis/\qquad |v|=v/e_{1}e_{0}$ a scalar .

$6/\qquad f_{0}e_{0}=\gamma [1+(f_{0}\wedge e_{0})/(f_{0}\cdot e_{0})]=\gamma (1+v)$

$7/\qquad 1=f_{0}e_{0}e_{0}f_{0}=\gamma ^{2}(1-|v|^{2})$

$8/\qquad \gamma =(1-|v|^{2})^{-1/2}$

Thus, as we guessed , $\gamma$ is in accordance we the definition of v .

It is now an easy task to deduce the Lorentz transformations and demonstrate the reciprocity.

Chessfan (talk) 17:26, 1 May 2019 (UTC)

Imagine a trajectory from (0,0)to $\tau f_{0}$ .What will be the time coordinate in the $(e_{0},e_{1})$ frame ? You simply project orthogonally the vector $\tau f_{0}$ on the vector $te_{0}$ and you find :

$9/\qquad t=\gamma \tau$

The reciprocity is obvious with (3) and the fact that we can imagine the $(e_{0},e_{1})$ frame moving backwards with velocity $(-v)$ .

--Chessfan (talk) 08:24, 2 May 2019 (UTC)

The Clifford (geometric) algebra approach to spacetime physics advocated by Hestenes, Baylis and others would be worth a separate article. There does exist a rather unsatisfactory Wikibooks presentation of the subject that would be a guide what NOT to do in writing a Wikipedia article on physics using geometric algebra. Prokaryotic Caspase Homolog (talk) 23:11, 11 May 2019 (UTC)

References

See Hestenes and Doran for free on : geocalc.clas.asu.edu and geometry.mrao.cam.ac.uk Chessfan (talk) 06:20, 12 May 2019 (UTC)

Corrected First Equation

The first equation reads

$t'=\gamma \cdot \left(t-{\frac {vx}{c^{2}}}\right)$

which, on inspection, is incorrect. Note the physical units within the parenthesis must evaluate to time, but the last term has physical units that evaluate to time over velocity. This is due to the squaring of the speed of light in the denominator of the last term, which is apparently a typo. I have removed that square.

-- motorfingers : Talk 08:10, 29 September 2020 (UTC)

The equation

$t'=\gamma \cdot \left(t-{\frac {vx}{c^{2}}}\right)$

is correct.

You are confusing with this equation:

$ct'=\gamma \cdot \left(ct-{\frac {vx}{c}}\right).$

See the literature. I have restored the correct equation. - DVdm (talk) 08:49, 29 September 2020 (UTC)

Whoops, it seemed so obvious in the middle of the night, and I did a quick web search at the time to verify the equation that seemed to validate what I thought at the time. I was reading $vx$ to be time, not the actual physical units of distance squared divided by time. In the morning, the error seems as clearly false as it seemed to be clear in the middle of the night. Sorry, and thank you, DVdm, for the quick reversion. -- motorfingers : Talk 15:15, 29 September 2020 (UTC)

Middle of the night is a fun place alright, but sometimes a dangerous one - DVdm (talk) 18:17, 29 September 2020 (UTC)

Section 6.2

The Tex for section 6.2 (contravariant vectors) is not showing the "-" on the "-1" in the exponents. It just looks like a " 1" — Preceding unsigned comment added by 2600:6C44:E7F:1100:F0B3:3663:E21A:A6FE (talk) 22:37, 2 February 2021 (UTC)

If you were using chrome, it is probably poor setups in handling MathML. See discussion in Pauli matrices. Cuzkatzimhut (talk) 22:46, 2 February 2021 (UTC)

I see it with Firefox under Windows. -- motorfingers : Talk 00:52, 3 February 2021 (UTC)

I just want to say that I love this picture :)

This picture!

— Preceding unsigned comment added by 98.128.172.242 (talk • contribs) 04:07, 25 April 2021 (UTC)

Same here mate! petite (talk) 10:42, 10 January 2022 (UTC)

First equation condition

I think it should be mentioned that the two frames have the same origin at t=0 for the first equation 31.164.189.193 (talk) 11:19, 13 January 2022 (UTC)

 Done. Not really needed, but won't do any harm: [1]. - DVdm (talk) 13:29, 13 January 2022 (UTC)

I came to this page to comment that the phrase "two frames with the origins coinciding at t=t′=0" is confusing, and found this comment thread.

The coordinates are for spacetime, so the origin is defined as (0, 0, 0, 0), which includes t=0. Saying two frames have the same origin at t=0 is like saying they have the same origin at x=0, or y=0, or z=0.

My concern is that the current text encourages people to think in terms of 3D space with time somehow separate, rather than thinking in term of 4D spacetime.

Maybe "... two frames with the origins coinciding at (0, 0, 0, 0)" would satisfy the original comment, while emphasizing spacetime. On the other hand, given that this is the definition of an origin it is an odd statement. Subbookkeepper (talk) 16:18, 14 September 2023 (UTC)

@Subbookkeepper: A very late reply, this, but I have solved it this way. - DVdm (talk) 18:51, 9 October 2024 (UTC)

Eigenstates

It is a simple matter to find the eigenvalues Sqrt[1-b)/(1+b)] and its inverse, where b is beta, and eigenstates ([1, 1] and [-1, 1]) of the [x, ct] Lorentz transformation. Some commentators on web forums claim that the eigenvalues are related to the Doppler effect. Should this be discussed in the article? Xxanthippe (talk) 22:52, 10 July 2024 (UTC).

Yes, a photon moving in the same direction as the implicit boost direction of the Lorentz transformation will have its energy and momentum scaled by the same multiplicative factor — an amount determined by the Doppler shift. Because it is the same scaling factor for the momentum and energy, that's saying that the photon's four-momentum is an eigenstate. Likewise for a photon with the opposite momentum — that is, with momentum antiparallel to the implicit boost direction of the Lorentz transformation — it will be red-shifted rather than blue-shifted but otherwise this too is a Doppler shift and an eigenstate. If we do mention any of this in the article, I'm thinking it should be exceedingly brief. —Quantling (talk | contribs) 15:33, 11 July 2024 (UTC)

The eigenstructure is a significant algebraic feature of any linear transformation, so it is odd that it is not mentioned in an article on the most important linear transformation in physics. This case, in particular, needs some discussion as it involves the apparent paradox (to Galilean thinkers) of transforming between frames that are both traveling at the same speed c, and the limiting processes that are involved because of that. However, I will leave additions to this topic to editors with more experience than myself. Xxanthippe (talk) 04:23, 12 July 2024 (UTC).

More of a technical question, but how does the eigenvector interpretation generalise to higher dimensions? For example, if we have two dimensions (space and time), we find two linearly independent eigenvectors whose direction (speed) does not change, c and -c. So far so good. However, if we add another spatial dimension, introducing the famous light cones, what would the independent eigenvectors be in this case? As far as I understand, an n-dimensional space can only have n linearly-independent eigenvectors (in this case n=3), which does not seem to be enough to represent the vectors where the speed does not change (light cones). Viktaur (talk) 22:36, 13 August 2024 (UTC)

A spatial vector orthogonal to the implicit boost direction will have an eigenvalue of 1. There will not be length contraction of this vector. —Quantling (talk | contribs) 16:22, 15 August 2024 (UTC)

Format errors

There are lots of Math output errors (Firefox 129.0.2), but renders OK with Chrome 128.0.6613.85. Could somebody identify the problem? I am reluctant to meddle myself due to inexperience. Xxanthippe (talk) 01:13, 29 August 2024 (UTC).

I don't see the render errors with Firefox. Perhaps refresh the page, restart your browser, etc. —Quantling (talk | contribs) 00:31, 30 August 2024 (UTC)

Many thanks for your advice. I have restarted Mac OS 14.6.1 and Firefox 129.0.2 but still get a dozen yellow banners with "Math output error" in red. The rest of the math renders OK. Chrome renders all the math OK but I get the same yellow banners on Safari. Best wishes Xxanthippe (talk) 01:08, 30 August 2024 (UTC).

A note to say that I got satisfactory rendering results after changing my Appearance Skin from legacy to default. Xxanthippe (talk) 07:14, 1 September 2024 (UTC).

General boost direction

I have added the expression for the general boost direction. Xxanthippe (talk) 06:20, 8 September 2024 (UTC).

If we keep it here, I have made some tweaks: proper link to given source, looks, remove wp:repeatlink, wording. I know, the trace is easily verified, but it's certainly beyond basic arithmetic as outlined in wp:CALC. - DVdm (talk) 12:53, 8 September 2024 (UTC)

I have restored the original formatting. Users can say which version they prefer. I have not yet mastered the alignment of matrices: maybe it can be done better by some clever person. Xxanthippe (talk) 02:32, 21 September 2024 (UTC).

I have restored the alignment, as was —admittedly tersly— explained in my edit summary and here in my message above, with the word "looks".

The way you had formatted, namely

$${\begin{bmatrix}ct'\\\\\\\\x'\\\\\\\\y'\\\\\\z'\end{bmatrix}}={\begin{bmatrix}\gamma &-\gamma \beta _{x}&-\gamma \beta _{y}&-\gamma \beta _{z}\\-\gamma \beta _{x}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}^{2}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{z}\\-\gamma \beta _{y}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}^{2}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}\\-\gamma \beta _{z}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{z}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{z}^{2}\\\end{bmatrix}}{\begin{bmatrix}ct\\\\\\x\\\\\\\\y\\\\\\\\z\end{bmatrix}},$$

has the 4x1 matrices (vectors) extremely elongated. Compare with

$${\begin{bmatrix}ct'\\x'\\y'\\z'\\\end{bmatrix}}={\begin{bmatrix}\gamma &-\gamma \beta _{x}&-\gamma \beta _{y}&-\gamma \beta _{z}\\-\gamma \beta _{x}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}^{2}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{z}\\-\gamma \beta _{y}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}^{2}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}\\-\gamma \beta _{z}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{z}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{z}^{2}\\\end{bmatrix}}{\begin{bmatrix}ct\\x\\y\\z\end{bmatrix}},$$

which looks better and is done along the standard as in Help:Displaying a formula. See other examples in articles such as Matrix multiplication and Transformation matrix.

In our same article in subsection Lorentz_transformation#Proper_transformations, we see

$$B(\mathbf {v} )={\begin{bmatrix}\gamma &-\gamma v_{x}/c&-\gamma v_{y}/c&-\gamma v_{z}/c\\-\gamma v_{x}/c&1+(\gamma -1){\dfrac {v_{x}^{2}}{v^{2}}}&(\gamma -1){\dfrac {v_{x}v_{y}}{v^{2}}}&(\gamma -1){\dfrac {v_{x}v_{z}}{v^{2}}}\\-\gamma v_{y}/c&(\gamma -1){\dfrac {v_{y}v_{x}}{v^{2}}}&1+(\gamma -1){\dfrac {v_{y}^{2}}{v^{2}}}&(\gamma -1){\dfrac {v_{y}v_{z}}{v^{2}}}\\-\gamma v_{z}/c&(\gamma -1){\dfrac {v_{z}v_{x}}{v^{2}}}&(\gamma -1){\dfrac {v_{z}v_{y}}{v^{2}}}&1+(\gamma -1){\dfrac {v_{z}^{2}}{v^{2}}}\end{bmatrix}}={\begin{bmatrix}\gamma &-\gamma {\vec {\beta }}^{T}\\-\gamma {\vec {\beta }}&I+(\gamma -1){\dfrac {{\vec {\beta }}{\vec {\beta }}^{T}}{\beta ^{2}}}\end{bmatrix}},$$

where the height of the RHS matrix is, depending on the number of columns, automatically adjusted, and not artificially "elongated" by inserting empty lines.

Comments by others are welcome. Cheers. - DVdm (talk) 12:25, 5 October 2024 (UTC)

There is an \align command but I have yet to master it. Until then. Xxanthippe (talk) 02:59, 8 October 2024 (UTC).

@Xxanthippe: Got it! We need to factor each vector component with a vertical phantom, an invisible element taken from the corresponding row of the matrix, using \vphantom{<MaxHeightElementOfRow>}: $${\begin{bmatrix}ct'{\vphantom {-\gamma \beta _{x}}}\\x'{\vphantom {1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}^{2}}}\\y'{\vphantom {{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}}}\\z'{\vphantom {{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}}}\end{bmatrix}}={\begin{bmatrix}\gamma &-\gamma \beta _{x}&-\gamma \beta _{y}&-\gamma \beta _{z}\\-\gamma \beta _{x}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}^{2}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{z}\\-\gamma \beta _{y}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}^{2}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}\\-\gamma \beta _{z}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{z}&{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}&1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{z}^{2}\\\end{bmatrix}}{\begin{bmatrix}ct{\vphantom {-\gamma \beta _{x}}}\\x{\vphantom {1+{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}^{2}}}\\y{\vphantom {{\frac {\gamma ^{2}}{1+\gamma }}\beta _{x}\beta _{y}}}\\z{\vphantom {{\frac {\gamma ^{2}}{1+\gamma }}\beta _{y}\beta _{z}}}\end{bmatrix}},$$ DVdm (talk) 10:00, 8 October 2024 (UTC)