Talk:Lorentz transformation

Question
So, I'm not too knowledgeable when it comes to science and math but I would like to know; did Lorentz actually create these formulas or where they simply named after him? — Preceding unsigned comment added by 2600:1700:69C1:2A00:CD3F:6125:720C:780B (talk) 21:59, 19 June 2018 (UTC)
 * Please sign all your talk page messages with four tildes ( ~ ) — See Help:Using talk pages. Thanks.
 * Better ask at the wp:Reference desk/Science. Here we discuss the article, not its content—see wp:Talk page guidelines. Good luck. - DVdm (talk) 10:15, 20 June 2018 (UTC)

Physical implications of time dilation.
I think that section gives a false idea of the Lorentz transformations, based on the fact that the boosted observer F’ is supposed to measure time intervals by observing a clock at rest in F !

But why should he do that ? He simply takes his own clock (wristwatch time) ! Which of course runs at the same rate than the clock at rest in F !

And by using his own clock he will measure the correct proper time in F’ !

The Lorentz transformations are an intrinsic Physical property of Spacetime! That is completely forgotten in the traditional presentations !

--Chessfan (talk) 07:42, 11 October 2018 (UTC)


 * Good point. I have reworded the paragraphs to avoid the ambiguous "boosted observer" and to make sure that "ticks" refer to single events, so that time intervals are measured "between two ticks". I also changed the order to remain closer to the standard sources. - DVdm (talk) 08:16, 11 October 2018 (UTC)

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 :

$$5 bis/ \qquad |v|=v/e_1e_0 $$ 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 $$t e_0 $$ 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)


 * 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:

c t' = \gamma \cdot \left( c t - \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)
 * ✅. Not really needed, but won't do any harm: . - 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)

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.  — Q uantling (talk &#124; 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).