Talk:Relativistic mechanics

Requested move

 * The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section. 

The result of the move request was: page moved.   A rbitrarily 0   ( talk ) 20:19, 30 October 2011 (UTC)

Relativistic Mechanics → Relativistic mechanics –

Per WP:CAPS ("Wikipedia avoids unnecessary capitalization") and WP:TITLE, this is a generic, common term, not a propriety or commercial term, so the article title should be downcased. Lowercase will match the formatting of related article titles. Tony  (talk)  11:58, 26 October 2011 (UTC)


 * Support. It's given in lower case in the Academic Press dictionary of science and technology. Kauffner (talk) 19:21, 26 October 2011 (UTC)


 * comment I think there is a strong case for simply redirecting this to special relativity, just like relativistic mechanics.TR 15:16, 28 October 2011 (UTC)
 * The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

transfer of content
Special relativity was too long and dense, there was a section in that article called "Relativistic mechanics", so I spilt the excess into here. Also modified the page ratings. =| F = q(E+v×B) ⇄ ∑ici 06:50, 4 June 2012 (UTC)

"the formula was correct as it stood "
I have no clue how to use this formula:


 * $$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} = \sum_{n=0}^{\infty} \prod_{k=1}^n \dfrac{(2k - 1)}{2k} \dfrac{v^2}{c^2} = 1 + \dfrac{1}{2} \dfrac{v^2}{c^2} + \dfrac{3}{8} \dfrac{v^4}{c^4} + \dfrac{5}{16} \dfrac{v^6}{c^6} + \ldots$$

Before you mention it I know product $$\prod$$ and summation $$\sum$$ notation, and for the series to have increasing powers of (v/c)2, would there not be any dummy variable in the summation/product? Where is the power dependence of (v/c)2 ? According to this layout is it not possible to factor out the (v/c)2


 * $$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} = \dfrac{v^2}{c^2} \sum_{n=0}^{\infty} \prod_{k=1}^n \dfrac{(2k - 1)}{2k}$$

like that?? Why is it easier to have a product for every term in the series when you can just calculate each coefficent and add them up like any series?

Here is my derivation using the binomial theorem:


 * $$\begin{align}

\gamma & = (1-\beta^2)^{-1/2} \\ & = 1 + \left(-\dfrac{1}{2}\right) (-\beta^2) + \dfrac{\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}-1\right)}{2!}(-\beta^2)^2 \\ & + \dfrac{\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}-1\right)\left(-\dfrac{1}{2}-2\right)}{3!}(-\beta^2)^3 \\ & + \dfrac{\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}-1\right)\left(-\dfrac{1}{2}-2\right)\left(-\dfrac{1}{2}-3\right)}{4!}(-\beta^2)^4 + \cdots \\ & \\ & = 1 + \dfrac{(-1)^1 \dfrac{1}{2^1}}{1!} (-\beta^2) + \dfrac{(-1)^2 \dfrac{1\cdot 3}{2^2}}{2!} \beta^4 + \dfrac{(-1)^3 \dfrac{1\cdot 3\cdot 5}{2^3}}{3!} (-\beta^6) + \dfrac{(-1)^4 \dfrac{1\cdot 3\cdot 5 \cdot 7}{2^4}}{4!} \beta^8 + \cdots \\ & \\ & = \dfrac{1}{0!}\beta^0 + \dfrac{1!!}{2^11!} \beta^2 + \dfrac{3!!}{2^22!} \beta^4 + \dfrac{5!!}{2^33!} \beta^6 + \dfrac{7!!}{2^44!} \beta^8 + \cdots \\ & \\ & = 1 + \sum_{n=1}^\infty \dfrac{(2n-1)!!}{2^n n!}\beta^{2n} \\ & = 1 + \sum_{n=1}^\infty \dfrac{\dfrac{(2n)!}{n!2^n }}{2^n n!}\beta^{2n} \\ & = \sum_{n=0}^\infty \dfrac{(2n)!}{(2^n n!)^2}\beta^{2n} \\

\end{align}$$

so the one I gave is correct. So what if the size of the coefficients becomes atronomical as the series progresses, its no worse than a product for each term. I will not replace this in the article or fight you on it, just curious on your version... F = q(E+v×B) ⇄ ∑ici 13:06, 21 June 2012 (UTC)


 * According to this edit, (v/c)2 is in the product, and is multiplied with the bracket (2k − 1)/(2k)? How is a reader supposed to know that, even if (v/c)2 its part of the fraction?


 * If it is multiplied with itself n times why can't it be written:


 * $$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} = \sum_{n=0}^{\infty} \left(\dfrac{v^2}{c^2}\right)^n\prod_{k=1}^n \dfrac{(2k - 1)}{2k} = 1 + \dfrac{1}{2} \dfrac{v^2}{c^2} + \dfrac{3}{8} \dfrac{v^4}{c^4} + \dfrac{5}{16} \dfrac{v^6}{c^6} + \cdots$$?


 * F = q(E+v×B) ⇄ ∑ici 13:30, 21 June 2012 (UTC)


 * The n=0 term is the empty product, 1. The n=1 term is
 * $$1 \cdot \frac{(2 \cdot 1 - 1) v^2}{(2 \cdot 1) c^2} = \frac{v^2}{2 c^2} \,.$$
 * The n=2 term is
 * $$\frac{v^2}{2 c^2} \cdot \frac{(2 \cdot 2 - 1) v^2}{(2 \cdot 2) c^2} = \frac{3 v^4}{8 c^4} \,.$$
 * And so forth. So this is correct.
 * The reader should use common sense. Neither the product nor the sum ends until there is something which indicates an end such as the "=" sign. And as you pointed out, it makes no sense if the velocity is not part of the product. JRSpriggs (talk) 13:36, 21 June 2012 (UTC)


 * I didn't point out it has to be part of the bracket. Based on your explaination (and my initial geuss) - the product is only for the coefficients, the (v/c)2 isn't part of the product, it is part of the sum, all that matters it that (v/c)2 has a summy variable n attatched so the reader does not have to geuss how to calculate each term.
 * It’s true the reader should use common sense, but its still unclear and inserting one power for sake of clarification is worth it. I have at least inserted the version just tweaked above so it is explicitly clear how to evaluate the sum... F = q(E+v×B) ⇄ ∑ici 14:11, 21 June 2012 (UTC)
 * It’s true the reader should use common sense, but its still unclear and inserting one power for sake of clarification is worth it. I have at least inserted the version just tweaked above so it is explicitly clear how to evaluate the sum... F = q(E+v×B) ⇄ ∑ici 14:11, 21 June 2012 (UTC)

Angular momentum
This article did not even mention angular momentum, so I transferred some content from Relativistic quantum mechanics. M&and;Ŝc2ħεИτlk 08:06, 17 April 2013 (UTC)

Ambiguity of "bivector"
We seem to be using "bivector" to mean two different things. I was using it as I have seen it used in some texts on general relativity &mdash; to mean two separate 3D vectors u and v construed as part of an antisymmetric tensor so:
 * $$\begin{pmatrix}

0 & - u_x & - u_y & - u_z \\ u_x & 0 & v_z & - v_y \\ u_y & - v_z & 0 & v_x \\ u_z & v_y & - v_x & 0 \end{pmatrix} \,.$$ Here u is an ordinary 3D vector and v is a distinct 3D axial vector.

Maschen appears to be using it to mean an axial vector with three components (such as the cross product of two vectors). JRSpriggs (talk) 15:01, 24 May 2013 (UTC)


 * No, I mean bivector as the plane element from the exterior product of x and p, with 6 components, not the pseudo/axial vector with three components from a cross product. The inclusion of angular momentum as a pseudovector was only for the classical case. I suspect you refer to the angular momentum section and the three components which may suggest they are the same?


 * "Three components in cyclic permutations are the familiar 3-orbital angular momenta:


 * $$M^{12} = L^3 = x^1 p^2 - x^2 p^1$$
 * $$M^{23} = L^1 = x^2 p^3 - x^3 p^2$$
 * $$M^{31} = L^2 = x^3 p^1 - x^1 p^3$$ ..."


 * I'll reword/rewrite. M&and;Ŝc2ħεИτlk 14:56, 25 May 2013 (UTC)


 * I think I caught up on what you mean, since the definition I inserted from Penrose
 * $$\mathbf{M} = 2\mathbf{x}\wedge\mathbf{p}$$
 * just has 3 independent components, while the matrix you have above has 6 altogether... I'll look for a better source soon. M&and;Ŝc2ħεИτlk 15:51, 25 May 2013 (UTC)

Indeed the correct definition is in Penrose - I just read it incorrectly (to be fair he used Latin indices for the components of the 4-vectors but it's no excuse). The angular momentum tensor is in fact:


 * $$\mathbf{M} = 2\mathbf{X}\wedge\mathbf{P}$$

where X is the 4-position and P the 4-momentum, and the components are:


 * $$M^{01} = -M^{10} = x^0 p^1 - x^1 p^0 = ct p^1 - x^1 E/c $$
 * $$M^{02} = -M^{20} = x^0 p^2 - x^2 p^0 = ct p^2 - x^2 E/c $$
 * $$M^{03} = -M^{30} = x^0 p^3 - x^3 p^0 = ct p^3 - x^3 E/c $$
 * $$M^{12} = -M^{21} = x^1 p^2 - x^2 p^1$$
 * $$M^{23} = -M^{32} = x^2 p^3 - x^3 p^2$$
 * $$M^{31} = -M^{13} = x^3 p^1 - x^1 p^3$$

Where M01, M02, M03, are the COM boosts and M12, M23, M31 for the orbital AM. I'll continue to look for sources, for now here's a quick few:, , and ,.

Thanks for noticing the error and helping to clear this up. M&and;Ŝc2ħεИτlk 16:25, 25 May 2013 (UTC)


 * P.S. Should close this section by saying the definition for AM as 2r&and;p is still a bivector in three dimensions, since two vectors are exterior-multiplied together. The discrepancy was really in the definition of relativistic AM, consequently the number of components. M&and;Ŝc2ħεИτlk 08:39, 27 May 2013 (UTC)

Separate article on relativistic angular momentum?
Since


 * the section in this article has become more about the mathematical description of AM in relativity, and seems out-of-style with the qualitative nature of this article with a minimum number of equations (a good thing), and
 * AM is a notable topic in both SR and GR (see refs in above section),
 * there's plenty more about AM in SR not in this article (which could or shouldn't be),
 * AM in GR looks sufficiently complicated to deserve it's own treatment (non-conservation in curved spacetime? or conserved currents under the relevant conditions? the relations to the energy-momentum tensor?, Komar quantities? etc. etc.), alongside the simpler SR description.

I think it could make it's own article. At least five articles
 * angular momentum
 * relativistic quantum mechanics
 * special relativity
 * general relativity
 * this article

would definitely link to it, and possibly:


 * two-body problem in general relativity?
 * gravitoelectromagnetism?
 * gravitational radiation?
 * Poincaré group?
 * Lorentz group?
 * Representation theory of the Lorentz group?

so there is no risk of it becoming an "orphan".

For this article, the section should be rewritten in qualitative prose, only quoting definitions and not the multitude of writing them down. Will do this soon unless there are strong objections (there shouldn't be). M&and;Ŝc2ħεИτlk 07:44, 26 May 2013 (UTC)


 * Done. Finally. Experts in general relativity would help there. M&and;Ŝc2ħεИτlk 23:41, 4 June 2013 (UTC)

Proof of Relativistic Energy and Momentum Equations
The bottom part shows the same interaction shown in the top part but from the point of view of a rocket moving toward the left at 0.999c. If you wish to double check the numbers just be careful not to do too much rounding off. Some of the results are extremely sensitive to even very tiny changes.



Just granpa (talk) 11:29, 2 July 2016 (UTC)

Possible Error on this page
I do not know how to write formulae on Wikipedia but this txt only version should be clear. The author of this "derivation" pop up box for the gamma^3 in Newton's Second Law is showing how/where this gamma^3 comes in, and he shows the parallel and perpendicular accelerations which he denotes by boldface letter "a" with subscripts for parallel or perpendicular. It is evident from one place in the derivation that "a" is taken to equate to "dv/dt" which is wrong. This is what I think is an error. I think this gamma^3 calculation [which everyone knows comes from time derivative of (gamma v) or from d(gamma v)/dt]. The result is that a=gamma^3 dv/dt wich implies a/gamma^3 = dv/dt. It is erroneous to write F = gamma^3 m a and it is correct to write F = gamma^3 m dv/dt.

Here is my discussion of this:

Is it F=gamma^3 m a

or

Is it F=gamma^3 m dv/dt

?

I think latter only not former. Here's why.

Isn't it true that in ground frame F=ma= m gamma^3 dv/dt (but not F=gamma^3 m a) so a = gamma^3 dv/dt or a/gamma^3 = dv/dt

In other words, the acceleration in ground frame is always constant for constant force, but what changes is only by how much dv/dt changes given that constant acceleration not the acceleration. (True or False?)

If true then integral a dt = integral gamma^3 dv/dt dt = gamma v -> this at = gamma v is found everywhere in textbooks on constant acceleration derivations

if False then integral a dt = integral dv/dt dt = v with no gamma at all. This is result if "a" is assumed as merely a label for "dv/dt" to "simplify" notation. But this is not true.

If true, this means that constant force always results in constant acceleration but that if gamma is large it just means that this constant acceleration results in smaller and smaller changes in velocity.

It does NOT mean that acceleration is non-constant or reduced in some way by gamma being large.

In fact, the product gamma^3 dv/dt is what is constant and in fact the acceleration is that constant.

I think it is an error to write F = gamma^3 m a because that is not true. F=ma is true but F= gamma^3 m dv/dt is true since a = gamma^3 dv/dt is true

How can this Wikipedia page be changed to reflect this? — Preceding unsigned comment added by Dukon (talk • contribs) 22:34, 28 December 2016 (UTC)


 * You are mistaken. The formula in the article is correct as is. JRSpriggs (talk) 19:43, 29 December 2016 (UTC)

Hi JRSpriggs. Thank you for taking the time to evaluate my remarks and make your comment.

Can you please explain how I get integrals to appear in this conversation text box? I am only txt typing them in but I'd like them to be nicely formatted in here. If you know how please share.

To the physics discussion at hand: If the formula in article is correct, then how do you define relativistic velocity to be gamma v? If you claim a = dv/dt ("a" is a label always for "dv/dt") this means there can be no gamma with the v, or no relativistic velocity since integral_s^t (a) dt = integral_s^t (dv/dt) dt = v always without gamma but relativistic velocity must have gamma with it. If you think formula in article is correct where does gamma in velocity come from in the integral_s^t (a) dt? The gamma^3 cannot be in front of (a) since then the left hand side would no longer equal a*t - a*s while right hand side is gamma_t v(t) - gamma_s v(s). This means you would not have result that a*t = gamma_t v(t) but that is relativistic result. Formula in article (considering "a" as label for "dv/dt") doesn't deliver correct relativistic velocity. Can you comment on this detail? Thank you for your input. Dukon (talk) 03:34, 30 December 2016 (UTC)


 * You need to be clear on which time you are differentiating with respect to. "t" is the time coordinate of the stationary observer. "&tau;" is the time as measured by a clock moving with the object in question, called proper-time (or co-moving time).
 * $$\frac{d \mathbf{v}}{d \tau} = \frac{d \mathbf{v}}{d t} \frac{d t}{d \tau} = \mathbf{a} \gamma $$
 * by the chain rule and the definitions of acceleration and gamma. JRSpriggs (talk) 20:23, 30 December 2016 (UTC)

neither sure how to just add my response to this nor whether you will be notified I replied but --- thank you for your posting additional information. However I asked a number of questions which I would kindly ask you to answer applying the new information you have supplied. Due to your not saying anything additional beyond your new information, I have no idea if you felt like just saying something true or you see how your new information answers my questions. If you do see how your new information answers my questions please state explicitly how it answers the questions. That would be helpful but contributing information about the chain rule which I would never object to and agree is true without applying it to dig out specific detailed answers to my specific detailed questions leaves my questions unanswered by your new information which we both agree is true. A little help? Thank you

Idea
Could Albert Einstein be talking about antimatter when he talks about invariant mass Tony Ratliff (talk) 19:03, 21 November 2023 (UTC)


 * No, see invariant mass, antimatter, and history of relativity Johnjbarton (talk) 19:18, 21 November 2023 (UTC)