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http://upload.wikimedia.org/math/6/d/c/6dc88e88a74eac0290cb0d575aa7a147.png There is an algebraic error in the above equation. The rightmost term should read: (c v1 + c v2)/(c + [v1 v2]/c). Basic unit analysis shows that the original quantity is incorrect.

Sorry for the botched image include. But the URL is there for reference. Cheers, Justin.

Right you are. I fixed it. —Keenan Pepper 02:52, 2 July 2006 (UTC)

## Problem of symmetry

Sorry folks, but are you sure that this formula is correct: $\mathbf{v_1} \oplus \mathbf{v_2}=\frac{\mathbf{v_1}+\mathbf{v_2}}{1+ \frac{\mathbf{v_1}\cdot\mathbf{v_2}}{c^2}} + \frac{1}{c^2} \cdot \frac{1}{1+\sqrt{1-\frac{v_1^2}{c^2}}} \cdot \frac{\mathbf{v_1}\times(\mathbf{v_1}\times\mathbf{v_2})}{1+\frac{\mathbf{v_1}\cdot\mathbf{v_2}}{c^2}}$

I have my doubts about that, since $\oplus$ is not symmetric. The result of this formula seems to be correct only, if $v_1$ and $v_2$ are collinear.

Can anyone confirm this? --141.33.44.201 10:20, 14 December 2006 (UTC)

The correct formula is only symmetric in the collinear case, as others have said below. The formula above is an attempt to correct for the fact that there is an extra time dilation factor of $\sqrt{1-v_1^2}$ for the perpendicular component in the case of non-collinear motions. The triple cross product is a three-dimensional way of selecting the component of $v_2$ perpendicular to $v_1$.
The triple cross product evaluates to $|v_1|^2$ times the desired component of $v_2$. The fraction in front is supposed to add with the $v_2$ from the first term, to produce the required $\sqrt{1-|v|^2}$. Since the first term contains the component perpendicular multiplied by 1, the second factor needs to equal $\sqrt{1-v_1^2}-1 \over |v_1|^2$ to work. It does, as can be seen by simplifying this fraction by multiplying top and bottom by $\sqrt{1-v_1^2} +1$. But because the algebraic manipulations were designed to work with the cross product, this is a confusing formula. Selecting perpendicular and parallel components should be done with the dot product, as in the rewrite. —Preceding unsigned comment added by 71.127.173.103 (talk) 02:35, 3 September 2007 (UTC)

## Velocity units

"When Velocity is expressed in metres per second, instead of as a fraction of the speed of light the equation becomes..."

Of course, this statement is true, but it would also be true if velocity is expressed in kilometers per hour, miles per hour, or knots too. The main thing is unit distance and time rather than fraction of C. Perhaps the statement should reflect this fact.

The formula IS NOT and DOES NOT HAVE TO BE symmetric in the component vectors. The problem is that the article does not explain the physical meaning of these components. The explanation should be added because many misunderstand the relativistic addition of velocities. (see p. ex. the award winning article "Speed of light". A bitter fight is going on because of the author's stubbornness. And guess what, they refer to THIS article as the source of their interpretation. Since there is no interpretatiton here the whole fight is futile.)

Here is the interpretatiton

Let us picture 3 observers 1, 2, and 3. Let v1 denote the volocity vector of 2 as observed by 1. Also, let v2 denote the velocity vector of 3 as observed by 2. Then if 1 observes 3, too, then they will see the velocity vector given by the formula. Is it clear now that the formula does not have to be symmetrical? The point is thet the roles of the observers are not symmetrical.

The rather suprising thing is that if 1, 2, 3, move along the same straight line then the formula becomes symmetrical.

IMPORTANT: SOMEBODY SHOULD INCLUDE THIS EXPLANATION ABOUT THE PHYSICAL MEANING OF THE COMPONENTS. I repeat, many misunderstand the Einstein formula. This is very dangerous. Relativity theory is already the hot bed of of stupidity and misunderstanding.

zgyorfi

Thanks. Meanwhile I looked the general lorentz-boost up in one of my physics books. And I (surprisingly) realised, that the formula indeed is not symmetric. --141.33.44.201 09:50, 29 March 2007 (UTC)

## cleanup tag or expert needed tag or what

the people here and a physicist elsewhere say this article needs work. I'm putting an expert needed tag since it seems likely to need smarties to clean it up.Rich 19:10, 28 April 2007 (UTC)

I did a complete rewrite. Likebox 02:45, 3 September 2007 (UTC)

## Two successive dot products in the first equation?

Why are ther two successive dot products in the first equation? Errors as simple as this give great cause for concern about the accuracy of the rest of the program. —The preceding unsigned comment was added by 194.200.70.203 (talk) 18:48, August 23, 2007 (UTC)

Those dots can be removed as what it on the lhs of each one is a scalar and not a vector. --EMS | Talk 19:40, 23 August 2007 (UTC)

## Doppler Stuff is Not Very Physical

I only kept it because the previous version had it as a reasonable addition formula. Maybe it should be erased entirely. At least the relativistic and nonrelativistic versions are correct now. Likebox 03:42, 3 September 2007 (UTC)

I figured out why it's a velocity addition law. But it is very peculiar--- its for the Doppler effects in a one-dimensional right-moving wave. Only then is there a group which gives this addition law. This has no relativistic analog. I don't know the context in which this addition law is used, if any. Likebox 23:43, 4 September 2007 (UTC)

I want to determine what we get by adding $c$ and $-c$ together. Because we are adding two relativistic speeds we should use the velocity addition formula, right? But it is undefined for $v_1 = -v_2 = c$. And things don't get any clearer if we consider the two limits $\lim_{v \rightarrow c} \frac{c - v}{1 - \frac{cv}{c^2}} = c$ and $\lim_{v \rightarrow c} \frac{v - c}{1 - \frac{cv}{c^2}} = -c$. It obviously is not a revokable singularity, so what is really going on here?

And if we set $c \oplus -c = 0$ we loose the associativity, since then we have $(v \oplus c) \oplus -c = c \oplus -c = 0$, and $v \oplus (c \oplus -c) = v \oplus 0 = v$. Can someone please explain what all the trouble is about? --Rndusr (talk) 00:02, 21 December 2007 (UTC)

The trouble is made clearer by taking a limit. If you have a particle moving with c-e where e is very small compared to c, and you add the velocity -(c-e) you get a particle at rest. If you add -(c-e/2) you get a particle moving with a speed c/3. If you boost by c-ae where a is an arbitrary number, you get an arbitrary result. There is no unique answer to adding c and -c because for speeds very close to light, the result of subtraction is any answer you want.Likebox (talk) 02:06, 21 December 2007 (UTC)

Rndusr, good question, yet it doesn't even matter. In the composition formula one of the "velocities to compose" is the velocity of one observer (object) w.r.t the other. This velocity cannot be c or -c. Light signals cannot be treated as observers, so your case (c,c) is out scope for the formula. DVdm (talk) 09:18, 21 December 2007 (UTC)

## Suggested change of name

I suggest that the name of the page is changed to 'velocity-composition formula' with a redirect from the original name. As has already been pointed out, relativity is a hotbed of misunderstanding and stupidity and use of the term 'addition' causes much confusion in that some people seem to think that in that relativity 2 + 2 can equal 3. Also, velocities can be added normally in some circumstances such as closing speeds. Martin Hogbin (talk) 12:29, 23 March 2008 (UTC)

## too complicated

The previous comment about renaming the article 'velocity-composition formula' makes sense, since it would not imply a simple 'addition' method. I was checking to see if the previous incorrect formula had been changed, and obviously it has. Now the article seems way too complicated with the 'long' discussion of doppler effects.Phyti (talk) 15:38, 17 May 2008 (UTC)phyti

I agree, the Doppler shift bit has no real relevance to the main subject and should be deleted from this article.Martin Hogbin (talk) 22:00, 1 July 2008 (UTC)

## Citation should be fine

First formula is the same as in "Zur Elektrodynamik der bewegter Koerper". Ann. Phys., 17, 891-921.
paragraph 5.
This should be appropriate reference;) AlexShkotin (talk) 15:43, 18 October 2008 (UTC)

## Gyrovector space??

Anyone know anything? --unsigned comment--

I added the link. I am currently constructing that page. Only colinear velocites are commutative and associative, but in general, addition of non-colinear velocities is non-associative and non-commutative. The set of admissable velocities forms a hyperbolic space. There have been various attempts in the past to use hyperbolic geometry to study special relativity but these have been unpopular due to lack of new results and hampered by the non-associativity, however gyrovectors changes all this. It "repairs" the situation with gyroassociativity and gyrocommutativity.

A book review is given here: http://www.springerlink.com/content/h7u2nq7mp8apunx9/fulltext.pdf Delaszk (talk) 21:04, 21 November 2008 (UTC)

Concerning gyrovectors someone might like to ponder the following problem:

It is known that mathematically $\mathbf{u} \oplus (\mathbf{v} \oplus \mathbf{w})$ is not identically equal to $(\mathbf{u} \oplus \mathbf{v})\oplus \mathbf{w}$. So that at most one expression can be correct. Which one is it and why?

JFB80 (talk) 20:46, 27 October 2010 (UTC)

This kind of pondering is not really on topic on a article talk page and should really be put forward at, for instance, wp:Reference desk/Science, but anyway...

I don't think that "correctness" is a relevant expression property here, but surely, when u is combined with the result of the combination of v and w, then the former is "correct", whereas when the combination of u and v is combined with w then the latter is "correct". DVdm (talk) 21:00, 27 October 2010 (UTC)

Since there is talk about non-associativity is it not good to be clear? Your reply avoids the problem which is obviously this: if there are points A,B,C,D having relative velocities u,v,w, which (if either) of the two formulae expresses the velocity of D relative to A? At most one can, so maybe the talk about non-associativity is meaningless JFB80 (talk) 08:28, 28 October 2010 (UTC)
It's worse than that - since it isn't commutative either, then all the following expressions are different: u+(v+w), (v+w)+u, (w+v)+u, u+(w+v), (u+v)+w, w+(u+v), w+(v+u), (v+u)+w
and it's even worse than that still - if you add more points E,F,... with relative velocites x,y,... then there are even more combinations of expressions every time you add a new point. 89.241.236.121 (talk) 11:30, 28 October 2010 (UTC)
The noncommutativity and nonassociativity mean that you have to be careful to match the expression to the physical meaning. If the velocity of D from A's perspective is given by u+(v+w) then you must be careful to use that expression in your calculations and not accidentally use (u+v)+w. But is the velocity given by u+(v+w) ? 89.241.236.121 (talk) 11:38, 28 October 2010 (UTC)
I don't think we can currently say that any of these expressions has physical meaning. How do we know the velocity composition law matches reality? There haven't been any experiments done to test the validity of the velocity addition law except when the velocites are colinear. Ungar says stellar aberration shows that the addition law is not a composition law and instead the addition should be given by what he calls the co-addition to the usual addition law. The co-addition law is symmetric (commutative) but it is still nonassociative. 89.241.236.121 (talk) 11:57, 28 October 2010 (UTC)
(ec) Ah yes, I see what you mean now. Tricky, although of course the magnitudes of the vectors are equal. Some of this is related to the fact that even $\mathbf{v} \oplus \mathbf{u} \neq \mathbf{u} \oplus \mathbf{v}$, so we don' even know the "real" relative velocity between A and C. The question is mentioned and treated in the cited source here but the essential pages (7 and 8) seem to be left out from the preview. If anyone has full access to it, they can add a remark in the text. DVdm (talk) 12:14, 28 October 2010 (UTC)
The content of those pages also appear in the paper The Relativistic Composite Velocity Reciprocity Principle, however in later works Ungar came to the conclusion that $\oplus$, should be replaced with the co-addition which is denoted $\mathbf{u} \boxplus \mathbf{v}$, for which we do have $\mathbf{u}\boxplus \mathbf{v} = \mathbf{v} \boxplus \mathbf{u}$. Also the associativity problem of choosing where to put the brackets can be solved in the case of the co-addition by, instead of composing $\boxplus$, deriving a formula for $\mathbf{u} \boxplus_3 \mathbf{v} \boxplus_3 \mathbf{w}$ which is symmetric in all 3 velocities, so co-addition $\boxplus$ is denoted $\boxplus_2$ to distinguish it from $\boxplus_3$, and this can be extended to derive a formula for $\boxplus_n$ which appears under the name gyroparallelepiped law in the books 1, 2. 89.241.230.124 (talk) 15:31, 28 October 2010 (UTC)
That's nice. Thanks for the link. As far as I'm concerned, please feel free to amend the article. DVdm (talk) 15:39, 28 October 2010 (UTC)
Are you suggesting that from now on we should use this new idea instead of the familiar one due to Einstein? Do you consider it established mathematics? JFB80 (talk) 21:39, 13 November 2010 (UTC)
I'm not sure what you mean with "we should use", but the thing is properly sourced and it does solve a problem in an obviously established mathematical way, so I don't see why it should not be briefly mentioned, as it is now. DVdm (talk) 21:33, 13 November 2010 (UTC)

## Euclidean rotation through a purely imaginary angle

The section discussing rapidity is obscure. While it is true that some authors introduce hyperbolic functions by using the imaginary variable in the classical circular functions sine and cosine, this approach makes the subject difficult for novices. Infinite series or reference to the length of sides of a right triangle on y=x from a point (u, 1/u) are more direct. Certainly the idea of Euclidean rotation is incorrect here.

Some modifications to the section using rapidity have been made.Rgdboer (talk) 00:41, 13 January 2009 (UTC)
Rapidity is not an angle, it is a relativistic velocity (scaled by the factor c). It did originate with Minkowski's (and Sommerfeld's) interpretation of the Lorentz transformation as a Euclidean rotation through an imaginary angle but then it was reinterpreted by Varicak in hyperbolic geometry as a velocity.JFB80 (talk) 21:11, 17 October 2010 (UTC)

## Constant accelerations

Frequently for space travel considerations, it is useful to know what the speed (relative to a stationary observer) over time is of a spacecraft undergoing a constant acceleration (in its reference frame, for instance if it were accelerating at 1g to simulate Earth gravity). This can be derived (below) from the velocity addition formula, and is v(t)=c tanh(at/c) (for a constant acceleration a). Is this worth including in the article?

Derivation: The relativistic addition of velocities formula is

$s = \frac{v+u}{1+(uv/c^2)},$

which means that if you are already traveling at a speed of v relative to an outside observer, and accelerate at a rate of a for a time interval $\Delta t$, your new speed is

$v+\Delta v = \frac{v+a\Delta t}{1+(va\Delta t/c^2)},$

so

$\Delta v = \frac{v+a\Delta t}{1+(va\Delta t/c^2)}-v=\frac{v+a\Delta t}{1+(va\Delta t/c^2)}-\frac{v+(v^2/c^2)(a\Delta t)}{1+(va\Delta t/c^2)}=a\Delta t\frac{1-(v^2/c^2)}{1+(va\Delta t/c^2)}.$

So we have

$\frac{\Delta v}{\Delta t} =a \frac{1-(v^2/c^2)}{1+(va\Delta t/c^2)}.$

Letting $\Delta t \to 0$, we obtain

$\frac{dv}{dt}=a(1-v^2/c^2)$.

This is a first order separable differential equation, so we can solve:

$\frac{c^2}{c^2-v^2}\, dv=a dt$
$\int \frac{c^2}{c^2-v^2}\, dv=\int a dt$
$c \tanh^{-1}(v/c)=at+C\,$
$v(t)=c\tanh(at/c)\,.$

skeptical scientist (talk) 02:48, 8 August 2009 (UTC)

Yes, this is a nice application of the composition formula. (See also my version - it is essentially the same). But unless we find an acceptable source for this, we couldn't really add it here.
By the way, I added one step and did some reformatting - hope you don't mind...
DVdm (talk) 20:03, 8 August 2009 (UTC)
A similar result was shown in the book: S.J.Prokhovnik The Logic of Special Relativity Cambridge U.P. 1967, Appendix 3, equation (3.8.6). More recently I have generalized it. JFB80 (talk) 20:01, 9 October 2010 (UTC)
A further comment: this problem goes right back to the beginning of the special theory with experiments of Kaufmann el al on the motion of an electron under constant acceleration in an electric field. These experiments first gave clear confirmation of the special theory (Bucherer Phys.Z 1908) So there is plenty of background if the topic is to be introduced into the article. The problem was discussed in Minkowski Raum und Zeit 1908 and Max Born Ann d Physik 1909 who gave the motion the name 'hyperbolische bewegung'. It would also fit in well with introductory comments on Galileo because it was he who first clarified the idea of constant acceleration with experiments on balls rolling down inclined planes. JFB80 (talk) 09:13, 11 October 2010 (UTC)
Yes, as I said it is a nice application of the formula. We could introduce the derivation in the article provided (1) the derivation can be found in these sources, and (2) other contributes agree that it is sufficiently important to be included. DVdm (talk) 15:32, 11 October 2010 (UTC)

## W. Kantor's 1972 paper on non-parallel velocities

I added the sentence "Experimental results have been reported that conflict with the formula's prediction for non-parallel velocities" about this paper: Nonparallel convention of light but it was reverted. Are you so sure that the paper is crackpot ? Charvest (talk) 19:46, 25 September 2009 (UTC)

Yes. That journal has published papers by Jack Sarfatti. Martin Hogbin (talk) 19:50, 25 September 2009 (UTC)

I haven't heard of him, but crackpot by association doesn't seem to be a watertight argument. Charvest (talk) 19:54, 25 September 2009 (UTC)
In this case it does. If I were you, I'd stay away from it afap :-) - DVdm (talk) 20:00, 25 September 2009 (UTC)
I've reverted my edits to other articles that included this ref. Upon closer reading, the paper even says that experiments contradict the co-linear case of the formula as well. Charvest (talk)

## Sommerfeld's paper: On the Composition of Velocities in the Theory of Relativity

I've removed the source because the accompanying commentary was wrong. The source does not say velocities are not cartesian. The source says "it apparently better corresponds to the meaning of the theory of relativity to calculate and (by consideration of the reality relations) to construct with rotation angles, instead of only using its tangents, the velocities." 89.241.225.124 (talk) 08:55, 12 December 2010 (UTC)

You were very quick to remove it. You should have thought about it a little because you are quite wrong. If you look at fig.2 you will see that Sommerfeld is combining arcs on the surface of a sphere. fig.1 is the Cartesian (Euclidean) figure. Just after fig.2 Sommerfeld said:
In summarizing we can say: For the composition of velocities in the theory of relativity, not the formulas of the plane, but the formulas of the spherical trigonometry (with imaginary sides) are valid. By this remark the complicated transformation calculus becomes dispensable, and can be replaced by a lucid construction on a sphere. Is not this clear enough? Please replyJFB80 (talk) 17:03, 12 December 2010 (UTC)
The adjective "cartesian" describes coordinate systems. It does not describe objects. For example complex numbers can be given in either cartesian or polar form. The word "cartesian" does not refer to a plane vs a sphere: Even a sphere can be described with cartesian coordinates. The equation of a sphere of unit radius at the origin in a cartesian coordinate system is x^2 + y^2 + z^2 = 1, but the equation of the same sphere in a spherical coordinate system is just r=1.
Fig 2 in 1 does combine arcs on a sphere but the arcs don't represent velocities, they represent angles. The source says "If we namely compose (Fig. 2) the rotation angles $\varphi_{1},\varphi_{2}$ as arcs on a unit sphere". Angles used in that paper are related to velocities by the equation near the beginning of the paper $\cos\varphi=\frac{1}{\sqrt{1-\beta^{2}}}$ where β=v/c.
The phrase "not the formulas of the plane, but the formulas of the spherical trigonometry (with imaginary sides) are valid." is pure rhetoric. Things can be represented all sorts of ways and as long as they give the right answers then no representation is more or less valid than any other. For example equations with trig functions can be rewritten with hyperbolic functions using the identities $\cosh x = \cos ix \,$ and $\sinh x = -i \sin ix \,$.
The set of admissable velocities is described by the cartesian coordinates { v=(x,y,z) such that |v|<c }. The addition of velocities is of course not the addition of R^3 but the velocity-addition formula. 89.241.239.164 (talk) 19:08, 12 December 2010 (UTC)
(1) Your remark about the use of 'Cartesian' is a quibble. Read 'Euclidean' for 'Cartesian' if you like. What are ordinary vectors, Euclidean or Cartesian? Anyone who wants to understand, can understand. (2) Spherical arcs correspond to Minkowski imaginary angles which then define rapidities. (3) Sommerfeld was an outstanding authority on special relativity. He did not indulge in rhetoric because he knew what he was talking about - refer to p.14 of S Walter's article [[1]] In my opinion Sommerfeld's paper was a fundamental contribution which, already 100 years ago, indicated a correct scientific solution to velocity composition currently discussed by the theory of gyro-vectors which is invading Wiki articles and monopolizing discussion. I would like to bring to notice the existence of an alternative - a good one. Can you object to that?JFB80 (talk) 17:35, 13 December 2010 (UTC)
You added to the article the Sommerfeld ref, and said "An alternative way to explain the effect known as gyro-rotation" with the note: "This approach recognizes that velocities are not Cartesian." The sentence begins with "This approach recognizes" which implies that what is being recognized is not recognized by the other approach. You say "Read 'Euclidean' for 'Cartesian' if you like.". In that case the sentence would become "This approach recognizes that velocities are not Euclidean." implying that the other approach (gyrovectors) treats velocities as Euclidean vectors, which is not true, gyrovectors don't treat velocities as Euclidean vectors. Perhaps you meant to say "This approach recognizes that velocities are not ordinary vectors." but again nobody is saying that they are ordinary vectors, so it is incorrect to write "This approach recognizes" as if the other approach doesn't. 89.241.237.134 (talk) 12:07, 14 December 2010 (UTC)
The gyrovector theory started off from Einstein's 1905 derivation generalized to 3 dimensions as described earlier in this velocity-addition article. The operation v(+)u is the velocity observed in a frame S of a point moving with velocity u in another frame S' which is moving relative to S with velocity v. So initially at least what you said (gyrovectors don't treat velocities as Euclidean vectors) was not true. Subsequently Ungar made an algebra out of this operation introducing the gyr angle and it became more and more abstract. But it is unclear whether these abstractions have any significant application other than the initial one for which ordinary mathematics should suffice (and is necessary anyway because it is difficult to locate and understand proofs of the gyro-formulae - for example, where is the formula for calculating gyr?). Hyperbolic velocities make their appearance via unproved statements about the Beltrami representation which in the article is not clearly related to the initial abstract definition of gyrovector.JFB80 (talk) 18:42, 15 December 2010 (UTC)JFB80 (talk) 07:05, 16 December 2010 (UTC)
The addition of Euclidean vectors is commutative. Relativistic velocity addition is not commutative in the 2d case discussed by Einstein, therefore it is nonsense to suggest the gyrovector approach treats velocities as Euclidean vectors. 89.241.234.151 (talk) 07:43, 16 December 2010 (UTC)
Please read more carefully - I said initially they were Euclidean vectors. In fact in his 1989 paper Ungar defined his variables u, v as belonging to R^3, the set of all 3-vectors in the Euclidean 3-space R^3 with magnitude smaller than the speed of light c in empty space. And in the 2 dimensional case, was'nt Einstein using ordinary Euclidean vectors? Who is talking nonsense? Reference: A.A.Ungar The relativistic velocity composition paradox and the Thomas rotation. Foundations of Physics vol.19, no.11, 1989 p.1386JFB80 (talk) 16:27, 16 December 2010 (UTC)JFB80 (talk) 16:35, 16 December 2010 (UTC)
Well the definition of a vector space (or gyrovector space) includes not just the set but also the addition operation. The set of relativistic velocities is a subset of euclidean vectors but they have different addition operations therefore they are not the same.
You make various remarks about unproved statements but Wikipedia is an encyclopedia not a textbook so proofs are usually not going to be included. There are plenty things the gyrovector space article could explain more fully but Wikipedia articles are works in progress. gyr[u,v] is a spatial rotation and its effect on a vector w is a new vector denoted by gyr[u,v]w the formula for which is given in this article. As a rotation gyr[u,v] can be represented as a 3x3 matrix, which can be calculated by finding the angle and plane between w and gyr[u,v]w and writing a rotation matrix for that angle and plane. 89.241.228.234 (talk) 18:21, 16 December 2010 (UTC)
Naturally complete proofs cannot be included in the Wiki articles (although references would be useful). But the gyro-speak would be clearer if more related to the mathematics we know. JFB80 (talk) 22:10, 18 December 2010 (UTC)

## velocity addition formula with velocity expressed as gamma

the velocity addition formula with velocity expressed as gamma is:

z = x y + sqrt(x^2 - 1) sqrt(y^2 - 1)
Just granpa (talk) 17:35, 14 December 2010 (UTC)

## Reverted edits by 96.237.182.21

I've reverted the changes made today. Note that this article writes the formula as v+u to match up with the rest of this article, but ungar writes the formula as u+v, therefore the formula as it appears in this article swaps u and v wherever they appear in Ungar. Also note that the coordinates of v were already in the next formula, but they were factored, i.e. v+(stuff)v = { 1 + (stuff) }v. 92.41.6.146 (talk) 20:26, 20 June 2011 (UTC)

## Small typo in 'Special theory of relativity' section

When it talks of the hyperbolic tangents it says

$s = {v+u \over 1+(vu/c^2)} .$

is the same as

$\tanh(\alpha + \beta) = {\tanh(\alpha) + \tanh(\beta) \over 1+ \tanh(\alpha) \tanh(\beta) }$

where

${v\over c} = \tanh(\alpha) \,\,\,\, {u \over c}=\tanh(\beta) \,\,\,\,\, {s\over c}=\tanh(\alpha +\beta) ,$

But unless I am missing something very obvious, the last term should just be $s =\tanh(\alpha +\beta)$. Washyleopard (talk) 15:17, 8 August 2012 (UTC)

Looks ok to me. - DVdm (talk) 15:35, 8 August 2012 (UTC)

Nevermind, I see it now. Forgot about the c's in the numerator. As I thought, obvious mistake on my part.Washyleopard (talk) 16:22, 8 August 2012 (UTC)

## Typo in the final expression for orthogonal velocities

The expression $S = \sqrt{V^2 + U^2 - V^2 U^2}$ should be as follows $S = \sqrt{V^2 + U^2 - V^2 U^2 / c^2}$

You can easily check it out. If inertial reference frame K' is moving with the speed V in the positive x direction relative to the frame K then

$v_x = {v'_x + V \over 1 + v'_x V / c^2 } \ , \quad v_y = {v'_y \sqrt{1 - V^2 / c^2} \over 1 + v'_x V / c^2 }$

For the case of orthogonal velocities:

$v'_x = 0 \ , \quad v'_y = U \ , \quad S = \sqrt{v_x^2 + v_y^2} = \sqrt{V^2 + U^2 (1 - V^2 / c^2)} = \sqrt{V^2 + U^2 - V^2 U^2 / c^2}$

MrPlough (talk) 08:36, 30 June 2013 (UTC)

In that context "units where c=1" are used. I.o.w. V=v/c and U=u/c. See the opening remark in the next subsection where "engineering units" are used. -DVdm (talk) 11:14, 1 July 2013 (UTC)