Talk:Velocity-addition formula

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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)

1 Problem of symmetry
2 Velocity units
3 ANSWER: Problem of symmetry
- 3.1 Answer for the Answer
4 cleanup tag or expert needed tag or what
5 Two successive dot products in the first equation?
6 Doppler Stuff is Not Very Physical
7 What about c-c?
8 Suggested change of name
9 too complicated

[edit] 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)

[edit] 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.

[edit] ANSWER: Problem of symmetry

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

[edit] Answer for the Answer

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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] What about c-c?

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)

[edit] 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)

[edit] 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

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