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)

Contents

[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 v1 and v2 are collinear.

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

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