Recent content by yre

This discussion has made me think I am misinterpreting what Ungar says about coaddition. It's not that the magnitude |u#v| gives the speed of C which I think is given by |u@v|=|v@u|. It's that neither u@v or v@u give the direction. The role of u#v is that when comparing the trigonometric...

Yes. The full 3D velocty addition formula is about directed velocities. This thread has gotten side-tracked into talking about translations and parallel movements, but what this discussion is about is the composition of two boosts, i.e. Lorentz transformations, i.e. the origin of B and C did...

Ok, yes, I was thinking C would be moving away from A in a straight line from A's perspective in which case I'd say there would be a boost from object A to object C. However in 3+1 dimensions C could also be moving across A's line of sight, so to speak, which wouldn't be a boost.

I'm taking A, B and C to be moving objects at rest w.r.t. inertial frames whose origin is at A, B and C.

You mean A,B and C are all traveling in the same direction, not necessarily on the same straight line, but all parallel to each other then do we get (u + v)/(1 + uv/c^2)? I think so, I should have been using the phrase parallel rather than colinear. and in this case there is no rotation and so...

We can rewrite B(u)B(v) as B(u@v)T(u,v). but we could also write B(u)B(v) as T(u,v)B(v@u). so B(u)B(v)=T(u,v)B(v@u)=B(u@v)T(u,v). u@v is not equal to v@u so there isn't a unique velocity of the Lorentz transformation.

We should be consistent about what is considered to be the "actual relative velocity" of things relative to A, which means fixing the coordinate frame of A, not rotating it. If B is boosted from A by B(u) and we say the relative velocity of B from A is u, then if the boost of C from A is B(w)...

Page 9 eq(22) gives the formula for "coaddition" u#v in terms of @ Page 10 eq(29) gives u#v in symmetric form. Page 13 says "Einstein velocity coaddition ... does give rise to an exact "velocity gyroparallelogram" in hyperbolic geometry." Page 23 poses the question "whether, in the...

Actually not so exactly. u@v and v@u have the same magnitude as each other but I don't know about the magnitude of Ungar's formula vCA=u@T(u,-v)v. I don't know how he got that formula.

According to Ungar the velocity vCA with the correct direction is given by vCA=u@T(u,-v)v

Not quite. You get a velocity with the same magnitude as u@v, but u@v doesn't give the correct direction of C relative to A.

Exactly. Same magnitude. Different direction.

The point is that the velocity addition formula u@v does not give the relative velocity of C relative to A. It gives the parameter of the boost of C relative to A. Therefore it is not a velocity composition formula in the sense that you'd normally expect something called a velocity-addition...

Let me repeat my last reply but with a different symbol so as not to confuse it with ordinary addition. Let's use @ for the velocity addition formula. The Lorentz transformation B(u) B(v) where B(u) and B(v) are the boosts of velocities u and v, is the same as B(u@v)T(u,v) where T(u,v) is the...

You are still not getting what I wrote. I said "u+v means the full non-colinear velocity addition formula." I should have used a different symbol.