Recent content by dragan

yes, precisely :) btw, we already have the solution, but the problem is interesting anyway. Most people usually say at first that it is a "standard" problem and it is enough to interchange the velocities (up to a minus sign) in the transformation formula.

Dear Tim, I think your answer is right. I got something a little different, and our formulas are probably equivalent (my formula also reduces to right answer in the special cases that you have mentioned). The only problem is how to use the formula when the given velocities are superluminal...

Dear George, you have written the same answer again (with an extra minus sign?) and I am sure again that it is wrong again. This is, however, the first answer that comes to one's mind. The right formula should be symmetric in respect to interchange of the arguments and the given formula is not...

kev, I offered the prize for a limited time, please check it up :) I already found a good answer for V, but it demanded going through SL(2,C) group. I also managed to revert the equation in a special case when one of the velocities has infinite magnitude. Both sulutions agree and I am happy :)...

Yes, precisely. I choose this function to be identity, so I define the problem in such a way, that O and O' use the same basis. Therefore the dot product makes perfect sense. Maybe I should have clarified this. Please note that the formula I shown is just a vector form of the usual formulas...

Well, here is a little proof :) From the definition: v' = f(V,v) (1) v = f(-V,v') (2) Function f must also have the property that changing the sign of its arguments changes its sign: f(-x,-y) = -f(x,y) (3) Now, let us assume that what you said is right: V = f(v',v) (4)...

George, Let us denote v' = f(V,v). From your suggestion it follows that the function f is symmetric and this is not true. This is obviously true for Galilean boost, but not for Lorentz transformation. The easiest way to find out that the transformation cannot be symmetric is choosing V...

jcds, Of course this is an issue, but it can be sorted out easily. All the velocities are defined in two inertial frames. V and v are defined in 3D vector space of observer O and v' is defined in 3D vector space of the observer O'. We have freedom of choice of both bases, so we simply choose...

Kev, I have been teaching Relativity for several years, the formula is part of a textbook I've written on the topic and no student has ever complained :)) Here is the formula in tex. c=1 for simplicity: v\prime = \frac{\sqrt{1-V^2}\left(v-\frac{v\cdot V}{V\cdot V}V\right)-V+\frac{v\cdot...

jcsd: All vectors v, v' and V always lie in one plane (you can show it by taking a vector product v'xV). Therefore it is reasonable to look for V in the form V = av + bv'. At least that is the way I tried :) I don't have the answer, but I am close, I believe. Btw, I was actually looking for a...

I would say that you just don't see physical implications. Of course they are, only I did not discuss them. Solution to the problem I posted is a small piece of a very general question which, is from the physical point of view very interesting. I just asked the question I couldn't answer...

George, Of course you are right in all that you said. This is exactly the reason for Thomas precession. The good news is that I think I already know how to determine V, but the calculations are surprisingly painful. What is also so funny, that the problem seems usually "obvious" to...

Ok, guys :) Kev, what you have shown is the first thing that comes to one's mind, but the formulas are not right. To find out, just plug them into the equation that I have given. Your formula are not solution of this equation. They only apply to v and v', not V. The reason they are wrong...

Thanks for the formula. There is no derivation given, but it is quite simple: one considers velocity four-vectors of B and C, calculates the product, which is invariant. Then one calculates the same product in an inertial frame attributed to B and equals the two. The formula for w^2 follows...

Yes, of course there is no single solution. And for some cases there is no solution at all. But the formula for velocity transform "works" not only for M being material particles or photons, but anything at all. It could be also superluminal phase velocity of a wave and the velocity...