Awesome. So I'll write out my final calculations. To correct for the cosmetic cancellation of 2, as you said, I would end up with: $$v = \frac {1 - \sqrt {1- \frac {u_2^2}{c^2}}}{ \frac {u_2}{c^2}}$$
Then, we get:
$$v = \frac {c^2 - \sqrt {c^4 - {c^2}{u_2^2}}}{u_2}$$
However, I would not know...
Thank you! I would say the solution with the minus would be correct as otherwise we would get a value over the speed of light which is not possible. Is this correct?
Thanks a lot!
The velocity of particle 1 in the primed frame will be ##u'_1 = \frac {u_1 - v} {1 - \frac {u_1*v}{c^2}}##
##u_1 = 0## as it is at rest in the regular frame. This way, the formula will be: $$u'_1 = \frac {0 - v} {1 - \frac {0*v} {c^2}}$$
This will end up in ##u'_1 = \frac {-v}{1} =...
That is my whole concern. The answer I get is non-relativistic but as we're doing special relativity right now, I suppose the calculation has to be relativistic as shown by my calculations. Therefore, I hope you can help me with trying to figure out where I went wrong.
Yes, it is in the very last line. v = 1/2 U. Plugging this back into the formula's with u approaching c would give me the velocity of particle 2 to be c and the velocity of particle 1 still to be -1/2 c, therefore no equal magnitude.
Once again, thank you for your swift response
Thank you for your response.
"v" indeed is not the velocity of particle 2 in the primed frame. (u-v)/(1-(uv)/c^2) however is. The velocity of the particle in the primed frame should be equal but opposite to the velocity of particle 1 in the primed frame. The velocity of particle 1 in the primed...
Homework Statement
In frame S particle 1 is at rest and particle 2 is moving to the right with velocity u. Now consider a frame S 0 which, relative to S, is moving to the right with velocity v. Determine the value of v such that the two particles appear in S' to be approaching each other with...