Special & General Relativity | Forum profile

Originally Posted by atyy I think the LHS is wrong, but the RHS is correct. Yes. But O does not conclude that the light strikes L' and R' at the same time. O concludes that the light strikes L' at t(t'(L')) and the light strikes R' at t(t'(R')). I would try t(t'(L')) = d/(2*λ(c+v)) and t(t'(R')) = d/(2*λ(c-v)) OK, I'm not sure about that, let me do some algebra...

Originally Posted by Doc Al Nope--Einstein makes no such assumption. As I've already stated, light does not 'belong' to any reference frame. What Einstein does assume is that all light moves at the same speed with respect to any reference frame. The equation I provided already takes into consideration the fact that the rod moves a distance vΔt track during the time it takes for the light to travel from one end to the...

Originally Posted by jmallett Einstein went forward with the assumption that light is relative to the frame of reference in which it exists, i.e. the train's light is not the track's light. Nope--Einstein makes no such assumption. As I've already stated, light does not 'belong' to any reference frame. What Einstein does assume is that all light moves at the same speed with respect to any reference frame. Certainly in...

Originally Posted by cfrogue OK, now your diagram has the center of the light cone at the origin of O and also at O', is this correct. Yes, you can see that the flash (the intersection of the yellow lines) is at the intersection of the x=0 and t=0 lines, and also at the intersection of the x'=0 and t'=0 lines. The apex of the light cone is therefore at the origin of both frames.

Originally Posted by atyy That's ok. You should take Matheinste seriously (he's very good), but we can have some fun anyway. You're too kind. You have obviously missed my many poor postings. Please, please please do not let me spoil your fun. For all my protests I am enjoying it too. Matheinste.

Originally Posted by cfrogue I did not say twin1 will elapse t/gamma in its proper time. If I did I meant twin1 will elapse t/gamma in the time of twin2. So you don't have a method to calculate the proper time of twin1 in his relative inertial motion phase, in such a way that you can compare his total aging to twin2's? Wasn't comparing their total aging the whole point of what you were asking in post 49 , which got this entire...

Originally Posted by cfrogue No..... Yup, that's wrong. Editing .... it's almost right ... ok, it's right. Calculation A t(t'(R')) = d/(2*λ(c-v) =d.sqrt(1-v^2/c^2)/[2(c-v)] =d.sqrt((1-v/c).(1+v/c))/[2(c-v)] =d.sqrt(((c-v)/c).((c+v)/c))/[2(c-v)] =d.sqrt((c-v).(c+v))/[2c(c-v)] =(d/2c).sqrt((c+v)/(c-v)) Calculation B For O', t'(R')=d/2c and L'=d/2 The Lorentz transformation is...

Originally Posted by cfrogue The only problem is that these two origins do not remain coincident because of the relative motion of O'. What problem? In each frame the sphere is centered around this frame's origin, and NOT AROUND BOTH ORIGINS IN ONE FRAME . Therefore It doesn't matter if they remain coincident. Originally Posted by cfrogue Does this make sense? Your 'problems' here? No, they don't make sense to me.

Originally Posted by edpell Yes a test of Lorentz invariance done on an object with mass. It seem like the Michelson Morley experiment nicely shows invariance (OK an invariance of speed not of rest mass) for a mass less photon but I would like to see results for an object with mass (for invariance of rest mass or further that there is no preferred frame). The fine structure constant looks like a good way to see if electric charge varies with...

OK here is a pendulum: A gravity clock consists of two spherical masses one large of rest mass M and one small of rest mass m. The smaller mass is suspended by a rigid frame, of negligible mass, at a height R above the center of the large mass. It is the bottom of a pendulum arm, of negligible mass, of length L. When displaced the pendulum has the period Given two observers A who will travel with the gravity clock and B who will remain behind...