My pleasure. Lets take this one step at a time. Given two clocks A and B at inertial rest to each other. Clock "B" is accelerated until it reaches 0.5c and then coasts for 5 minutes according to its clock. That would be 5.7735 minutes according to A's clock because gamma equals 1.1547. Agree? Yes/No. After this time period "B" now accelerates once more until it has achieved a relative velocity of 0.866c and then goes inertial. From this description I would think you have no arguement with respect to "B" having a velocity of 0.866c relative to A. Yes/No? Once we agree here I'll go to the next step.
OK so far. No. In the final configuration of the scenario you have the A frame, the C frame and the BD frame. v = Vac, u =Vbc=Vdc . So A views B and D as having the same velocity.
Then I'll skip your following response and continue my presentation. Now that you agree B has a velocity of 0.866c relative to A, lets move forward. Going back to the beginning we now stipulate that simultaneous with the launching of B that another craft (C) is launched simultaneous with B from A and remains perfectly side by side in a comoving direction. I would assume therefore that you agree they are also side by side during the 5 minute inertial break in the trip. And that they are a relative rest to each other. Yes/No?
You almost got away with that, MacM, but nobody was fooled, really. You can't back up your velocity addition claims, can you?
Let's not reuse C. We'll only confuse the issue. So we launch a ship E from A so that it is comoving (at rest wrt C,B and D)? In other words at rest wrt the C frame?
You are pathetic James. I am in the middle of walking you and others through that very fact. Now back off. Stuff you BS.
We have not gotten to D yet. Do you agree B and C are at relative rest co-moving at 0.5c relative to A during the 5 minute inertial interval? Yes/No?
C has no connection with B. B is the same craft as in the first scenario. It independantly accelerated in two steps to 0.866c relative to A. Now indeed C has aboard it another craft "D" which is identical to "B". At the end of the inertial period do you agree that for B to accelerate it is equally valid to compute X fuel, X thrust and X time to achieve 0.866c relative to A regardless if you view it as re-accelerating from A or just accelerating from B? Yes/No?
Great. So we are in agreement that B accelerating from its momentary inertial period relative to A, using X fuel, X thrust in X time achieves 0.866c relative to A. That is a change in velocity of 0.366c from the 0.5c during the inertial period. Now since B and C & D are at relative rest, "D" accelerates simultaneous with B away from C. D being an identical craft to B also uses X fuel, creating X thrust and runs side by side with B for X time until B reaches 0.866c relative to A. We agree A has a velocity of 0.866c relative to A. Do we agree that physics are the same in every frame? Yes/No.
I forgot for a moment that B was a separate craft. So after the "intermission" B accelerates to its final inertial frame, .866c wrt A. If you mean that B expends some fuel to accelerate to C and then some more to reach .866c or alternately one large burn from A to .866c, then yes. Don't ask me to compute the energy, thrust or time. They are frame dependant and inverse hyperbolic trig functions give me a headache. Please Register or Log in to view the hidden image!
No B and C are seperate crafts and each have accelerated to 0.5c and went inertial simultaneously. Agreed actual figures are not necessary. Please Register or Log in to view the hidden image! Now since B and D are identical and since it really doesn't matter if you are looking at A or C as the initial referance, D and B are identical in all respects. That is D MUST have a velocity of 0.866c relative to A; whereas according to SRT. C = v = 0.5c, D = u = 0.366c, w = v + u / (1 + vu/c^2) w = 0.732c not 0.866c. One must either deny ultimate velocity is equal for consumption of equally applied energy or that the differential velocity between B an C vs D and C is not the same even though they have been side by side the entire trip from A.? Care to justify that? Please Register or Log in to view the hidden image!
So B and D which are now both traveling at .5c wrt A and are now (during the intermission) identical (mass fuel etc. B having burned off it's extra fuel) fire their engines for identical burns and accelerate together to .866c. They are now in a single frame (which I called BD), moving at a velocity wrt to C (the rest frame of the intermission) which satisfies the relativistic velocity addition. As I said v = V c wrt a, u = Vb wrt c =Vd wrt c and Vb wrt A = Vd wrt A . What's the problem?
I found your problem, you are already measuring the velocity from A's frame, you don't need to transform it anymore. it moves from 0c, to .5c, to .866c all WRT A, there is no more math needed. the only way you would need more math would be if you measured some interval WRT some other frame than A. QED
I think I just said that. Please Register or Log in to view the hidden image! >>edit At least I think I did. :m: