You said it better anyway. I tend to go into excessive detail sometimes. Please Register or Log in to view the hidden image!
It doesn't satisfy VA. According to VA D is only going 0.732c relative to A. While B is going 0.866c. The hint should be that there is no physical justification to treat them differently as I have shown. That is B makes the trip if C and D are there or not. D's velocity change does not have any relationship to A, B or for that matter even C. The changes in velocity are basic physics and are identical, yet SRT claims different terminal velocities wrt A. B = 0.866c D = 0.732c Had I claimed C was a shuttle craft moving relative to the mother craft "A" and had fired a missle "D" you would claim 0.732c velocity between the missle and the mothership. Yet it can be seen that B made the same trip using the same energy and flew side by side with D and yet is claimed to be going 0.866c.
No Mac, they are both at .866c wrt A. You don't seem to grasp that ship C expended extra fuel boosting D up to .5c wrt A invalidating your argument. Assuming perfectly efficient engines exactly as much as B needed to boost to .5c wrt A.
talking about the fuel does not matter, the bottom line is that you transform it when you don't need to. you define all of the velocity changes WRT A, so there is no need for: which is where your problem lies.
True, but I suspect fuel consumption and the like are at the heart of Mac's conceptual troubles. I don't like to speak for someone else though.
Incorrect. http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html A shuttlecraft moving at 0.5c relative to its mothercraft, fires a missle at 0.366c from the shuttle. SRT claims the missle is going 0.732c relative to the mothership. You stand corrected. Both "B" and "D" (according to SRT) are not going 0.866c relative to A.
It is not I that have problems here. Look again. Explain just where you see any differance in my A, C, D case vs the referanced link case. According to the velocity addition formula D is moving 0.732c wrt A, not 0.866c.
no, you are wrong, you define all of your velocity changes WRT A. if you defined the second velocity change (of .366c) WRT the craft that were already moving WRT A at .5c, then your argument would hold water. however, you would be adding that .366c to 0c and then it would not be co-moving WRT them. I am sorry, but you have fallen into the same trap I have seen many others here fall in, you mix up your frames.
I think this is where you problem is. if D accelerates to a speed 0.366c faster than B and C, then it will not be going .866c WRT A. if it goes 0.366c faster WRT A, then it will going 0.866c in your argument you define it as going .366c faster WRT A, which would mean it does not need to be transformed, because all of you other speed measurements are WRT A also.
Yes, but that isn't the original problem. Originally, B and D accelerate from .5c wrt A to .866c wrt A. This is not the same as accelerating to .366c wrt C and gives a different answer as it is different scenario. The entire point of the addition formula is that relativistic velocities don't add linearly. Vab != Vac + Vcb >>edit Cato, I know I'm just restating but I thought it might be a little clearer.
Well, I'm not going to waste much more time on this. If you merely look at the link I posted you can sse you are absolutely in error. A is the resting observer. C is the moving observer and D is the missle. The velocity wrt A is (according to SRT) 0.732c. Now if you don't know how and where to apply Velocity Addition, I hardly think you are in a position of saying somebodyelse is wrong. You seem totally confused by this simple example where B and D have co-moved the entire trip but have different velocities per SRT.
No you missed the point entirely. I stipulated that either A or C as referance still means you must claim veloicity wrt A OR delta velocity wrt CD to be different when in fact B and D remained side by side co-moving the entire trip. I haven't missed anything. I gave both views. Now B and D hve co-moved the entire trip. Explain how they have different terminal velocities wrt A.
This is almost laughable. You just repeat what I know and have demonstrated but say that is what the formula is for. I know what the formula is for. The issue is you must justify how when B and D have co-moved the entire trip wrt A they can possibly have different terminal velocities.
the problem you did and the problem you linked to are not the same. you never change POVs so you never need to do a transform. you say B speeds to .5c WRT A, then .866c WRT A, then ask how fast it is going WRT A. the answer is obviously, as defined by you in the set up, .866c. let me ask you this, at what rate does B leave CD WRT CD? what rate does B leave CD WRT A?
No, you completely missed the point. When A measures Vab = .866c and Vac=.5c, C and B measure Vbc != (not equal) .366c . Specifying Vab wrt A (aka as measured by A)=.866c and Vac wrt A=.5c is not the same as when Vac wrt A= .5c and Vcb wrt C =.366c. A, B and C will agree that two distinct sets of final frame velocities are being descibed. They won't agree on what those velocities are but they able to calculate what each other measure with the velocity formula.
You are partially correct but off topic. B is not the same as the linked case but D is. It is the comparison of B to D that is my point. There is no physics differance and consequently it shows that the link (SR Theory) that fails because B and D are identical in all physics respects yet SRT treats them differently simply to preclude v > c.
You are almost there but not quite. B and D are identical in all physical details. The point I am making is the different results based on SRT. There is simply no justification to claim the different velocity between B and C and D and C. BC is 0.366. We agree on that but DC is only 0.166 yielding a net 0.732c between DA. Yet B and D have accelerated and consumed equal energy in a side by side change in velocity. SRT (Velocity Addition) is false. FYI: It has nothing to do with transformations between frames as you and Cato have implied. It is a direct calculation of compound velocity designed strictly to preclude v > c for the purpose of maintaining the merger of time and space into time-space. It violates all sorts of physics.
you are almost there, but not quite. no, we do not agree with that. I will say it again. you define all of your velocity changes WRT A's frame. YOU DO NOT NEED TO DO ANY MATH IF YOU DO NOT CHANGE POV's
Look I am only comparing POV's. During the inertial break B and C are at rest to each other and both are in co-moving equal motion wrt A. When B accelerates once again there is no bias or physics justification to declare the acceleration from C is any different than the acceleration from A. The velocity addition Formula only applies to D which launches from C, just as B launced from C and does not use the VAF. The physics are identical but you are treating the velocity achieved based on identical consumption of fuel, etc and a side by side change in velocity yet are claiming different terminal velocities. IT CANNOT BE JUSTIFIED BY ANY ACTUAL PHYSICS. IT IS MERELY AN AD HOC MATHEMATICAL CONTRIVANCE IN SRT TO KEEP TEPMPORAL SYNCHRONIZATION WHEN MERGING TIME AND SPACE INTO TIME-SPACE BY NOT ALLOWING V > C TO OCCUR. You must recall that we have taken this step by step. The AB trip results in a 0.866c relative velocity. You can write a scenario where BC are at rest and B accelerates using X fuel, X thrust in X time and you will claim rightfully that the relative velocity BC is 0.366c. That is the same fuel, thrust and time used when doing the DC trip. There is not differance. D is the absolute same as B in terms of physics. Yet it is only when you declare that D is launched from C which has a relative velocity wrt A that the VAF becomes employed and distorts the change in velocity of D when physically it can be seen that it in reality MUST be identical to B.
