MacM's Claims

Discussion in 'Physics & Math' started by Rosnet, Aug 5, 2005.

  1. MacM Registered Senior Member

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    10,104
    Speak english. You are making no sense what-so-ever.
     
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  3. MacM Registered Senior Member

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    10,104
    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.
     
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  5. kevinalm Registered Senior Member

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    993
    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.
     
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  7. MacM Registered Senior Member

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    10,104
    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?
     
  8. James R Just this guy, you know? Staff Member

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    39,397
    You almost got away with that, MacM, but nobody was fooled, really. You can't back up your velocity addition claims, can you?
     
  9. kevinalm Registered Senior Member

    Messages:
    993
    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?
     
  10. MacM Registered Senior Member

    Messages:
    10,104
    You are pathetic James. I am in the middle of walking you and others through that very fact. Now back off. Stuff you BS.
     
  11. MacM Registered Senior Member

    Messages:
    10,104
    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?
     
  12. kevinalm Registered Senior Member

    Messages:
    993
    During the time before C launches B and D they are all moving at 0.5 c wrt A, yes.
     
  13. MacM Registered Senior Member

    Messages:
    10,104
    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?
     
  14. cato less hate, more science Registered Senior Member

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    2,959
    I think everything is fine so far.
     
  15. MacM Registered Senior Member

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    10,104
    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.
     
  16. kevinalm Registered Senior Member

    Messages:
    993
    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.

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  17. cato less hate, more science Registered Senior Member

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    no, typo, A should be at rest WRT A right? do you mean "D" and "B" move at .866c WRT A.
     
  18. MacM Registered Senior Member

    Messages:
    10,104
    No B and C are seperate crafts and each have accelerated to 0.5c and went inertial simultaneously.

    Agreed actual figures are not necessary.

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    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?

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  19. MacM Registered Senior Member

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    10,104

    Clarify?
     
  20. kevinalm Registered Senior Member

    Messages:
    993
    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?
     
    Last edited: Aug 25, 2005
  21. cato less hate, more science Registered Senior Member

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    2,959
    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
     
  22. kevinalm Registered Senior Member

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    993
    I think I just said that.

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    >>edit At least I think I did. :m:
     
  23. cato less hate, more science Registered Senior Member

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    2,959
    yeah, I must have been typing my message when you posted.
     

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