Pure SR, Relativistic Mass, and its Gravity?

Discussion in 'Physics & Math' started by Neddy Bate, Jun 5, 2018.

  1. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,940
    E doesn't refer to proper energy, that's one mistake you make. The proper energy, neglecting independent atomic motions in each rocket, is always \(mc^2\). Secondly, there's no reason to assume conservation of x-momentum in the x-moving rocket, because there's an equal and opposite force decreasing the relativistic mass and therefore x-momentum of whatever is causing it to accelerate.

    In order to assume constant acceleration, you automatically have to assume there's some room on the velocity y-axis to accelerate before light speed is actually reached.
     
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  3. Q-reeus Valued Senior Member

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    3,417
    I was in error there and indeed E is the total energy γmc² (γ here the net Lorentz boost wrt rest frame of m), and is inclusive of any 'atomic motions' i.e. thermal energy.
    I take back my take back in #100 - that in #93 re conserved x-component momentum is actually correct.
    The correct relativistic 3-momentum component transforms are as for boxed set of 4 equations here:
    https://hepweb.ucsd.edu/ph110b/110b_notes/node54.html
    And invariance of transverse momentum components shown there demands a decrease in x-component transverse velocity, re current scenario, to compensate.
    That confusing piece makes no sense to me. There are no x-component forces acting. Conservation of x-momentum therefore holds, irrespective of y velocity, notwithstanding my erroneous statement to the contrary in #100.
    That is a weak argument easily refuted. Suppose we start with equal and opposite x-component velocities, as seen in inertial frame K, of magnitude 0.8c. Now have y-component velocities build up till y-component momentum exceeds the x-components in magnitude. Nothing prevents that in principle. Clearly we must have magnitude of y velocities exceeding that of x component velocities seen in K.
    Try fitting (0.8c)² + (v_y)² < c², with |v_y| > |v_x = 0.8c|. NO CAN DO!
     
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  5. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    Something is pushing the rocket and making it accelerate. If the rocket acquires some additional momentum along the x-axis as a result, then a corresponding opposite momentum is transferred to whatever pushed it, or equivalently the source of the push loses some of its momentum along the x-axis equal to whatever the rocket gains. If rocket exhaust is creating the thrust, the exhaust has momentum along x that must be factored in, and it carries mass away from the rocket which also compensates for momentum changes. The x-momentum of the rocket on its own, neglecting exhaust and external forces, is not conserved even though its x-velocity is. Including exhaust and external forces, obviously conservation applies.

    For a guy who claims to know Relativity so much more than others, you reach some truly utterly absurd conclusions about uniform motion.

    Like I said, the problem as posed in the OP only makes sense if the y-velocity is limited to well under light speed. I thought it was obvious enough that I didn't need to explicitly state that the x-velocity must be similarly limited so that the Pythagorean sum doesn't exceed \(c\).
     
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  7. Q-reeus Valued Senior Member

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    Of course but as consistently specified throughout this thread, only along y axis!
    Irrelevancies. All that matters is we have specified some initial relative x-axis velocities. THEN we boost exclusively along y axis via rocket thrust. Exhaust plumes etc. are of no concern to this thought experiment that is perfectly entitled to neglect practical problems of rocket engineering. Leave those to NASA & Space-X etc.
    I never claimed to know 'so much more than others', just more than some here who need to learn. I made one or two mistakes myself, but still come out ahead of you. More on that next post.
     
  8. Q-reeus Valued Senior Member

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    3,417
    Cop-out. My simple example given in #102 refutes your claim x component velocities will be constant seen in K. Try and actually prove otherwise.
     
  9. Q-reeus Valued Senior Member

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    3,417
    Referring to: http://farside.ph.utexas.edu/teaching/em/lectures/node110.html
    Frame S' moves at x-axis velocity v wrt lab frame S, as (1360) there makes evident. A transverse particle velocity u2 in S transforms to u2' in S' according to (1362).
    Taking the case particle in S has zero x-axis velocity u1 = 0, we have the simple result u2' = u2/γ. This nicely illustrates that a boost along one axis implies a reduction of velocity along an orthogonal such axis.

    Translated into K frame scenarios here, it's inevitable that a boost along y axis, with zero force thus zero momentum change along x-axis, implies a reduction in x axis velocity. In fact by the factor 1/γ, where γ here has v in γ = 1/√(1-v²/c²) the instantaneous y axis velocity.
     
  10. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,940
    And that something which does the pushing, such as rocket fuel, is co-moving with the rocket and therefore has an x-momentum of its own which changes as its velocity changes from ignition. The total x-momentum is still conserved.

    If you want to talk about momentum conservation along x, you need to factor in the relativistic mass carried off by the rocket exhaust. Basic centuries-old physics here, equal and opposite reactions- that doesn't change with Relativity.

    This whole argument is utterly irrelevant in light of what I just discussed above. Your x-momentum conservation argument doesn't apply to the rocket without factoring in the momentum of whatever's pushing it. Go learn about basic rocket physics before you mouth off to others.

    You're claiming that if a stationary observer sees two objects, one of which is stationary along x and another having velocity v along x, and they both accelerate strictly along the y-axis at a constant rate as seen by that observer, somehow magically this same observer starts seeing the x-velocity changing even though there's no acceleration there. You're not wiggling out of this one, it's basic physics, you're in the wrong and you ought to apologize for throwing insults at me over it. You know what they say about keeping silent and letting everyone think you're an idiot, rather than opening your mouth...
     
  11. Neddy Bate Valued Senior Member

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    1,873
    Some inertial frame S can certainly launch a rocket straight up its own y axis.

    Some other inertial frame S' can certainly launch a rocket straight up its own y' axis.

    In the standard configuration, there is a constant velocity v in the x-direction between the two frames.

    Thus each frame measures the x-component of the other frame's rocket's consistently as either v or -v. There can be no doubt about that.
     
    Last edited: Jun 17, 2018
  12. Neddy Bate Valued Senior Member

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    1,873
    I agree with you here, Q-reeus.

    But your "x-components of velocity must decrease" idea cannot be right, as exemplified simply in my previous post (#108, just above this post).

    On the other hand, the x-component of the velocity of a treadmill belt inside one of the rockets would decrease, as you helpfully reminded me to take into consideration pages ago.
     
    Last edited: Jun 17, 2018
  13. Q-reeus Valued Senior Member

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    3,417
    Let's dispose of this useless distraction now. Evidently you never figured our attention has always been to use 'rocket' or 'spaceship' as simply a source of acceleration, thus accompanying velocity change along the acceleration axis. Only the 'payload' is being considered - your fool insistence on further muddying the waters by bringing in 'practicalities of rocket engineering' merely introduces needless complication. That in no way alters the essential physics Neddy and I are interested in.
    We might just as easily, for the sake of this 'does transverse velocity change' bit, have been using a charged particle coasting along x-axis, suddenly entering between the parallel plates of a charged capacitor, whose applied E field is along y-axis. Same basic physics. Basic physics you keep getting wrong.
    Yes the total x-momentum is conserved. As is the x-momentum of our payload - the 'capsule' or 'capsules' variously containing our test mass & scales and/or treadmill. What is of real interest.
    See above. Again - stop this continual muddying the waters.
    Wrong - see above.
    Yes, and given you have well and truly nailed your colours to the mast on this matter, it's you, not me, that will have to back down and apologize. And look the idiot.

    I gave a simple scenario, last para in #102, that plainly shows you must be wrong. Or you truly believe that x-axis velocity there will only alter once y-axis velocity hits some magic value - some mystical 'wall'?! He he. Relativity doesn't work like that. Logically, there must be reduction of x velocity for any finite y velocity owing to a strictly y directed boost along that y axis.
    And the velocity transforms used in #106 make that perfectly clear. Your best bet is to simply cease posting here. It's a tactic I've seen many times from many members when shown up as wrong. Your choice.
     
  14. Q-reeus Valued Senior Member

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    3,417
    Sorry Neddy but it is right. Please ponder my response above. Those relativistic velocity transforms I linked to in #106 are correct and, properly applied, demand x-component velocity reduction, given the strictly y-axis rocket boosts. One needs to invert the axes from the S/S' case there, to the K frame case here, because we are making a 'virtual' translation into a K' frame moving along y-axis wrt K. That transition from K to K' is really just a transition in time in K frame. At t = 0, K' frame coincides with K frame. At some later time, thanks to the y acting boost, K' now has relative velocity -v_y wrt the 'old' K.
    Same basic transforms apply, just a different specific. Ponder my example last para #102. There MUST be reduction in velocity along x-axis. Once that is seen and internalized, I think we are finally done here.
    [PS: Viewing the treadmill belt as slowing down due to ever-changing time dilation (as seen in K frame), as done early on this thread, is a complimentary not contradictory pov to what those transforms linked to in #106 reveal.]
     
    Last edited: Jun 18, 2018
  15. Neddy Bate Valued Senior Member

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    1,873
    Q-reeus,

    The inertial reference frames S and S' are not attached to the accelerating rockets themselves. The two inertial frames maintain a constant, uniform, relative velocity of v, at all times, by definition (SR). The inertial reference frames do not accelerate.

    If the rockets are launched straight up the y and y' axes respectively, then they are always located at X-COORDINATES x=0 and x'=0 at all times. The X-COMPONENT of their relative velocity can be calculated from that alone.

    You are way over-thinking this.

    Rearranging the basic Lorentz transform:
    x' = γ(x - vt)
    x'/γ = x - vt
    (x'/γ) + vt = x
    x = (x'/γ) + vt

    Substituting x'=0
    x = (x'/γ) + vt
    x = (0/γ) + vt
    x = vt
     
    Last edited: Jun 18, 2018
  16. Q-reeus Valued Senior Member

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    3,417
    Neddy, do you or do you not accept my example in #102 proving x-axis coasting velocity must decrease if there is a boost along an orthogonal axis e.g. y?
    Do you or do you not accept that the relativistic velocity transforms as used in #106 confirm precisely the above?

    If the answer is yes to both, what are you now saying?
    And btw your manipulation of Lorentz transforms used in #112 is referring to displacements, not velocities. Additional factors enter when differentiating to get velocities. That gamma factor has to be differentiated for starters (and gets still more involved when further differentiating to get accelerations). Which matters are all taken care of in those final expressions dealing with velocities in (1362):
    http://farside.ph.utexas.edu/teaching/em/lectures/node110.html

    Further, your last displacement result x = vt is trivial as it ends up back in the original frame.
     
  17. Neddy Bate Valued Senior Member

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    1,873
    Q-reeus,

    I am talking about SR, not anything else such as GR.

    The gamma factor is simply
    γ = 1 / √(1 - (v²/c²))
    Where v is the unchanging velocity between frame S and frame S'. Why would you differentiate? I can only guess you are attaching the reference frames to the accelerating rockets? The inertial frames do not accelerate, ever.

    The rocket launched straight up the y' axis of frame S' always (at all times) has the x' coordinate of x'=0
    The end result of my derivation, x=vt, tells you the x coordinate of that rocket, regardless of its position on any y or y' axis.
    The x-component of its velocity is simply v.
     
  18. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,940
    I'm glad to see you and I in agreement on this very basic issue, and your explanation of the resolution is more than adequate. Do you also now see how the situation of the two rockets isn't symmetric, that the picture would look different if the x-velocities were reversed by a reference frame shift?
     
  19. Q-reeus Valued Senior Member

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    3,417
    So am I.
    Because your displacement transform expressions are not velocity transforms. Those displacement expressions must be correctly differentiated to go from x's to dx/dt's. OK?
    That would have been the best thing to stick to all along. Those earlier results in #13 are easily arrived at by sticking with accelerated frames - yes. But you thought one could 'simplify' by going to an inertial frame(s). As I repeated over and over, it's more complex not simpler to do so. Hence we are stuck in this seemingly endless loop!
    Recall your enthusiastic picking up on pervect's use of 'momentarily co-located inertial observer'? You can't see the parallel here? Groan.
    Yes, trivially true in S', but:
    A: You have shifted back to making, in effect, K' frame now displaced along x-axis wrt K. That's a different situation entirely to my use of a boost along y-axis. OK?
    B: Your displacement result x = vt assumes a scheme where orthogonal accelerations are entirely absent. It has no bearing on what I have shown to be necessarily so in #102.
    Sorry, incorrect thinking. Again - please state whether you accept my simple example in #102. Yes or no?
    If only we had agreed to stick to using accelerated frames where lab and/or mass+scales on treadmill belt were stationary, this would be done and dusted back in page 1. Sigh.
     
  20. Neddy Bate Valued Senior Member

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    1,873
    You mean like this?

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    Yes, that is SR.

    You don't need to boost a reference frame just because a rocket is travelling straight up the y' axis. Is this new to you?

    If you were located at a NASA site where they were launching a rocket, would you start with a coordinate system attached to the rocket, and disregard any coordinate system attached to the ground?
     
  21. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    Anyone who doesn't understand that "no acceleration" on an axis means no velocity change on that axis... should not be talking about how they've got all the GR experts stumped.
     
  22. Q-reeus Valued Senior Member

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    3,417
    Once again - do you agree with my simple finding in #102? Yes or no? No further diversions please. Just answer that one question.
     
  23. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,940
    Post #102 refers to a source which shows that passengers in the rocket moving along x will agree with the inertial observer in the primed frame that their momentum along the x' axis is 0. That means they remain aligned at x'=0, which means their x-velocity remains constant. That has nothing to do with momentum conservation.
     

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