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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    5,762
    Why do you keep insisting on applying boosts to a static observer? The passengers in the rockets will see a change in \(v_x\), not the observer on the ground. Please note that Neddy and I are talking about invariant \(v_x\) in the case of a ground observer watching two rockets take off with one having a horizontal velocity at the start, we're not talking about the onboard belt here.
     
    Last edited: Jun 19, 2018
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  3. Q-reeus Valued Senior Member

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    3,052
    In addition to having painted yourself into a corner in insisting, over many postings here, that transverse velocities are unaffected by an orthogonal axis boost, as seen in an inertial frame, you fail miserably to understand that explained in #111. Too bad. Doing a desperate shifty and shifting to accelerated frames won't save you from embarrassment. All I have to do is point to your many posts which are referenced to Neddy's inertial K, K' frames.
    Now, let's have that 'momentum example' that will somehow 'prove' a boost along y-axis has no effect on x-axis speed.
    Well? Where is your magnificent momentum example?
     
    Last edited: Jun 19, 2018
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  5. Neddy Bate Valued Senior Member

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    Before replying I just want to reiterate that I am using a pair of pre-defined inertial frames which do not change throughout the experiment. I am not changing the inertial reference frames constantly in order to approximate the acclerating frame of the rocket, as I suspect you might be doing.

    With that said, you had previously agreed that the rocket ascending the y' axis remains at x'=0 regardless of its upward motion, because the rocket never veers off of the y' axis which is fixed at x'=0.

    So no 'transverse' upward motion even needs to be considered, since it does not change the x' coordinate of the rocket at any time. We can simply substitute x'=0 into the appropriate equation and obtain x=vt. This should be so surprise, as it is the very definition of the standard configuration in SR. The x-velocity of the y' axis as measured by the unprimed frame is defined to be v.

    I suppose you mean this part of #102:

    No, I do not agree that your link applies to a pair of pre-defined inertial frames which do not change throughout an experiment wherein a rocket ascends the y' axis of the primed frame, and the unprimed frame measures its x-component velocity to be v at all times.

    Look at it this way. The launch pad in the primed frame reports the launch is a success, because the rocket remains on its course, which is the y' axis, constantly located at x'=0. The observer driving by in their car must agree that the launch is a success and that the rocket remains on course, even though he finds the speed shown on his speedometer to be the lateral speed of the rocket. The Q-reeus explanation requires that the rocket launch is not a success, because it veers off course, but only in the car frame. Meanwhile the launch remains a success in the primed frame. Contradiction!
     
    Last edited: Jun 19, 2018
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  7. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    And here's where Q-reeus' attempt to apply momentum conservation goes wrong and leads to absurd conclusions:

    Suppose I stand on a train with horizontal speed \(v_x\) and throw a baseball with rest mass \(m\) straight up into the air (in my frame of reference). Q-reeus stands stationary on the ground watching. Before the toss, he assigns to the ball and train a gamma factor \(\gamma_1=\gamma(v_x)\), and immediately afterward when the initial y-velocity is added to the ball's trajectory, he assigns to the ball a factor of \(\gamma_2\) while the train's gamma factor remains unchanged.

    Gain in x-momentum of the ball, as seen by Q-reeus: \((\gamma_2-\gamma_1)mv_x\)
    Energy transferred to the ball by the toss, as seen by Q-reeus: \(\Delta E=(\gamma_2-\gamma_1)mc^2\)
    Decrease in rest mass of train + me due to energy transferred to the ball: \(\Delta M=\frac{\Delta_E}{\gamma_1c^2}\)
    Loss in x-momentum of train + me, once again as seen by Q-reeus: \(\gamma_1\Delta Mv_x=(\gamma_2-\gamma_1)mv_x\)

    Does Q-reeus see the ball's momentum along x conserved when it's tossed? No.
    Does he see conservation of total x-momentum of ball + train + me? Yes.
    Do equal and opposite reactions matter when discussing momentum conservation? Yes, it's not just a problem for Elon Musk to worry about.
    Does the ball's x-velocity relative to the train change when it's thrown? No.
     
    Last edited: Jun 19, 2018
  8. Q-reeus Valued Senior Member

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    3,052
    No, your example is irrelevant and wrongly analyzed wrt momentum.
    That much is ok.
    At this point you go wrong. There is no such gain. You can't just add gammas like that. Excising from #102:
    "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."
    Correct but misleading and irrelevant.
    Wrong - for the same reason as given above. The actual relativistic momentum transforms refute your notion and misuse of gamma factors.
    Actually, he does. Again - as per those momentum transforms linked to in #102.
    Trivially true, and irrelevant to what really matters to basic theme of this tortuous thread.
    Again, trivially true and irrelevant. The correct observation, relevant to my #139, is that baseball vertical velocity is less by factor γ1 as seen by me, relative to as seen by onboard thrower. That's all there is to the train example. Which is thus a train wreckage for your failed 'refutation' of that shown in #139 and many other posts.
     
  9. Q-reeus Valued Senior Member

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    3,052
    No, I said simple *example*. There is only one simple example in #102, last para there. I first asked you in #113, and, just like CptBork, you keep dodging a straight answer.
    Well you should agree, even with that part you quoted. It's established SR after all.
    Only in your head. There is no 'veering off course' involved. Let's adapt that simple *example* in #102 to your above. The rocket is a super-duper one. So is the car. Let the rocket attain a y-axis velocity of 0.8c at some given moment, seen in the launch pad frame. You in that super car are driving past the launch pad at that moment, with an x-axis velocity also 0.8c relative to the launch pad. Now tell me, what is the total velocity of rocket as seen by you in that car? Can it be compatible with it still having y-axis and/or x-axis components of velocity = 0.8c?
     
    Last edited: Jun 20, 2018
  10. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,762
    Q-reeus, I've had more than enough experience with internet trolls for many years to know that some people won't change until they decide for themselves that it's time for a change. I'm not going to keep repeating the same arguments ad infinitum. I think any reasonable person can see that you don't understand the concepts you're trying to explain and justify here, and ultimately Neddy and I will probably have to simply bypass you in order to reach further conclusions.

    The observer watching the train pass by doesn't boost, they remain fixed in place. The ball has gamma factor \(\gamma_1\) before the toss, \(\gamma_2\) immediately afterward. We are interested in calculating the ball's changes in energy and x-momentum as a result of the toss, and verifying that they are equal and opposite to the changes in energy and x-momentum experienced by the train and its passenger. There's nothing mentioned about adding gamma's here.
     
  11. Q-reeus Valued Senior Member

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    3,052
    Adding with a minus sign attached to quantity is still addition. Your momentum expressions have the wrong general form. Evidently you cannot face the fact, established by the accepted SR maths I linked to, that momentum components transverse to a given boost axis, are unaltered by that boost. You have dug yourself into a deep hole.
    If Neddy wishes to embrace your fantasy hybrid Newtonian/SR world, so be it. Life's too short to waste on a futile endlessly looping saga. I wait only to see what Neddy has to say to my last post.
     
  12. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,762
    What's the point in linking to equations if you're not even going to apply them properly? You need to rethink that part instead of pointing to the same link over and over and using equations that apply to a completely different situation (even after converting boost axes).

    And of course we're taking a difference in gamma's! The ball's rest mass doesn't change, its x-velocity doesn't change, so the only thing left to change is \(\gamma\). How else do you calculate a diffence from before and after?
     
  13. Q-reeus Valued Senior Member

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    3,052
    Chaff tossing. You must know better, surely. Apparently not.
     
  14. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,762
    I gave you multiple opportunities to pretend that you simply misunderstood the problem and thus save some face. Then you could have blamed me for poorly communicating, or something like that. Your loss.
     
  15. Q-reeus Valued Senior Member

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    3,052
    Say what you want. Believe what you want. I'm done arguing with you. Savvy?
     
  16. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,762
    Sure, whatever you want is fine with me.
     
  17. Q-reeus Valued Senior Member

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    3,052
    After some further thinking on this matter, ....

    Case 1:
    Suppose a particle of charge q and proper mass m moves at constant x-axis velocity \(v_x\) = 0.866c -> \(\gamma_x\) = 2 in inertial frame K. It next enters one end of a parallel-plate capacitor exerting a fapp uniform y-axis oriented E field. Regardless of motion in K frame, force qE on the charge is constant, and acts strictly along the y-axis. Suppose the KE gain \(\Delta_y\)qE in moving a transverse distance \(\Delta_y\) between the plates, equals that of the initial x-axis KE. At no time does a force act along x-axis, so the initial x-axis momentum magnitude of \(\gamma_x v_x m\) = \(2 v_x m\) is preserved.
    Conservation of energy requires net KE \(\gamma mc^2\) = \(4mc^2\). Giving in turn a net velocity acting 45 degrees between x & y axes, of magnitude
    v = \(\sqrt(1-\frac{1}{gamma^2})\)c = 0.9375c.
    Which has a component along both x & y axes of 0.663c. Clearly there has been a reduction in initial 0.866c x-axis velocity. Without any K frame x-acting force.

    Seen in x-axis boosted frame K' where, initially \(v_x\) is zero, there exists a transverse \(B_z\) field, of equal magnitude and opposite sign to that described below in Case 2. It results in a magnetic Lorentz force acting in K' to slow the charged particles \(v_x\) seen in K, as \(v'_y\) builds in value.

    Case 2:
    As above, except instead of particle moving into a fixed-plates capacitor, the capacitor moves with the particle along x-axis, it's applied \(E_y\) field being suddenly switched on at some instant. The crucial distinction with case 1 is the presence of a transverse \(B_z\) field seen in K frame. As the particle gains velocity along y-axis, it feels an increasing magnetic Lorentz force \(F_x = v_y\times B_z\), which acts in the direction of motion in K. Exactly countering the purely kinematical reduction in \(v_x\) applying to Case 1.

    This situation then, where the source of transverse impulse moves with the particle as seen in K, is where there is no change in net \(v_x\).
    While the impulse imparted is strictly transverse to x-axis as seen in particle initial rest frame K', there is a non-zero impulse imparted along x-axis as determined in K.

    The parallel to Case 2 is the train and thrown baseball set out in #144. I will now concede there is an increase in x-axis ball momentum when thrown vertically in the train frame. Owing to the fact there must be a hidden transverse impulse acting along x-axis as determined in the platform frame. It would be more difficult to pin down, but non-simultaneity will play an essential role - along the lines as discussed in #37.

    This should also apply to the rocket/spaceship situation of say #143. Leading edge of the rocket nozzle fires up earlier than the trailing edge will, as seen in a frame K having coasting x-axis motion wrt launch pad. So in effect I would now say the rocket does in fact tilt slightly seen in K, such that \(v_x\) remains constant throughout the y-axis speed boost. Actually Lorentz contraction demands a tilt will occur as seen in K.

    In summary:
    For a situation as per Case 1, initial coasting velocity decreases according to #139.
    Whereas as per Case 2 and analogues, initial coasting velocity is constant.
    In both cases, the velocity induced by some impulse acting orthogonal to a coasting velocity seen in some frame K', reduces wrt impulse rest frame K, according to #139.
     
    Last edited: Jun 20, 2018
  18. Neddy Bate Valued Senior Member

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    1,671
    Well that's good, so the rocket remains on course going straight up the y' axis. Now tell me what you think the x-component velocity of the y' axis would be, as measured from the unprimed frame? It should be v, surely?

    Since the gamma factor of 0.80c is 5/3 the total velocity of that rocket would be the composite of its x-component velocity 0.80c and its y-component velocity 0.80c/(5/3)=0.48c. Note the 5/3 there comes from the given v=0.80c of the car, which measures the clock on the launch pad as time dilated, and thus the rocket's ascension slowed by that time dilation.

    And, yes, it is compatible with it still having an x-axis component of velocity = 0.80c.

    From Pythagorus:
    (total velocity)² = (x-velocity)² + (y-velocity)²
    total velocity = √ ( (x-velocity)² + (y-velocity)² )
    total velocity = √ ( (0.80)² + (0.48)² )
    total velocity = √ ( 0.64 + 0.23 )
    total velocity = √ ( 0.87 )
    total velocity = 0.93c
     
  19. Neddy Bate Valued Senior Member

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    1,671
    By the way, you can do the same thing with light. Aim a laser beam up the y' axis instead of the super-rocket. The launch pad frame measures the light speed as 1.00c. From the car frame, the total velocity of that light would be the composite of its x-component velocity 0.80c and its y-component velocity 1.00c/(5/3)=0.60c. Note the 5/3 there comes from the given v=0.80c of the car, which measures the clock on the launch pad as time dilated, and thus the light's vertical-component motion slowed by that time dilation.

    And, yes, it is compatible with it still having an x-axis component of velocity = 0.80c.

    From Pythagorus:
    (total velocity of light)² = (x-velocity)² + (y-velocity)²
    total velocity of light= √ ( (x-velocity)² + (y-velocity)² )
    total velocity of light = √ ( (0.80)² + (0.60)² )
    total velocity of light = √ ( 0.64 + 0.36 )
    total velocity of light = √ ( 1.00 )
    total velocity of light = 1.00c
     
  20. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,762
    If a particle is traveling at 0.866c along the x-axis, you can't accelerate it strictly along the y-axis to reach a 45-degree angle. The particle's inertial mass would approach infinity as it approached lightspeed, and the Pythagorean velocity sum would always remain below c with the x-velocity remaining constant. If you were to instead boost the lab frame along y without affecting the particle, the x-velocity component would now be reduced in this boosted frame and you could pick a frame in which the particle is now traveling at 45 degrees. You can't boost the observer's reference frame when you want to talk about conserved quantities, you have to pick a fixed inertial frame and leave it in place.
     
  21. Q-reeus Valued Senior Member

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    3,052
    The inertial frame K is fixed in Case 1 scenario. A shift into an x-axis boosted K' frame was done later simply to point out what is seen there. You are not seeing that in K frame, the charged particle gains a transverse KE that could easily be made to match or exceed the initial coasting x-axis one. Hence also the respective momenta. Net velocity in K must reflect that fact. For an electron, the potential difference needed to achieve a transverse KE -> \(\gamma_y\) = 2 is not much over half a million volts. Easy. That's when there is zero \(v_x\). To achieve a net \(\gamma\) =4 in K frame requires a PD of twice that i.e. not much over a million volts.

    I'm now satisfied with the general explanations given in #154. Without recourse to the case of particle and capacitor co-moving in an inertial frame, it would have been far more difficult to understand the fundamental difference between Case 1 & Case 2 type scenarios. This has shifted much from the original query. I'm hoping it's done now. Hoping.
     
    Last edited: Jun 21, 2018
  22. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,762
    Energy doesn't have a directional quantity to it. You can't say anything about kinetic energy along one axis vs. another. If an acceleration is applied along a given axis, there is a limit to how long this acceleration can be sustained as the particle's mass approaches infinity, but there is no reason why such an acceleration requires any change in velocity along any other axis.

    If you consider the case of a force constantly applied in the y-direction, as in the case of a charged particle moving through a uniform electric field, the direction of acceleration only matches the direction of force when that force is either applied parallel to the direction of the particle's motion, or transverse. As soon as the particle starts to deflect, there is going to be some off-axis acceleration due to the electric field. There is nonetheless no reason to expect it will be traveling at 45 degrees w.r.t. the original axis of motion, I think the actual answer will be somewhat more complicated.

    In the case of the rockets as Neddy and I have been discussing, a constant acceleration along y as seen by the static observer will lead to the appearance of horizontal forces acting on the rocket which coasts along x, along with equal and opposite forces acting on its propellant, but there will be no change in \(v_x\), unlike "Case 1" of the charged particle, in which the force is specified to apply along y, but not necessarily the acceleration. Forces don't transform as 4-vectors the way positions and momenta do.
     
  23. Q-reeus Valued Senior Member

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    3,052
    True but it implies for Case 1 a corresponding momenta split as I explained.
    See above.
    Again true but not relevant.
    Yes there is. For Case 1, applied force qE is strictly along y-axis. No change in transverse, x-axis momentum occurs. Which requires a change in \(v_x\).
    Agreed! And that contradicts your previous statement in red highlight. There is indeed a negative acceleration along x-axis. Just enough to keep \(p_x\) constant.
    For the specific values I gave, equal momentum is attained along x and y. Hence the angle is 45 degrees.
    You are there agreeing with my assessment in #154.
    Indeed, but it's fundamental electrodynamics that strictly y-axis acting F = qE applies at all times in Case 1. As a result, there is an increasing negative acceleration along x-axis as velocity along y-axis picks up. Again, just enough to have \(p_x\) remain constant. Ponder the difference between Case 1 and Case 2 in #154. No transverse B in K, vs a transverse B in K. Magnetic Lorentz forces!
     

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