Pure SR, Relativistic Mass, and its Gravity?

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

  1. Q-reeus Valued Senior Member

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    3,411
    Nonsense. That single example in #102 knocks your claim dead. That in #106 seals it with the established maths. You are simply too egotistical to concede. And stop this butting-in in an attempt to confuse Neddy with ever more muddying of the waters. Go elsewhere and rant on politics which is your evident 'expertise'.
     
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  3. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    I just pointed out what your example actually shows. The formula you cite is comparing the x'-momentum seen by the inertial observer at x'=0 to what's seen by observers onboard the associated rocket, which will be 0 for both. That says nothing about what the inertial observer at x=0 sees.

    Please stop defending your mistakes and trying to look like the resident expert. And go learn something about rockets while you're at it.
     
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  5. Q-reeus Valued Senior Member

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    No. It shows by simple logical necessity any boost along y will result in a decrease in x-axis coasting velocity. All seen in the same inertial frame. Something you are heavily invested in denying.
    Wrong again. The correct transposition from their S, S' coordinate system to the K, K' one I used is fully explained in e.g. #111. And you continue to avoid dealing - properly - with #110.
    Yes yes, just keep making such assertions, insinuations, and irrelevant misdirections. No, actually don't. Just quit now and salvage some self-respect.
     
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  7. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    #110 relies on #102, which relies on a source whose formula is incorrectly applied to conclude that the moving rocket's x-momentum is conserved in the stationary observer's frame. The moving rocket's x-momentum is not conserved on its own without factoring in the source of the acceleration.

    Neddy makes a very simple use of the Lorentz transform to show the same thing I've been saying. Rather than blathering incorrectly about momentum conservation, show what's mathematically wrong with Neddy's application.
     
  8. Q-reeus Valued Senior Member

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    3,411
    Doubly wrong. #110 referred back to both #102, and #106 - whose linked to formulas, correctly applied, dovetails nicely with that simple refutation in #102 of your basic conceptual error. One you keep refusing to concede on.
    Still nonsense. Still trying to muddy the waters despite my dealing with all such fluff in #110. Pathetic.
    Done already - #111, #113, #116. Once Neddy answers my single question posed in #119, this issue should be over. Hopefully the entire saga. No thanks to your unwelcome and confused intrusions.
     
    Last edited: Jun 18, 2018
  9. Neddy Bate Valued Senior Member

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    1,870
    I will answer "yes or no" to that question, but only if you first promise to answer "yes or no" to this simple question:

    Question: You have already agreed that a rocket fired straight up the y' axis of inertial (never boosting) frame S' would always be located at x'=0 in that frame, so do you also agree that Lorentz transforming x'=0 to the inertial (never boosting) frame S produces x=vt as I derived?
     
  10. Neddy Bate Valued Senior Member

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    1,870
    It's not quite so simple as that. Consider my rocket which is accelerating straight up the y' axis so as to produce a constant "1 gee" on board the rocket. Also consider a treadmill inside that rocket which has a constant x-component velocity as measured by the rocket itself.

    As the y'-component velocity increases hyperbolically relative to frame S', frame S' measures the x-component velocity of the treadmill belt as slowing down due to increased time dilation. Likewise, frame S will measure the treadmill belt as slowing down even more.

    It is easy to mistakenly generalize that result to the rocket's x-component velocity, as measured by frame S. I made that mistake earlier in the thread. I don't know if Q-reeus is making that same mistake though, (as it seems he has some other reasoning).
     
  11. Neddy Bate Valued Senior Member

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    1,870
    Yes, I'm glad we agree also. And I am also glad you find my explanation satisfactory.

    I see the situation as symmetric, because of the following reasons:

    1a. Frame S launches a rocket straight up its y axis in such as way as to produce "1 gee" on board, by definition of the scenario.
    1b. Frame S' launches a rocket straight up its y' axis in such as way as to produce "1 gee" on board, by definition of the scenario.
    2a. We could choose to view the situation from frame S and say that the rocket on the y' axis is moving with x-component equal to v.
    2a. We could choose to view the situation from frame S' and say that the rocket on the y axis is moving with x'-component equal to -v.
    3a. Regardless of how we choose to view the situation, the passengers aboard both of the rockets experience "1 gee" on board, by definition of the scenario.
    3b. We could even view the situation from some other inertial frame S'' where both rockets have equal and opposite x''-component velocities, and the passengers aboard both of the rockets experience "1 gee" on board, by definition of the scenario.

    But if you mean it is asymmetrical that frame S can say that one rocket has an x-component velocity of zero, and the other rocket has an x-component velocity of v, then I suppose that is asymmetrical. But that asymmetry does not change what the passengers feel on board the rockets.
     
  12. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    The asymmetry I was referring to involved a static observer in frame S watching both rockets accelerate at the same rate, as seen from that observer's POV. If each rocket accelerates so as to feel 1g, static observers in either S or S' wouldn't see them both accelerating at the same rate. Does that make sense to you?

    P.S. As to Q-reeus' arguments, he's mistakenly concluding that the rocket moving horzontally w.r.t. S must have a fixed x-momentum, neglecting the changes in x-momentum applied to whatever is causing the rocket to accelerate (fuel, a guiding hand, whatever). Obviously for x-momentum to be fixed, the x-velocity would have to change, which is why he derives his absurd result. I'm planning to give a simple example to demonstrate his misconception, but that will have to come later.
     
  13. Neddy Bate Valued Senior Member

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    1,870
    Yes, now I understand. I had not meant to imply that both rockets would be ascending at the same rate in the 'asymmetric' arrangement. It was a given that the passengers in both should feel a constant "1 gee."

    Given that, in the asymmetric arrangement, frame S measures the rocket on its own y axis to be ascending faster than the laterally moving rocket ascending the y' axis. And likewise, frame S' measures the rocket on its own y' axis to be ascending faster than the laterally moving rocket ascending the y axis. This is simply time dilation applied to the upward rate of the rocket as measured from the frame in which it has no lateral movement.

    Oh, I didn't realize that. Well hopefully you can clear that issue up, and then finally all will be well with the world again.
     
  14. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    Ok that all sounds good so far. Now another thing you need to account for is that the gravitational acceleration on Earth is 9.81m/s^2 for a stationary observer on the surface, but it will look different to other observers in motion relative to the first. Sound sensible?
     
  15. Neddy Bate Valued Senior Member

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    1,870
    If you mean that someone stationary on earth can drop a test mass and measure its acceleration as 9.8 m/s/s, but that someone flying at relativistic speed laterally to that experiment would measure the drop as time dilated, then yes that sounds sensible. But that does not make the person who is stationary on earth feel anything different. If they stand on a spring scale, they still measure their "usual" weight, despite someone else flying past them.

    What I have been trying to isolate is a case where a mass on a spring scale would measure something other than its "usual" weight. But I wanted to do so without using any actual gravity, but rather just Einstein Elevator scenarios.

    That is why, in all of the scenarios I have been describing in this thread, there are only inertial frames which have no gravity. The only gravity, per se, is what the passengers feel on board the rockets, spaceships, or even the laboratory (lab) which was conveniently equipped with a rocket engine underneath.

    In every scenario, it was given that the rockets, spaceships, and lab all should accelerate straight up the y-axis of an inertial frame in such a way that passengers at rest with respect to their vessel should feel 1 gee, (hyperbolic acceleration up the y axis). Furthermore, I wanted the inertial frame to remain unchanged, not constantly redefined to follow the rocket upward. I thought it would be useful to keep the inertial frame defined so that we could trace out the rocket's / spaceship's / lab's coordinates relative to an inertial frame, if we needed to.

    From there, I am very much interested in using one of these scenarios to show that a mass on a spring scale would measure something other than its "usual" weight.

    Toward that end, we really only need one scenario, which is the lab with the rocket engine underneath. I have equipped that lab with a treadmill-type-device which provides constant motion relative to the lab. We can have a spring scale on the floor of the lab as a control, and we can put another spring scale on top of the belt, and another one underneath the entire treadmill. Of course we would have to zero out the scale so as to neglect the weight of the treadmill itself.

    Then we could weigh a test mass on all three scales and figure out the results. I think Q-reeus concluded that the result would be that the scale on the floor of the lab weighs the test mass at 1 gee, the scale underneath the treadmill weighs the test mass at 1*gamma gees, and the scale on top of the belt weighs the test mass at 1*gamma*gamma gees. (Where gamma is found using only the constant velocity of the belt relative to the lab.)
     
    Last edited: Jun 19, 2018
  16. Neddy Bate Valued Senior Member

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    1,870
    By the way, the reason I went to all the great lengths described above was because I mistakenly believed that the belt could be shown to be co-moving with some other accelerating frame experiencing 1 gee, and thus the belt should also weigh according to 1 gee. But in the meantime I have realized that I was most likely mistaken about that.
     
    Last edited: Jun 19, 2018
  17. Neddy Bate Valued Senior Member

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    1,870
    Now I'm thinking about the belt of the treadmill...

    It has a "top" belt, and a "bottom" belt, both with equal speed in opposite directions.

    I am also thinking Q-reeus wants to attach a continuum of inertial reference frames to the upward accelerated objects.

    So, if we just compare the reference frames of the top and bottom belts, I think we must conclude neither one measures the other as having an ever-diminishing x-component velocity. Surely?

    Does that help? lol...
     
    Last edited: Jun 19, 2018
  18. Q-reeus Valued Senior Member

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    3,411
    As soon as I realized your Lorentz transforms were merely displacement not velocity ones, little attention was further paid to their exact form.
    You are out of your depth in SR, as is the vengeful stalker CptBork, who is only here to spite me. And is encouraging you to remain in a hybrid Newtonian/SR fantasy world.
    Anyway short answer is NO. How did you fail to see those transforms were dealing purely with x directed motions?! Where is there any transverse motion? There isn't any.
    Within here is given the generalized version, that includes transverse motion: http://www.mathpages.com/home/kmath188/kmath188.htm
    Nut out for yourself how to apply the relevant expressions there, and discover, lo and behold, one gets a result consistent with what I have been continually pointing out to you in vain.

    Now, either give me that straight answer, or I leave you to become the disciple of vengeful stalker CptBork. , whereupon you will continue in both of your fantasy hybrid Newtonian/SR worlds.
     
  19. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    I'm not here to stalk you. I used to post regularly in this section until arrogant know-nothings like yourself started watering down the discussions. It seems like Neddy's starting to fit the pieces of the puzzle together, so leave if you don't feel like making a positive contribution.

    If you understood Relativity as well as you pretend to, you'd realise that your incorrect application of momentum conservation is the type of argument typically used to disprove Relativity, not to reinforce its validity.
     
  20. Q-reeus Valued Senior Member

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    3,411
    Liar. Of course you are. Smarting and fuming over my posting in that other thread. For which you only had rhetoric and foul words to respond with.
    How unfortunate for those you managed to confuse way back then, similarly no doubt as you are confusing Neddy here.
    Innuendo is cheap. You know full well that simple scenario in #102 destroys your continued fool claim x velocity is unaffected by a y boost, but dance around that fact owing to conceit and malice.
    No Neddy is perpetually confused thanks in no small part to your intrusions here.
    Bunkum disposed of back in #110 and elsewhere. Too bad the head honchos here know too little physics themselves to meaningfully intervene and straighten you out.
     
  21. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    I came here with positive intentions Q-reeus, and right away you started your bullshit. It's obvious your only purpose here is to try and look like you know how to do physics, so you can push pointless alt theories that have some kind of emotional appeal to you. You're looking very stupid here and it's about to get worse when I get around to posting my momentum example, just leave while you can still pretend that you misunderstood the problem.

    BTW go ahead and invite the mods to have a look at this thread. I'm sure James R will have a field day explaining to you why velocities can't change along an axis where there's no acceleration.
     
  22. Q-reeus Valued Senior Member

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    3,411
    One last try. Going back to the appeal to time dilation re slowing of belt speed, first discussed in #37, this can actually be adapted to the quite restrictive Lorentz transformations given in #112. We transpose the coordinates so that y's are used in place of x's, and vice versa. Before a boost along y in inertial frame K, velocity of say a particle purely moving along x is v_x.
    Now boost along y to give a γ = 1/√(1-v²/c²), with v the y-axis boost velocity. We know the transverse distances wrt y-axis e.g. along x-axis, are unaffected by the y-axis boost. But there is a time dilation by factor γ in K' (moving along y-axis wrt K, NOT along x-axis ).
    Hence motion along x axis must be slowed by that factor such that dx'/dt' = v'_x = (v_x)/γ.

    Which is also the result by using the velocity transforms directly as per #106.
    While the vengeful stalker is quite clearly too deeply invested in denying that unavoidable result to ever concede, it's hoped Neddy doesn't succumb to the same psychological impediment. Because it's psychology i.e human nature not physics which is the real determinant by this stage.
    Pride or progress - make a choice Neddy. And DO honour your promise to answer my simple question posed in #119!
     
    Last edited: Jun 19, 2018
  23. Q-reeus Valued Senior Member

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    3,411
    Bluff won't work. Please, by all means devastate me with your promised momentum example. I so look forward to your rock-solid proof 1 = 2 or equivalent nonsense. DO IT! More fun for me.

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