Pure SR, Relativistic Mass, and its Gravity?

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

  1. Q-reeus Banned Valued Senior Member

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    4,695
    Good then, and yes it's not simple.
    In any frame, inertial or not, having a non-zero x-axis speed component wrt lab i.e rocket frame. Thus in both the belt frame and K frame.
    My reading of your 2nd, treadmill scenario in #30, which is what we have been dealing with, is that observer at rest in K has an x-axis velocity such that lab, and mass-on-scales on moving belt, have equal and opposite x-axis velocities seen in K. Which remains true even when y velocity becomes appreciable. But there will be corresponding equal and opposite x-component decelerations.
    Good, there is some progress now. Time-rate-of-change of y velocity yields in turn time-rate-of-change of overall time dilation factor, which in turn means uniform motion in lab frame is non-uniform i.e. accelerated in say K frame, or any other inertial frame having an x-component of velocity wrt lab/rocket frame.
    You have that basically right. I specifically chose a dilute gas for the reasons given there. In the case of a solid or liquid, there is essentially equipartition of molecular/atomic/lattice thermal energy between KE and PE since one there has a system of excited oscillators. Which is different to the case of a gas at typically encountered pressures. The observed net mass increase will be the same in all cases, but with a different split as to the detailed contributions.
    I do suggest to ponder that example in #15 some more - it's simple way of tying relativistic mass and Lorentz contraction together to show how that γ² factor arises naturally.
    Groan - more punishment for me.
     
    Last edited: Jun 11, 2018
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  3. Neddy Bate Valued Senior Member

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    It's not punishment, you are just a good person helping another good person understand something. So, if karma is a thing, you should get a reward in this life, and maybe even one in the next!

    Does this mean that the spaceships also have x-component decelerations? In other words, do the spaceships remain co-located with the lab & belt, or not?
     
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  5. Q-reeus Banned Valued Senior Member

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    4,695
    Thanks for those kind sentiments!
    From what I read in 1st scenario in #30, yes they do. It would be better though to have started with relative x-component velocities, then have them simultaneously (as seen in K) fire up and accelerate along y. Anyway the best way to handle all of this is to take my earlier hint - work in a frame that most simplifies. Either the belt frame or lab frame. Then apply the appropriate SR transforms to get corresponding values in any other frame. A reading in one frame e.g. weight read by scale, MUST be the same in any other frame. It's an invariant. Must run.
     
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  7. Neddy Bate Valued Senior Member

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    If that is the case, then unfortunately I am back to thinking there should be no difference between the measurements made by the spaceship A & lab floor, and there should be no difference between the measurements made by the spaceship B & the belt. The only reason I thought I was starting to see a difference was when I thought the time dilation of the belt was causing it to not remain co-located with spaceship B.

    Would that make a difference? If so, what difference? The belt on the treadmill could be considered zero-friction so that we could start it with a push of the hand, and it would run forever. Then we could fire up of the rocket engine underneath, and it would still become time dilated more and more as its speed increased. So I am not seeing a difference? Please elaborate when you have time. Thank you.

    Of course the weight read by scale would be an invariant, yes. And we already agreed the two spaceships get the same weight reading as each other, which all frames must agree is the case.
     
  8. Q-reeus Banned Valued Senior Member

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    4,695
    While that last sentence is correct, it merely reflects the fact your twin spaceships scenario is NOT equivalent to the accelerated treadmill one. Something I have been at pains to emphasize on several earlier occasions.
    Honestly, I do strongly recommend at this point you take up my suggestion and field your K frame scenario re treadmill at e.g. PhysicsForums. There are any number of relativity experts there who are used to dealing with such apparent conundrums, though probably not your particular scenario. Again, please post a link here to any such started thread.
     
  9. Neddy Bate Valued Senior Member

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    2,548
    Q-reeus,

    I do believe that you understand the maths on this subject better than I do, so in the end, you might end up being correct. That would be fine with me, but only if there would be some accompanying explanation that would make sense. For example, when I thought you were saying that spaceship B and a point on the treadmill belt do not remain co-located, but apparently that is not the case after all.

    So, as it stands, we have spaceship B who is always co-located with a point on the treadmill belt, (at all times!), and yet you say somehow they end up with a different measurements of identical phenomenon. All I can say is that under pure SR, with everything being about relative velocity, I just have to err on the side of caution, and assume there must be a mistake there someplace. I do not think it is a mistake in SR, but possibly some miscommunication on my part, or something else.

    I'll take a break from the thread now, but first I want to I thank you again. If I ever do think of some way to get us past this, I will post here, and maybe we will pick this up at a later time.
     
  10. Q-reeus Banned Valued Senior Member

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    4,695
    What you are missing Neddy is that in the treadmill case, there is lateral belt motion wrt the actual source of acceleration - the rocket. By equivalence principle, that #15 example with the long rod mass shows the crucial aspect of relative motion wrt a proper frame source of g = -a.
    Your twin spaceships scenario is merely a mirror image situation which guarantees symmetric i.e. identical scale readings.
    OK and again thanks for a thumbs up. Still - by fielding your scenario - specifically the accelerated treadmill seen in K frame - at some other quality forum like PhysicsForums, you are bound to get some fresh takes on it that may provide just the breakthrough you need.
     
  11. Neddy Bate Valued Senior Member

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    2,548
    In post #30, I describe the the belt and the lab floor as having equal lateral motion with respect to the origin of frame K, just in opposite directions. I thought it was implied that the "actual source of acceleration - the rocket" was evenly distributed laterally, sort of like the distribution forces along a bridge. So, if I put my "actual source of acceleration - the rocket" at x=0 moving up the y axis, does that make both scenarios equivalent?
     
  12. Q-reeus Banned Valued Senior Member

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    4,695
    The fundamental situation is this: whenever there is constrained i.e. not in free-fall motion, transverse to some proper-frame source of acceleration, be it gravitational or an equivalent accelerating rocketship etc., the 'g force' experienced in the moving frame is amplified by that factor γ² derived earlier by various means. And it will show as a scale reading.
    Your twin spaceships scenario does not satisfy that requirement.
    Again - take it to PhysicsForums. Get some different approaches (or maybe just the same) on it there.
     
  13. Neddy Bate Valued Senior Member

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    2,548
    So, let me just make sure I understand what I think you are saying. Correct me if I'm wrong.

    In the two-spaceship scenario, frame K sees two separate rockets, each causing purely upward acceleration. One is attached to spaceship A and the other is attached the spaceship B.

    But in the lab-belt scenario, frame K sees only one rocket, but it is moving laterally because it is attached to some point on the lab? And even though the end result is the belt and spaceship B are always both co-located, the two different rocket locations cause differences in scale readings of weight?
     
  14. Q-reeus Banned Valued Senior Member

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    4,695
    I don't follow that last highlighted bit, but anyway with the twin spaceships, acceleration source and scale in each ship are always co-located with no relative motions. So obviously only a proper mass reading is possible in each ship. Which is simply not the situation with accelerating treadmill. A point on the belt IS in relative motion to the source acceleration which is in lab frame.
    AGAIN - take it to PhysicsForums!
     
  15. Neddy Bate Valued Senior Member

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    2,548
    In the last highlighted bit, "the two different rocket locations" refers to the location where the rocket engine itself is mounted. In the case of the spaceship B, the location where the rocket engine itself is located is that it is mounted to spaceship B itself. In the case of the lab/belt the location where the rocket engine itself is located is that it is mounted to the lab itself.

    And in the last highlighted bit, "the differences in scale readings of weight" refers to your claim that the reading on the scale in spaceship B would be different than the reading on the scale on the belt, even though the two test masses are identical. Thus the difference is caused by the two different mounting locations of the rockets, with one being at rest with respect to the scale and the other being in motion with respect to the scale.

    Yes, that is what I was finally putting my finger on. In the frame of spaceship B, the rocket is stationary, so you say its scale weighs normally. But in the frame of the belt, the rocket is in lateral motion, so you say its scale weighs differently by a factor of gamma squared.

    I guess I understand now. So if I had chosen to mount the rocket to the treadmill belt itself, then the scale on the belt would weigh normally, but the scale on the lab floor would weigh differently by a factor of gamma squared.

    So what happens if we mount a rocket to the lab itself and we mount a rocket to the belt itself, and we adjust the thrust of both so that the whole lab still moves up the y axis of frame K in the same way as before?

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    Why now, when we have finally gotten to the bottom of it? lol...
     
    Last edited: Jun 12, 2018
  16. Q-reeus Banned Valued Senior Member

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    4,695
    Yes. And by accumulating understanding from post #1 onward, it hopefully has become clear why. Similarly why there is a single gamma factor for case of scales under the treadmill.
    That's an ugly situation to consider. You are subjecting the entire system to a time-changing centre of thrust. Your rocket will be tilting thus not flying straight up. I'm out on that one!
    See above!
    I truly, truly, truly hope so!!!

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  17. Neddy Bate Valued Senior Member

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    Thank you for your help, Q-reeus. I will leave it at that.
     
  18. Q-reeus Banned Valued Senior Member

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    You're welcome Neddy. And great news!

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  19. Q-reeus Banned Valued Senior Member

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    I hesitate to post this, but here goes....
    I recommended several times to field at PhysicsForums the initially baffling case of accelerating treadmill seen in inertial K frame. Well I just noticed a thread there that treats the same overall physics but with a different and much simpler to analyze scenario:
    https://www.physicsforums.com/threa...ore-when-going-at-relativistic-speeds.948690/

    It was amusing to see a couple of seasoned GR pros making initially silly assessments, then covering their tracks and back-tracking later on. It's only when one gets to p2 #42 that the correct relation is presented, albeit using somewhat advanced maths. So much fluff and confusing irrelevancies were bandied about in that thread, I take back my initial recommendation!

    Still, from #42 onward there, things settle mostly. pervect is the go-to man there, and his #49 briefly touches on the difference between weighing with scales still and mass moving, vs weighing with both scales and mass moving - i.e. your scales static under the treadmill vs moving on the treadmill. His #50 And #54 are also worth a read, if you can follow the maths. None of them introduce the concept of 'relativistic weakening' as I did, but overall it all tallies with my own analysis, but what a saga to have to plough through.

    I want to make this clear - above is for your private study - NOT to restart discussion here! Any issues bugging you in that thread should be raised over there! OK! Hopefully though it's all now sufficiently clear. SR is always consistent - only wrong application causes issues.
     
    Last edited: Jun 13, 2018
  20. Q-reeus Banned Valued Senior Member

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    Slight correction to #56. On a re-read of that linked-to PF thread, pervect actually first introduced the correct result in #33 there, but without any derivation. His #36 linked to a fairly long-winded GR-based derivation in another thread. Other than that...THE END.

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  21. Neddy Bate Valued Senior Member

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    Hmmm, interesting. Thanks!
     
  22. Neddy Bate Valued Senior Member

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    Sorry, just one last comment. In that other thread, post #50, the author uses a "dropped mass" to measure the acceleration, and states that the dropped mass is just an instantaneously co-moving inertial observer. If spaceship B is always co-moving with the belt, then each one's instantaneously co-moving inertial observer is always identical to the other's. This does not seem to support the idea that spaceship B and the belt measure different accelerations.

    If you would be kind enough to give me a super-quick explanation, I will end the thread on that. Thanks again.
     
  23. Q-reeus Banned Valued Senior Member

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    4,695
    Sure. Once again - scales/mass in spaceship B are at rest wrt acceleration source proper-frame: spaceship B. Logically then only the proper mass can be read by the scales. For scales/mass on the moving belt, there is relative transverse motion wrt lab-frame/rocket - the proper-frame source of accelerated motion. Or, if it were gravity causing the "g's", a 'stationary' planet. Therefore mine and pervect's gamma squared findings apply there.
    THE VERY END.

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    Last edited: Jun 14, 2018

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