Pure SR, Relativistic Mass, and its Gravity?

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

  1. Neddy Bate Valued Senior Member

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    1,870
    Thank you for finally seeing that if spaceship B and the belt were to be co-moving, then it would be very problematic for anyone to claim they experience different gees of acceleration.

    Yes, it was one of my original premises that spaceship B and the belt would be co-moving. But I started to question that, back when you had me thinking about how frame K would find the belt to be constantly slowing down its horizontal speed, due to increasing effects of time dilation caused by its increasing speed relative to the y axis. Around that time, I asked you if they would remain co-moving or not, and you told that they would. So we both share some of the blame for this confusion.

    Wait, now you are talking about two spaceships. We are supposed to be talking about spaceship B and the treadmill belt running in the lab. It's probably just a mis-typing, but we really should try to be careful here. We had already agreed on the twin spaceship scenario, so if you are now changing that, I need you to specifically tell me that, please.

    I had long ago given up on the idea that spaceship A and spaceship B would always remain at the same y coordinate in each of their own frames. I am relying on inertial frame K to set up the scenario where spaceship A and spaceship B always remain at the same y coordinate as each other according to frame K. Surely we can still agree on that, right?

    If spaceship B and the belt are not co-moving, then I have no problem with them experiencing different gees of acceleration. But I will not be fully happy until I understand how it can be possible that spaceship B and the belt could not be co-moving. It still seems to me that both are equivalent inertial frames, and so it seems they should be co-moving in both the x and y directions of inertial frame K.

    I suppose that if they are not co-moving in the x direction, then that could lead to them also not being co-moving in the y direction, and that seems to be what you are saying above. But how can it be possible that spaceship B and the belt could not be co-moving in just the x direction of frame K? Let's start with that.

    So, are you now saying that frame K would find the belt to be constantly slowing down its horizontal speed due to increasing effects of time dilation caused by its increasing speed relative to the y axis, but that would not be the case for spaceship B??

    By the way, please do not try to throw a curve ball and say that that spaceship B, being a spaceship, could have rocket thrust in the x direction, whereas the belt, being a treadmill, has no rocket thrust in the x direction. I specifically want all these motions in the x direction of frame K to be inertial, please.
     
    Last edited: Jun 16, 2018
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  3. Q-reeus Valued Senior Member

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    More specifically, if they remained co-moving - which they don't. Big difference.
    Then yes. But the whole thing has been mired in your choosing scenarios that make it harder not easier. My original analyses, strictly in either belt or lab frames made it easier.
    It's not your separate twin spaceships scenario. The treadmill-in-lab is being accelerated by a spaceship/rocket. Hence let's call it spaceship C. Whatever, there are two spaceships involved spaceship B and C - but not the symmetric twin spaceships A & B in that other case you introduced.
    See above.
    So think about everything I have pointed out earlier. In the end, findings in #13 have tallied with that in e.g. #49 in the PF thread. Again - are pervect and I totally mistaken? Or you just continue to see things from a confusing perspective? Bets?
    It's as I stated it, in qualitative terms, back in #37. The #35 ref to Lorentz contraction was a quick stab and not really pertinent. Non-simultaneity plus rate of change of time-dilation of belt seen in K are the crucial determinants. And to repeat - it's a far more difficult situation to analyze. Moral - always choose a scenario simplifying things!!!!!
     
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  5. Neddy Bate Valued Senior Member

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    Okay, there was never a spaceship C, so I just wanted to be sure. Let's just stick with calling it the lab. You say you want to simplify things, but then switch from talking about the belt & spaceship B to the lab frame. I think you are confusing yourself.

    Anyway, the scenario is as simple as possible. The belt is moving horizontally at constant speed relative to the lab. Spaceship B is moving horizontally at constant speed relative to the lab. Thus spaceship B is co-moving with the belt at all times. Which part do you think is wrong?

    Or, even better: Which one, belt or spaceship B, would you say is not moving at constant speed relative to the lab frame? Please answer "Belt," or "Spaceship B," or "I don't know."

    Are you now saying that frame K would find the belt to be constantly slowing down its horizontal speed due to increasing effects of time dilation caused by its increasing speed relative to the y axis, but that would not be the case for spaceship B? Please answer "Yes," or "No," or "I don't know."

    Clear answers to those questions will let us wrap this up, finally. Thank you.
     
    Last edited: Jun 16, 2018
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  7. Q-reeus Valued Senior Member

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    3,411
    That from the belt frame, spaceship B has only 1/4 the acceleration experienced in belt frame! And 'at all times' is impossible since the treadmill is only so long, and finite-sized spaceships B & C are passing each other in opposite x-directions.
    As per #37.
    From spaceship B's pov the belt has essentially constant speed. I say 'essentially' because it depends, owing to both opposite x-axis motion of spaceships and ever changing y-axis speeds, on which instant that is evaluated. When exactly 'co-located' along x-axis, we can say it's exactly constant. At some other instant, any observed change in belt speed seen in B frame will be a 2nd-order thing not worth the effort to pick over in detail. Trust me there or not.
    Whereas in K frame there is a ready made large initial x-axis speed of belt, which combines then with y-axis acceleration to give that as per #37.
    Can I really go home now?
     
  8. Neddy Bate Valued Senior Member

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    1,870
    That is begging the question. You need to conclude that, not state it as a premise.

    "At all times" (during the duration of the experiment in question) is just a way of saying it remains so the whole time spaceship B is moving by the belt, not after it reaches the end of the belt. Though the "frame" of spaceship B would still be co-moving with the belt, even after spaceship B itself passes the end of the belt.

    Spaceships A and B are never co-moving, at any time. What are you talking about?

    The question was whether spaceship B moves at constant speed relative to the lab frame, and you answer from the frame of spaceship B and what it measures about the belt. It's almost like you are gaslighting me.

    Let me try to make it as easy as possible:

    1. I assume the belt moves at constant velocity relative to the lab. Do you see any problem with that?

    2. I also assume spaceship B moves at the same constant velocity relative to the lab. Do you see any problem with that?

    Surely something must be wrong, because if there are no problems above, then spaceship B and the belt must both experience the same gees as each other.
     
    Last edited: Jun 16, 2018
  9. Q-reeus Valued Senior Member

    Messages:
    3,411
    It's already been shown. Same over in PF thread. Simply adapt to your specifics.
    You quoted B & C but then refer to A & B. Different scenarios - remember? And anyway, in that A & B scenario, they were give opposite x-axis velocities, so how could they be co-moving?!
    Of course not. It was a given postulate from the start. Nowhere have I even suggested otherwise.
    AS determined from either ship A or C lab frame, they do not continue to determine constant relative speeds. It's as I explained before in say #24 & #26. Remember, they are passing each other laterally while also accelerating upwards. It becomes a remote determination case once the ships are not co-located. A square grid laid out in that inertial K frame is an ever-changing locally squashing grid seen in either ship's frame. And will be curvilinear on a remotely determined basis.
    In addition to those transformation equations for acceleration I linked to back in #12, here's the ones for velocity:
    https://en.wikipedia.org/wiki/Lorentz_transformation#Transformation_of_velocities
    If you take the trouble to plug in the relevant inputs, for two different instants in your scenario - co-located vs not co-located - my claims will be verified. Don't ask me to do the sums for you. Exert some effort on it yourself and learn by directly working with the maths.
    Yes unfortunately for both of us you have a really stubborn comprehension issue. Once more - join up at PF, and attack the relativity expert folks there. Please!
     
  10. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    All this argument shows is that the relativistic mass of an object doesn't affect the gravity it feels from a planet that's at rest with respect to that object, when you can treat that planet's gravity as approximately uniform. Whether I'm at rest with respect to Earth or whether I see it moving away from me at near light speed, the equivalence principle shows that residents on the surface will feel the same force, as measured in their own frame of reference.

    If you try to go further and make detailed arguments based on Newton's gravitational model, you'll get inconsistent results in SR, which is why Einstein formulated GR in the first place.
     
  11. Neddy Bate Valued Senior Member

    Messages:
    1,870
    Okay I finally figured it out. Thanks for all of your help, everyone.

    The quick explanation is that the belt and the spaceship are not co-moving. The reason is that the x-component velocity of the belt slows down due to ever-increasing time dilation in the lab (as measured by inertial frame K), but the x-component velocity of the spaceship remains constant (as measured by inertial frame K).

    As long as the belt and the spaceship are not co-moving, I have no problem with them experiencing different gees of acceleration.


    -------------------------------------------------


    MORE RIGOROUS EXPLANATION:

    Let there be two inertial frames, one called K and the other called L. Let frame L be in uniform relative motion along the x axis of frame K, at constant velocity v. The way I choose to picture it in my mind is from the point of view of frame K. While frame K appears stationary, frame L appears to be moving in a rightward-direction at constant velocity v.

    Let there be a laboratory (lab) initially located at the origin of frame K, and let there be a spaceship initially located at the origin of frame L. Within the lab is a treadmill type device running in a rightward-direction at a constant velocity relative to the lab. We shall choose to make that velocity equal to v, which is the relative speed between the two frames. The way I choose to picture it, from the point of view of frame K, this would appear to make the belt and the spaceship co-moving.

    However, as we shall see in the following paragraphs, the belt and the spaceship will not be co-moving after we introduce an upward acceleration to both the lab and the spaceship. For convenience, the lab has a rocket engine mounted under it, as does the spaceship.

    At the time when the two origins are co-located, both of the rocket engines are turned on. (Side note, both of the rocket engines provide a "one gee" environment.) The lab accelerates directly up the y axis of frame K, and the spaceship accelerates directly up the y axis of frame L.

    It is important to note that as the lab accelerates upward, frame K measures the x-component of the velocity of the belt to be decreasing due to increased time dilation inside the lab caused by its ever-increasing upward velocity. However, frame K does not measure the x-component of the velocity of the spaceship to be decreasing at all, because the relative velocity between the two frames is constantly v, which is the constant relative velocity between frame K and frame L.

    Even though there is increased time dilation inside the spaceship caused by its ever-increasing upward velocity, it does not cause frame K to measure the x-component of the velocity of the spaceship as decreasing. Frame K and frame L always maintain constant velocity v between them regardless.

    Thus the belt and the spaceship are not co-moving, and my earlier claims have no standing. As long as the belt and the spaceship are not co-moving, I have no problem with them experiencing different gees of acceleration.
     
  12. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    I think you're overcomplicating it a lot. If a stationary observer watches two identical rockets accelerating at the same rate in space along the y-axis, and there's a velocity along the x-axis between the two rockets, the relative velocity along x stays constant. However, if one of the rockets is stationary along the x-axis (eliminating a possible symmetry in the problem), the accelerations measured by passengers on the two rockets will be different as will the thrusts required to achieve those accelerations and the resulting forces felt by those passengers.
     
  13. Neddy Bate Valued Senior Member

    Messages:
    1,870
    Yes, which is what I said in my previous post. But what I realized is that is not so for a treadmill, running in the x direction, which is being carried by one of the rockets. Its velocity in the x direction would be slowing down continuously, due to increased time dilation, (unlike your twin rocket scenario).

    Just take your original arrangement, where I presume you agree that the passengers all feel the same forces, and view it from an observer who is driving in a car at the same x-component-velocity as one of the rockets . From the car frame, one of rockets would be stationary along the x axis, but the other rocket would be moving along the x axis. And yet the passengers would still all feel the same forces. The forces they feel don't change just because you look at them from a moving car.

    I think you're over-simplifying it a lot.
     
    Last edited: Jun 17, 2018
  14. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,910
    The passengers in the two rockets will feel different accelerations and forces. If I get in a car and chase the rocket moving along the x-axis, I will now see the two rockets passing a given point on the y-axis at different times, so the situation isn't symmetric. You can't treat either rocket as being at rest along the x-axis with equal results, because you've already specified that an observer sees them both accelerating along the y-axis at the same rate while the x-position of one of the rockets is fixed.

    Passengers in both rockets would feel the same force if they both had, for instance, the same velocity along the x-axis relative to the static observer, in opposite x-directions. However, that's not the situation you've specified, and what you've specified can't be morphed into this situation by a change of inertial observers.
     
    Last edited: Jun 17, 2018
  15. Q-reeus Valued Senior Member

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    3,411
    An empty statement since by definition relativistic mass (strictly, energy) = proper mass when relative v = 0.
    Another confusing statement. The equivalence principle more specifically WEP (weak equivalence principle) merely states that locally i.e. ignoring tidal effects, uniform rectilinear acceleration and gravity are indistinguishable. Further, radial motion wrt Earth is a totally different case to that discussed here, which is strictly about transverse motions wrt a source of a or g. Stop muddying the waters.
    For the limited situation discussed by OP, where only intensive not extensive effects are relevant, it works out just fine. As shown here and over at PF thread.
     
  16. Q-reeus Valued Senior Member

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    3,411
    Wrong. You also don't understand SR. In stationary i.e. inertial observer frame, as y velocities increase, x velocities MUST start decreasing. It all comes out by consistent application of those relativistic transformations for acceleration I gave way back in #18. Again: https://en.wikipedia.org/wiki/Acceleration_(special_relativity)
    It can be understood as satisfying conservation of 3-momentum. In particular the x-axis components, given a continuously increasing transverse mass owing to y velocities.
    More confusing fluff. WHY would the passengers experience different proper accelerations?! Just because we choose a rest frame where one rocket has zero x-axis velocity? Nonsense thinking. Equal proper accelerations in both rockets was specified! Butt out of here - you only came in in the first place out of spite over my posting in another thread. Well instead of your intended stealing the show, you've simply made a fool of yourself. Repeatedly.
     
  17. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    You're a moron and you should go annoy people with your lack of knowledge somewhere else. As seen by the stationary observer, the x-velocities are fixed, the only acceleration they see is in the y-direction, as defined by the problem. We're not talking about the x-velocities as seen by the passengers in the rockets, and I'd appreciate it if you'd check your ignorance at the door before berating other posters or trying to appear knowledgeable.
     
  18. Q-reeus Valued Senior Member

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    3,411
    It's you that only tries to appear knowledgeable. I gave you a basic reason why, from an inertial observer's frame, x-components of velocity must decrease. Instead of name-calling, deal with the actual relativistic physics involved. Plug in the correct values in either the 3-acceleration or 4-acceleration formulas I linked to. My claim will be confirmed. That's SR. To repeat, there MUST be deceleration of x-component velocities seen in inertial frame. All without there being any proper x-axis accelerations experienced in either spaceship.
     
  19. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    5,910
    Don't complain about insults, you're the one who started trash talking, and you've been doing that with Neddy for several pages now. You don't apply coordinate transformations to determine the x-velocity in the inertial frame, that's already been specified. The inertial observer sees two rockets accelerating strictly along the y-axis, that means the same observer sees no change in velocity along x. Very basic definitions here.
     
  20. Q-reeus Valued Senior Member

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    3,411
    Rubbish. Unlike your foul-mouthed self, very evident over in that other thread and similar ones, I never resorted to name-calling insults with Neddy.
    Revealing your ignorance again. The x-axis velocities are an initial t = 0 spec. The actual physics requires those initial values to change with time as y velocities change.
    Too basic, and wrong. Instead of using the relativistic acceleration transforms, which clearly you are reluctant to try, one can simply examine the SR 4-velocity or 4-momentum expressions. Given their respective norms are invariants, it's easy to see that any change in 3-velocity or 3-momentum components along one direction necessarily effects any non-zero values in the other ones. Basic stuff - IF one actually understands SR that it. Again - butt out and stop further confusing Neddy with misinformation.
     
  21. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,910
    Using basic acceleration formulas, I conclude the same \(\gamma^2\) factor as you do. You're just terrible at deriving and explaining it, and need to stop berating other posters. There's no confusion here, an inertial observer sees two rockets accelerating at a constant rate along the y-axis with an initial relative velocity along x. That initial x velocity doesn't change in the inertial observer's frame of reference. It does change in the frames of the two rockets, but that's not what I've been discussing.
     
  22. Hayden Registered Senior Member

    Messages:
    110
    There are endless number of paradoxes in SR. It is kinda GIGO, Garbage in Garbage out.
     
  23. Q-reeus Valued Senior Member

    Messages:
    3,411
    Still wrong. Why not admit, and let's stick for now with 4-momentum P = (E/c, p). Since |P| is an invariant, and proper energy E is also here, it follows any increase in one component of 3-momentum p must result in a decrease in any other non-zero component that has zero proper force acting on it. Specifically, increasing p_y has to be compensated by a decreasing p_x. Which, given that the overall relativistic mass γm is increasing, demands v_x is decreasing.
    What I will have to take back was that appeal to conservation of x-component momentum given in #93. It had ignored the counter-intuitive implications of 4-momentum invariance.

    Just the fact that c is a limiting speed in any frame should be enough to see that as y-axis speed increases continually, x-component must at some point reduce just to satisfy
    (v_x)² + (v_y)² < c².
     
    Last edited: Jun 17, 2018

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