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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    Regardless of whether it feels right to you, that 4x boost in belt frame is correct. It's just how the maths of SR works out. As I have tried to show, one can tackle it from multiple pov, in either lab or belt frame, and as long as SR is consistently applied, the end result always agrees.
    SR stands for special relativity. Relative to the belt frame, the lab frame is undergoing a vertical acceleration 4x that measured by a stationary accelerometer in the lab frame. As the maths shows. This is not Galilean relativity. It's not Newtonian physics. But SR has been experimentally and observationally confirmed to very high precision in countless situations. So we accept it as true.
     
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  3. Q-reeus Valued Senior Member

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    It will entirely depend on which particular reference frame one picks to evaluate the overall scenario. You specified only a relative velocity, not which reference frame to evaluate from. If ship A is chosen, from that pov, ship B, at the moment of coincidence with ship A along x axis, has an instantaneous acceleration 1/γ² = 1/4 times that experienced in ship A. That follows from the inverse transformation eqn given in (1c) of that Wiki article I gave earlier. Now, from ship A's pov, the mass in ship B weighs more by factor γ = 2. Further, from ship A's pov, the scale in B is 'relativistically weakened' by that same factor of 2. Multiplying those three factors together, one has that the scale reading in ship B is identical to that in ship A. And of course from ship B's pov, the above will reciprocally apply to ship A. All is well in SR land!
     
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  5. Neddy Bate Valued Senior Member

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    That was part of the point I was trying to make. The spaceships' relative velocity sideways along the x axis is purely relative, so, since it is given that the on-board scales read the same as they would on earth, there would be no way to choose which of the two spaceships would need to travel at a different rate upwards along the y axis, if that were to be necessary for some reason.

    Thus, we can conclude that the two spaceships would not need to travel at different rates upwards along the y axis, and therefore they must be co-moving with respect to the y axis. The distance between them along the y axis never changes over time, even though the distance between them along the x axis does constantly change over time.

    Okay, all of that may well be the case, as long as the two scales end up displaying equal values, and as long as you agree that the two spaceships are co-moving along the y-axis.

    If you agree, then I would ask you to please direct your attention back to my original scenario where the entire lab frame, including the treadmill device, accelerate upward at one uniform rate, in order to produce one "gee" of force within that frame. That means the spaceship scenario is a perfect analogy, and the scale at rest on the belt physically displays the same weight that is displayed on the scale which is on the floor of the lab. Correct?

    As for my revised scenario, where the scale is underneath the whole treadmill device, that scale might display a different value, I don't know. But the one on the belt and the one on the floor display the same value, right?
     
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  7. Q-reeus Valued Senior Member

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    That above reasoning and conclusion only holds from an inertial reference frame where velocity along axis x of ship A is equal in magnitude and opposite in direction to that of ship B. Move away from that symmetrically positioned reference frame, and your reasoning above fails. Moreover, the evaluation I gave in #22 applied only to the instant the two ships crossed paths.
    At a later situation as measured in either ship's frame, y acceleration results in a y component of velocity, and that makes the whole evaluation more complex. Still gives results consistent with SR, but with considerably more involved maths.
    See above. Co-moving from which pov, from which frame of reference, and/or at which instant in time?
    No. See above. My summary in #13 remains correct.
    No. See above. The scale under the treadmill case is as given by point 2: in #13.
     
  8. Neddy Bate Valued Senior Member

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    Sorry but that is incorrect. The distance between the two spaceships (as measured along the y axis only) never changes, at any time, in the reference frame of ship A, and it also never changes in the reference frame of ship B either. We would have to be able to agree on this in order to both be discussing the same topic.
     
  9. Q-reeus Valued Senior Member

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    In various versions of the twins paradox, the concept of a distant 'now' is a somewhat arbitrary inferred thing, and relative age is only strictly determined when the two twins meet again at a common spacetime event. Likewise, if you try and do the actual maths from either spaceship's frame, the inferred y displacement of the other spaceship at a given distant 'now' will not only differ from the inferred y displacement of your spaceship, but you must deal with an ever changing curvilinear 'x-y' grid. Owing to the accelerated motions.
     
  10. Neddy Bate Valued Senior Member

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    1,575
    Look at what we were calling the lab frame, with its treadmill-style device sitting on the floor. The belt always moves along the x axis of both frames (parallel to the floor). The belt never moves along the y axis of either frame, not at any moment in time, not anywhere along its length. Now, factor in the acceleration, and you still have every event on the belt having the same y coordinate, regardless of the time, so even relativity of simultaneity can't change that. Unless you can explain to me how the "ever changing curvilinear 'x-y' grid" shifts for the conveyor belt scenario?
     
  11. Q-reeus Valued Senior Member

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    The treadmill scenario is conceptually quite distinct from your two spaceships one. I was obviously dealing with the latter in #26. Even in the treadmill case, there is a very mild distortion of coordinates occurring but in a way different from the two spaceships case. One required by the equivalence principle.
    Having wasted a fair amount of time searching for a suitable online resource, the best I have found not too heavy on advanced maths is:
    https://web2.ph.utexas.edu/~gleeson/RelativityNotesChapter6.pdf
    Try reading it - maybe something will click.
     
  12. Neddy Bate Valued Senior Member

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    1,575
    My argument is that the distance between the spaceships (as measured along the y axis only, perpendicular to the relative velocity v) can remain constant at all times in both of their reference frames, as well as the third reference frame that you had mentioned (in which they both have equal x-velocities in opposite directions). The spaceships can be co-moving along the y axis of that third reference frame, in such a way that they can both feel one "gee" due to their upward acceleration. If you disagree, then tell me why two identical spaceship which both feel one "gee" at all times cannot both be accelerating upward at the same rate?

    Once you see that the distance between the spaceships (as measured along the y axis only, perpendicular to the relative velocity v) can remain constant at all times, you will see that the spaceship scenario is identical to the treadmill scenario. This is because the distance between the treadmill and the floor (as measured along the y axis only, perpendicular to the relative velocity v) also remains constant at all times. I don't know why you can't see that yet, but be that as it may, we are stuck there until you do.

    Thanks for the link, I will give it a try when I have time.
     
  13. Neddy Bate Valued Senior Member

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    1,575
    Perhaps it would help if we build the spaceship scenario entirely from an inertial reference frame. We will call our inertial frame K.

    Let two identical spaceships move together up the y axis of frame K in such a way that they both feel one "gee" on board, similar to being on earth. So far, the spaceships have no relative velocity, because we have not set them in relative motion yet. But please note that the two spaceships do not separate from each other along the y axis. They maintain the same upward movement as one another, according to frame K.

    At some moment of our choosing, we let both spaceships start to separate from each other in the x direction, while still maintaining their identical upward motion. We give spaceship A an x-component to its velocity which we could call -u where the negative sign means it moves to the left or the -x direction. We give spaceship B an x-component to its velocity which we could call +u where the positive sign means it moves to the right or the +x direction.

    By maintaining symmetry in this way, it should be pretty easy to see that the spaceships still maintain the same upward movement as one another in frame K. This holds true even though their x coordinates are not the same over time. The x coordinate of ship A would change according to x_A=-ut, and the x coordinate of ship B would change according to x_B=+ut. All of this is measured in frame K.

    On board each of the two spaceships, the one "gee" environment lets them weigh their identical test masses with their identical spring scales, and they both get the same reading as if they had weighed them on earth.

    ------

    Now you might say that frame K is the only frame in which the spaceships maintain the same upward motion as each other, and perhaps that is true. But what is more important is that we can build the treadmill scenario in frame K, and we get the exact same result by analogy, as follows:

    ------

    Let there be a treadmill on the floor of a lab, and let the whole lab move up the y axis of frame K in such a way that they both feel one "gee" on board, similar to being on earth. So far, the belt and the floor have no relative velocity, because we have not set them in relative motion yet. But please note that the belt and the floor do not separate from each other along the y axis of frame K. They maintain the same upward movement as one another, according to frame K.

    At some moment of our choosing, we let a point on the belt and a point on the floor start to separate from each other in the x direction, while still maintaining their identical upward motion. We give the floor an x-component to its velocity which we could call -u where the negative sign means it moves to the left or the -x direction. We give the belt an x-component to its velocity which we could call +u where the positive sign means it moves to the right or the +x direction.

    By maintaining symmetry in this way, it should be pretty easy to see that the belt and the floor still maintain the same upward movement as one another in frame K. This holds true even though their x coordinates are not the same over time. The x coordinate of the point of the floor would change according to x_A=-ut, and the x coordinate of the point on the belt would change according to x_B=+ut. All of this is measured in frame K.

    On the floor, and on the belt, the one "gee" environment lets them each weigh their identical test masses with their identical spring scales, and they both get the same reading as if they had weighed them on earth. Note that this is all perfectly analogous to the spaceship scenario.
     
  14. Q-reeus Valued Senior Member

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    That both experience proper accelerations of 1g does not translate into either determining the other's acceleration as 1g. The maths says otherwise - as already covered.
    No they are quite different in character.
    I never claimed otherwise wrt that treadmill case - but that situation doesn't translate holus-bolus into the two spaceships scenario. The latter is more complicated and I suggest to go back and ponder only what I wrote regarding the treadmill scenario. And that applies to your #30 also - stick to the original treadmill case. Mixing up SR and Galilean relativity only confuses and creates false paradoxes.
    Hope you do.
     
  15. Neddy Bate Valued Senior Member

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    1,575
    Please read my post #30 where I start with a completely inertial (non-accelerating) reference frame K, and then use it to describe the spaceship scenario. After that, I use the exact same wording and equations to describe the treadmill scenario. They are identical. They both involve purely inertial relative motion in the x direction, and accelerated motion in the y direction. I don't understand how you can claim they are different.

    At first post #30 might appear to be rather long, but the second half is practically a word-for-word repeat of the first half. The first half is the spaceship scenario, and then the second half is the treadmill scenario.
     
  16. Confused2 Registered Senior Member

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    487
    Stunning thread - thank you.

    Not my thread but can I suggest a slightly different approach using geometry?

    Staying with the treadmill in a lab..(I thought OP worked well)

    Before starting...

    We mark the start and end of the treadmill in inertial frame K as A and B.

    So (obviously) the treadmill is of length AB.

    We place a clock on the treadmill. At time t0 and t'0 we accelerate the lab upwards at (say) a and start the belt going at a velocity v that gives a γ = 1/√(1-v²/c²) of (for convenience) 2.

    In the lab frame it takes t=AB/v for the clock to travel along the belt. While the clock has been travelling the lab has advanced in the inertial frame and the end of the treadmill is now (Newton's Laws of Motion) at (say) C where BC=0+(1/2)at². Due to time dilation wrt the lab frame the clock on the belt records a time t' which is (at the chosen velocity) t'=0.5t.

    It is apparent that the path AC is a hypotenuse so longer than treadmill length AB but I think we can ignore this if v >> a.

    Since both the lab and the clock have advanced wrt K by the same amount we have BC=(1/2)at²=(1/2)a't'² giving a'=4a. *IF* and only if we can ignore length contraction in the treadmill frame wrt to the lab frame. Can we ignore it? I'm not smart enough to know.
     
  17. Q-reeus Valued Senior Member

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    bad post
     
  18. Q-reeus Valued Senior Member

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    OK so I actually studied your #30 this time. Concerning the inertial K frame evaluation of treadmill scenario:
    At first I was agreeing with you (hence my last 'bad post'), but since the results in #13 are valid, something has to be wrong with your reasoning here.
    While it's true in K frame a mass on the lab floor will have equal and opposite instantaneous x-axis velocity to one on the belt, hence equal relativistic mass and scale stiffness applies in K, we must continue to have (belt scale reading) = γ² (lab floor scale reading). Consequently the respective accelerations must vary by that factor. How?

    The source of acceleration is the rocket therefore in the lab frame. One must then apply the relevant transformations, the middle pair given in (1c) here:
    https://en.wikipedia.org/wiki/Acceleration_(special_relativity)
    Of course you have to know what the value is in lab frame already having an x-axis velocity in K. Anyway, for sure when the correct values are inserted, the resulting non-symmetric wrt K frame values will be consistent with those given in #13.
    How to explain it 'physically'? Well given there is a time-changing y velocity happening, that in turn effects the x-components of velocity seen in K. Which gives an overall somewhat messy situation where there is time-rate-of-change of Lorentz contractions happening - and that implies 'unexpected' additional accelerations as seen in K.

    Moral: make it easy on yourself. Always try and evaluate forces, accelerations etc. in a given proper frame, then just apply the relevant SR transforms to determine the required values in any other frame.
     
  19. Neddy Bate Valued Senior Member

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    Thank you for studying #30, and I am encouraged that you were agreeing with me for awhile.

    I still question how the results in #13 can be justified to be applied to the belt, but not applied to the floor. The frame K analysis shows them as two equivalent inertial frames, in exactly the same way that the two spaceships are two equivalent inertial frames. Whatever justification you could offer for applying the results in #13 to the belt but not to the floor, would also justify applying the results in #13 to one spaceship but not the other. Which would clearly be completely arbitrary since the spaceships are identical.

    Using frame K, could you please elaborate on the differences you perceive there to be between the spaceship scenario and the belt/floor scenario?

    If all of that is correct for the belt but not the floor, then it must also be correct for one spaceship and not the other. It just cannot be right, as far as I can tell. I swear, I am not doing this to prove you wrong, I don't take any pleasure in that. I would like us to agree on the equivalence of the two scenarios so that we can move on the UPDATED scenario in #12, where a scale is placed underneath the entire treadmill.

    Maybe it will help if I explain what lead me to make that update. I was thinking that if there is no relative motion between a scale and a mass, then the scale must weigh the rest mass and not the relativistic mass. That's all there was to it. You are talking about the acceleration being affected, and not even talking about the relativistic mass. Considering the spaceship scenario, I can only see the accelerations as being the same for both.
     
  20. Q-reeus Valued Senior Member

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    Your whole thinking remains embedded in Galilean relativity, or at best some hybrid of that and SR. I get the strong impression you are out to prove SR is wrong. It aint. Only wrong application of SR yields inconsistencies. When done in either the belt or lab frame, the exercise is comparatively easy and no paradoxes arise. As I have shown. Your 'simple' K frame scenario is actually quite a deal more complex to evaluate. You only think it's straight-forward owing to a Newtonian/Galilean-relativity worldview. To qualitatively expand a bit on what I wrote last time:

    In lab frame, at say t = 0, the rocket fires and there is a uniform upward acceleration a. In belt frame, the upward acceleration is not only amplified by factor γ², but is initially non-uniform wrt along x-axis - owing to non-simultaneity. One end of the belt initially accelerates upward before the other end, after which both ends accelerate uniformly. But that initial differential means the belt has acquired a slight tilt. (If that sounds crazy well sorry but it's just how it is in SR.) Which given the mildness of acceleration a is of no consequence in belt frame.

    However, in inertial frame K, non-simultaneity is also present wrt rocket acceleration, and in consequence the belt has a tilt induced there, of a magnitude somewhat less than that in belt frame. What is fundamentally different in K is that in addition to a tilt, the belt speed is decelerating owing to the upward acceleration. Note it's not an ordinary mechanical effect but purely relativistic - horizontal belt speed remains zero in belt frame.
    I'm not prepared to labour with the maths, but am confident that when worked through in detail, you will find the combination of tilt and deceleration of belt speed seen in K will provide that necessary final result consistent with those given in #13. Yes - it really is a complicated mess to work out in detail!

    As for your belief that the weight of a mass sitting on a scale is always the proper one regardless of speed in a given frame, I ask you to go back and ponder the example given in 2nd main para in #15. If you reject that example then I quit as it means you flat out reject SR.

    Still, there are other avenues for you to take. Why not field your K-frame example in a forum with a far greater physics talent pool - say PhysicsForums?
    [If you do take that route, please post here a link to any such thread. I would be interested to see how others handle it.]
    Or do an online search for articles dealing with acceleration in SR, and email query the author(s)? I gave a link to one such in #28.
     
    Last edited: Jun 10, 2018
  21. Neddy Bate Valued Senior Member

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    I think you are redefining the inertial frame you are calling K to be one where the whole lab only moves upward, and has no movement along the x axis of that inertial frame. If so, then I would suggest calling it inertial frame K' because I had already defined what I called frame K to be an inertial frame where the lab moves to the left, while the top of the belt on the treadmill moves to the right, both at the same speeds. I did this for symmetry, in an attempt to get Q-reeus to see the equivalence between that scenario and the spaceship scenario.

    But admittedly, it would be simpler to let the lab accelerate upward on the y axis of an inertial frame such as K' so that the whole lab has no horizontal movement, even though the belt on the treadmill does. The hypotenuse of your triangle and the time dilation of the your clock on the belt could certainly be calculated, with respect to frame K'. But we would have to also consider that the lab itself is accelerating upward through frame K', so the lab clocks would not match the clocks of frame K'. But at least frame K' and the rest frame of lab itself (not including the moving belt) could be compared with regard to their mutual x axes, and we would not have to worry about length contraction or relativity of simultaneity along that axis.
     
    Last edited: Jun 11, 2018
  22. Neddy Bate Valued Senior Member

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    I assure you that I am not out to prove SR wrong. I was hoping to make a simple scenario that would help me understand relativistic mass a little better. But perhaps, as you say, it is not as simple as I thought.

    This is something I had not thought of before, so please help me understand. In which frame does the belt become tilted?

    In frame K measurements, my chosen point on the belt is always co-located with the spaceship of what I consider (perhaps incorrectly) to be an equivalent scenario. Regardless of the belt's tilting, the point on the belt and the spaceship are co-moving to the right along the +x axis of K, while also co-moving upward along the y axis, in which direction they both accelerate. I think you agreed to that earlier? If not, please tell me what I am missing.

    Okay, this is another thing I had not thought of before. I think you are telling me that the spaceship and my chosen point on the belt would not be co-moving after all, because in frame K, the belt is slowing down due to increased time dilation of the lab frame. Wow, that is a new twist. Okay, I am starting to see that I might be in way over my head! I'll have to think about that some more.

    No, I do not reject SR. I simply have not come to terms with how relativistic mass works. I don't understand what you are saying in #15 but earlier you were saying something along the lines that two identical test masses would weight differently if one were much hotter than the other. The way I see it is that the atoms in the hotter mass are all moving fast relative to the scale underneath it. In the much colder case, the atoms could theoretically be stationary with respect to the scale underneath it, and so no relativistic mass increase would show up. That is all I meant, and seems consistent with what I was saying.

    I didn't think it would be this complicated, so I don't think I'm up for that task. What I want to do next is see if I can update the spaceship scenario (which I think you and I agree on) to be one where relativistic mass could be weighed. That will be tricky, of course.
     
    Last edited: Jun 11, 2018
  23. Confused2 Registered Senior Member

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    487
    As an interested but not knowledgeable observer of the thread..
    The treadmill is inherently a 'large interval of time' which creates complications .. my intuition (usually wrong) suggests you can get relativistic mass with a short belt. I don't have the knowledge or the time to go for it right now.
     

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