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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    I am trying to posit a simple way to use pure Special Relativity (SR) to approach relativistic mass. And from there, I think we might be able to make some conclusions about whether so-called "relativistic mass" gravitates or not.

    We start with a simple laboratory (lab) reference frame, and let it be inertial for now. We put a treadmill device at rest on the floor of the lab, and then turn on the treadmill so that the belt is moving horizontally at constant speed, parallel with the floor of the lab. Let the bed of the treadmill be arbitrarily long, like a long conveyor belt, so that we can treat an arbitrarily short segment of the belt as an inertial frame. We will call the inertial frame of this segment the belt the "belt" reference frame.

    At this point there is no acceleration in the scenario, and since this is pure SR, there is no gravity. So, a spring scale with a mass on top of it would have no reading, because the mass would be essentially "weightless" on the scale. This would be the case regardless of whether the scale & mass are at rest on the floor of the lab, or whether they are at rest on the belt. In both cases the scale reads the same, in this case a zero value.

    Now we let the whole lab frame accelerate upward at a constant rate. The floor of the lab and the belt now both have an acceleration similar to gravity, and the spring scale now displays a reading for the weight of the mass. There is still no difference whether the scale & mass are at rest on the floor of the lab, or whether they are at rest on the belt. In both cases the scale reads the same value, in this case a non-zero value.

    Thus we conclude that any so-called relativistic mass added to the mass due to it moving through the lab frame on the belt, does not contribute to its gravity, as measured by the weight of the mass. Is that conclusion warranted, or am I missing something?
     
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  3. NotEinstein Valued Senior Member

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    The scale gives a reading for the weight of the object (in Newtons), not its mass (in kilograms). You have to divide by the acceleration to get to the mass. I don't see how the weight of the object has anything to do with the relativistic mass in this scenario?

    (Also, it's dangerous to compare this scenario to any with gravity, because you've explicitly set gravity to zero at its start.)
     
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  5. Michael 345 Looking for Bali in Nov Valued Senior Member

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    I think I am reading this correctly but I'll put it into my words

    The acceleration 90° up at 1G I get

    You are questioning if the load moving horizontal to the vertical adds gravity

    If you consider the horizontal movement at 1G the combined result would be at 45°

    Since the result of 45° movement, in my thoughts, would slow the 1G vertical acceleration (or are you maintaining 1G regardless of horizontal movement?)

    Perhaps all I have put is BS - well so be it

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  7. Q-reeus Valued Senior Member

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    Yes you are doubly wrong. It never occurred to consider whether adding thermal energy to a dilute gas (such that inter-molecular or container-wall/molecule collisions occupy a negligible fraction of the average time molecules are in flight) would obviously raise it's gravitational/inertial mass? And that increase in turn must be associated with increased mean molecular KE? Specific calculations will bear that out precisely - I'm assuming you have enough grasp of SR to perform such calculations unaided.

    And further, that you can then figure that the 'relativistic mass' measured in (accelerating) belt frame is greater than when measured in the (accelerating) lab frame by factor γ = 1/√(1-v²/c²).
    Where v is the horizontal belt speed wrt lab frame, and in both measurements the mass is sitting on the moving belt. If the mass is sitting on the lab floor, there can obviously only be a measurement of the proper i.e. invariant mass - in that lab frame.
     
    Last edited: Jun 5, 2018
  8. Confused2 Registered Senior Member

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    A clock on the belt and a clock in the lab frame might be interesting.
    Edit... Is the acceleration the same in the belt and lab frames?
     
  9. Q-reeus Valued Senior Member

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    3,044
    No. As per #4, it's measured value in belt frame is higher by that factor γ = 1/√(1-v²/c²) γ*γ. A trivial application of (specifically, weak) equivalence principle.
    [Correction: In an earlier thread I had deduced the vertical acting gravitational acceleration, experienced by a passenger in a train moving horizontally wrt Earth's surface, must be greater than for a stationary observer by the factor γ*γ = 1/(1-v²/c²), not γ alone. That must, by equivalence principle, also apply to current scenario. Hence above is likewise a correction to what I wrote in #4.]
     
    Last edited: Jun 5, 2018
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  10. Neddy Bate Valued Senior Member

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    Agreed. I don't think I suggested otherwise, at least I did not intend to.

    The idea is to use pure SR and the equivalence principle to simulate a gravitational field. Or, if you are uneasy about the generalization to a gravitational field, just to test the behavior of relativistic mass in an elevator accelerating upward at a constant rate.

    The speed of the belt relative to the lab can be considered arbitrarily fast, so as to impart a relativistic mass increase in the test mass being weighed on the belt. The relativistic mass does not appear to show up in my scenario since, as you say, we simply have a mass, an acceleration, and a measurement of weight.

    But Q-reeus seems to think otherwise. I will reply to him next.
     
    Last edited: Jun 5, 2018
  11. Neddy Bate Valued Senior Member

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    1,596
    Okay now we are getting somewhere. This is what I wanted to know. So you are saying the spring scale should show a larger reading due to the relative velocity between the lab and the belt. So an observer on the lab floor says the belt is moving, and so he thinks the scale on the belt should show a greater reading than his own. But an observer on the belt says the lab is moving, and so he thinks the scale on the floor should show a greater reading than his own. Which one is correct?
     
  12. Neddy Bate Valued Senior Member

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    1,596
    An observer on the lab floor says the belt is moving, and so he expects the clock on the belt to be ticking at a slower rate than his own. But an observer on the belt says the lab is moving, and so he expects the clock on the floor to be ticking at a slower rate than his own. Both are correct, as strange as it sounds. Relativity of simultaneity ties up any loose ends.

    The whole lab, including the treadmill device, accelerate upward at a constant rate. This imparts a gravity-like sensation whereby the spring scales can show a measurement of the weight of the test masses.
     
  13. NotEinstein Valued Senior Member

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    Ah, I misunderstood your scenario, my mistake. You're saying that if the on-belt object's mass increases (due to the movement on the belt), it should also have a larger weight, and thus the scale should reflect that. Interesting scenario...

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  14. Neddy Bate Valued Senior Member

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    Yes, but I don't think the relativistic mass increase shows up in my given arrangement. I don't see how it can. Hopefully Q-reeus will be able to explain it.
     
    Last edited: Jun 5, 2018
  15. Neddy Bate Valued Senior Member

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    UPDATE:

    Thinking more about this, I have decided to put the whole treadmill device on top of a spring scale. The spring scale can be set to zero with the treadmill running at speed so that the weight of the treadmill itself is negated. Then a test mass can be placed on top of the treadmill belt, and weighed. This ensures the test mass is moving relative to the spring scale. In my previous arrangement there was no relative movement between the test mass and the spring scale, so this change fixes that issue.

    As a control, there can be an identical spring scale and an identical test mass at rest on the floor of the lab.
     
  16. Q-reeus Valued Senior Member

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    3,044
    Your new scenario is actually what I had in mind when stating in #4:
    "And further, that you can then figure that the 'relativistic mass' measured in (accelerating) belt frame is greater than when measured in the (accelerating) lab frame by factor γ = 1/√(1-v²/c²).
    Where v is the horizontal belt speed wrt lab frame, and in both measurements the mass is sitting on the moving belt."

    And btw I retract that part of my correction in #6 stating "Hence above is likewise a correction to what I wrote in #4."
    I had forgotten scenario in #4 had the mass riding on the belt in both scale readings, whereas in #6 the contrast was between weighing the mass in the lab frame and stationary wrt lab frame, vs weighing in the belt frame with mass on the belt.

    So to summarize, and given your 'new' scenario in #12:

    1: For the control with mass stationary in lab frame and weighed in lab frame, we simply have scale reading = -ma, where m is the inertial mass and a = |a| the magnitude of vertical acceleration.

    2: Mass on belt and scale stationary wrt lab frame and under the belt - then scale reading is -γma, with as before, γ = 1/√(1-v²/c²).

    3: Mass on belt and scale also on belt - then scale reading is -γ²ma.

    You can confirm all cases by working through the SR force transformation eqns, e.g. http://www.sciencebits.com/Transformation-Forces-Relativity

    A short cut is to note the transverse relativistic mass (obsolete terminology but still useful) is just γ times the proper value. Since every Eotvos type experiment ever conducted has confirmed that inertial mass = passive gravitational mass, 2: above follows directly by recourse to equivalence principle.

    A further transformation with scale on belt - 3: above - then follows from 2: by application of the relativistic transverse force expression found in that linked article above.
     
  17. Neddy Bate Valued Senior Member

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    Yes, I think my new scenario in #12 is the change that was needed to make the relativistic mass show up.

    1. Okay.
    2. I can accept that if there is relativistic mass, it would show up in the gamma factor here.
    3. Your answer here is something I find hard to understand or accept. You seem to be saying that the scale and the mass are both at rest with respect to each other in the inertial frame of the belt, so it seems to me that the speed of the belt with respect to the lab should not factor into it in that case?
     
  18. Q-reeus Valued Senior Member

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    3,044
    Yes, which is just in keeping with your revised scenario. And was my thinking in #4.
    As stated last post, it follows by applying a further transformation of force when scale is moved into the belt frame. To put it briefly f' = γf, where f' is the transverse force component measured in the belt frame, and f = -γma the transverse force - of the mass moving on the belt - measured in lab frame. So one finishes up with f' = -γ²ma as per 3:.

    Another way of seeing it is via equivalence principle. Suppose instead of acceleration a, an equivalent local gravitational field g = a is induced by a very long, stationary rod of uniform mass density. With the rod in motion at speed v along the rod's length direction, there is both an increase in overall rod mass by factor γ, AND a length contraction of the rod by factor 1/γ. Thus the rod mass density has been raised by factor γ*γ, and this will reflect in a consequent rise in locally observed g by that same factor.
    You should be able to see the essential equivalence to your scenario with acceleration and moving belt.

    A still further way of seeing it is to note that a moving object e.g. the spring in the scale on belt, is materially weakened wrt it's elastic modulus, relative to a stationary such scale, by that factor γ.
    Hence it will deflect more by factor γ than one might naively expect. I won't go into proving here what may be termed 'relativistic weakening' but it follows from considering how a spring oscillator frequency transforms, when in uniform motion vs when stationary in a given inertial frame. When I raised various consequences of such 'relativistic weakening' some years ago in PhysicsForums, it threw the resident GR expert - he and other pros there declined to respond. Wasn't an exercise in any of their textbooks I guess.
     
  19. Neddy Bate Valued Senior Member

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    The revised scenario has a scale at rest in the lab frame, underneath the entire treadmill device. The entire belt is in motion relative to the lab frame, thus that scale is not at rest with the belt.

    On the other hand, it was the original scenario had a scale at rest in the inertial reference frame of the belt. The scale was sitting on the belt, underneath the test mass, which was also at rest in the belt frame. So I do not see any reason for that scale to measure anything other than the weight of the proper rest mass, surely?
     
  20. Q-reeus Valued Senior Member

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    3,044
    In #15 three different but entirely complementary explanations for why that conclusion is wrong were given. Analyses carried out either in the lab frame or belt frame have to yield a consistent final result. I'm off on other things for a while, but in the meantime, do try and think through what I wrote last time.
     
  21. Q-reeus Valued Senior Member

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    3,044
    The example of 2nd para in #15 was meant to show that, by appeal to equivalence principle, the acceleration experienced in belt frame must be γ²a, not just a experienced in lab frame. To directly confirm that, simply apply the appropriate relativistic transform for accelerations given in e.g. (1c) here: https://en.wikipedia.org/wiki/Acceleration_(special_relativity)
    Do it right, and one has full consistency with everything in #15 and my earlier posts on this topic, as it must be. Still not convinced?
     
  22. Neddy Bate Valued Senior Member

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    So if the lab frame feels one "gee" of gravity due to its upward acceleration on its y axis, and if the belt is moving at a constant velocity of v=0.866c along the x axis of the lab frame, yielding gamma=2.000, you are saying the gravity felt on the belt would be 2*2=4 gees of gravity on the y axis? That doesn't seem right. You are saying inertial motion on the x axis somehow increases the acceleration on the y axis by a factor of 4?

    The belt frame is an inertial frame, and a test mass at rest on the belt has a velocity of 0.000c in that frame. Likewise, a scale at rest on the belt has a velocity of 0.000c in that frame. In that frame, the velocity of v=-0.866 describes the motion of the lab frame, which does not pertain to a mass on the belt or a scale on the belt. I can't see any reason to use v=0.866 when the scale and mass are both at rest on the belt.
     
    Last edited: Jun 6, 2018
  23. Neddy Bate Valued Senior Member

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    Imagine two spaceships in deep space, one passing the other at a relative velocity of v=0.866c along the x axis. Now let there be two identical spring scales and two identical test masses, each one at rest on each space ship. The scales measure the rest mass on each ship, and because they are in deep space they measure zero. Now let each space ship accelerate upward in such a way to produce 1 gee of gravity on board, while still maintaining a relative velocity of v=0.866c along the x axis. Now each scale measures the weight of the test mass as if it were on earth. Wouldn't the upward accelerations of the two spaceships have to be identical?
     

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