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Free fall acceleration in SR

  1. Feb 25, 2013 #1
    This is a spin-off of a parallel discussion, starting from:
    https://www.physicsforums.com/showthread.php?p=4281037#post4281037

    The question is what SR predicts that an accelerometer in free-fall will read. This issue may be simply due to different people using a different meaning of "SR", but it could have a deeper cause.

    A basic reference for this discussion:
    Einstein 1905, http://www.fourmilab.ch/etexts/einstein/specrel/www/
    and another one for context:
    Langevin 1911, http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time

    SR uses the inertial frames of classical mechanics; in my opinion it's obvious that SR doesn't predict that an accelerometer in free fall will indicate a large acceleration. That conflicts with the known laws of physics, even of classical mechanics.

    Arguments in favor of both opinions may help to clarify this issue.
     
  2. jcsd
  3. Feb 25, 2013 #2

    Dale

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    SR predicts that an ideal accelerometer reads:[tex]a^{\mu}=\frac{d^2 x^{\mu}}{d\tau^2}[/tex]Where x is the worldline of the accelerometer in a SR inertial frame and τ is the proper time along that worldline.

    If you claim that a frame can be treated as a SR inertial frame, then the above formula is what SR predicts for the proper acceleration of objects.

    That question cannot even be addressed by SR since SR does not handle gravitation. You cannot generally establish a self-consistent SR inertial frame in the presence of gravity. This question is outside the domain of applicability of SR.

    The real question is when is it appropriate to use SR as an approximation in a scenario where there is gravity. The answer to that question is that it is appropriate to do so when the resulting errors for the measured quantities in the scenario are small. That is not the case in a gravitational turn-around twin scenario.
     
    Last edited: Feb 25, 2013
  4. Feb 25, 2013 #3
    Your definition does not appear in the provided references which don't even use that term; I can find no reason for expecting such a flagrant error in SR. The first reference defines SR wrt the inertial frames of classical mechanics; wrt such frames the laws of Newton hold in good approximation. An ideal accelerometer that in free fall reads zero would be a colossal erroneous modification of Newton's mechanics. Please provide a reliable reference to back up that claim.
    SR was always assumed to be applicable on Earth (of course, in good approximation!); for example CERN uses it in the presence of gravity.
    That's certainly not the topic of this thread, and the preliminary discussion was from about here: https://www.physicsforums.com/showthread.php?p=4281150. I hold that by design the resulting error for that case must be small whatever the details of the turn-around, as it was meant to be of sufficiently short duration to be neglected. It's certainly worthy of its own thread, with a detailed calculation.
     
    Last edited: Feb 25, 2013
  5. Feb 25, 2013 #4

    stevendaryl

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    You can say that that's true by definition of "ideal accelerometer", but you can't prove that there is any actual device that is an ideal accelerometer.

    The canonical example of an accelerometer that I always use is a cubical box, with a ball suspended in the center by 6 identical springs connected to the center of each wall of the box. You measure acceleration by the deflection of the ball from the center.

    But if there were a force that pulled springs, ball and box in proportion to their masses, then acceleration due to this force would not be measurable.

    Another alternative for an accelerometer is to use light beams: if the light beam travels straight, then the device producing the beam is not accelerating. If the light beam curves in one direction or another, then that indicates acceleration. Maybe this can be used to measure acceleration? It all depends on whether light itself is affected by forces.
     
  6. Feb 25, 2013 #5

    Mentz114

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    Did you mean to write this ? You must mean something completely different from what I understand to be 'free fall'.
     
  7. Feb 25, 2013 #6

    Dale

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    See these references:
    http://physicspages.com/2011/05/25/acceleration-in-special-relativity/
    http://www.mth.uct.ac.za/omei/gr/chap2/node2.html and http://www.mth.uct.ac.za/omei/gr/chap2/node4.html
    http://en.wikipedia.org/wiki/Four-velocity#Definition_of_the_four-velocity and http://en.wikipedia.org/wiki/Four-acceleration

    The formula is correct. If you disagree, then please provide the formula you believe is correct.
     
    Last edited: Feb 25, 2013
  8. Feb 25, 2013 #7

    Dale

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    That is correct. The only such forces within the domain of applicability of SR are fictitious forces, and fictitious forces are not measured by accelerometers. The only other such force in any mainstream theory is Newtonian gravity, which is incompatible with SR and therefore outside its domain of applicability.

    I believe that one consequence of the first postulate is that your mechanical and your optical accelerometers will read the same.
     
  9. Feb 25, 2013 #8

    Mentz114

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  10. Feb 25, 2013 #9

    WannabeNewton

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    I would like to know what he/she meant by that too because it certainly isn't my understanding of free fall.
     
  11. Feb 25, 2013 #10

    stevendaryl

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    I wasn't sure about that, either, but he might mean that according to Newtonian physics, an object in freefall is accelerating under the force of gravity. So an "accelerometer" that measures zero in freefall isn't correctly measuring the Newtonian notion of acceleration. But then, that just means that there is no (localized) device that can measure acceleration in the Newtonian sense.
     
  12. Feb 25, 2013 #11

    Mentz114

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    Yes, I guess that's what he means. In the other thread GregAshmore mentions 'unbalanced forces'. I suppose an accelerometer can be said to measure resistance ( reaction ) to an applied force ( speaking in Newtonian terms ).

    It shows how tricky gravity is an why it makes sense to geometrise it.
     
  13. Feb 25, 2013 #12

    PeterDonis

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    You should have included the bolded phrase; as DaleSpam pointed out, it's crucial. SR can't deal with curved spacetime; the spacetime in the "gravitational turnaround" scenario, where the traveling twin is in free fall the whole time, is curved. It has to be, otherwise the traveling twin couldn't be in free fall the whole time.
     
  14. Feb 25, 2013 #13

    PeterDonis

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    You left out a phrase here too, and again it's crucial. The CERN people don't analyze experiments including the effects of gravity; they analyze experiments in a local inertial frame in which the effects of gravity are negligible. That's why they can use SR for the analysis: because SR is still valid within a local inertial frame.
     
  15. Feb 26, 2013 #14
    Obviously I did not, thanks for pointing that out! Indeed, I meant the inverse, as I stated earlier and even in my first post here. An ideal [mechanical] accelerometer that in free fall doesn't read zero would be a colossal erroneous modification of Newton's mechanics.
     
    Last edited: Feb 26, 2013
  16. Feb 26, 2013 #15
    I checked your first two references, which according to my browser do not even contain the word "accelerometer". I didn't look further.

    ADDENDUM: I overlooked that you asked me to state the obvious. You suggest(ed?) that SR proposed laws that were known to be erroneous. All parts of an accelerometer fall at the same rate in a homogeneous field; to suggest that according to SR this known fact would not be true not only makes no sense to me, it also doesn't follow from either the postulates or the Lorentz transformations.

    BTW, I suddenly notice in post #2 a change of opinion about the SR predicted accelerometer reading; that's good, it means that such discussions have merit. :smile: It's very well possible that I'll similarly change my opinion, but I need to see some substantial argument.
     
    Last edited by a moderator: May 6, 2017
  17. Feb 26, 2013 #16
    That could be a lead; please elaborate.
     
  18. Feb 26, 2013 #17
    With "fall" I mean the standard meaning of "falling" in SR and English, of a freely moving object in a gravitational field. That is in contrast to "inertial".
     
  19. Feb 26, 2013 #18
    The surface of the Earth is not a local inertial frame, but as discussed earlier we fully agree on the fact that we work with approximations. You appear to hold that SR doesn't make any prediction in a gravitational field. However, CERN certainly is in a gravitational field.
     
    Last edited: Feb 26, 2013
  20. Feb 26, 2013 #19

    PeterDonis

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    No, I said that SR can't explain how the traveling twin in the Langevin version of the twin paradox doesn't feel any force during the turnaround. You can still derive the difference in proper time between the stay-at-home twin and the traveling twin using SR, by making the inertial legs very long compared to the turnaround; but that doesn't explain how the turnaround can be made in free fall. And you can't finesse this by saying "well, it's all in a local inertial frame", because the free-fall turnaround cannot be made within a single local inertial frame. This is a crucial difference between that case and the CERN case; CERN experiments can be analyzed within a single local inertial frame because they happen so fast and over such a short span of distance compared to the size of the Earth.
     
  21. Feb 26, 2013 #20

    Dale

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    The definition I gave and the pages I linked to give the formula for the four velocity and explain it's relationship to proper acceleration and the coordinate acceleration in the momentarily comoving inertial frame. I am sure that you are aware that proper acceleration is the acceleration measured by an accelerometer.

    What I asked you to state is the formula that you believe SR uses to predict the reading on an accelerometer, which, with characteristic evasiveness, you have failed to do.

    This has become a pattern with you. I refer to some unambiguous mathematical expression, you claim it is wrong, I ask you to provide what you believe to be the correct mathematical expression, and you fail to do so, over, and over, and over, ...
     
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