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A Theories that respect one form of the equivalence principle but violate another

  1. Apr 29, 2016 #1
    This extract is taken from Sean Carroll's textbook. It talks about a theory that respects the Weak Equivalence Principle but violates the Einstein Equivalence Principle.

    we could imagine a theory of gravity in which freely falling particles begin to rotate as they moved through a gravitational field. Then they could fall along the same paths as they would in an accelerated frame (thereby satisfying the WEP), but you could nevertheless detect the existence of the gravitational field (in violation of the EEP). Such theories seem contrived, but there is no law of nature that forbids them.

    Now, the Weak Equivalence Principle has any one of the following forms:

    1. the inertial mass is equal to the gravitational mass
    2. there exists a preferred class of trajectories through spacetime, known as inertial or freely-falling trajectories, on which unaccelerated particles travel - where unaccelerated means "subject only to gravity"
    3. the motion of freely-falling particles are the same in a gravitational field and a uniformly accelerated frame, in small enough regions of spacetime

    and the Einstein Equivalence Principle has the following form:

    In small enough regions of spacetime, the laws of physics reduce to those of special relativity; it is impossible to detect the existence of a gravitational field by means of local experiments.


    In the theory mentioned in the extract, what does it mean for the particles to fall along the same paths as they would in an accelerated frame and how does that satisfy the WEP?
     
    Last edited: Apr 29, 2016
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  3. Apr 29, 2016 #2

    PeterDonis

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    Imagine that you were in an rocket accelerating at 1 g (meaning an object at rest in the rocket would feel weight equivalent to a 1 g acceleration) in empty space, and you were standing on the "floor" of the rocket (the "floor" here would actually be the end of the rocket where the exhaust was being ejected), and you dropped an object. It would accelerate straight downward (towards the floor) at 1 g. And this behavior would be indistinguishable from the behavior of an object dropped by a person standing in an enclosed room on the surface of the Earth, feeling weight equivalent to a 1 g acceleration. A frame of reference in which the rocket, or the enclosed room, is at rest would be an accelerated frame, and the paths of the objects in both cases, relative to that frame, would be the same. This would satisfy the WEP.
     
  4. Apr 29, 2016 #3
    Thanks for the explanation!

    I now understand that the free-falling particles in both the accelerated frame (with acceleration ##\textbf{a}=-\textbf{g}## relative to some inertial frame (inertial as defined in special relativity)) and the gravitational field (with field strength ##\textbf{g}##) fall downwards in a helical motion. This respects the WEP.

    But how could we nevertheless detect the existence of the gravitational field?
     
  5. Apr 29, 2016 #4

    PeterDonis

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    Huh? That's not what I said. I said they will both fall straight downward.

    To test the hypothetical theory of gravity Carroll describes, you would look for whether the free-falling particles started to spin as they fell. If they did, they must be falling in a gravitational field (in this hypothetical theory); if they didn't, they must be falling in an accelerating rocket in flat spacetime.

    Of course, in actual experiments, we see no difference--the free-falling particles don't spin in either case. That tells us that hypothetical theories of the sort Carroll describes are ruled out.
     
  6. Apr 29, 2016 #5
    Well, the Weak Equivalence Principle states that the motion of freely-falling particles are the same in a gravitational field and a uniformly accelerated frame, in small enough regions of spacetime.

    Does this not mean that, if WEP is to be satisfied, then if the particle rotates as it falls freely in a gravitational field (in the hypothetical theory), it also ought to rotate as it falls freely in a uniformly accelerated frame in flat spacetime?
     
  7. Apr 29, 2016 #6

    stevendaryl

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    As I understand it, a particle with intrinsic angular momentum, such as an electron, will experience a very tiny spin-dependent force due to the rotation of a massive planet. This effect is analogous to the force between two magnetic dipoles in electromagnetism. If you take this tiny force into account (I think it's called the Lense-Thirring effect), then the path of an electron in freefall will not be a geodesic. So it seems that General Relativity does not actually satisfy the strong equivalence principle. Even in a very small region of spacetime, the motion of an electron would be different in the presence of a massive spinning planet than it would in flat spacetime.

    Maybe somebody who knows more than I do can clarify this: Does GR satisfy the strong equivalence principle, when you take into account spin?
     
  8. Apr 29, 2016 #7

    PeterDonis

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    No. The WEP only talks about the particle's trajectory. It does not talk about any other degrees of freedom of the particle. Rotation is a separate degree of freedom from the particle's trajectory, so the WEP doesn't say anything about it.
     
  9. Apr 29, 2016 #8

    PeterDonis

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    I don't think this is correct. A particle with spin will still follow a geodesic; it will just be a different geodesic than the one a particle with no spin, but all other attributes the same, would follow.

    Also, I don't think the Lense-Thirring effect (which Gravity Probe B was testing for) is limited to particles with spin; it's due to the angular momentum of the massive object, not any angular momentum of the test object, and is present for spinless test objects. But I believe there is an extra effect something like spin-orbit coupling, over and above the usual Lense-Thirring effect, which comes into play for particles with spin orbiting a spinning massive object.

    An electron moving in an electromagnetic field will not follow a geodesic; but that's because the EM field is exerting a force on it which results in it having nonzero proper acceleration. The Lense-Thirring effect and its relatives (see above) is a "gravitational" effect and does not cause nonzero proper acceleration. At least, as I understand it.

    Also, AFAIK there is a different effect due to spacetime curvature which can cause nonzero proper acceleration; it arises from different parts of an asymmetrical object interacting with spacetime curvature (tidal gravity) in different ways. But an object like this is not a "test object", and all versions of the EP only apply to test objects (objects that do not cause any spacetime curvature and have no internal structure). So the EP doesn't apply to this scenario.

    AFAIK, yes, as above.
     
  10. Apr 29, 2016 #9

    stevendaryl

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    I suppose that depends on how you define "geodesic", but if the path is spin-dependent, then it certainly is not a geodesic in the sense of extremizing [itex]\int d\lambda \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}[/itex]. But the main point is that it violates the equivalence principle, in that two electrons floating in empty space will not experience spin-dependent deviation, while the same two electrons in freefall will.
     
  11. Apr 29, 2016 #10

    PeterDonis

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    Yes, I see what you mean. I'll have to look up the actual math on this; I might be misremembering since it's been quite a while since I looked at it.
     
  12. Apr 29, 2016 #11

    PeterDonis

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    Ok, a quick look at the Gravity Probe B website reminded me of at least one thing I misremembered. The Lense-Thirring effect does not change the trajectory of an object; it causes the spin axis of a spinning test object ("gyroscope") orbiting a spinning massive object to precess. (More precisely, it adds an additional precession to the geodetic precession that is present for a gyroscope orbiting any massive object, spinning or not.) But the precession is not observable locally; you have to have a distant reference in a fixed direction (such as a distant star) to measure it. So the L-T effect does not violate the EP. (Also it does not affect spinless objects, so I misremembered that too. :oops: )

    Here is the Gravity Probe B website's page on the L-T precession (it also discusses other relevant effects like geodetic precession):

    https://einstein.stanford.edu/SPACETIME/spacetime4.html

    There is another issue involved here as well, to do with the effect of frame dragging on radial trajectories, but I'll save that for a separate post when I have more time.
     
  13. Apr 30, 2016 #12

    stevendaryl

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    Hmm. I misunderstood, then. Let's consider two massive circular rotating disks. What I thought was the case was that the force between the disks was different, depending on whether the disks were rotating in the same direction or opposite direction.

    In Wikipedia: (https://en.wikipedia.org/wiki/Gravitoelectromagnetism#Higher-order_effects)
     
  14. Apr 30, 2016 #13

    PeterDonis

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    Yes, this is true. But neither disk is a test object, since both are producing significant spacetime curvature; one way of thinking of the "force" produced is as an interaction between the spacetime curvatures produced by the two objects. What I said about the trajectory not being affected by the L-T effect was only intended to apply to test objects. Sorry if that wasn't clear.
     
  15. May 2, 2016 #14
    Even before considering rotation of the test objects or of the planet below them, in that case, aren't there are some little things to account for already?

    - a pair of test objects released at rest with respect to one another not subject to an external gravitational influence (deep space) will accelerate toward each other

    - a pair of test objects released at rest with respect to one another subject to an external gravitational influence (planet) will fall toward the planet's center of mass, so as to approach each other, and will further approach each other due to mutual gravitation

    - a pair of test objects released at rest with respect to one another subject to an acceleration (elevator) will have parallel vectors due to that acceleration, and also approach each other due to mutual gravitation

    The test objects' net mutual approach seems different in these three cases, least for deep space, most for planet, elevator in the middle (same as deep space up to relativistic energy increase from elevator perspective)?
    If the gravitational case is not a planet with a center of mass but a gravitational field without convergence, won't the source center of mass be either at an infinite distance or an infinite plane of some thickness at a finite distance... both impossible (and identical in effect to the accelerating elevator)?

    In the case of the planet, if the test objects are connected by a rigid rod, does the resistance to the various path components acting to laterally displace the test objects toward each other influence their vertical trajectories (does gravity care whether it's efforts to bring the test object's paths together toward the center of the planet's mass are being effective or wasted)?
     
    Last edited: May 2, 2016
  16. May 2, 2016 #15

    timmdeeg

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    http://arxiv.org/pdf/1206.7093v1.pdf
    According to Mathisson a spinning particle in a gravitational field performs a helical motion. Shouldn't this allow us to distinguish acceleration from gravity?
     
  17. May 2, 2016 #16

    PAllen

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    This effect is not inconsistent with the principle of equivalence. In reading the paper, note the discussion from the bottom of page 2 to the top of page 3. There is a minimum size of spinning body in GR if you don't want local violations of SR (FTL flows; captured in the dominant energy condition). This means that a spinning body is not really a test particle, as it is interacting with the curvature over a finite size. This is made more explicit in the following pater, which cites the above:

    http://arxiv.org/abs/1501.04879

    where (see the top of page 4), the effect is noted as precisely due to curvature over scale of the body - parallel transport around a closed loop not preserving the vector. Thus, this effect is due to the relative motion of parts of the body interacting with the tidal gravity over the scale of the body. This makes it irrelevant for evaluating the equivalence principle, which explicitly applies only on scales where tidal gravity can be ignored.
     
  18. May 2, 2016 #17

    stevendaryl

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    Yes, I thought about that--even if a spinning body as a whole doesn't follow geodesics, it could be that the little bits of matter that make it up could satisfy the equivalence principle. But what about particles with intrinsic spin?
     
  19. May 2, 2016 #18

    PAllen

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    In classical GR, there can be no such thing. Any spin rate for radius below some size implies FTL motion. Also note that the second paper notes that a spinning 'particle' will follow a geodesic if the there is no curvature detectable over a closed loop of characteristic size. That is, even with a spinning particle idealization (which this paper uses), the effect is only due to interaction with tidal gravity - thus irrelevant to equivalence principle.
     
  20. May 2, 2016 #19

    stevendaryl

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    My question isn't really about GR, but about people's best guess as to whatever extension of GR would be needed to allow particles such as electrons that have intrinsic spin. If it's really a point-particle, then I would think that tidal forces couldn't be important.
     
  21. May 2, 2016 #20

    PAllen

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    The paper in my post #16 suggests that a particle with intrinsic spin does detect tidal gravity within a characteristic radius determined by its spin parameter; and if tidal gravity is minimal within that radius, the particle would act no differently than a non-spinning particle. That is consistent with the tests of falling neutrons - they behave no dfferently than expected for a particle without spin (due to weakness of tidal gravity near earth). In particular, these experiments have shown that the free fall is unaffected by the direction of the neutron's spin axis.
     
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