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Is my intuition correct?

  1. Nov 8, 2011 #1

    Matterwave

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    So, when we are talking about geodetic precession, my intuition tells me that the angular momentum vector of the spinning particle should be parallel transported along it's orbit. Because parallel transport of a vector around a closed loop doesn't reproduce the same vector, then the parallel transported vector will not point in the same direction as the original, vector, giving precession. Is this the correct intuitive understanding of geodetic precession?

    Also, I just finished reading Wald's chapter 4, and in the problems he introduces the Lense-Thirring effect. In the homework you show that a space-like vector S at the origin inside a spherical shell of mass rotating at angular frequency omega parallel transported along the geodesic (purely time-like, i.e., you are "at rest" at the origin) will precess due to the rotation of the outside spherical shell. I understand if the vector is like, e.g. an angular momentum vector and it would precess, but what if I just put a solid rod "free falling" (so it's really really short) at the center? Would the rod start to get dragged along with the rotation? (i.e. if I placed it somehow "at rest" w.r.t. the outside stars who's gravitational fields are negligible inside my bubble, would it start to rotate with the bubble?)

    Would this be a confirmation of Mach's principle, in that the exterior gravitational field determines what it means to rotate or not, rather than there being absolute rotation?
     
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  3. Nov 8, 2011 #2

    Dale

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    Yes, that is correct. This is the basis of gravity probe B.
     
  4. Nov 8, 2011 #3

    Matterwave

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    Thanks, how about my intuition for the Lense Thirring effect?
     
  5. Nov 9, 2011 #4

    Dale

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    I don't know. I was hoping someone else would.
     
  6. Nov 9, 2011 #5

    Matterwave

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    I'm hoping someone would tell me too hehe.
     
  7. Nov 9, 2011 #6

    pervect

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    My intuition also says that the center of a short rod should follow a geodesic, and that a vector drawn from the center of the rod to the end should be parallel transported.

    I tend to think of the rod as being a Born Rigid rod, and it's impossible to start one of those rotating, in the sense of having angular momentum, (at least in flat space-time) so one thing I'm lacking is a proof that Born rigid rods actually exist in the space-time in question. I suppose you'd also need a better formulation about what "not rotating" means and a proof that it works in curved space-time as well as flat.

    As far as what Mach's principle means, there's a lot of debate on the issue. My current feeling is the problem is that it means different things to different people. It's really up to someone to define and interpret the principle solidly enough that it can actually be tested experimentally. The current situation is that the words "Mach's principle" are associated with concepts that are similar , and related, but different, depending on one's interpretation.

    I don't see a lot of point in arguing semantics, so it's better to talk about what you really mean by "Mach's principle" and go on from there, rather than to argue about what the words mean.
     
  8. Nov 9, 2011 #7

    Matterwave

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    I guess my question is better phrased as such:

    If I were inside such a rotating spherical shell of mass which was rotating with respect to very distant stars (for convenience, I introduce the distant stars to help me define "rest"), and this shell was completely opaque to all EM radiation so that I could not ever detect the distant starts, could I create some experiment which would tell me that this shell was rotating?

    For example, suppose I could shoot/pin a dot (or several) to the interior of this shell to show me one (or several) specific location(s) on this shell, and then track that dot with a laser, then this laser beam could define my definition of a comoving frame. If there were absolute rotation, in the sense that Newton would describe a rotating frame as non-inertial, then I should be able to feel centrifugal forces beginning to arise. If I put a speck of dust in my comoving frame slightly displaced from the origin, then this speck of dust should, due to centrifugal force, move farther and farther away from the origin correct?

    If there was no such thing as absolute rotation, then, the comoving frame is, as far as anyone is concerned, an inertial frame of reference without any centrifugal forces. In this scenario then, if I put a speck of dust in this frame displaced form the origin, it should stay at constant radius.

    Is my thinking flawed in this sense? If not, which scenario would play out in actual physical reality (i.e. which scenario is favored by Einstein's GR)?
     
  9. Nov 9, 2011 #8

    PeterDonis

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    If you are inside a rotating spherical shell, spacetime is not flat. It's only flat inside a *non*-rotating spherical shell. I would have to dig out my copy of Wald to check my memory, but I believe the metric in the homework problem in question turns out not to be flat. (If a geodesic at the origin inside the shell precesses when parallel transported, I don't see how the metric *could* be flat.)
     
  10. Nov 9, 2011 #9

    Matterwave

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    So I've been thinking about this and it seems that we could solve the problem analytically if we knew the metric (and curvature) inside the shell.

    If we took coordinates in which the shell is rotating, and we place a particle at the center at rest (4 velocity (1,0,0,0)), and another particle at epsilon coordinate distance in the "r" direction "at rest" (i.e. it's 4 velocity in our coordinate system is (1,0,0,0), then we can use the geodesic deviation equation to see if the proper distance between these two particles increases or not right. If they do increase, then it seems that our coordinate system is actually non-inertial in the sense that an apparent centrifugal force arises.

    We can also examine if we gave that second particle initial velocity matching the rotation of the sphere, i.e. it's still placed a coordinate distance epsilon in the "r" direction but has initial 4-velocity (~1,0,omega*epsilon,0) where the 3rd slot is the phi direction (longitudinal direction), and see if the proper distance in "r" increases or not.

    Perhaps the solution which does not increase the proper distance r is in the middle of these two situations, with some acceleration in the phi direction?

    I'm not sure how to get the metric inside this sphere, so I'm not sure how to carry out these calculations.
     
  11. Nov 9, 2011 #10

    PeterDonis

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    The Wikipedia page on frame-dragging...

    http://en.wikipedia.org/wiki/Frame-dragging

    ...gives a formula for the "acceleration due to the Lense-Thirring effect" inside a rotating spherical shell. It would appear from the formulas that the "acceleration" referred to is relative acceleration of geodesics, i.e., apparent centrifugal and coriolis forces. From the article:

    There are two papers referenced, but unfortunately they're behind a paywall so I haven't looked at them to see details behind the formula.
     
  12. Nov 9, 2011 #11

    Matterwave

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    In that formula, I see some terms that look extremely like centrifugal and coriolis force. But are they using the coordinates that rotate with the body or coordinates which are "fixed" in some sense? It seems they are using "fixed" coordinates since for coordinates rotating with the body, omega would be 0 right?

    Would this confirm then, that a particle in this fixed coordinate system fixed at small epsilon distance from the origin would accelerate out to towards the sphere, and that therefore the rotation of the sphere really does determine (at least somewhat) what a "rotating frame" really means (i.e. our fixed frame is actually a "rotating frame" because these centrifugal forces are present)?

    The first term looks like coriolis, the second term looks like centrifugal, but what about the last term? It has no omega dot, so it doesn't look like the Euler force.
     
  13. Nov 9, 2011 #12

    PeterDonis

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    Google Books has an excerpt from a book by Cuifolini that talks about the metric inside a rotating spherical shell (hope this ultra-long link works):

    http://books.google.com/books?id=gG...nepage&q=metric inside rotating shell&f=false

    I won't post equations from the book, but two key features of the metric he presents are:

    (1) Test particles experience forces "formally similar to centrifugal and coriolis forces". In fact, if you look at his equations for the coordinate acceleration of a test particle geodesic, it is nonzero even on the *axis* of rotation of the shell. An object at the exact origin of coordinates does not accelerate, but an object on axis but not at the origin will oscillate about the origin, along the z-axis (the axis of rotation), with a period that depends on the angular velocity, mass, and radius of the shell. (The period is the *same* regardless of the amplitude of the oscillation in z, i.e., it is perfectly linear, so you could set up a bunch of particles all along the z axis, start them all off at mutual rest at some instant of time, and they would all oscillate in perfect step, all passing by the origin at the same instant.)

    (2) All inertial frames inside the shell are "dragged" (i.e., gyroscopes will precess) at a constant "angular velocity" proportional to that of the shell. (The proportionality constant depends on the mass and radius of the shell.)
     
  14. Nov 9, 2011 #13

    PeterDonis

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    That's the way it looks to me. But I think the Cuifolini book explains things better. (Of course, if the paper the Wiki page is quoting from were freely available, it would be easier to see how it explains things.)

    Yes, I think so, but see below for a further comment. The Cuifolini book excerpt appears to show the same thing for off-axis motion. (For on-axis, see my previous post.)

    I think one could define "zero angular momentum observers" or ZAMOs in a way similar to how it's done for Kerr spacetime, yes; and such observers would be "rotating" with respect to the asymptotically flat background spacetime "at infinity".

    As far as I can tell, it's that last term that causes objects off the axis of rotation to move outward towards the sphere. The other terms are all cross products so they would not cause a particle that had no initial radial velocity to acquire one, as far as I can see.
     
  15. Nov 9, 2011 #14

    Matterwave

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    the [itex]-\bf \omega\times(\omega\times r)[/itex] term is in the radial direction I believe.

    That last term would seem to suggest what you described above, which is that particles on the axis would oscillate up and down along the axis! Indeed that is a very weird force and does not appear in classical rotational frames of reference. (Perhaps the Euler force is missing because constant omega has been assumed)
     
  16. Nov 10, 2011 #15

    PeterDonis

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    Ah, you're right. That's the radial motion term; the third term, as you say, is the on-axis motion term.
     
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