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Is my understanding of the equivalence principle correct?

  1. Jul 11, 2012 #1
    I have been reading my books section on the weak equivalence principle over and over again, I think I understand it now, this is my understanding.
    Since all particles are accelerated by gravity at the same rate, no matter what they're composition or mass are, only a frame free falling with said particles can be considered to be inertial to them.
    However, because gravity fields are nonuniform from place to place and time to time, we can only construct approximate inertial frames over a small region in space-time.

    Please correct me if I am wrong
     
  2. jcsd
  3. Jul 11, 2012 #2
    Yes, you're right!
    Try to read the beginning of chapter 4 of Lecture Notes on General Relativity by Sean Carrol (you can find it at arXiv:gr-qc/9712019v1).
    It is impossible to disentangle the effects of a gravitational field from those of being in an accelerating frame, simply by observing the behavior of freely falling particles.
     
  4. Jul 11, 2012 #3
    Not all gravitational fields are non-uniform. If one has a flat spacetime then the gravitational field is uniform. The weak equivalence principle states
    What you referred to is the strong equivalence principle which says the same thing except that the free-fall frame is limited in spatial size as well as limited in time too. I.e. one is restricted to a small enough region in spacetime. In that sense you are 100% correct.

    Note: All partilces don't fall at the same rate. Particles fall at a rate which is independant on the mass of the particle. The rate at which a particle falls is dependant on the velocity in which it falls.
     
  5. Jul 11, 2012 #4

    PeterDonis

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    Correct.

    This is kind of a weird way of saying it, so I'm not sure if you are really expressing the weak equivalence principle here. The terms "inertial frame" and "freely falling frame" are just two different terms for the same thing; that has nothing directly to do with the WEP, it's just a matter of definition. See further comments below.

    The Wikipedia page appears to me to capture standard usage reasonably well:

    http://en.wikipedia.org/wiki/Equivalence_principle

    The first version of the WEP looks a lot like what I quoted above: "The trajectory of a point mass in a gravitational field depends only on its initial position and velocity, and is independent of its composition." But that says nothing about "frames". The WEP is independent of frames (at least I would say it is). What you say about frames is certainly *consistent* with the WEP, but I would not say it "is" the WEP. The WEP is just the physical statement that the motion of freely falling bodies is independent of their composition. It is true that this physical fact is what *allows* us to construct local inertial (freely falling) frames in which freely falling objects are at rest.

    You go on to say:

    This is correct, but it's something different from the WEP (at least I would say it is). This may be more of an issue of terminology than physics; it appears to me that you have a good grasp of the basic physics involved.
     
  6. Jul 11, 2012 #5

    PeterDonis

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    In flat spacetime there is no gravitational field--zero. I suppose that counts as "uniform", but I'm not sure it's what you meant.

    I'm not sure what you mean here. "Fall at the same rate" refers to the *acceleration*, not the velocity. More precisely, it refers to "coordinate acceleration", the "acceleration" that a freely falling, i.e., weightless, object will be observed to have by an observer who is accelerated, i.e., who feels weight. This acceleration is independent of the particle's velocity.

    The actual *trajectory* of the particle does depend on its *initial* velocity; obviously the trajectory of a rock dropped from rest at the top of a cliff will be different from that of a rock thrown upward from the same cliff. But the observed coordinate acceleration of both rocks, as seen by an observer at rest at the top of the cliff, will be the same.
     
  7. Jul 11, 2012 #6
    Although that's the popular way of interpeting the gravitational field, its not one I prescribe to. And it's not the way Einstein viewed it. According to Einstein, if you are in an inertial frame of reference in flat spacetime, you can transform to a new set of spacetime coordinates, corresponding to a uniformly accelerating frame of reference, and the spacetime will be indistinguishable from a uniform gravational field. That's Einstein's Equivalence Principle. In his paper The Foundation of the General Theory of Relativity, Albert Einstein, Annalen der Physik, 49, 1916 he wrote in section 2
    Spacetime curvature and tidal gravity are precisely the same thing expressed in different languages. What you said implies that there is a gravitational field present if and only if there are tidal forces present. Instead of that Einstein had the same viewpoint that if you were at rest in a frame of reference and you dropped an apple then the apple will accelerate to the ground and that action is to be interpreted as the action of a gravitational field. What you said denys that. In you viewpoint only a field with tidal forces in it has a gravitational field. If so then nobody has ever experienced a gavitational field because people have never experience tidal forces acting on their bodies.

    A uniform (aka homogeneous) gravitational field is a gravitational field where (1) The Reimann tensor vanishes and (2) not all of the Christoffel symbols vanish. The metric of a uniform gravitational field is components which vary with the spacetime coordinates. Its easier to percieve this if you stick to Cartesian spatial coordinates. Otherwise its easy to confuse accelrations due to the curvilinear coordinates rather than acceleration due to gravitational acceleration. The following is an example of a metric from a uniform gravitational field. Its identical to the metric for a uniformly accelerating frame of reference.

    ds2 = c2(1 + gz/c2)2dt2 + dx2 + dy2 + dz2

    Contrary to what you asserted, the existance of a gravitational field is not determined by the non-vanishing of the Reimann tensor, but by the non-vanishing of all the Christoffel symbols. I.e. no [tex]\Gamma[/tex]'s means no "gravitational field" ...

    This also means that the variability of all metric components determines the existance of a gravitational field. They are synonymous with the Newtonian gravitational potential. This is the reason why the metric is often referred to as the tensor potentials.

    The acceleration of a body in freefall is a function of the bodies velocity as well as the particles acceleration. Most GR texts point this out. Take a look at the geodesic equation. If you set it up right then on the left side there will be the acceleration and on the right you'll have the Christofel symbols contracted with the 4-velocities of the particle. This means that the acceleration is velocity dependant.
    Consider Newtonian gravity for a moment. If you dropped a stone from rest at t = 0 then the accleration due to gravity would not depend only on the initial speed. However, if you were throw the stone downwards then the gravitional acceleration would still not depend on the stones speed. Now lets use general relativity. If you through a stone downwards then the gravitational acceleration would depend on the velocity at which the stone was thrown. This is true in general throughout relativity, i.e. force is velocity dependant.

    This is easy enough to prove. Simply calculate the gravitational acceleration of a body in freefall first by using zero initial velocity and then use a non-zero velocity. I've done this myself. If I knew how to use Latex better I could show you ... when I'm not in a lazy mood that is. lol!
     
  8. Jul 11, 2012 #7

    atyy

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    Yes, that is eg. how Rindler says it. So more properly, the vanishing of the Riemann tensor corresponds to no "tidal" gravitational field. Since acceleration is fake gravity, many consider tidal gravity to be real gravity, ie. that which cannot be mimicked by acceleration, and so it is also common terminology to say that the existence of a gravitational field means that spacetime is curved.
     
  9. Jul 11, 2012 #8

    PeterDonis

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    It's true that the term "gravitational field" is ambiguous. I should have said (as atyy said) that flat spacetime corresponds to zero *tidal* gravity--i.e., zero curvature tensor. You are using "gravitational field" to refer to the connection coefficients, which I agree can be nonzero in flat spacetime. Sorry for the misinterpretation on my part.

    Locally, yes. But the term "uniform gravitational field" is problematic if you try to extend it beyond a local inertial frame. Two key issues:

    (1) The acceleration of "uniformly accelerating" objects is not "uniform" in the sense of being the same everywhere--at least, not if you are going to set up a coordinate chart in which the "uniformly accelerating" observers are all at rest. This chart is the Rindler chart, and the acceleration felt by observers at rest in the Rindler chart varies with position; it is not the same everywhere. So if you interpret the Rindler chart as describing a "gravitational field", the field is not "uniform"--it varies with position.

    (2) The Rindler chart does not cover all of flat Minkowski spacetime; it only covers "region I", the "right wedge" containing all hyperbolas with x^2 - t^2 > 0 and x > 0 (where x, t are the standard Minkowski coordinates, and I have suppressed the other two space coordinates). So the "gravitational field" described by this chart can't be interpreted as existing "everywhere", only in the region covered by the chart.

    Ah, I see what you were referring to. Yes, this is true, although it's a negligible effect unless the stone is thrown with a relativistic velocity. And the effect is still independent of the mass and composition of the stone, so the WEP still holds.
     
  10. Jul 12, 2012 #9
    That's also what it says in Gravitation by Misner, Thorne and Wheeler.

    I don't see it that way myself. It see it as being just as real as any other gravitational field.

    Yeah. I'm aware of that. I just disagree.
     
  11. Jul 12, 2012 #10
    Yes. I'm aware of all that too. Thanks. There are two ways to think of a gravitational field as being uniform in Newtonian gravity (1) the gravitational acceleration is the same everywhere in the field and (2) the tidal force tensor is zero everywhere. If we trtanslate to GR and keep the the #2 definition then the relativistic form of the tidal force tensor is the Reimann tensor and that is zero everywhere in a uniform graitational field.
     
  12. Jul 12, 2012 #11
    I forgot about an article that I have in my personal data base. It directly addreses your question. The article is What is the principle of equivalence?, Hans C. Ohanian, Am. J. Phys. 45(10)), October 1977. The abstract reads
     
  13. Jul 12, 2012 #12

    stevendaryl

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    The coordinates you are talking about are Rindler coordinates for flat spacetime, and they really AREN'T uniform. The magnitude of the "gravitational field" varies linearly with distance: it is stronger to the rear (in the direction the "field" points) and weaker to the front.
     
  14. Jul 12, 2012 #13
    If we have a uniform gravitational field, isn't that a pointless concept as everything will be accelerated the same, hence we won't be able to detect it
     
  15. Jul 12, 2012 #14

    stevendaryl

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    But the claim that the tidal force is zero is a fact about the curvature tensor, and not a fact about the "gravitational field", if you interpret the latter to be equivalent to the connection coefficients [itex]\Gamma^{\mu}_{\nu \lambda}[/itex]. So it seems a little schizophrenic to interpretation "nonzero gravitational field" to mean [itex]\Gamma^{\mu}_{\nu \lambda}[/itex]≠ 0 and to interpret "uniform gravitational field" to mean [itex]R^{\mu}_{\nu \lambda \tau}[/itex] = 0.
     
  16. Jul 12, 2012 #15

    stevendaryl

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    If you're going to talk in terms of a gravitational "field", then I think it makes sense to use a definition of "gravitational field" that reduces to the Newtonian concept of a "gravitational field" in the appropriate limit. The curvature tensor really doesn't.

    Of course, General Relativity doesn't actually require that you talk about the "gravitational field" at all; the curvature tensor is all you need (well, plus a lot of mathematics to extract observables from that).

    This is sort of nit-picky terminology, but in my opinion, it makes sense to say that "gravity" is present if and only if there is a nonzero curvature tensor, but to say that a "gravitational field" is present if and only if there is a nonzero connection coefficient (which is a coordinate-dependent fact).
     
  17. Jul 12, 2012 #16

    A.T.

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    That's the whole point of the EP. You cannot detect the difference between a uniform gravitational field, and an uniformly accelerated reference frame.
     
  18. Jul 12, 2012 #17
    I disagree. You seem to have in mind the Newtonian definition of uniform where the gravitational acceleration is the same everywhere. That is not the wy I see it. In Newtonian gravity a uniform gravitational field is a field which the Newtonian tidal force tensor is zero. This is precisely how it transferst to general relativity, i.e. the curvature tensor is zero in a uniform gravitational field. That's the definition used in journals on the subject, i.e. they start with the definition of uniform/homogeneous as having a zero Reimann tensor.
     
  19. Jul 12, 2012 #18
    I disagree. The gravitational field is said to be uniform/homogeneous if and only if the tidal force tensor is zero. That is the fact about the "gravitational field".

    The presence of a gravitational field exists if not all the Christoffel symbols don't vanish. The presence of tidal forces is dictated by the non-vanishing of the Reimann tensor.

    Yes. That is how its defined in the GR literature. e.g. Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173
     
  20. Jul 12, 2012 #19
    I agree. That's why I don't think of gravity as a curvature in spacetime since you can have a gravitational field without spacetime curvature. The vacuum domain wall is a great example of that.

    I don't understand that comment. Theories don't require people talking. However one has to know what it means for a gravitational field to be present. There was an article on the subject in the American Journal of Physics where the author showed that a charged paticle falling in a uniform graviational field radiates. The author had to know what a uniform gravitational field was and its expression in order to make his calculations.

    Not if someone wishes to know the equivalence principle as Einstein stated it.
     
  21. Jul 12, 2012 #20
    That was beautifully stated sir, beautifully!! :biggrin:
     
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