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Inertial frames in SR and GR

  1. Jan 28, 2010 #1
    How are inertial frames defined in Special and General Relativity? In Newtonian physics, an inertial frame is usually defined as one in which N2 holds. Clearly this cannot be the same definition as for SR. In GR an inertial frame is one in which SR holds (I think). However, there is now a distinction between globally and locally inertial, and frames which are apparently accelerating towards each other can still be locally inertial.
     
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  3. Jan 28, 2010 #2
    In both SR og GR, an inertial frame is one which is moving along a geodesic in space-time. In SR this reduces to nonaccelerating frames as inertial.
     
  4. Jan 28, 2010 #3
    First of all, I concur with the reply above.

    Yes, in SR an intertial frame can be said to fill the entire space. N2 would be valid for all movements of particles as seen from this frame, no matter how far away.

    In GR this needn't be the case There could be tidal forces present, if the frame is located in a non-uniform gravitational field. So in GR it could be the case that the inertial frame is only exact in an infinitesimal volume, since deviations from N2 appear as soon as a particle is not in the origin of the inertial frame.

    The fact that each planet experiences tidal forces is a manifestation of this. They do not follow N2 because their size is finite.

    In GR, two intertial frames can accelerate towards each other. Just consider two frames, each freely falling towards a massive object, from opposite sides.

    Torquil
     
  5. Jan 28, 2010 #4
    Isn't the Earth moving on a geodesic? Would someone standing on a non-rotating Earth be in an inertial frame? And the geodesics depend on the frame we're working in, but it still holds that someone travelling along one will be in an inertial frame?
     
  6. Jan 28, 2010 #5

    sylas

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    Yes.

    No. They are kept on the surface by normal forces. A geodesic at the surface is the line traced by a falling object which otherwise does not interact with matter. The geodesics proceed downwards, through the surface. You can tell it is not inertial because you can feel weight.

    The Earth itself, however, is falling unimpeded around the Sun... a geodesic.

    In GR, a geodesic is a frame which is freely falling. As I understand it, in GR, a frame is really only local to the observer falling along the geodesic.

    Cheers -- sylas
     
  7. Jan 28, 2010 #6

    Ich

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    Earth's center is moving approximately on a geodesic.
    Geodesics at the surface are accelerated inwards by earth's gravitation, so someone standing at the surface actually accelerates away from a geodesic.
    Further, and maybe more pertaining to your question: a point on the surface is moving alway parallel to earth's center, but, due to tidal effects, experiences different gravitational "forces" from moon and sun. So even in the absence of earth's gravitational field, it doesn't move on a geodesic. That's how inertial frames are valid only for small regions in curved spacetime.
    No, geodesics depend on initial conditions (velocity) and the curvature of spacetime. They are not coordinate dependent.
     
  8. Jan 28, 2010 #7
    "No. They are kept on the surface by normal forces. A geodesic at the surface is the line traced by a falling object which otherwise does not interact with matter. The geodesics proceed downwards, through the surface. You can tell it is not inertial because you can feel weight."

    It seems a little strange that the Earth is moving along a geodesic and I'm moving along with the Earth yet I'm not moving along a geodesic. I suppose Ich is right that if I was in the centre of the Earth I would be moving along a geodesic.

    "No, geodesics depend on initial conditions (velocity) and the curvature of spacetime. They are not coordinate dependent."

    From the point of view of a man in freefall, he sits at the origin the whole time, whereas from the point of view of the Earth he traces out a geodesic. Or from the point of view of someone on a roundabout, a geodesic might be a spiral. They do depend on the frame of reference - that's where the affine connection come's into the geodesic equation.
     
  9. Jan 28, 2010 #8

    sylas

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    No; the geodesics are the same in all cases. All that is different is the co-ordinates of the geodesic. The geodesic itself is a set of points in spacetime... a path or a locus. The same set of points is a geodesic independent of what co-ordinate system you use to identify the points.

    Cheers -- sylas
     
  10. Jan 28, 2010 #9

    Ich

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    Approximately. An extended object is always subject to tides, so no point of it moves exactly on a gedesic. But the center is as close as possible.

    And sylas beat me to your second statement. There is only one geodesic, but infinitely many ways to assign coordinates to it.
     
  11. Jan 28, 2010 #10

    George Jones

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    Yes.
    This confuses a function with the image of a function, but I'm probably being overly pedantic here.
     
  12. Jan 28, 2010 #11

    sylas

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    Not at all! I'd much prefer to be as correct as I can manage, and I don't always get the terminology right. But I'm not sure of the issue here. A pedantic explanation of the best way to express it would be much appreciated!

    Cheers -- sylas
     
  13. Jan 28, 2010 #12

    George Jones

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    A geodesic depends not only on a set of points in spacetime, but also on how ("fast") the points are traversed. A curve in a spacetime [itex]M[/itex] is a mapping

    [tex]\gamma : I \rightarrow M,[/tex]

    where [itex]I[/itex] is an interval (often closed) of the real line. A curve [itex]\gamma[/itex] is a geodesic if the tangent vector [itex]\dot{\gamma}[/itex] is parallel transport along the curve, that is, if

    [tex]\nabla_{\dot{\gamma}} \dot{\gamma} = 0.[/itex]

    An observer is a future-direct timelike curve [itex]\gamma[/itex] parametrized by proper time. This means that

    [tex]g \left( \dot{\gamma} , \dot{\gamma} \right) = 1[/tex]

    all along the curve. In this case, the tangent vector "field" is the observer's 4-velocity, often denoted by [itex]\mathbf{u}[/itex], the observer's 4-acceleration is

    [tex]\mathbf{a} = \nabla_{\mathbf{u}} \mathbf{u},[/tex]

    and a freely falling observer is a geodesic observer curve.

    Sometimes pinning down things so precisely confuses, and sometimes it helps with understanding.
     
  14. Jan 28, 2010 #13

    sylas

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    Ah! Thank you. I think I have been thinking of a geodesic as what should more properly be called the image of a timelike geodesic. Given the image of a timelike geodesic, I guess there is only one unique way to give a geodesic that is parameterized by proper time, and so I didn't worry about the map.

    Quite so! But if I understand the definitions better, I may be able to improve my own descriptions, even if not given in full detail, so that they remain correct when made more precise; which would be nice. Not to mention that I simply wanted to know it better myself.

    Thanks -- sylas
     
  15. Jan 28, 2010 #14
    Mathematicians mostly use M as a n-dimensional embedded manifold in a (n+1)-space and they of course clarify what topology the manifold is equipped with. In this case (basically in GR), since M is a smooth manifold and of course equipped with the metric topology in a Riemannian space(time), i.e. the local coordinates are curvilinear, one has to notice that

    [tex]\gamma : I \rightarrow M,[/tex]

    depends, above all else, on the metric itself and then when it comes to the dynamical properties of the moving particles, mass points, on geodesics, considering the fact that [tex]\nabla_{\dot{\gamma}} \dot{\gamma} = 0[/itex], it is said to be a geodesic depends on how fast the points are traversed. This means that the geodesics are split into various categories just by studying the dynamics of particles moving on them. This is the reason why Mathematicians are not that much acquainted with, for instance, light-like geodesics.

    So I would see that both sylas and George look at one thing from two different angles which each calls for its own viewpoints.

    AB
     
  16. Jan 28, 2010 #15

    George Jones

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    As usual, Altabeh, I think I disagree with what you have written, but I am sufficiently confused by what you have written to be sure that I disagree.:wink::biggrin:

    Also, as usual, work intervenes, and I don't know when I'll get back to this thread. IF I do get back to this thread, I want to address some technical issues, but first a simple example. Take R^4 to be Minkowski spacetime, with g suitably defined using the standard global chart as an inertial coordinate system.

    Let \gamma : [0, 1] -> R^4 be given by \gamma(t) = (t, 0, 0, 0), and let \alpha : [0, 1] -> R^4 be given by \alpha(t) = (t^2, 0, 0, 0).

    What is the relevance to this example of what you have written.
     
  17. Jan 28, 2010 #16
    I'm so happy that you disagree again with me this time on this issue and I hope you are not going to not be back to this thread as you once left my post https://www.physicsforums.com/showthread.php?t=371614" without an answer.

    Let's say you wanted to declare that geodesics are realized by two qualities:

    1- A geodesic is a set of points in spacetime that is defined to be locally the shortest path between points in the spacetime. (From a mathematical point of view)

    2- The instantaneous derivative of a curve wrt the proper time s at each point and the change in the direction of this derivative along the curve must be taken into account so a little analysis on these two things is required when studying geodesics. For example, [tex]\dot{\gamma}=a[/tex] with a being a constant, represents a straight line or the shortest path between two points in a Minkowski spacetime and thus [tex]\dot{\dot{\gamma}}=0[/tex] for every line; this is the shortest because the curvature vector of [tex]\gamma[/tex] is null.

    In mathematics, the second part of the "quality" 2 is indeed a manifestation of the first formula of Frenet formulae,

    [tex]R=|\dot{\dot{\gamma}}|[/tex], (1)

    where R is the magnitude of curvature vector.

    Which in physics and specially GR plays a significant role as the equation of geodesic. So that one can see that it just falls upon the equation of geodesic to determine how fast geodesics are traversed. All these things happen to exist on metric spaces so the first thing that a geodesic depends on is the metric of that space(time) not anything else. You say it is the first part of "quality" 2 which can along with "quality" 1 distinguish a geodesic from other curves. But I'm saying that it

    not how fast geodesics are traversed. The difference between my idea and yours is that I don't talk about a special M but you take M for granted to be a given metric (I think). So you can see that it is so simple-minded to even think that we are in a disagreement with each other.

    See that you specify R^4 or the Minkowski spacetime. Now you must get my idea: through the use of equation (1), [tex]|\frac{d^2}{dt^2}\alpha(t)|=R_\alpha=2,[/tex] which means [tex]\alpha[/tex] is not a geodesic in R^4. But this could be if we considered another spacetime with another metric.

    AB
     
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  18. Jan 29, 2010 #17

    George Jones

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    Okay, now I have some idea what you mean, but in this thread, I think that is best to take spacetime to be an arbitary but fixed pair [itex]\left( M , \mathbf{g} \right)[/itex], where [itex]M[/itex] is a 4-dimensional differentiable manifold and [itex]\mathbf{g}[/itex] is a Lorentzian "metric", so that a metric-compatible torsion-free connection is fixed.

    I think that box 10.2 form Misner, Thorne, and Wheeler hints at a physicist's version of what you mean. I don't seem to have any math references that talk about this. Could you give some?

    Like always, I am hard-pressed to find time for this, as I'm preparing to go out of town today, and much of my weekend will be spent doing stuff with my wife and three-year-old daughter. Maybe Monday. I want to discuss some of the technical details.

    For example, while it is true in a Riemannian mainfold that shortest distance (if it exists) between two points is a geodesic, it not true that
    Easy counterexamples to this statement exist! For example, consider [itex]S^2[/itex] with its metric inherited by considering it to be a surface in [itex]\mathbb{R}^3[/itex]. Think about geosdesics on the surface of the Earth!
     
  19. Jan 29, 2010 #18

    atyy

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    Could the difference between George's and Altabeh's definitions (the acceleration def, not the length def) is that for George, a geodesic is an affinely parameterized geodesic, while for Altabeh, a geodesic can be a non-affinely parameterized curve traversing the same path?
     
  20. Jan 29, 2010 #19

    Dale

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    The concept of inertial frames doesn't really apply to GR. An inertial frame is a coordinate system in which the laws of physics take the standard form, and in GR the laws of physics are expressed in terms of tensor quantities in order to ensure that they take the same form in all coordinate systems.

    The comments about geodesics describe inertial objects, which is not the same as describing inertial reference frames.
     
  21. Jan 29, 2010 #20

    Fredrik

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    You can define an inertial frame in GR too. Consider a point p on the world line of an object. We take p to be the origin. The tangent of the world line at p is by definition a geodesic through p. We take that to be the time axis. The time coordinate at a point q on the time axis is defined to be the proper time along the time axis from p to q. The x,y,z axes are chosen from a hypersurface through p that's orthogonal to the world line. They're either chosen arbitrarily or chosen to match how the object is oriented in space. And, uh, I realize that I don't fully understand the whole construction myself (in curved spacetimes). I'm going to have to think about it, or look it up.

    If spacetime is flat, the result is the co-moving inertial frame at p. If spacetime is curved, the procedure may not define a coordinate system that covers the entire manifold. Hence the term "local inertial frame".
     
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