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What is geodesic deviation

  1. Jul 24, 2014 #1
    Definition/Summary

    Where two particles very close together have the same velocities, their two geodesics are parallel, though only instantaneously, and so the gap (a 4-vector, of time and distance) between them has zero rate of increase, but has non-zero acceleration.

    The acceleration (a 4-vector) of this gap is the rate of deviation between the geodesics.

    The acceleration, for very small gaps, depends, linearly, both on the velocity of the particle and on the the direction of the gap.

    The linear equations relating this acceleration velocity and direction (all 4-vectors) combine to make a 4th-order tensor equation, the geodesic deviation equation.

    Equations

    Geodesic deviation equation:

    [tex]\frac{D^2\,\delta x^{\alpha}}{D\tau^2}\ =\ -\,R^{\alpha}_{\ \mu\beta\sigma}\,V^{\mu}\,V^{\sigma}\,\delta x^{\beta}[/tex]

    for 4-velocity [itex]V[/itex] and gap [itex]\delta x[/itex]

    where [itex]\tau[/itex] is proper time and [itex]R[/itex] is the Riemann curvature tensor

    Extended explanation

    Geodesic:

    A geodesic is the world-line (a path in four-dimensional spacetime, or 4-curve) "followed" by a free-falling body, or by light.

    For example, a rocket in orbit follows a geodesic, but a rocket using its engines does not, nor does a rocket on the ground.

    There are also "faster-than-light" geodesics.

    Technically, a particle does not move along a world-line: the world-line is a curve in spacetime, representing the whole history of the particle.


    Proper time:

    The gap (between two particles on very close geodesics) is measured with respect to the metric coordinates (including coordinate time).

    But it is measured between them, not at the same coordinate time, but at the same proper time, shown on identical (imaginary) steady clocks moving with each particle.

    And the acceleration of the gap is also measured not with respect to coordinate time, but to the proper time on either clock.

    Light "follows" a null geodesic, along which "separation" is zero, and proper time is constant: so, for null geodesics, the geodesic deviation equation cannot be written in the form given above.

    Riemann curvature tensor:

    This is a 4th-order (4 indices) tensor. It converts 3 vectors (two of them are identical velocity vectors, and the third is a position vector) into a fourth vector (an acceleration vector).

    Although it appears to have 64 parameters, it has various symmetries (including the first Bianchi identity), resulting in only 20 independent parameters.

    Example of how geodesics deviate:

    Imagine two clocks attached to the ends of a shelf. The shelf is "horizontal", and is free-falling directly towards the Sun.

    So the centre of the shelf is moving directly towards the centre of the Sun, but the two clocks are not … obviously, they would miss it by half the length of the shelf! :biggrin:

    So they are not "following" geodesics.

    However, they are moving parallel to each other, and, obviously, the rate-of change of the gap between them is zero.

    Disconnect them from the shelf: they now are "following" geodesics, and those geodesics start parallel, but will get closer as they fall nearer the Sun.

    The rate-of change of the gap between them started at zero, but will increase. The rate of this increase, or the acceleration of the gap, is the geodesic deviation for an initially horizontal gap.

    In this case, the acceleration is in the same direction as the gap itself: it is simply a multiple of the gap. But we will now see that that is not generally the case …

    Do the same thing for the shelf being "vertical": after they are disconnected, the lower clock will accelerate faster, and the two clocks will go at a different rate.

    So clock time and coordinate time will get out of step: when the two clocks show the same time (their proper time, [itex]\tau[/itex]), they will be at different coordinate times ([itex]t[/itex]).

    So although the gap initially was only in the vertical direction, and had no time component, its acceleration does have a time component.

    Do the same thing for the shelf being at any angle, and moving (without turning) at any velocity (not necessarily parallel or perpendicular to the shelf).

    Again, generally, the acceleration of the gap will not be parallel to the gap: so our general experiment converts an input 4-vector (the gap) into a completely different 4-vector (the acceleration).

    That is why we need a tensor, and that is also how the tensor works. :smile:

    If we "make the shelf infinitesimally small", this tensor becomes the Riemann curvature tensor.


    Geodesic deviation in Newtonian spacetime:

    In a uniform Newtonian gravitational field (acceleration = constant), bodies at different positions experience the same acceleration, and so at all times their relative acceleration is zero.

    But in the usual central Newtonian gravitational field (acceleration proportional to 1/r), with usual coordinates, relative acceleration is non-zero, and so there is geodesic deviation, even without considering either the special-relativity difference between [itex]t[/itex] and [itex]\tau[/itex] or general relativity.

    However, there will be no deviation in the "time direction", and so the curvature tensor will have only 6 independent parameters.

    Electromagnetic comparison:

    By comparison, the world-line deviation equation between world-lines followed by two charged particles with the same charge/mass ratio freely moving (in flat Minkowski spacetime) in an electromagnetic field is:

    [tex]\frac{D^2\,\delta x^{\alpha}}{D\tau^2}\ =\ \frac{q}{m}\,F^{\alpha}_{\ \mu\,;\,\beta}\,V^{\mu}\,\delta x^{\beta}[/tex]

    where [itex]q[/itex] is charge, [itex]m[/itex] is mass, and [itex]F[/itex] is the electromagnetic tensor

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
  2. jcsd
  3. Jul 2, 2017 #2
    Within a closed FLRW geometry with constant curvature, geodesic deviation is not zero. Does this lead to a frequency shift of photons? (If photons have a non zero spatial extension the geodesic deviation could lead to a shape deviation of the photon)
     
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