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Geodesics & Constant Velocity

  1. Oct 8, 2013 #1
    "The geodesics in R n are the straight lines parametrized by constant velocity".

    This can be proved with the geodesic equation:
    [tex] \ddot{x^a} + \Gamma^a_{bc} \dot{x^b} \dot{x^c}=0 [/tex]
    Locally we can find a coordinate system such that [tex] \Gamma =0 [/tex], and thus:
    [tex] \ddot{x^a} =0 [/tex]
    So along a geodesic at some point P, acceleration = 0 and there is constant velocity.

    But geodesics are the paths that freely falling objects will take, and freely falling objects accelerate according to Newton's law F=ma. But from the geodesic equation we get that a=0 and thus F=0 ... what gives?

    Also, a geodesic always a local path (extending off a point), right?? I.e. a freely falling rock dropped off a cliff will traverse many different geodesics on its way to Earth vs. a single geodesic on its way to Earth?

    Thanks,
    Ari
     
  2. jcsd
  3. Oct 8, 2013 #2
    Hi Ari, I am not an expert, but here is my 2 cents worth until the real experts log in ;)
    In GR gravity is not a force in the Newtonian sense. Free falling objects simply follow a geodesic. It takes a force to stop them moving along a geodesic. For example an object sitting on a high platform will experience proper acceleration. If the platform collapses, the object follows a geodesic and experiences no proper acceleration. When it hit the ground it once again experiences a force that prevents it following a geodesic.

    The local path between two points that has the greatest proper time interval is a geodesic (or at least part of one). This does not mean that the geodesic has to be local, i.e it does not start and stop and those two points. The geodesic path can be extended into the infinite past and future. Your freely falling rock dropped off a cliff will follow many geodesic paths if and only if it bounces of other objects on its way down the cliff. If it is freely falling it will follow one unique geodesic.
     
    Last edited: Oct 8, 2013
  4. Oct 8, 2013 #3

    Mentz114

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    Where did you read that ? It's obvious that geodesics do not necessarily have constant velocity, certainly not if there is gravity about.
     
  5. Oct 8, 2013 #4

    WannabeNewton

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    In addition to yuiop's excellent comments, let me just elucidate a minor point. In GR, the equations of motion for a freely falling particle with 4-velocity ##u^{\mu}## are ##u^{\nu}\nabla_{\nu} u^{\mu} = 0##. Notice here that there are no distinct gravitational and inertial terms; the idea is that both gravity and inertia are codified in an inseparable manner into the metric associated derivative operator ##\nabla_{\mu}## as a consequence of Einstein's equivalence principle. There is no unique way to decompose the motion of a freely falling particle into gravitational and inertial components; this is the covariant framework of free fall motion in GR.

    If we were to write down the equations of motion in the Newtonian limit, after having chosen a fixed background global inertial frame ##(t,x^i)##, we would get something of the form ##\frac{\mathrm{d} ^2 x^{i}}{\mathrm{d} t^2} = - \Gamma ^{i}_{tt} = -\frac{\partial \varphi}{\partial x^{i}}## where ##\varphi## is a quantity given in terms of the ##h_{tt}## component of a perturbed metric tensor ##h_{\mu\nu}## (perturbed in the sense that it describes "small" deviations from flatness). Notice that we can write this more suggestively as ##\vec{a} = -\nabla \varphi##. What we have done in the Newtonian limit, relative to that choice of fixed background global inertial frame, is decompose ##\nabla_{\mu}## into gravitational and inertial components exactly like the equations of motion for free fall in Newtonian mechanics.
     
  6. Oct 8, 2013 #5

    WannabeNewton

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    He/she said the geodesics of ##\mathbb{R}^{n}##, which are certainly straight lines of constant velocity.
     
  7. Oct 9, 2013 #6

    pervect

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    The geodesics of interest for a freely falling object are not the geodesics of R^n. The geodesics of R^n would be for example the geodesics of a straight line in Euclidean space.

    The geodesics of interest for a freely falling object are the geodesics of the Lorentzian signature space-time near the surface of the Earth, which is not the same as R^n. If you use an earth-center inertial coordinate system and ignore the effects of the Earth's rotation (a very good but not perfect approximation), you can approximate the metric near the surface of the Earth with a Schwarzschild metric. You can then go on to write down the geodesic equations (which will be considerably more complex than the ones you just wrote) and get an approximately correct trajectory for a falling object.
     
  8. Oct 11, 2013 #7

    A.T.

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    The proper acceleration of free fallers is zero. That's what a free falling accelerometer measures.

    The geodesic is determined by the starting point and the initial direction. The initial direction in space-time is determined by the initial velocity. If you let it go from rest, the initial direction is along the time dimension, like shown in this animation.

    https://www.youtube.com/watch?v=DdC0QN6f3G4
     
    Last edited: Oct 11, 2013
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