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Gravitational Acceleration

  1. Jan 30, 2014 #1
    If this question has an obvious answer, please excuse my ignorance. I'm still very new to the world of physics relative to most of you. But my question is simple. In Newtonian physics I know the rather simple explanation along with the corresponding formulas but in modern physics I'm a little confused. If gravity is just a distortion in spacetime which causes bodies to travel in a geodesic path along spacetime, why does that body accelerate? If a body just wants to remain traveling in a straight line at constant velocity, than why does "bending" that straight line result in acceleration. From my, very limited, understanding the body is still traveling in a locally straight path so what has actually changed that causes the acceleration? Thank you in advance.
  2. jcsd
  3. Jan 30, 2014 #2


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    A freely falling body undergoes no proper acceleration; proper acceleration is what you measure with an accelerometer. Acceleration due to gravity is, in a very loose sense, analogous to inertial forces from Newtonian mechanics such as the centrifugal and Coriolis forces. More precisely, imagine we have a non-rotating spherically symmetric star generating the gravitational field of interest. An observer, let's call him ##O##, hovers outside of the star with a rocket. Another observer ##O'## is freely falling and at an event ##p## passes right by ##O##. Now by definition ##O'## has no proper acceleration because he is freely falling. On the other hand ##O## does have a proper acceleration because he is using the thrusters of his rocket to counter the gravitational field so that he can just hover in place; this proper acceleration is the magnitude of an acceleration vector ##\vec{a}##. Say now we go to the rest frame of ##O##. When ##O'## falls past him at ##p##, ##O## will attribute to ##O'## the acceleration ##\vec{g} = -\vec{a}## because we're in the rest frame of ##O## and at ##p## we have ##O'## freely falling down past ##O##. This ##\vec{g} = -\vec{a}## is the gravitational acceleration.

    To make it a bit more concrete, consider the uniform gravitational field within the Earth. You're standard on the ground so the ground is pushing you up with a proper acceleration ##a = 9.81 m/s^2##. A person that's freely falling towards the ground after being dropped from the sky (we ignore air resistance) has no proper acceleration by definition. However you, in your rest frame, say that the person is accelerating down towards the ground at a rate ##g = -9.81m/s^2##. This is what you normally call the gravitational acceleration but as you can see it's in a loose sense like an inertial force arising from the fact that you have an acceleration ##\vec{a}## of which the proper acceleration exerted by the ground on your feet is the magnitude and in your rest frame this manifests itself as an acceleration ##\vec{g} = -\vec{a}## of a freely falling object.

    EDIT: I would recommend that you, for the time being, ignore the fact that freely falling observers locally have truly straight world-lines (we say that they are locally inertial observers). This will be a hard thing to understand until you have more math under your belt.
    Last edited: Jan 30, 2014
  4. Jan 30, 2014 #3


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    If a body follows a geodesic that happens to bring it closer to another body, then its velocity wrt to that thing can increase, so ##d^2x/dt^2 \neq 0 ##. This is called coordinate acceleration because the first body feels no forces acting on it.

    [WN beat me to it]
  5. Jan 30, 2014 #4


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    Your mashing classical ideas and general relativity ideas together and using classical frames of reference to compare them.

    Classically when we see an object move with constant speed along a curved trajectory or accelerate along straight trajectory we say that a force is acting on it to make it move. We would be standing on the ground observing the motion. So objects that fall follow these kinds of trajectories and so we say that a gravitational force is acting on them.

    Relativistically if you were in the object falling toward the Earth along a curved trajectory then you would feel weightless because you and the object are falling at the same rate. Feeling weightless means no external forces are acting on you at all.

    Einstein combined these ideas together to say that a geometrical model could be used to describe gravity that of space being locally flat but being curved in the larger picture. In this geometrical picture straight lines are geodesic lines. All falling trajectories follow geodesic lines. If you were to use a rocket to change your falling trajectory then your new trajectory wouldn't be a geodesic and you'd feel a force pushing you.
  6. Jan 30, 2014 #5
    So in Layman's terms, there really isn't an acceleration due to gravity. The acceleration observed of the falling object by a person on the ground is due to mechanical forces from the ground and in truth, the gravity never actually caused any acceleration of the falling object. Correct? To the falling object, there would be no acceleration?
  7. Jan 30, 2014 #6


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  8. Jan 31, 2014 #7


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    Because in distorted space-time a geodesic path changes its direction with respect to the space & time dimensions. You can visualize this like this:


    You have to distinguish between:
    - coordinate acceleration = dv/dt in some coordantes
    - proper acceleration = what an accelerometer measures

    No proper acceleration, because a free falling accelerometer measures zero. As you see in the video above, there are no interaction forces acting on the falling apple in Einsteins model. But in non-inertial reference frames (like the rest frame of the surface) it has coordinate acceleration (it gains speed).
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