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If one considers the free fall of an object of mass $m$ from a hight $h$ in the Earth's gravitational field (neglecting air friction) from the point of view of GR, what would be the main corrections to the Newtonian result for, say, total time until the object reaches the ground?

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  • $\begingroup$ This is essentially the question: "How long does this take in GR to fall to the ground?", right? (That question would be homework-like, in my opinion) $\endgroup$
    – ACuriousMind
    Mar 13 '15 at 14:18
  • $\begingroup$ Well, I am not interested in the actual calculation, I just assume this a well know result in GR and was curious as to the order of magnitude of the difference from the classical (Newtonian) result. $\endgroup$
    – Andrei
    Mar 13 '15 at 15:28
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There isn't a simple closed form expression for the distance as a function of time in general relativity. However if you're just interested in how big the difference is I think there is a nice way to see this.

For simplicity let's take a falling object with zero total energy. In the Newtonian case this means the kinetic energy is equal to the potential energy i.e. that the object notionally starts at rest at infinity. Equating the kinetic and potential energy gives:

$$ \tfrac{1}{2}mv^2 = \frac{GMm}{r} $$

from which we get the Newtonian equation of motion:

$$ \frac{dr}{dt} = -\sqrt{\frac{2GM}{r}} \tag{1} $$

The derivation of the corresponding equation in general relativity is somewhat involved so I'll skip the details and just quote the result. If you're really, really interested the gory details are given in this article. Anyhow, the result is:

$$ \frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right) \sqrt{\frac{2GM}{r}} \tag{2} $$

where $r_s$ is the Schwarzschild radius.

The difference between the Newtonian expression (1) and the GR expression (2) is that factor of $1 - r_s/r$, so calculating the factor is a good way to see how big the difference is. Let's do this for the Earth's surface. The Schwarzschild radius is:

$$ r_s = \frac{2GM}{c^2} $$

and if we feed in the mass and radius of the Earth we get:

$$ \frac{r_s}{r_e} = 1.39 \times 10^{-9} $$

So GR predicts the velocity when the object reaches the Earth's surface will be smaller than the Newtonian expression by a factor of $0.9999999986$.

The more eagle eyed amongst you will have spotted somethng odd - equation (2) tells us that when $r = r_s$ the factor of $1 - r_s/r$ goes to zero, so the velocity is zero at the event horizon. This is the reason for the notorious claim that nothing can fall into a black hole, because all falling objects tend asymptotically to a velocity of zero at the horizon.

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  • $\begingroup$ Is this the non simple closed form equation? goo.gl/photos/e4Gfz7fWtksGY7g47 The time it takes for a particle to fall from r2 to r1. The equevelant of t=sr(2h/g) in Classical Physics $\endgroup$
    – Bill
    Oct 3 '16 at 17:51
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The difference is often due to the addition of the Lorentz factor, which you can apply to equations of motion using a Lorentz transformation. Dr. Wolfgang Rindler was kind enough to write a small paper on this on scholarpedia. The Lorentz factor ($\gamma$) is often defined as: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

This changes basic momentum from $p = mv$ to $p=\gamma m v$. The wikipedia article actually shows a chart about the value of the Lorentz factor depending on speed. The Lorentz factor tends to not matter at low speeds ($v<<c$), but becomes increasingly important as the speed increases, in a very non-linear manner.

When you say how long does it take for an object to fall, do you mean as measured by the object falling or an on the surface who is not moving? According to the summary of a talk given by Benjamin Knorr here, fast-moving objects experience an (constant) acceleration in proportion to the cube of the Lorentz factor when compared to the stationary acceleration. Specifically: $$a_{moving} = \gamma ^3 a_{stationary}$$

which means the fast-moving object will experience a much larger force, which helps it conform to the stationary observer's observations of the object. This means the fast-moving object takes less time to hit something according to its perspective than the stationary observer sees.

So the correction to Newton's falling apple will go from $$\Delta x = v_0 + at^2$$ to: $$\Delta x_{apple} = \gamma v_{0, observer} + \gamma^3a_{observer}t_{apple}^2$$ where variables with the "apple" subscript are relative to the apple's perspective and variables with the "observer" subscript are the observer's. The observer will be considered stationary. For our observer, the apple falls like normal.

If the apple fell at mach 1, the Lorentz factor is about 1.000 000 000 000 644, which means the apple does not experience much time dilation. (I did it quickly on wolframalpha.com, and they didn't give me enough places to see a difference beyond the micro-second range.) For the observer, nothing really changes if they're using Newtonian or Relativity-Adjusted Newtonian mechanics.

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  • $\begingroup$ So why did John in the answer above say that there isn't a simple closed form equation in GR? You can very easily solve for time in your equation and whoila!! $\endgroup$
    – Bill
    Sep 26 '16 at 19:03
  • $\begingroup$ @Bill I think I'm coming at it from a simpler perspective than John... Honestly, I need to take time to read that paper he linked to find out why he didn't go for the Lorentz factor. $\endgroup$
    – PipperChip
    Sep 26 '16 at 19:37
  • $\begingroup$ Now I get it. The Lorentz factor differentiates Classical Physics with Special Relativity. The question is about General Relativity which explains free fall with curved spacetime $\endgroup$
    – Bill
    Oct 8 '16 at 10:29

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