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So in General Relativity you can solve the geodesic equation for my body and find that it's "trying" to travel a path through the center of the earth due to local curvature. I remain on the ground as it is continually intercepting me from this path.

This explanation doesn't leave any room for my personal mass to play a role, and so it appears to fail in explaining the obvious fact that it's much more difficult to lift a car away from the ground than myself.

So what's going on here? Obviously I generate a gravitational influence which will change the geodesics slightly, but I estimate that the difference in this effect (between me and the car) cannot account for the difference in weight. Maybe I'm wrong? If so, how exactly does my gravitational curvature result in the difficulty one has in lifting me?

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It's not typical to talk about the equivalent of Newtonian dynamics in general relativity, because one typically talks about the geodesic equation:

$${\ddot x}^{a} + \Gamma^{a}{}_{bc}{\dot x}^{b}{\dot x}^c = 0$$

But, if we really want to get into it, the "Newton's second law" way of looking at this would be:

$$m\left({\ddot x}^{a} + \Gamma^{a}{}_{bc}{\dot x}^{b}{\dot x}^c\right) = F^{a}$$

And in this picture, the four-force applied by the ground is the thing that keeps you from moving in a geodesic, and the mass is supplied in the obvious place.

To make this even more super simple, consider a person sitting stationary on the surface of a static planet of radius $R$. Then, the metric is ${\rm diag}\left((-\left(1 - \frac{2M}{r}\right), \frac{1}{1-\frac{2M}{r}}, r^{2}, r^{2}\sin^{2}\theta\right)$, and $x^{a} = N(t, R, 0, 0)$ (note that $t$ varies, while $R$ does not, it is a constant), where $N$ is some normalization factor that guarantees ${\dot x}^{a}{\dot x}_{a} = -1$, so we have $N^{2} = \frac{1}{1-\frac{2M}{R}}$.

Then, $\ddot x =0$, and the only nonzero $\dot x$ is the $t$ component. The only nonzero $\Gamma_{tt}{}^{a}$ is $\Gamma_{tt}{}^{r} = \frac{M}{r^{2}}\left(1-\frac{2M}{r}\right)$

Substituting, we find:

$${\vec F} = {\hat r}\,N^{2}m\frac{M}{r^{2}}\left(1-\frac{2M}{r}\right)$$

or, putting factors of $N$ back in, and setting $r=R$ due to the location of our person,

$${\vec F} = {\hat r}\,\frac{Mm}{R^{2}}$$

Which should be familiar once we recognize that we are in units of $G = c = 1$

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In GR, it's much more natural to quantify the strength of a gravitational interaction by the acceleration it induces on a body than by the force, because the acceleration is the same regardless of the body. This is basically another way of stating the equivalence principle. Under this quantification scheme, you and a car are actually being attracted toward the center of the earth by the exact same amount - by $9.8 \text{ m/s}^2$ (in a frame in which the Earth's surface is a rest).

The reason it's more difficult to lift a car than a person is because when you do so, the force pushing on the car/person is electrostatic rather than gravitational in origin, and for electrostatics, force is what's the same across all bodies of equal charge, not acceleration - so force is the natural way of quantifying the strength of electrostatic interaction, unlike in the case of gravity. If you tried using gravity to pick up a car or a person, you would find that it's equally easy to lift up either.

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    $\begingroup$ Sounds like an awesome answer. I'll have to think about it for bit. $\endgroup$ – DPatt Jun 19 '17 at 20:46
  • $\begingroup$ @DPatt Here's another way to think about it. You intuitively think that gravity is "pulling harder" on a car than on a person. I claim this is wrong, and really it's more natural to think of gravity as "pulling equally hard" on both. To see, this imagine you're in a free-falling frame falling towards the Earth's surface and watching the person and car. In this frame, gravity is clearly "pulling equally hard" on both - because its pull is zero on both! The difference is that in this frame, the Earth seems to be pushing on both the car and the person to keep them accelerating at ... $\endgroup$ – tparker Jun 19 '17 at 23:33
  • $\begingroup$ ... $9.8 \text{ m/s}^2$, via electrostatic repulsion between the ground and the objects. Clearly it has to push harder on the car than on the person, because the car has more inertia. If they're on soft dirt or mud, the electrostatic repulsion might not be strong enough to give it the full acceleration, and the car will sink into the ground. So in this frame, it's easy to see that the pull of gravity is the same on the car and the person - the difference is that the electrostatic force from the ground is pushing harder on the car. $\endgroup$ – tparker Jun 19 '17 at 23:36
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The metric in hyperbolic coordinates gives us: $$d\tau^{2}=r^{2}d\omega^{2}-dr^{2}\tag{1}$$ For a non-relativistic frame, pick a proper acceleration to be $A=\frac{c^{2}}{R}=g=9.81m/s^{2}$. We observe that for the proper acceleration to be as such, we have to pick a trajectory (hyperbola) which is $R=\frac{c^{2}}{g}\approx{10^{16}m}$ away from the origin. Consider another trajectory just a bit away from the $r=R$ hyperbola such that $r=R+x$. Now substitute the above into the metric and multiply & divide the first term by a factor of $R^{2}$. The final metric will be in the following form: (with $R\omega=t$) $$d\tau^{2}=\left(1+\frac{2x}{R}\right)dt^{2}-\frac{dx^{2}}{c^{2}}=\left(1+\frac{2xg}{c^{2}}\right)-\frac{dx^{2}}{c^{2}}=\left(1+\frac{2U(x)}{c^{2}}\right)-\frac{dx^{2}}{c^{2}}\tag{2}$$

The geodesic equation states that: $$\frac{d^{2}x^{n}}{dS^{2}}=-\Gamma^{n}_{mr}t^{r}t^{m}\tag{3}$$ If the $n$ component is $x$, then, as long as the particle moves slowly, propertime=coordinate time. Thus, we can re-write equation $(3)$ as: $$\frac{dx^{2}}{d\tau^{2}}=-\Gamma^{x}_{mr}\frac{dx^{r}}{d\tau}\frac{dx^{m}}{d\tau}\tag{4}$$. When $m$ & $r$ are time components, $m=r=t$ $$\frac{dx^{2}}{dt^{2}}=-\Gamma^{x}_{tt}\tag{5}$$ From equation $(5)$, we observe that the geodesic equation simplifies to show that the Christoffel symbol is nothing but the gravitational field and equation of motion of a particle in such a field is completely independent of its personal mass. (Christoffel symbols are proportional to a the gradient of the metric tensor aka the potential along a particular direction).

We also notice that, from equation $(2)$, the action can be constructed as follow: $$\mathcal{S}=-mc^{2}\int{\sqrt{1+\frac{1}{c^{2}}\left(2U(x)-\left(\frac{dx}{dt}\right)^{2}\right)}}\tag{6}$$ We can observe that the Lagrangian for this action is: (after binomial expansion) $$\mathscr{L}=-mc^{2}-mU(x)+\frac{m\dot{x}^{2}}{2}\tag{7}$$ Solving the Euler-Lagrange equation, we obtain: $$m\ddot{x}=-m\frac{\partial{U(x)}}{\partial{x}}\tag{8}$$ Hence, we obtain the Newtonian equation of motion: $$\ddot{x}=-\frac{\partial{U(x)}}{\partial{x}}\tag{9}$$ Hence this equation is completely independent of personal mass.

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