How is the Earth pushing "up"/"outwards" at $9.8\text{ m/s}^2$ in general relativity? [duplicate]

Question

So I've been trying to wrap my head around GR for a bit now, but one thing that keeps me down is the idea that while earth bends spacetime around us, the earth is also accelerating towards us at $9.8\text{ m/s}^2$, which is why objects accelerate towards it at that rate. However, I haven't been able to source this claim, it comes up independently in the information I am consuming. One claim by "ScienceClic English" claims that the geological forces of the earth itself is expanding the earth at a rate of $9.8\text{ m/s}^2$ while the curvature of spacetime keeps earth the same size. You can imagine my difficulties sourcing this as all results on "expanding earth" in any variation returns debunking of the expanding earth theory itself. The other claims do not go into any detail, they merely claim that the earth is accelerating "up" at $9.8\text{ m/s}^2$.

What is the deal with this? I feel a force acting upon me, so already there's the natural intuition towards gravity existing. Yet we know gravity isn't a real force, I am just experiencing the curvature of spacetime, but what else? What governs the law that things fall towards the earth at $9.8\text{ m/s}^2$ on earth?

Answer 1

A crucial issue here is what we mean by acceleration. In Newtonian physics, objects which are not under the influence of external forces move in straight lines at constant speed; deviation from this behavior is called coordinate acceleration because it refers to the rate at which the object's coordinates change.

On the other hand, in GR there is a subtly different concept called proper acceleration, which measures the deviation of an object from a free-fall trajectory. A pencil sitting on a table is not moving, so it does not possess any coordinate acceleration. However, since a free-fall trajectory (on the surface of the earth) would be accelerating downward at 9.8 m/s$^2$, the proper acceleration $\mathbf a_\text{proper} = \mathbf a_\text{coordinate} - \mathbf a_\text{free fall}$ is directed upward at 9.8 m/s$^2$.

In GR, one typically refers to force as the things which induce proper acceleration - essentially, $\mathbf F = m\mathbf a_\text{proper}$. It is in this sense that gravitation is sometimes said not to be a force - under the influence of gravity, objects are in free-fall and therefore their proper acceleration is zero.

On the other hand, if you multiply $\mathbf a_\text{free fall}$ by an object's mass and move it to the other side of the equation, you get $$\mathbf F + \underbrace{m\mathbf a_\text{free fall}}_{\text{Gravitational force?}} = m\mathbf a_\text{coordinate}$$

Ultimately, the debate over whether to call the gravity a force comes down to whether you want $m\mathbf a_{\text{free fall}}$ to be on the left or right hand side of the equation of motion, and is therefore in some sense a semantic point. There are subtleties, of course - the facts that non-tidal gravitational forces are not measurable and that gravitational and inertial masses are observed to be universally proportional, along with the beautiful geometry of general relativity, seem to suggest very strongly that we should adopt the proper acceleration POV. But if you're only interested in doing calculations, it doesn't really matter.

Answer 2

The short answer is that Earth doesn't expand upward. Instead, a massive object like Earth curves space-time. Objects with no forces on them follow straight paths at constant speed in flat space-time. In curved space-time, those paths are not straight. Two objects that start out near each other can follow different paths that accelerate away from each other. This is the cause of gravity.

To get a start on seeing how this all works, see Why can't I do this to get infinite energy?. This talks about a simple thought experiment that led Einstein to General Relativity. The upshot is that some straightforward properties of the universe, like conservation of energy, lead to the conclusion that time must run slower near the earth.

A consequence of slower time near Earth is space-time is curved. To get an understanding of what this curvature means, see my answer to Intrinsic Curvature of a Cylinder.

A consequence of curved space-time is that light is deflected toward the earth as it passes by. To see this, consider light from a distant star. The light is a plane wave. Part of the wave passes near the Earth, where time runs slower. That part of the wave travels slower and gets a little behind. The wavefronts are no longer planar. They have been tilted a little toward the Earth. Light propagates perpendicular to its wavefronts. Light is deflected a little toward the Earth.

The same thing happens as starlight passes near the Sun. This is the famous solar eclipse experiment that provided the first evidence for General Relativity.

However, the outcome was twice as much deflection as Einstein predicted. Einstein went back to the drawing board. After some thought, he concluded that in addition to a distortion of time, there is also a distortion of space. The distance to the center of the Earth is slightly longer than you would expect from measuring the circumference of a distant orbit. From this, it is possible to show that the trajectory of matter is also deflected.

Without curvature, an object sitting at rest on the surface of the Earth would float there. Because space-time is curved, it is deflected toward the Earth. The ground is rigid, and resists being pushed into. It pushed back hard enough to hold the object still. This is the upward force you are thinking of.

Answer 3

the earth is also accelerating towards us at 9.8m/s2, which is why objects accelerate towards it at that rate. However, I haven't been able to source this claim

You don’t need to source the claim, you can measure it experimentally with an accelerometer. For example, you can use the one in your cell phone. Simply use an app that lets you read the accelerometer, place the phone on the ground, and you will see that the ground is accelerating upward at $9.8 \text{ m/s}^2$.

This type of acceleration is called proper acceleration. It is physical and independent of the coordinates. The other type of acceleration is coordinate acceleration, which depends on the reference frame. In non inertial frames the coordinate acceleration differs from the proper acceleration, and the difference is described by introducing “fictitious forces” or “inertial forces”.

So, if the surface of the earth is accelerating then how does it stay the same size? This is due to curvature. Consider the surface of a sphere. If you walk straight forward on the surface of a sphere you will travel in a great circle. The equator is a great circle, but other latitude lines are not. To stay on any other latitude line you cannot walk straight forward but must continuously turn towards the nearest pole. In other words, a person walking due east on the 1 deg N line is continually turning away from a person walking due east on the 1 deg S line, although the distance between them is constant. Geometrically this is nearly identical to how the curvature of spacetime works.

Answer 4

Without external forces, we move along geodesics defined by the geodesic equation:

$$ \ddot x^m = - \Gamma^m_{ij}\dot x^j\dot x^i$$

where $\Gamma^i_{jk}$ are the Christoffel symbols.

In the approximation that the Earth is not rotating near the speed of light, we can use the Christoffel symbols from the Schwarzschild metric, but before we do that, remember the we are not moving near the speed of light either, so that our 4-velocity is:

$$ \dot x^{\mu} = \gamma(c, v_x, v_y, v_z) \approx (c, 0,0,0) = (x^0,0,0,0) $$

so that the geodesic equation is:

$$ \ddot x^m = -\Gamma^m_{00}x^0x^0 = -\Gamma^m_{00}c^2$$

and we only need (with $c=1$) a few Christoffel symbols:

$$ \Gamma^t_{00} = 0$$ $$ \Gamma^r_{00} = \frac{GM}{r^2}(1-\frac{2GM}r)^{-1}$$ $$ \Gamma^{\theta}_{00} = 0$$ $$ \Gamma^{\phi}_{00} = 0$$

That means only the velocity in the $r$ coordinate changes, and the non-zero part of the geodesic equation is:

$$ \ddot r = -\frac{GM}{r^2}(1-\frac{2GM}r)^{-1}$$

Since the Earth is not near being a black hole, $$ r \gg 2GM $$

so:

$$ \ddot r = -\frac{GM}{r^2}$$

which is exactly Newton's Law of gravity: you accelerate radially inward at:

$$ \frac{GM}{R_{\rm Earth}^2} \approx 9.8\,{\rm m/s^2} $$

It's remarkable that an equation with so much going on reduces to the simple form we're used to (and we used the solution that predicted black holes, even).

Answer 5

"The earth is also accelerating towards us at 9.8m/s2, " What? That sentence is totally wrong. Let's look at the Newton's third law.

$F_{you \; put \; on \; earth} = - F_{earth\; puts\; on\; you}$

This equation is about force and not acceleration.

$m_{earth}a_{of \; the \;earth \;moving \;toward \; you} = - m_{you} a_{of \;you \;moving \;toward \; the \; earth}$

$a_{of \;you \;moving \;toward \; the \; earth}$ is $9.8\;m/s^2$ , mass of the earth is $5.79\times 10^24 \; kg$. Use your mass and calculate the acceleration at which the earth move toward you.

Before understanding GR you need to understand Newtonian gravity very well.