Why doesn't planet Earth expand if I accelerate upwards when standing on its surface?

Question

According to General Relativity I am being accelerated upwards by planet earth while writing this question. But a curious person on the the other side of the planet relative to me would have the same experience. That means we are accelerated in opposite directions, although earths diameter do not seem to increase. How can this be?

Answer 1

Spacetime curvature makes this possible. Here's an analogy. There are two paths on opposite sides of the equator, at a constant distance from it. Someone walking east along the path north of the equator will have to continually turn slightly left to stay on the path. (If that isn't obvious, imagine it's so far north that it visibly circles the pole.) Likewise, someone walking east on the path south of the equator will have to turn right. Two people walking side by side along the paths will stay the same distance apart, even though they're constantly turning away from each other. This wouldn't be possible on the Euclidean plane, but it's possible on a curved surface. That's what happens in general relativity, but the direction they're walking is the time direction, and the turning is acceleration.

Answer 2

According to General Relativity I am being accelerated upwards by planet earth while writing this question.

According to general relativity you are being accelerated upward by the normal force. This is exactly what happens in Newtonian mechanics.

One difference between the two is that Newtonian mechanics deems gravitation to be a real force while general relativity does not. A frame based on a person standing still on the surface of a non-rotating rogue planet is very close to being an inertial frame in Newtonian mechanics. The person is standing still because the upward normal force and the downward gravitational force cancel one another.

An inertial frame in general relativity is comoving with a stream of falling apples. A person standing still is accelerating upward from the perspective of a stream of falling apples. This upward acceleration must necessarily be the result of a real force, which is the normal force.

But a curious person on the the other side of the planet relative to me would have the same experience. That means we are accelerated in opposite directions, although earths diameter do not seem to increase. How can this be?

Another key difference between Newtonian mechanics and general relativity is that inertial reference frames span the universe in Newtonian mechanics but are local in general relativity. Mathematically, "local" means infinitesimally small. The concept is a bit more expansive in physics, where it means small enough that instruments cannot detect accelerations due to differential gravity (e.g., tidal effects).

Nowadays, Einstein's elevator car thought experiment doesn't quite cut it as instruments capable of detecting the differential gravity across an an object the size of an elevator car have been developed; this was the basis of the European Space Agency's Gravity field and Ocean Circulation Explorer (GOCE) satellite. A relativistic inertial frame with its origin at a person's center of mass standing still on a planet does not extend to a person standing still on the other side of the planet.

Answer 3

Disregarding the earth rotation for being too slow, we can use the Schwarzschild metric as a good approximation:

$$c^2d\tau^2 = \left(1 - \frac{2GM}{c^2r}\right)c^2dt^2 - \frac{1}{\left(1 - \frac{2GM}{c^2r}\right)}dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2$$

For a body at rest on the earth surface, $$dr = 0,\, d\theta = d\phi = 0\implies\left(\frac{dt}{d\tau}\right)^2 = \frac{1}{\left(1 - \frac{2GM}{c^2r}\right)}$$

The second covariant derivative of $r$ with respect to $\tau$, is:

$$\nabla^2_{\tau}r = \frac{\partial^2 r}{\partial \tau^2} - \sum{\Gamma_{ij}^r\frac{\partial X^i}{\partial \tau}\frac{\partial X^j}{\partial \tau}}$$

Most of the terms of the summation are zero because the body is at rest. The first term of the right side is the conventional acceleration, that is also zero for the same reason. The non zero terms are:

$$\nabla^2_{\tau}r = \left(1 - \frac{2GM}{c^2r}\right)\left(\frac{GM}{r^2}\right)\left(\frac{\partial t}{\partial \tau}\right)^2 = \frac{GM}{r^2}$$

That is our $g$.

Answer 4

Benrg has a nice geometrical answer, but I'll add a somewhat hand-wavey and less technical analogy. Consider a spaceship that's falling from orbit. When it's still some distance from the ground, it turns on its rockets and decelerates, eventually coming to a stop and then hovering just a few metres above the ground. The people in the rocket will feel 1g of acceleration, and the rocket will be consuming fuel to keep the spaceship in the air. Yet the spaceship isn't going anywhere, it's just hovering.

The forces experienced by bodies on the surface are kind of similar. The acceleration upwards is exactly counterbalanced by the tendency of the Earth to want to contract (to follow its natural path in spacetime).

Answer 5

It is accelerated from YOUR referential. In free fall you will follow the space-time geodesics. But the earth's ground prevents you from falling toward the center of mass of the earth. So in your referential, you are accelerated upward by the ground.

Answer 6

Because of Earth's gravity (and rotations, but we will focus on gravity since that seems to be the point of your question) you are in an accelerated frame of reference, not an inertial one. Near the surface of Earth g is about 9.8 meters per second squared. This will make your weight be about the same on Earth as it would seem to be if you were accelerating at 9.8 meters per second squared out in space, far enough away from any other body for gravity to be negligible.

Answer 7

It comes down to a definition of acceleration. Acceleration is most universally appreciable as a force application contradicting an object's natural position or trajectory. Notice that this does not require that the object move -- only that it is being affected by a force, as in 'experiencing pressure'.

So by this definition an object at apparent rest on a table top is being forced by the solid table surface, and feels the pressure of this force throughout its form, and so on.

It also helps to appreciate gravity as an electromagnetic phenomenon, as the definition of acceleration also applies to (ferro-)magnetic forces. When you see two strong magnets pulling or pushing each other, it appears that they are exerting a force, as though expending energy...

But to the magnets their unhindered relative motion represents a state of rest given their natural atomic states. Energy expense is only experienced by the person holding the magnets apart/together against the natural tendency, and by the magnets when they are prevented from their natural relative motion (including if/when they impact).

Answer 8

That is because we are just accelerating radially outwards and not "moving" radially outwards. This case is analogous to circular motion where there is radial acceleration but no radial movement.

You can refer to my article in the link below for detailed explanation: https://paribeshregmi.medium.com/a-soft-intro-to-general-relativity-aa46da221747

Answer 9

The only way I can understand Albert's warped space-time as acceleration, is if space is like the old ghostly ether, and it is flowing and collapsing into bodies of mass. If mass pulls spacetime into itself, then it makes sense to me why at the event horizon where space-time, water-like ether, begins flowing into the black hole at a rate equal to light. Since solid matter cannot accelerate to light-speed,then there is simply no escape. And light, should form a shell where it is breaking even trying to escape as fast as space-time is contracting inwards, and essentially hovering. If spacetime is just bent but static, then it would simply be a matter of changing angular direction to get out. Spacetime distortion only makes common sense if it is flowing or collapsing into concentrations of mass or energy, just as it is expanding elsewhere in the empty parts of the universe. How is that different from the old concept of the 'Ether'?

Answer 10

All I see here is a bunch of requests for special pleadings. No referencing in practical reality. Bouyancy ceasing to exist in freefall proves that any tangible force in reality is null. There absolutely can not be motion without a force or an effect of that force. It would have to be constant velocity in order for our density layers not be affected or mass not increasing on either end of the line. It is the object decelerating from Earth's reference frame that gives the illusion of acceleration. We know this to be true because man has never experienced an environment void of matter. There will always be resistance. Back to the ol drawing board.

Answer 11

What is asserted is that gravity isn't a force at all; rather, it's acceleration such that if I sit on a chair here in California, the acceleration is upward by Earth against me and my chair. What about my friend in China sitting on his chair? What about everyone else, Earth-wide, sitting on their chairs. I don't know if gravity is force or not; I do know that gravity is still unexplained. I took a lot of physics in college but I'm no scientist. Admittedly, I don't completely get space-time which is probably where gravity's explanation lies. But I'm pretty sure it's not fabric.

Einstein was brilliant for figuring it out. But he's dead and now we need another genius to explain.