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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?

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    $\begingroup$ Gravity pulls down, not up. Unless you are floating into the sky at an ever-increasing rate, you are not accelerating upwards. $\endgroup$ – Nuclear Hoagie Oct 9 '20 at 18:52
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    $\begingroup$ @NuclearWang You're referring to Newton's picture of gravity. The question is about General Relativity's picture of gravity, in which the surface of the earth is indeed accelerating (in the absolute sense) away from the center, even though its distance from the center is not changing. It's a valid question. $\endgroup$ – Chiral Anomaly Oct 9 '20 at 21:40
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    $\begingroup$ @ChiralAnomaly That is incorrect. In Newtonian mechanics, a frame fixed with respect to a person standing still on a non-rotating rogue planet is very close to an inertial frame. The person doesn't move because the normal force keeps the person from sinking into the planet by pushing the person upward. From the perspective of general relativity, that frame is non-inertial. There's a fictitious acceleration in general relativity, directed downward, that almost exactly matches the Newtonian gravitational acceleration. In both cases, the normal force is a real force that pushes the person upward. $\endgroup$ – David Hammen Oct 10 '20 at 0:10
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    $\begingroup$ @DavidHammen I understand. The point of my comment wasn't that the previous commenter was using Newton's picture, but that the previous commenter was not using the GR picture used in the question. By "Newton's picture," I meant that before GR people would have described two forces on the standing person: the upward force of the ground and the downward pull of gravity -- which just happens to pull equally on all equal-mass parts of the person's body, so only the upward force from the ground is "felt." Of course we can also describe Newton's model using the GR picture, as in your comments. $\endgroup$ – Chiral Anomaly Oct 10 '20 at 2:41
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    $\begingroup$ Relevant Veritasium Video $\endgroup$ – Paŭlo Ebermann Oct 10 '20 at 14:34
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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.

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  • $\begingroup$ Question: The person is accelerating due to the normal force, but the normal force is (at least in the Newtonian picture) a contact force - a short distance force arising from the electromagnetic interaction between surfaces. If we ignore the person, though, the ground itself is accelerating with respect to the stream of falling apples, and it doesn't seem quite right to say that this is due to the normal force. I suppose what really resolves the apparent paradox is that the word "accelerating" has a somewhat altered meaning within the formalism of GR (as compared to the Newtonian picture)? $\endgroup$ – Filip Milovanović Oct 10 '20 at 10:25
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    $\begingroup$ @FilipMilovanović The ground is accelerating upward with respect to a stream of falling apples because of contact forces with matter below. Newtonian mechanics is well validated in regimes where velocities are extremely low with respect to the speed of light and distances are extremely large with respect to the Schwarzschild radius. GR has to be compatible with Newtonian mechanics in such regimes, and it is. There is little, if any, measurable difference between GR and Newtonian mechanics in such regimes. $\endgroup$ – David Hammen Oct 10 '20 at 10:38
  • $\begingroup$ I'm not saying it's incompatible - I'm just wondering if there's a conceptual difference here in the meaning of the word acceleration compared to how it was used in the pre-Einstein world. Regarding the contact forces: I understand what you're saying; in GR, if you're not following a geodesic, then there's some force acting on you, and in this particular scenario, it's the normal force. Now, you've said that a ground layer is accelerating due to contact forces with the layer below - which is fine, but you can do that layer by layer until you run out of layers. $\endgroup$ – Filip Milovanović Oct 10 '20 at 11:35
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    $\begingroup$ The horizon is a conceptual border, not a real object. A rocket that maintains itself just outside the horizon is accelerating. Even though it goes nowhere. So yes, the word has acquired a slightly different meaning now. Acceleration is no longer just the second derivative of displacement; it's now fundamental, since it's something observers can calculate and agree on, while displacement, duration and velocity are relative. $\endgroup$ – Ross Presser Oct 10 '20 at 15:18
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    $\begingroup$ @RossPresser while I appreciate the simplicity and elegance of benrg's answer (and wish I could upvote it a dozen times), I feel that this observation gets at the heart of what non-physicists like myself need to know about GR and acceleration. Would be great to see it in a standalone answer. $\endgroup$ – Lawnmower Man Oct 11 '20 at 0:34
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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.

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    $\begingroup$ Thanks! Very interesting analogy! (A casual reader could be confused by the analogy mentioning the earth in a way "unrelated" to the question, but that's okay once it's understood.) $\endgroup$ – Stéphane Gourichon Oct 10 '20 at 9:25
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    $\begingroup$ So, is this what people mean when they say that spacetime is "flowing into" a black hole? As the object moves through time, the coordinate grid (i don't know if that's the right term) appears to sort of flow into it, so locally it looks like the boundary is accelerating in a spacetime diagram? Not sure if I worded that correctly, but, am I close when it comes to the general idea? $\endgroup$ – Filip Milovanović Oct 10 '20 at 10:12
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    $\begingroup$ I'm curious, then: what is causing the outward acceleration, and why does the spacetime curvature effect balance it exactly? $\endgroup$ – The_Sympathizer Oct 10 '20 at 14:19
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    $\begingroup$ The acceleration is caused by the solidity of the earth. If the earth were permeable, you'd fall freely toward the center. In General Relativity, anything other than free fall is acceleration. So it balances because you are not moving. $\endgroup$ – Ross Presser Oct 10 '20 at 15:13
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    $\begingroup$ @HTNW I was kind of thinking that, just wanted to be sure. So basically the EM forces are constantly driving the Earth apart, but it "doesn't get anywhere" because spacetime doesn't let it. $\endgroup$ – The_Sympathizer Oct 11 '20 at 1:38
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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.

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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 $g$.

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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.

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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).

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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

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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.

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