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It is fine to say that for an object flying past a massive object, the spacetime is curved by the massive object, and so the object flying past follows the curved path of the geodesic, so it "appears" to be experiencing gravitational acceleration. Do we also say along with it, that the object flying past in reality exeriences NO attraction force towards the massive object? Is it just following the spacetime geodesic curve while experiencing NO attractive force?

Now come to the other issue: Supposing two objects are at rest relative to each other, ie they are not following any spacetime geodesic. Then why will they experience gravitational attraction towards each other? E.g. why will an apple fall to earth? Why won't it sit there in its original position high above the earth? How does the curvature of spacetime cause it to experience an attraction force towards the earth, and why would we need to exert a force in reverse direction to prevent it from falling? How does the curvature of spacetime cause this?

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. So we cannot just say that it fell to earth because its geodesic curve passed through the earth. Why did the spacetime curvature cause it to start moving in the first place?

Qmechanic
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user1648764
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    I have always wondered about this (and related). This is so brushed aside in populist explanations! – GreenAsJade Mar 11 '14 at 12:06
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    This exact question was puzzling me on my drive home yesterday, and my kids were wondering why I was tracing curves in the air over the steering wheel :) – Charlweed Jan 20 '18 at 22:57
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    Space-time curvature doesn't "cause" gravity. Space-time curvature is gravity. – EvilSnack Apr 13 '20 at 16:35
  • In agreement with EvilSnack, I have to suggest that the OP should look at the 8th to 10th minutes of the video at https://www.youtube.com/channel/UCpMcsdZf2KkAnfmxiq2MfMQ , which illustrates the fact that it's the revolving, around the sun, of the ORBIT of Mercury which provided the 1st confirmation of General Relativity. Newtonian gravity had sufficed to explain the orbiting of planets and moons, but the explanation of the revolving of an orbit took GR, as it conflicted with the notion of gravity as an anthropomorphic "force", however subtle and faint the curvatures may seem at close range. – Edouard Dec 06 '20 at 03:13
  • As the OP has visited this site very recently, I should mention that I've had to withdraw an earlier comment of mine which included the remark that "our motion thru time is entirely passive", because we can accelerate our motion, and any acceleration causes time dilation, although I doubt that the infinitesimal dilation caused by the acceleration of their motion by any person walking is perceptible by any means currently available. – Edouard Dec 11 '20 at 15:17

6 Answers6

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To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation.

Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" through spacetime is called a four-velocity and it is always equal to the speed of light. Spacetime in gravitation field is curved, so the time axis (in simple terms) is no longer orthogonal to the space axes. The apple moving first only in the time direction (i.e. at rest in space) starts accelerating in space thanks to the curvature (the "mixing" of the space and time axes) - the velocity in time becomes velocity in space. The acceleration happens because the time flows slower when the gravitational potential is decreasing. Apple is moving deeper into the graviational field, thus its velocity in the "time direction" is changing (as time gets slower and slower). The four-velocity is conserved (always equal to the speed of light), so the object must accelerate in space. This acceleration has the direction of decreasing gravitational gradient.

Edit - based on the comments I decided to clarify what the four-velocity is:

4-velocity is a four-vector, i.e. a vector with 4 components. The first component is the "speed through time" (how much of the coordinate time elapses per 1 unit of proper time). The remaining 3 components are the classical velocity vector (speed in the 3 spatial directions).

$$ U=\left(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau}\right) $$

When you observe the apple in its rest frame (the apple is at rest - zero spatial velocity), the whole 4-velocity is in the "speed through time". It is because in the rest frame the coordinate time equals the proper time, so $\frac{dt}{d\tau} = 1$.

When you observe the apple from some other reference frame, where the apple is moving at some speed, the coordinate time is no longer equal to the proper time. The time dilation causes that there is less proper time measured by the apple than the elapsed coordinate time (the time of the apple is slower than the time in the reference frame from which we are observing the apple). So in this frame, the "speed through time" of the apple is more than the speed of light ($\frac{dt}{d\tau} > 1$), but the speed through space is also increasing.

The magnitude of the 4-velocity always equals c, because it is an invariant (it does not depend on the choice of the reference frame). It is defined as:

$$ \left\|U\right\| =\sqrt[2]{c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2} $$

Notice the minus signs in the expression - these come from the Minkowski metric. The components of the 4-velocity can change when you switch from one reference frame to another, but the magnitude stays unchanged (all the changes in components "cancel out" in the magnitude).

mpv
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    Can you comment on why the four-velocity is always the speed of light? – GreenAsJade Mar 11 '14 at 12:07
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    @GreenAsJade: The four-velocity is commonly defined as $u^i = \frac{dx^i}{d\tau}$. Written as a four vector, it looks like $\vec{u} = \gamma (c, \mathbf{v})$, with $\gamma = (1-v^2/c^2)^{-1/2}$. Its length squared is then $\gamma^2 (c^2-v^2)$, which is equal to $c^2$. – Javier Mar 11 '14 at 14:19
  • Could I ask for clarification whether the velocity referred to in this answer is proper velocity or not? Initially at t=0, the apple is at rest, and I believe one measure of velocity will equal c, whereas the other will not. In the latter case, it would seem that the four-velocity does, in fact, not equal c. Perhaps I should submit a separate question, since this is more detailed than the OP's intention. – Alan Rominger Mar 11 '14 at 14:37
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    Does this mean that particles passing through space in the speed of light does not pass through time? Do electrons not have an age? – Pål GD Mar 11 '14 at 23:35
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    @PålGD: Correct, particles moving at the speed of light do not age. Thus the "time delation" mentioned in space travel things, where they travel a long way but not much time passes. – Mooing Duck Mar 12 '14 at 00:36
  • @AlanSE: "proper" velocity is three dimensional, this post deals almost entirely with (what he called) "four-velocity". The "four-velocity" is always equal to the speed of light. At t=0, the apple is moving "at the speed of light" through time. – Mooing Duck Mar 12 '14 at 00:39
  • @MooingDuck When you say that it is moving through time at the speed of light, what reference frame is this according to? Any? That wouldn't be consistent with time dilation. So this still isn't resolved. – Alan Rominger Mar 12 '14 at 02:12
  • As I understand things, light moves at "the speed of light" in all reference frames, which is part of why (general? special?) relativity is so strange. It breaks the normal rules. However, this is right on the edge of my understanding of physics, so that could very well be 100% incorrect. – Mooing Duck Mar 12 '14 at 02:23
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    @AlanSE The apple is moving through time at light speed only in the reference frame where the apple is at rest (spatially). In some other reference frame (where the apple has some spatial velocity) its speed through time is slower. The four-velocity is a vector which has 4 components. All these components can vary between frames, but the magnitude of this 4-vector remains unchanged (always equals c). – mpv Mar 12 '14 at 06:00
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    @PålGD Particles moving at the speed of light (in space) indeed do not age, because their proper time is zero (due to time dilation). But this does not apply to electrons (which you mention in your comment), because electrons do not move at the speed of light. It applies to photons. – mpv Mar 12 '14 at 06:25
  • @mpv You make the statement that the 4-velocity has a magnitude of c in all reference frames. If the time component is not equal to c, then that means a spatial component must be non-zero. In the case of the apple, which? Don't all reference frames agree it to be at rest (using schwarzschild metric for far away observer)? – Alan Rominger Mar 12 '14 at 12:32
  • @AlanSE I'm not sure I understand the problem. This is basic transformation. If the time component is less than c, the spatial components will compensate for that to make the magnitude exactly c. Which particular components? It depends on the chosen reference frame. In a frame which is moving at c/2 along y axis, the y component of the apple 4-velocity will be -c/2. In another frame it will be other components. The apple is not at rest in all reference frames. There is a lot of frames where the apple is moving in various directions at various speeds. – mpv Mar 12 '14 at 14:13
  • Surely you must have $\frac{dt}{d\tau} > 1$ in a frame where the apple has non-zero velocity? So the proper time measured by the apple is less than the elapsed coordinate time. – Simon Woods Mar 12 '14 at 14:29
  • @mpv You've now established that an observer at r=infinity in schwarzschild coordinates perceives the apple to have a velocity. But these aren't moving reference frames. Let's say we have A,B,C, which are Newton, the apple, and aliens. The aliens are stationary relative to Earth and Newton. At t=0, B is stationary relative to A. So if I accept your position, B must be non-stationary to C, but this is clearly wrong. Even if the surface of the Earth was non-stationary to a far away observer, there is some upward/downward velocity where it is both stationary and gravitationally time dilated – Alan Rominger Mar 12 '14 at 14:58
  • @AlanSE I'm not sure how you come to such conclusions. If A,B,C are all stationary with respect to each other, they are not moving with respect to each other. That is elementary. How is my answer suggesting something else? All I'm saying is that if you establish a reference frame which is moving with respect to the apple, then the apple has a spatial velocity in such a frame. That is still elementary. I'm not talking about observers at infinity, just reference frames. I suggest you enter that as a separate question, because it seems difficult to clarify in a comment. – mpv Mar 12 '14 at 15:51
  • @SimonWoods You are correct. I edited the answer to reflect that. Thanks for the catch! – mpv Mar 12 '14 at 16:15
  • @mpv Indeed, I may submit such a question. I see the point about moving observers, and I agree the 4-vector always has a length of c in that special relativity sense. My problem is that a falling object at the top of its parabolic trajectory seems to not have that same property. Right now, my hunch is that the resolution lies deeper than what we've discussed here, and is contained in the non-orthogonality of the time axis you discussed - which leads to observers disagreeing on what constitutes the top of the apple's arc. But this still doesn't resolve the other objection I mentioned. – Alan Rominger Mar 12 '14 at 16:21
  • If I'm explaining to kids, is it correct to say there are 4 dimensions, and gravity causes movement along the 4th (time) to be transferred to the other three, conserving overall velocity. In infinite gravity, movement through time stops, and falling velocity is the speed of light. – Charlweed Jan 20 '18 at 23:10
  • Great answer +1. The only thing to possibly add is that what we call "the speed of light" is actually the speed of time. This makes it easy to understand why we always move in spacetime with the speed of time - because we move in time in our rest frame. This also gives the intuition to the velocity time dilation in SR - the closer we move to the speed of time, the slower we move relative to the movement of time. – safesphere Jul 08 '19 at 17:01
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    "Time flows slower" and "time gets slower and slower" suggest that time itself is equivalent to passage thru it. Wouldn't it be more correct to say that "the duration of passage thru time increases" and "the duration of passage thru time increases more and more", so as to preserve the concept of time as a dimension (rather than a substance or energy) more clearly? – Edouard Nov 28 '20 at 03:50
  • @mpv if the magnitude of the 4-velocity is always the speed of light, then how can the apple in your example as you say - "in this frame, the "speed through time" of the apple is more than the speed of light $$dt/dτ>1$$, but the speed through space is also increasing - have a speed through time that's greater than $$dt/dτ>1$$? – Tivity Jun 14 '22 at 12:50
  • @mpv If the magnitude of the 4-velocity is always the speed of light, then how can the apple in your example as you say - "in this frame, the "speed through time" of the apple is more than the speed of light? – Python House Oct 30 '23 at 15:01
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When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve.

Even when at rest in space, the apple still advances in space-time. Here is a visualization of the falling apple in distorted space-time:

http://www.youtube.com/watch?v=DdC0QN6f3G4

answerman
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  • I've always visualized it as the apple being held by the stem at the top of a valley, the bottom of which is the Earth's gravity well / center of mass. The stem breaks and the apple rolls down the hill into the "valley's" gravity. The animation is very good, but I'm left wondering why space-time is curving away from the direction of earth's gravity. (I think I'm interpreting that correctly because the stem/branch force arrows point opposite from the center of the earth, and that direction is shown to be the center of the curve of the spacetime graph.) – Patrick M Mar 12 '14 at 16:56
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    This is a great video... – Earth is a Spoon Mar 13 '14 at 02:49
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    @PatrickM It's because you've been misled. The gravity well picture is good for showing (classical) gravitational potential and geodesics. It does not represent how spacetime is curved which would be difficult to show for any spatial dimension > 1. Hence the apple video is the only legitimate example I've seen of what curvature really means. Notice here the spatial dimension is not curved at all, but the time dimension is -- that is close to how it is in reality under GR. – Nimrod Sep 16 '18 at 22:20
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As to the first paragraph, gravity shows up as geodesic deviation; initially parallel geodesics do not remain parallel.

Since, for a freely falling particle, the proper acceleration (the reading of an accelerometer attached to the particle) is zero, it is correct to say that a particle whose worldline is a geodesic has no proper acceleration.

But it is not correct to say that a freely falling particle has no coordinate acceleration.

Regarding the second paragraph, if a particle's wordline is not a geodesic, the particle will have a proper acceleration, the particle's accelerometer will not read zero. Two particles that are preventing from falling towards one another will have weight.

Regarding the third paragraph, I think you need to sharpen your conception of worldines and geodesics. If a particle exists, it has a worldline and the worldline of a particle that is free to fall is a geodesic even if the particle is momentarily stationary.

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Not everything needs to follow geodesic Spacetime curvature available to it. With external force, you can prevent a particle from following Spacetime curvature. Only "freely" falling particles follow Spacetime curvature available to them. So, when you see a stationary object not following Spacetime curvature, it's because an external force is preventing it from going to it's inertial trajectory... Means, it's not in "Free Fall".

Come to Apple: In terms of Spacetime, nothing is in rest. An Apple, when attached with tree, is also in motion. But, the motion exist fully in time with zero space component. This motion is NOT according to Spacetime curvature available to it because external forces holding root of Apple oppose it at microscopic level. When these external forces stop working, Apple starts to follow Spacetime curvature which converts time component of motion to space component. That's why Apple's acceleration is merely inertial motion. You can see removal of time component of motion in Gravitational Time Dilation.

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Imagine you are on the northern hemisphere on Earth (assuming it's a perfect sphere).

Now go north with constant speed: you can just go straight north you don't need to steer.

Now go east with constant speed: this is something different now, in order to remain on the same latitude circle you must steer northwards constantly. If you don't see why, try to imagine you are doing it on the 89° latitude circle. If you stop steering, you begin to go "straight" along a geodesic and "fall" towards the equator.

This correction force depends on where you are and which direction you are going (and wish to remain on a "coordinate-straight" path), it's a linear map that maps your velocity into force. It's called the Christoffel-symbols. It's a property of your chosen coordinate system and the geometry of the space-time.

Now in reality on Earth you are in a coordinate system where the coordinates given by the latitude, longitude and altitude and time. Your 4-velocity in space-time is constant $c$. If you stand still, you go straight in the time direction. But in order to keep this four velocity, you feel an upwards force from the floor, this is the effect of the Christoffel-symbols. If you lose the floor your trajectory in space time will be a geodesic and you will fall.

Calmarius
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It seems that there is a common misunderstanding about Einstein's general relativity theory: general relativity is a theory of space and time. From general relativity, the presence of masses causes curvature in spacetime. When other masses move nearby, their trajectory will be affected by the curvature of spacetime which is determined by the geodesic equation. But we can't see the curved spacetime, we can only see its effects on bodies' motions, and these effects can be understanded as if bodies exert forces on other bodies, which is Newtonian view point. So technically, Newton's law of universal gravitation is a theory of effects of the curved spacetime, it's not perfect, but already precise enough to describe motions of planets in our solar system. Einstein's general relativity is a theory of the causes of the curved spacetime, it's a theory of why planet is moving in the way it is. G.R is more precise than Newton's theory when it comes to strong or time varing gravitational fields.

  • This A doesn't answer the OP's final Q: "Why did the spacetime curvature cause it to start moving in the first place?". – Gerold Broser Aug 27 '22 at 02:11
  • Because this is a fundamental thesis of general relativity: when bodies do not subjec to any forces, they follow a geodesic path. Since gravity is not a force, it's the manifesto of curved spacetime. When an apple detaches from a tree, it's not moving through space, but it's moving through time. The geodesic equation describes motions in four dimensional space, not only three dimensional space. Since everything has to move through time, it's trajectory will be described by the geodesic equation and falls towards Earth. – YuJuchong123 Aug 27 '22 at 02:59