Why does an object initially at rest begin to fall?

  • #1
Grasshopper
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Summary:
Is there a relationship between gravitational time dilation and falling bodies? Does the fact that time is relative with respect to altitude have anything to do with why a body will fall? Is so, what is the relationship? I have no conceptual explanation for why this should happen.
Motivation for why I'm asking this:
I'm trying to better understand why an object that is initially at rest with respect to a massive body will fall simply by virtue of the curvature of spacetime. If it were moving through space with respect to the Earth initially, I could see the curved path warping it to the Earth, but if it's at rest in space, what is the impetus to begin the motion through space?

To remedy this ignorance, I began looking at some math that is a bit above my head. I found a derivation that starts at, after some approxiations, the Schwartzchild metric for a weak gravitational field. This is that source:

https://aapt.scitation.org/doi/10.1119/1.4972045

There is one part in this derivation I did not understand at all (the killing vector). However, after the article got passed that step, I was able to follow along (my understanding was decent as of section (6.) ), which, by considering some obvious approximations (including slow speed), led to this:

##\frac{dx}{dt} = (2ax)^{1/2}##

And from there it's easy to see how you'd get an object falling with the familiar x = (1/2)at^2 equation.


It thus seems to me that simply having that curved spacetime metric automatically results in a body falling.



My major conceptual problem:


BUT, this is a mathematical explanation and doesn't help my intuition that is demanding the at rest apple have some sort of push to begin motion through space. So I started thinking about time dilation and wondering if its a transformation between coordinate system of the apple and the earth wherein hides the conversion of motion only through time to motion through space and time. Unfortunately, it then occurred to me that one of the assumptions made in the above derivation was that dx/dt is approximately dx/dτ (because of the non-relativistic speed of the object). If that's true, the gravitational time dilation shouldn't have much of an effect, I wouldn't think.


So I'm thinking that conceptual explanation for why the object at rest begins to fall is faulty.



If you have some insight on what would cause the initially at rest apple to fall to Earth, please post it. I listed this as intermediate, but all levels of math are welcomed as long as an explanation comes with them. Thanks, as always.
 

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  • #2
PeroK
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The motion (any motion) between the ball and the surface of the Earth is relative. You can take a coordinate system where the ball remains at rest; you can take a coordinate system where the Earth remains at rest; or, you can take a coordinate system (relative to the Sun, say) where neither remains at rest.

The first issue is to accept that there is nothing absolute about the ball's motion. Whether the ball moves all depends on your choice of coordinates.

That leaves us needing an explanation for the relative motion of the ball and, say, an object on the surface of the Earth. The ball, when dropped, has no forces on it. The object on the surface of the Earth has a real upward force on it, and hence undergoes real (proper) acceleration. In which case, whatever local coordinates we use the ball and the surface of the Earth must relatively accelerate towards each other. If you choose a coordinate system where the Earth is at rest, then the ball must move in that system.

In summary, the local relative motion results from the real force on the surface of the Earth and the ball being free of any force.
 
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  • #3
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I'm trying to better understand why an object that is initially at rest with respect to a massive body will fall simply by virtue of the curvature of spacetime.

You're looking at it wrong. The object was not "initially" at rest with respect to the massive body; it was momentarily at rest with respect to the massive body. A little bit before it was at rest, it was rising. (Assuming, of course, that the object was in free motion--no forces acting on it--the whole time.) So you need to look for an explanation of the object's entire trajectory, not just the "falling" part.

The explanation is that, if no forces are acting on the body, its trajectory through spacetime is the one that locally maximizes its proper time. In other words, its trajectory is a timelike geodesic of spacetime. This is the spacetime analogue of, for example, a great circle trajectory on the surface of the Earth. If you start out on the Earth and just walk in "as straight a path as you can", your path will be a great circle. In other words, that path is the "default" path that you take if you don't try to take any other path. A geodesic of spacetime is the same kind of thing: it's the path an object takes when it's not "trying" to take any other path, where "not trying to take any other path" means "no forces are acting on the object". (Note that gravity itself is not a force in GR.)

What really requires explanation, then, is not why the individual object's path rises, comes to rest, and then falls, relative to the massive body, but the global configuration of all such paths--why, for example, an apple above Australia has a path that behaves oppositely, relative to the Earth, from an apple in England ("rising" and "falling" are opposite directions, globally, for the two apples). That global configuration of the spacetime geometry is what GR explains using the Einstein Field Equation, which gives us the Schwarzschild metric as the spacetime geometry outside a massive body like the Earth. But you don't need all that to understand the trajectory of a single apple: the simple fact that the apple's trajectory is a geodesic is enough.

If it were moving through space with respect to the Earth initially

It is. It rises before it comes to rest. See above.

Also, "moving through space" is the wrong way to look at it. It is moving through spacetime, along a trajectory in spacetime. All objects do that; it's impossible for anything to "stand still" in spacetime. So asking why any particular object moves in spacetime is pointless; all objects do.

What particular motions through spacetime are also "moving through space" depends on your choice of coordinates. In local inertial coordinates, the apple is not moving through space: the Earth is. But both are moving through spacetime.

if it's at rest in space, what is the impetus to begin the motion through space?

It doesn't need any. "At rest in space" depends on your choice of coordinates, and no choice of coordinates can affect the actual physics. The physics is motion through spacetime, as above.
 
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  • #4
Grasshopper
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Thanks for the replies.

PeroK:
The acceleration of the Earth is one thing I didn't consider, although that one I do not fully understand (I have seen an equation that shows why it must be true, but it is not intuitive to me). I do understand the acceleration force felt upon the surface. This is basically just the normal force, is it not? What I don't understand is how it essentially propels the Earth to the apple. Unless it's just an application of the equivalence principle. If that's the case it's so subtly obvious that it's not obvious.

Then the other problem would be why doesn't the acceleration on the other side of the Earth keep it steady with respect to the apple. But, I can imagine a couple ways this could be explained: (1) Maybe these things do not apply globally, but only apply locally; (2) Maybe what matters is the differential in acceleration, and since the apple is closer to the nearside of the Earth than the far side, it is the nearside acceleration of the Earth that wins. I'm sure the real explanation is much more satisfying.



PeterDonis:

When I was saying the apple was "at rest in space," I meant that in the apple's frame, all its spacetime motion is through time. Is that incorrect?

Aside from that, it does appear that you are saying that there is no real cause other than the fact that the spacetime metric naturally encodes this motion. It's a geodesic through spacetime, and spacetime is curved in such away that unless the speed of the apple is the escape velocity, it will eventually come into contact with the Earth. Alternatively I suppose you could say the same thing is true of the Earth, could you not? The Earth moves through spacetime along a geodesic, and that geodesic includes an event where the apple and the Earth collide. How far off is this?
 
  • #5
PeroK
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Re the acceleration at the Earth's surface, that is Newtonian physics! The equivalence principle says that locally we have Special Relativity and at low speeds we have Newtonian physics ##F = ma##. But without the fictitious Newtonian gravitational force.

Note this is local. Where GR does come in is in preventing you extending this local analysis - in particular extending it to the far side of the Earth is not possible.
 
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  • #6
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When I was saying the apple was "at rest in space," I meant that in the apple's frame, all its spacetime motion is through time.

That's correct, but it's correct for all of the apple's motion, not just for the instant that the apple is at rest relative to the Earth. So there is no need for any "impetus" to cause the apple to start "moving in space" in this frame, since in this frame it is always at rest; the Earth is what moves.

it does appear that you are saying that there is no real cause other than the fact that the spacetime metric naturally encodes this motion.

Yes. But "motion" is a confusing word here. The apple's worldline is a geodesic of the spacetime geometry. That is true regardless of which frame you adopt: the apple's frame, in which the apple is always at rest, or the Earth's frame, in which the apple is only momentarily at rest at the maximum altitude of its trajectory.

The Earth moves through spacetime along a geodesic

The center of the Earth does, but not the rest of the Earth. The Earth is a large massive body held together by its own gravity, so it is in hydrostatic equilibrium; that means points on the Earth's surface are being pushed outward by the layers of Earth underneath them. That means points on the Earth's surface are not traveling on geodesics.
 
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  • #7
Grasshopper
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Okay, so here's a related question:

How would a reference frame above the Earth, watching two apples approach the Earth from opposite ends of the planet, describe the events? Surely it could not describe the Earth as accelerating toward both apples, right? But both apples would describe the Earth as accelerating to them.
 
  • #8
PeroK
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Okay, so here's a related question:

How would a reference frame above the Earth, watching two apples approach the Earth from opposite ends of the planet, describe the events? Surely it could not describe the Earth as accelerating toward both apples, right? But both apples would describe the Earth as accelerating to them.

If we assume each apple is close to the Earth, then in each case we can use a local frame to describe the local events. Neither local frame, however, can be extended to a global frame. Not using the equivalence principle.

If you want a single frame covering both events (you have a wide choice) but perhaps the simplest is to use the Schwarzschild coordinates. That is a global coordinate system and the physics cannot be approximated by SR or Newtonian mechanics. You must, in that case, use GR.

At that point you have to face up to the fact that you need GR and the equations of motion for GR. If we imagine the apples are initially held at rest relative to the surface of the Earth, then the explanation is:

1) The apples and the surface of the Earth are initially subject to real forces that prevent their following a geodesic. I.e. initially there is no relative motion.

2) When the apples are released, they both follow geodesics - and given the global spherical symmetry of spacetime curvature means they both fall towards the Earth (and in a global sense towards each other).

You can analyse each individually in terms of local physics and local forces. But, as soon as you look at the global system, you must bring the global curvature of spacetime into the equation. The global behaviour requires GR and curved spacetime.

A final point: if you try to assume that they remain at rest relative to the Earth after release, you have the absurd conclusion that the local force holding them in place is doing nothing. That the local motion is the same with and without the restraining force. And that is physically absurd.
 
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  • #9
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So, if the apple is feeling a force (because it's being held in place), then it by definition is not following a geodesic, which means it is not taking the shortest path through spacetime? In that case, since it's spatial coordinates are not changing, I must assume that its path through time is different than what it would be if it was not being held up by a force. I'm not sure if the time component interval would be longer or shorter since the Minkowski distance formula has that negative sign, though.

Regardless, if that's the case, then I would start to see a bit of the connection between gravity and time (which relates to the thread I mead a few weeks ago).


Would you or another be kind enough to tell me how far off the mark that is? Thanks again. I know you guys work hard outside of here, so it's nice to get your insight on the forums.
 
  • #10
PeroK
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So, if the apple is feeling a force (because it's being held in place), then it by definition is not following a geodesic, which means it is not taking the shortest path through spacetime? In that case, since it's spatial coordinates are not changing, I must assume that its path through time is different than what it would be if it was not being held up by a force. I'm not sure if the time component interval would be longer or shorter since Minkowski distance has that negative sign, though.

Regardless, if that's the case, then I would start to see a bit of the connection between gravity and time (which relates to the thread I mead a few weeks ago).


Would you or another be kind enough to tell me how far off the mark that is? Thanks again. I know you guys work hard outside of here, so it's nice to get your insight on the forums.

It's not off the mark, but I'm not sure how helpful it is. The apple held in place is not on a geodesic, that's true. What that means is quite subtle in terms of spacetime paths. But, if you grind through the maths you get spacetime geodesics that are almost the same as in Newtonian gravity. Especially in the case of an apple close to the Earth's surface.

I've seen some attempts to explain this in terms of gravitational time dilation, but personally I'm not convinced. Note that in the Newtonian case there is no way really to get elliptical orbits except by grinding through the equations. This is a fact of life in physics. Why ellipses? Do the maths!

Anyway, you have the same in GR: you grind through the Schwarzschild solution and out come approximately the Newtonian equations. Although, you can sort of piece it together as we have tried to do in this thread by an appeal to the equivalence principle and local physics and its inevitable Newtonian approximation.

And, of course, if you want to prove that GR really does produce approximately Newtonian gravity, then you have to do the maths. I don't believe there is any way round this.

In any case, being stationary is not on a geodesic, so the apple can't do that. Not easy to prove, though!
 
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  • #11
jbriggs444
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which means it is not taking the shortest path through spacetime?
For time-like geodesics it is the longest path -- path which experiences the greatest elapsed proper time between two specified events.

A zig-zag path will have a shorter elapsed proper time. Down toward a limit of zero.

Note that the presence of a real force does more than make the path deviate from being a geodesic. It changes which event the path ends at. If you want to compare apples with apples (or grapes with grapes), you need to compare the path of a person sitting at the dinner table with the path of a grape that he is throwing vertically and catching in his mouth. That is a comparison of a geodesic path and a non-geodesic path between the same two events.
 
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How would a reference frame above the Earth, watching two apples approach the Earth from opposite ends of the planet, describe the events?

Read this again:

The apple's worldline is a geodesic of the spacetime geometry. That is true regardless of which frame you adopt

What does it imply about the global frame you describe?
 
  • #13
A.T.
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Is there a relationship between gravitational time dilation and falling bodies? Does the fact that time is relative with respect to altitude have anything to do with why a body will fall?
Yes, the gradient of gravitational time dilation is directly related to the free fall (coordinate)acceleration in a hovering frame.

If you have some insight on what would cause the initially at rest apple to fall to Earth, please post it.
It follows a geodesic in space-time, which has distorted coordinates:



How would a reference frame above the Earth, watching two apples approach the Earth from opposite ends of the planet, describe the events?

On the global scale spacetime is intrinsically curved (as shown below). It cannot be rolled out like the cone in the video above, which approximates just a small radial range:

gravity_global_small-png.png


The red path is the geodesic world-line of a free falling object, that oscillates through a tunnel through a spherical mass. Note that the geodesic always deviates towards the "more stretched" proper time, or towards greater gravitational time dilation. Gravitational time dilation has an extreme point at the center of the mass (gradient is zero), so there is no gravity there (but the maximal gravitational time dilation).

Surely it could not describe the Earth as accelerating toward both apples, right? But both apples would describe the Earth as accelerating to them.
You have to differentiate between:

- proper acceleration: frame invariant, what an accelerometer measures, deviation from geodesic path in space-time

- coordinate acceleration: frame dependent, 2nd derivative of position in some coordinates
 
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What does it imply about the global frame you describe?
How would a reference frame above the Earth, watching two apples approach the Earth from opposite ends of the planet, describe the events? Surely it could not describe the Earth as accelerating toward both apples, right? But both apples would describe the Earth as accelerating to them.
About the global frame to use to describe apples' paths, do you mean for instance the Schwarzschild's coordinate system from the center of the Earth ?
 
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  • #15
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Read this again:



What does it imply about the global frame you describe?
I imagine it would have to mean that the path would remain a geodesic regardless, and same with the apple on the other side. I am skeptical that a frame can be arbitrarily large and still be valid for every relevant point, though.

Related question: But where does velocity come into play here?

Something moving slow and something moving fast have to travel through the same curved spacetime, but something moving fast is obviously moving through a larger interval of space per unit time. Do space and time in this case balance out so the path through spacetime is the same length, even if different components are of different size? (neglecting a situation, if it is possible, where velocity is so great that it actually affects spacetime curvature)
 
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PeroK
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I imagine it would have to mean that the path would remain a geodesic regardless, and same with the apple on the other side. I am skeptical that a frame can be arbitrarily large and still be valid for every relevant point, though.

Related question: But where does velocity come into play here?

Something moving slow and something moving fast have to travel through the same curved spacetime, but something moving fast is obviously moving through a larger interval of space per unit time. Do space and time in this case balance out so the path through spacetime is the same length, even if different components are of different size? (neglecting a situation, if it is possible, where velocity is so great that it actually affects spacetime curvature)
Geodesics are very much paths through spacetime and not space. For example, the orbit of the Earth is only a geodesic at the speed the Earth is moving. If you wanted to do an Earth orbit faster or slower, then that would not be a geodesic and you would need a force to control your motion.

In particular, the curvature involves both the time and space dimensions.

Things get a lot more complicated when you consider particles that themselves influence the spacetime geometry. But that's really a detail.
 
  • #17
jbriggs444
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Do space and time in this case balance out so the path through spacetime is the same length, even if different components are of different size?
In spacetime, the endpoints of the path are events. They specify both a position and a time. If you go faster, you are almost certainly changing the end event on the path. So you are no longer comparing two paths between the same two endpoints.

But it is possible for two timelike geodesic paths between the same two endpoint events to have different elapsed times.
 
  • #18
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I imagine it would have to mean that the path would remain a geodesic regardless, and same with the apple on the other side.

Yes.

I am skeptical that a frame can be arbitrarily large and still be valid for every relevant point, though.

Certainly a global frame can. Schwarzschild coordinates cover the entire vacuum region of spacetime outside the Earth.

An inertial frame can't be arbitrarily large in a curved spacetime, but the global frame just described is not an inertial frame.

where does velocity come into play here?

Velocity relative to what?

Something moving slow and something moving fast have to travel through the same curved spacetime

Assuming we have defined what "moving slow" and "moving fast" are relative to, then yes, but they will travel on different worldlines through the same curved spacetime.

Do space and time in this case balance out so the path through spacetime is the same length

"The same length" by what criterion? We're talking about different curves in the same geometry, so "the same length" by itself has no meaning. You have to specify some criterion for how the "length" along each curve is to be determined--between what pair of points on each curve the "length" is to be.
 
  • #19
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I am skeptical that a frame can be arbitrarily large and still be valid for every relevant point, though.
An inertial frame cannot be arbitrarily large in curved space time. The diagram of curved spacetime in post #13 shows non-inertial coordinates (except for the center of the mass, which is inertial locally)

Something moving slow and something moving fast have to travel through the same curved spacetime
No. Different velocity means different direction in spacetime. So even when starting at the same point in spacetime (event), objects with different velocity will not move through the same parts of curved spacetime,
 
  • #20
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Okay, so, it's clear I've come close to maximizing the information you can get without actually learning the math of general relativity, which leads me to at least two more question (the first is kind of a dumb one, honestly): All of you clearly have been studying general relativity for a long time, and you've gained all this intuitive understanding of it along the way. So here it is:

Q1: When did you gain this intuitive understanding along the way?

I ask because many times when I've been learning basic physics I've been able to solve a problem without actually having the faintest clue what was going on, just by remembering the structure of the math.

Q2: I have a copy of Essential Relativity by Rindler, and I plan to watch a lecture series on linear algebra and then jump in. Would you consider this a reasonable place to start? It's the only one on the topic I have. (also, would you say undergraduate linear algebra would be sufficient math background, or should I touch up on something else?)
 
  • #21
PeroK
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Q2: I have a copy of Essential Relativity by Rindler, and I plan to watch a lecture series on linear algebra and then jump in. Would you consider this a reasonable place to start? It's the only one on the topic I have. (also, would you say undergraduate linear algebra would be sufficient math background, or should I touch up on something else?)
A bit of linear algebra never hurts, but you need to focus on vectors, tensors and calculus (inclucing vector calculus) and differential equations.

The other thing is that knowing physics helps as well: not just specific knowledge, but the way physics is done. Specifically, unless you know SR well, it's unlikely you'll make much headway with GR. It's too big a step. And generally, the more you pick up how physicists throw stuff around the better: the liberal use of Taylor Series; the understanding of a potential function; the use of coordinate transformations etc.

The risk is that you get started with GR and then get stumped by something like a simple coordinate transformation. GR is an advanced undergraduate/graduate level subject. Any author is bound to assume a working knowldege of a lot of physics and mathematics. They won't (and indeed cannot) explain how quadratic equations are solved etc. The risk is that you see something come "out of the blue" that a well-prepared student would have seen 20 times before.

There is an excellent lecture series from MIT here:

https://ocw.mit.edu/courses/physics...ction-and-the-geometric-viewpoint-on-physics/

But, ultimately, it is a graduate-level subject.
 
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  • #24
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I suppose there’s a world of difference between knowing how to manipulate an equation and understanding the math. At least that’s been my life. I can solve really easy PDEs (like a 1D heat equation) by just following the cookbook, but conceptualizing what it means is another story.
 
  • #25
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I suppose there’s a world of difference between knowing how to manipulate an equation and understanding the math. At least that’s been my life. I can solve really easy PDEs (like a 1D heat equation) by just following the cookbook, but conceptualizing what it means is another story.

John Archibald Wheeler put it like this:

“Never make a calculation until you know the answer. Make an estimate before every calculation, try a simple physical argument (symmetry! invariance! conservation!) before every derivation, guess the answer to every paradox and puzzle. Courage: No one else needs to know what the guess is. Therefore make it quickly, by instinct. A right guess reinforces this instinct. A wrong guess brings the refreshment of surprise. In either case life as a spacetime expert, however long, is more fun!”

https://www.goodreads.com/quotes/988735-never-make-a-calculation-until-you-know-the-answer-make
 

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