Earth orbits the Sun because the Sun's mass curves spacetime. But the Sun is 150 million kilometers away from here; how can mass curve spacetime that it's not actually in? Is that a form of action at a distance?
Asked
Active
Viewed 3,633 times
11
6 Answers
32
The curvature of spacetime can be separated mathematically into two components, Ricci curvature and Weyl curvature. They are locally independent, but their joint variation over spacetime is constrained by mathematical relations (the second Bianchi identity).
General relativity says that the Ricci curvature is determined by the local matter density (stress-energy), but there is no direct constraint on the Weyl curvature.
So, in vacuum regions (Schwarzschild field, gravitational waves, etc.), the Ricci curvature is zero while the Weyl curvature can be nonzero. The physical value of the Weyl curvature is determined by the mathematical curvature relations and the boundary conditions.
The Weyl curvature represents the propagating degrees of freedom of the gravitational field, which can exist without matter. This spreads the influence of gravity beyond the immediate location of matter, but does not represent action at a distance because it still acts causally (limited by the speed of light).
Typically we solve for the gravitational field of the Sun as a steady state, which makes it seem like a global result that appears all at once. However, if we pose an initial-value problem containing a central mass (thus determining Ricci curvature) with a different initial configuration of Weyl curvature, the "extra" Weyl curvature would break up into gravitational waves and ultimately disperse to large distances, leaving the steady-state (Schwarzschild) solution.
That is, roughly speaking, the Weyl curvature is indirectly determined by matter, as the effect of Ricci curvature on Weyl curvature propagates outward at the speed of light.
nanoman
- 3,810
-
Love that Weyl curvature (which is central to Penrose's Conformal Cyclic Cosmology, one of my personal favorites), but it's often been described as a hypothesis, I guess because its subtle mathematical connections to reality (which your answer describes well) are not easily subjected to astronomical or experimental verification. Maybe the rather recent actual detection of gravitational waves has changed that. – Edouard Sep 22 '20 at 17:22
16
The curvature extends far away from the mass creating it, becoming progressively gentler as the distance increases, and it never completely goes away. The same curvature becomes stronger as the distance decreases, making all the effects of gravity more powerful the closer you get to the mass.
These are the characteristics of a field which extends throughout space, and upon which things like matter and electrical charge can act. For all known fields like this, there is a finite speed with which disturbances can travel through it and be felt later by distant objects and that speed is c, the speed of light.
niels nielsen
- 92,630
-
As the curvature extends itself into, say, the space around an identical mass equidistant from the 1st, do their curvatures cancel each other out so as to result in the "flat space" often attributed to inflationary cosmology, or do they superimpose themselves upon each other, like different forms of energy? (I'm trying to figure out whether or not your description corresponds to Guth's occasional description of gravity as "negative energy" in his more popular works, which doesn't seem to find much formal support in "mainstream physics".) – Edouard Sep 02 '20 at 04:16
-
1the curvatures do not cancel and the space between them is not "flat". Furthermore, an object placed at the midpoint between them will gravitate towards one or the other if perturbed even a tiny amount (there is no stable equilibrium point between them). – niels nielsen Sep 02 '20 at 06:39
5
This is not a perfect analogy, but imagine a rubber sheet, held flat in a frame. Now pinch as small circular region in the rubber (eg, push the sheet through a small ring). The small pinched region causes the rubber to stretch, with greater stretching near to the ring, becoming progressively less away from it. The pinched ring is analogous to a gravitating mass, and the stretching of the sheet is analogous to the scaling distortion of a map of space surrounding a gravitating body (bearing in mind that what we mean by curvature mathematically is actually scaling distortions of maps, not something curved in the usual sense of the word).
Of course it is a bit more complicated if one considers spacetime rather than just space, but one cannot distort space without also distorting spacetime.
Charles Francis
- 11,546
- 4
- 22
- 37
-
Ok, so does that mean that the curvature of a tiny piece of empty space can be computed from the curvature of its neighboring pieces, without considering distant masses? That would be what the rubber sheet analogy would seem to imply. – Doradus Aug 29 '20 at 20:32
-
3@Doradus, Essentially, yes. The properties of any region of empty space (tiny or large) can be computed (at least in principle) from Einstein's equation together with the boundary condition. This is exactly how we compute Schwarzschild geometry, from Einstein's equation together with boundary conditions of a central mass and flat space at infinity. – Charles Francis Aug 29 '20 at 20:39
-
-
4
-
1@Doradus It's also how gravity "escapes" a black hole so the gravity of a black hole can influence stuff outside of it. Although, I must admit I have no idea how this idea is supposed to work when you treat gravity as gravitons. – DKNguyen Aug 31 '20 at 00:21
3
In reality it is worth mentioning that it is not mass, but stress-energy that causes spacetime curvature.
Actually, we are inside the Earth's static gravitational field, that is why the Moon is orbiting Earth, and the Earth (together with the moon) is inside the Sun's static gravitational field, that is why the Earth-Moon system is orbiting the Sun, and the Sun (together with the Earth and the Moon) is inside the static gravitational field of the Milky way (everything in the Milky Way orbits around the galactic barycentre, whose location closely coincides with Sagittarius A), that the Sun is orbiting around, you get the idea.
https://en.wikipedia.org/wiki/Sagittarius_A*
https://en.wikipedia.org/wiki/Barycenter
But it is not so simple, because all objects inside the Milky way (and outside it) do have an effect on every single object everywhere in the universe. Of course, when you do calculations, then certain effects become negligible with distance.
What we call a static gravitational field, does extend infinitely. Does this mean that here on Earth we are affected by the gravitational field of all objects in the whole (not just observable) universe? In theory yes.
https://en.wikipedia.org/wiki/Gravitational_field
You are asking " Is that a form of action at a distance?", in reality it is not.
Gravitational influences do propagate at the speed of light, not instantaneously.
How fast does gravity propagate?
Does this mean that changes in these gravitational fields affect us instantaneously? No, because as everything, the speed of these changes in the static gravitational field have to follow one simple rule, nothing, not even the changes in the static gravitational fields can travel faster then the speed of light.
Árpád Szendrei
- 28,452
2
From one perspective, spacetime has characteristics similar to a physical medium in this sense. In a medium, any given small portion is connected to those immediately next to it. Hence, if you disturb that small portion, it will disturb those around it, and those will in turn disturb the ones around them, and so forth. This results in the generation of a ''wave'' ahead of which the medium is in its undisturbed state, and behind which the medium is in its disturbed state.
Likewise, spacetime acts as such a "medium" in that it works the same way. If you disturb a small area, it must disturb those around it and those around that and so forth. So once you gather up mass in an area, the disturbance generated there must be propagated to the surrounding spacetime.
Now, when we consider an object travelling through spacetime, its motion is determined by the characteristics of that in its immediate vicinity. Thus if that spacetime has already been disturbed by a distant large mass, then it will follow an altered course.
From another perspective, which is perhaps closer to how it is described in the mathematical formulation, and which avoids the notion of always imagining a piling-up process to have occurred prior, one can say that any given mass distribution ''intrinsically'' has a "halo" of spacetime distortion that must invariably come with it. The reason it must be a "halo" is because that if it were entirely localized to a sharp border with the object, the spacetime would fail to be continuous - continuity, by definition, implies a gradual change, not an abrupt one.
From the viewpoint of general relativity alone, continuity of spacetime appears to be a fundamental property without a further "reason", though then again general relativity is almost surely not the complete story of gravity, as one may have heard.
The_Sympathizer
- 20,136
-4
There is no contradiction. The Sun's mass is quite humble and not able to curve spacetime by very much at all; hence the almost unnoticeable deviances in Mercury's orbit required for GR to overthrow Newtonian.
Mass curves spacetime and gravity keeps planets in orbit.
If the motive is questioning a TOE, which is not here yet, it may help to know how space differs from spacetime.
-
3Mass curving spacetime is gravity. Gravity is the curvature of spacetime. – OrangeDog Aug 30 '20 at 22:00
-
1Thanks for the Sabine Hossenfelder link! Just discovered her channel recently. – Doradus Sep 16 '20 at 00:33
-
@OrangeDog, the Earth's orbit is not the result of the curved spacetime from the Sun's mass. The small deviances in Mercury's existing orbit are. This is to my knowledge the largest discrepancy between Newtonian gravity and GR in our solar system, and the discrepancy is many orders of magnitude smaller than orbital capture. It's clear two things are at work, not one as you insist. – Henrik Erlandsson Sep 18 '20 at 14:24
-
1@HenrikErlandsson you are wrong. General Relativity provides a complete description of gravity and planetary orbits. The spacetime curvature caused primarily by the sun is responsible for the orbits of both Earth and Mercury. It is not some "correction" that is applied to Newtonian mechanics. Newton/Kepler equations are just a simplified version that gives increasingly incorrect answers in more extreme conditions. – OrangeDog Sep 18 '20 at 15:51
-
1@HenrikErlandsson - I'm not sure where you are getting this from, but you're mistaken. – Doradus Sep 22 '20 at 15:09
-
@Doradus, I'm sorry, but I seem to be the only one answering the question as asked. Earth's orbit is perfectly predictable using Newtonian physics, without the need for action at a distance from GR. We would still be using it, if not for the small deviances in the orbit of Mercury. – Henrik Erlandsson Sep 25 '20 at 22:48
-
1@HenrikErlandsson - Your answer is flawed in several ways. First, my question was about curved spacetime, which is not a feature of Newtonian gravity, so Newtonian gravity is no more relevant than Ptolemaic epicycles. Second, Newtonian gravity is entirely based on action at a distance, so it obviously does not resolve that part of the question. Third, and most important, your answer is factually incorrect, so even if it addressed the question (which it doesn't) it would still merit the downvote. – Doradus Sep 29 '20 at 11:03
However, the original problem with action at a distance is that it was instantaneous and broke relativity.
– Toby Peterken Aug 29 '20 at 16:45Also, the view that things exchange photons is a bit complicated. EM is mediated by the EM field. Photons are ripples in that field and it turns out we can approximate a complicated field configuration by looking at lots of different ripples.
– Toby Peterken Aug 30 '20 at 08:50