<p>Yes. In General Relativity, the positive spacetime curvature due to a gravitational source means that a ball of radius <span class="math-container">$r$</span> has a smaller volume than that given by the usual Euclidean value of <span class="math-container">$\frac43\pi r^3$</span>. This is measured by the <a href="https://en.wikipedia.org/wiki/Scalar_curvature" rel="nofollow noreferrer">scalar curvature</a>, (aka the Ricci scalar). From Wikipedia:</p>
<blockquote>
<p>Specifically, the scalar curvature represents the amount by which the volume of a small geodesic ball in a Riemannian manifold deviates from that of the standard ball in Euclidean space.</p>
</blockquote>
<p>The <a href="https://en.wikipedia.org/wiki/Riemann_curvature_tensor" rel="nofollow noreferrer">Riemann curvature tensor</a>, which expresses the total curvature of spacetime, is composed of the <a href="https://en.wikipedia.org/wiki/Weyl_tensor" rel="nofollow noreferrer">Weyl curvature tensor</a> and the <a href="https://.m.wikipedia.org/wiki/Ricci_curvature" rel="nofollow noreferrer">Ricci curvature tensor</a>.
Paraphrasing slightly from the Weyl tensor article, the Weyl curvature measures the tidal distortion that a body feels when moving along a geodesic, but it's volume-preserving, that is, it doesn't convey any information on how the volume of the body changes.</p>
<p>The Ricci curvature tensor contains all the the information about how volumes change in the presence of tidal forces.</p>
<blockquote>
<p>In general relativity, the Weyl curvature is the only part of the curvature that exists in free space—a solution of the vacuum Einstein equation—and it governs the propagation of gravitational waves through regions of space devoid of matter.</p>
</blockquote>
<hr />
<p>When discussing gravity, it's helpful to consider the motion of test particles, that is, particles that respond to gravity but which have negligible mass, so they don't cause any gravity themselves.</p>
<p>Let's say we have an ideal spherically symmetrical gravitation source, eg a planet, and a collection of test particles initially arranged as the surface of a small sphere at some distance above the planet.</p>
<p>As the sphere of test particles free-falls down towards the planet it will begin to distort, under the influence of the Weyl tensor. It will get stretched slightly in the vertical direction because the acceleration is slightly smaller at the top than it is at the bottom (as we know from Newton's inverse square law). And it'll get slightly squashed in the horizontal direction because the &quot;down&quot; vectors aren't exactly parallel, they're pointing towards the centre of the planet. So the sphere becomes an ellipsoid. However, the volume of the ellipsoid is identical to that of the original sphere, and it remains identical, no matter how close it gets to the planet.</p>
<p>In contrast, if we create a large sphere of test particles surrounding the planet, and perfectly centred on it, then the Ricci tensor acts, causing the volume of the sphere to reduce as the test particles fall towards the planet. The shape of the sphere remains spherical because all of the test particles are at the same distance from the centre of the planet.</p>
<hr />
<p>Unfortunately, it's difficult to give further details without getting deeply into linear algebra or tensor calculus. So articles on this topic are either fairly superficial (like this answer), or are incomprehensible to people without a suitable tertiary mathematics education. Penrose explains a bit more detail in <em>The Emperor's New Mind</em> (which I've read a coyple of times), and I expect he gives even more detail in <em>The Road to Reality</em> (which I haven't read).</p>

CC BY-SA 4.0
2021-08-02T08:28:45.440Z
Great answer. And you make the difference between the two curvature tensors very clear! This is actually the first time I understand their difference. Thanks for that! 
Deschele Schilder
CC BY-SA 4.0
2021-08-02T10:22:46.527Z
I cant find the right words, so...♩
Deschele Schilder
CC BY-SA 4.0
2021-08-02T12:01:38.463Z
The problem is that the question was apparently asking if a *continuously contracting* metric exists near massive objects (in contrast to the actually real continuously *expanding* metric of intergalactic space in cosmology) -- which is not the case.
Peter Erwin
CC BY-SA 4.0
2021-08-04T17:57:06.563Z
@PeterErwin If an object approaches another, it will see space contracting as gravity grows stronger. It is not an explicit time dependence of course but only an implicit one.
Thomas
CC BY-SA 4.0
2021-08-05T10:18:41.813Z
@Thomas And if an object moves away from another, it will space "expanding"... But the question (as clarified in the comments) was asking whether space near a massive object is contracting in the same way as intergalactic space is expanding -- that is, with an explicit time dependence.
Peter Erwin
CC BY-SA 4.0
2021-08-02T07:21:05.243Z
2021-08-02T10:05:38.140Z
2021-08-02T10:05:38.140Z
PM 2Ring
<p>Space expands on a cosmic scale. It makes objects accelerate away from each other, as observed.</p>
<p>Can we say, by the same token, that space contracts around massive objects in space, with the consequence that objects accelerate towards each other instead of accelerating away?</p>
<p>The difference being that expansion of space is driven by other stuff (dark energy) than the contraction (mass/energy)?</p>

CC BY-SA 4.0
2021-08-02T10:52:17.037Z
"Space expands on a cosmic scale" = space is continually expanding. Do you mean 1) "Is space near massive objects continually contracting?" (the answer is "No"). Or do you mean 2) "Is the space near massive objects more compact (but not changing with time) than space far from a massive object?" (in which case PM 2Ring's answer applies).
Peter Erwin
CC BY-SA 4.0
2021-08-02T11:03:05.810Z
@PeterErwin I mean if its continuously contracting near a mass. Why cant the same mechanism that accelerates masses *away* from each other be used to make them accelerate *towards* each other?
Deschele Schilder
CC BY-SA 4.0
2021-08-02T11:58:53.940Z
OK, so the answer is "No". (Objects "accelerating towards each other" is what standard GR gives you with a static metric.)
Peter Erwin
CC BY-SA 4.0
2021-08-02T12:14:57.737Z
@PeterErwin But accelerating away can be described with a negstive curvature like a saddle. Objects on such a surface  have diverging geodesics, like the galaxies wrt each other. On a static nonexpanding metric.
Deschele Schilder
<general-relativity>
Can we say space contracts around a massive object in space?
Deschele Schilder
# Can we say space contracts around a massive object in space?

(Asked by: username_0 on 2021-08-01)

Space expands on a cosmic scale. It makes objects accelerate away from each other, as observed.

Can we say, by the same token, that space contracts around massive objects in space, with the consequence that objects accelerate towards each other instead of accelerating away?

The difference being that expansion of space is driven by other stuff (dark energy) than the contraction (mass/energy)?

**Tags:** general-relativity
> **Comments:**
> **username_1**: "Space expands on a cosmic scale" = space is continually expanding. Do you mean 1) "Is space near massive objects continually contracting?" (the answer is "No"). Or do you mean 2) "Is the space near massive objects more compact (but not changing with time) than space far from a massive object?" (in which case PM 2Ring's answer applies).  
> **username_0**: @PeterErwin I mean if its continuously contracting near a mass. Why cant the same mechanism that accelerates masses *away* from each other be used to make them accelerate *towards* each other?  
> **username_1**: OK, so the answer is "No". (Objects "accelerating towards each other" is what standard GR gives you with a static metric.)  
> **username_0**: @PeterErwin But accelerating away can be described with a negstive curvature like a saddle. Objects on such a surface  have diverging geodesics, like the galaxies wrt each other. On a static nonexpanding metric.  

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## 1 answers

### Answer by username_2 on 2021-08-02. Score: 1

Yes. In General Relativity, the positive spacetime curvature due to a gravitational source means that a ball of radius $r$ has a smaller volume than that given by the usual Euclidean value of $\frac43\pi r^3$. This is measured by the [scalar curvature](https://en.wikipedia.org/wiki/Scalar_curvature), (aka the Ricci scalar). From Wikipedia:

> Specifically, the scalar curvature represents the amount by which the volume of a small geodesic ball in a Riemannian manifold deviates from that of the standard ball in Euclidean space.

The [Riemann curvature tensor](https://en.wikipedia.org/wiki/Riemann_curvature_tensor), which expresses the total curvature of spacetime, is composed of the [Weyl curvature tensor](https://en.wikipedia.org/wiki/Weyl_tensor) and the [Ricci curvature tensor](https://.m.wikipedia.org/wiki/Ricci_curvature).
Paraphrasing slightly from the Weyl tensor article, the Weyl curvature measures the tidal distortion that a body feels when moving along a geodesic, but it's volume-preserving, that is, it doesn't convey any information on how the volume of the body changes.

The Ricci curvature tensor contains all the the information about how volumes change in the presence of tidal forces.

> In general relativity, the Weyl curvature is the only part of the curvature that exists in free space—a solution of the vacuum Einstein equation—and it governs the propagation of gravitational waves through regions of space devoid of matter.

---

When discussing gravity, it's helpful to consider the motion of test particles, that is, particles that respond to gravity but which have negligible mass, so they don't cause any gravity themselves.

Let's say we have an ideal spherically symmetrical gravitation source, eg a planet, and a collection of test particles initially arranged as the surface of a small sphere at some distance above the planet.

As the sphere of test particles free-falls down towards the planet it will begin to distort, under the influence of the Weyl tensor. It will get stretched slightly in the vertical direction because the acceleration is slightly smaller at the top than it is at the bottom (as we know from Newton's inverse square law). And it'll get slightly squashed in the horizontal direction because the "down" vectors aren't exactly parallel, they're pointing towards the centre of the planet. So the sphere becomes an ellipsoid. However, the volume of the ellipsoid is identical to that of the original sphere, and it remains identical, no matter how close it gets to the planet.

In contrast, if we create a large sphere of test particles surrounding the planet, and perfectly centred on it, then the Ricci tensor acts, causing the volume of the sphere to reduce as the test particles fall towards the planet. The shape of the sphere remains spherical because all of the test particles are at the same distance from the centre of the planet.

---

Unfortunately, it's difficult to give further details without getting deeply into linear algebra or tensor calculus. So articles on this topic are either fairly superficial (like this answer), or are incomprehensible to people without a suitable tertiary mathematics education. Penrose explains a bit more detail in *The Emperor's New Mind* (which I've read a coyple of times), and I expect he gives even more detail in *The Road to Reality* (which I haven't read).

> **Comments:**
> **username_0**: Great answer. And you make the difference between the two curvature tensors very clear! This is actually the first time I understand their difference. Thanks for that!   
> **username_0**: I cant find the right words, so...♩  
> **username_1**: The problem is that the question was apparently asking if a *continuously contracting* metric exists near massive objects (in contrast to the actually real continuously *expanding* metric of intergalactic space in cosmology) -- which is not the case.  
> **username_3**: @PeterErwin If an object approaches another, it will see space contracting as gravity grows stronger. It is not an explicit time dependence of course but only an implicit one.  
> **username_1**: @Thomas And if an object moves away from another, it will space "expanding"... But the question (as clarified in the comments) was asking whether space near a massive object is contracting in the same way as intergalactic space is expanding -- that is, with an explicit time dependence.  

---

