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In general relativity, if a volume of particles moves unrestricted through spacetime, is their volume always conserved?

Say we let a collection of particles at rest wrt each other, fall freely in a gravitational field. Will tidal forces keep the volume they occupy constant? My intuition says yes, but how do we prove that? If the metric varies it seems reasonable that the number of small cubes stays the same (while the shape varies,).

Assume the small particles don't attract each other.

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  • $\begingroup$ It depends on what solution of the einstein equation you are talking I think $\endgroup$ May 20 at 19:49
  • $\begingroup$ Why would it? Wouldn't the particles attract due to gravitational effects? $\endgroup$
    – Mauricio
    May 20 at 20:05
  • $\begingroup$ @Mauricio I forgot to mention that their mutual gravity can be ignored. Good point! Tanx. Ill edit. $\endgroup$
    – Felicia
    May 20 at 21:36
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    $\begingroup$ Even in special relativity, the volume of a body is observer-dependent. Anyone moving with respect to the body will see one linear dimension compressed and the other two remain the same (compared to someone at rest with the body). $\endgroup$
    – RC_23
    May 21 at 0:05
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    $\begingroup$ I think you'll also need the constraint that the particles are in a vacuum if you want the volume to be unchanged, otherwise the Ricci tensor is not necessarily 0. $\endgroup$
    – Eric Smith
    May 21 at 0:10

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In general no, the volume is not conserved. To take an extreme example, consider a shell of particles surrounding the Earth. As they freely fall, their volume will definitely decrease!

EDIT: my example is extreme, of course. Baez at https://math.ucr.edu/home/baez/einstein/node5.html points out that the volume of a "small" sphere of test particles in vacuum does not change, and I'm sure he's right about that. But in my example there's a mass at the center, so the whole volume is not in vacuum.

EDIT: to put some numbers on this, I'll calculate the volume of a 1 km thick shell falling near the Earth's surface (so the inner radius is about the same as Earth's radius, 6372000 meters, and the outer radius is 6373000 meters). Near Earth's surface Newton's equations are an excellent approximation for gravity, so the radius of a falling shell as a function of time is given by the differential equation $r'' = \frac{GM}{r^2}$, where $GM$ is approximately $4 \times 10^{14} m^3 / s^2$ for Earth. Unfortunately there isn't a nice closed form solution for this equation, but we can solve it numerically using Wolfram Alpha with something like

solve [y'' + (4*10^14) / (y-6372000)^2 = 0, y(0)=0, y'(0)=0] from 0 to 10 using r k f

That is for the inner surface of the shell, and gives a delta of -492.569, which is reasonable (we know gravitational acceleration is about 9.8 m/s^2 near earth's surface, so we expect an answer near 490 m for a 10 second fall).

Doing the same for the outer surface gives a delta of -492.415, which again is reasonable (it falls slightly less because it starts further away).

Finally, calculating the volumes gives us a starting volume of $4/3 * \pi * 1.21826269 * 10^{17}$ and a final volume (after 10 seconds) of $4/3 * \pi * 1.21826197641753715846882634 * 10^{17}$, so the volume of the shell has indeed decreased as expected.

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    $\begingroup$ But the tidal force pulls the particles away from each other in vertical direction. $\endgroup$
    – Felicia
    May 20 at 21:31
  • $\begingroup$ Yes, but tidal forces are a second order effect, whereas the regular gravitational force decreasing the radius of the shell is first order. That is, the shell's radius is shrinking due to gravity much faster than its thickness is increasing due to tidal forces. $\endgroup$
    – Eric Smith
    May 20 at 22:14
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    $\begingroup$ The shell is made of smaller volumes that stay approximately constant. There's a big difference... it's similar to how we say that spacetime is flat locally at each point, and yet globally it can be curved. In any case I'll update my answer with some calculations. $\endgroup$
    – Eric Smith
    May 21 at 12:02
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    $\begingroup$ Couldn't you argue that the "volume of the shell" in this example is zero, since it has zero thickness? And if you try to follow the same line of argument with a "thick shell", then you run into the fact that the thickness of the shell will increase as its radius decreases. $\endgroup$ May 21 at 14:51
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    $\begingroup$ So you were; mea culpa! I assume that you carried out the calculations so that the difference (one part in $10^6$, or so) is meaningful and not just rounding error? $\endgroup$ May 21 at 14:57
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It depends on where they fall. Suppose a tunnel from South to North pole, and a bunch of test masses falling down from one of them. At each moment, the particles farer to the center of earth have bigger acceleration downward, and that closer to tunnel walls, bigger sideways (to the axis) component of acceleration.

From the frame of the particles, if measuring only their relative distances, ignoring the earth, all that happens is a gradual approximation, decreasing the volume.

From the frame of the Earth, the particles also move like an harmonic oscillator, besides getting closer and closer.

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  • $\begingroup$ I don't see why the volume doesn't stay constant. $\endgroup$
    – Felicia
    May 20 at 21:40
  • $\begingroup$ The particles are attracted to the center of the Earth, so there is a component of acceleration to the axis of the hole for all of them. And as $g$ increases with $r$, the downward acceleration for that more far from the center of the earth is greater. Both effects compress the bunch. $\endgroup$ May 20 at 21:47
  • $\begingroup$ But how do you know the volume of a ball of particles does not stay the same? There is expansion in the vertical direction and compression in the horizontal. $\endgroup$
    – Felicia
    May 20 at 23:03
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    $\begingroup$ Expansion in the vertical direction outside the earth, but not inside my tunnel, where there is compression also in the vertical direction. $\endgroup$ May 20 at 23:07
  • $\begingroup$ Ah! Because gravity decreases? $\endgroup$
    – Felicia
    May 21 at 0:25
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In general relativity, if a volume of particles moves unrestricted through spacetime, is their volume always conserved?

Say we let a collection of particles at rest wrt each other, fall freely in a gravitational field. Will tidal forces keep the volume they occupy constant? My intuition says yes, but how do we prove that?

You can not prove it, despite your intuition, because your intuition is wrong, in general.

The other two answers have already supplied two counter examples. One of the counter examples is a volume with a hole in it, where an attractive mass is inside the hole.

The other counter-example provided can be thought of as a long thin cylinder that is inside a slightly wider cylindrical hole drilled all the way through the earth. The particles undergo simple harmonic motion about the center of the earth, so they can be started off in such a way that the volume compresses than expands, etc.

The proof on Baez's website that a small volume (a small blob of coffee dropped near the surface of the earth in his example) remains constant is not generally true for any volume. In his case the compression in the "width" is offset by the expansion in the "length," but this clearly does not always happen since we already saw two counter-examples.

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  • $\begingroup$ In an answer by Dale I read: "If you start with a spherical ball of coffee grounds then as it falls the ball will be stretched vertically and compressed horizontally into an ellipsoid, but the volume of the ball will remain the same" the examples in the two answers are not proven. Just stated. Its just intuition. $\endgroup$
    – Felicia
    May 20 at 23:39
  • $\begingroup$ I read in comment on an answer I gave (which is why I asked thidms question), after this question: "The volume is preserved because the Weyl tensor only affects the shape, not the volume. The Ricci tensor affects the volume, but in free space (eg, in the vicinity of a ball of marbles of small mass, so the ball has negligible effect on the curvature) the Ricci tensor vanishes." What must I think? $\endgroup$
    – Felicia
    May 20 at 23:47
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    $\begingroup$ The Ricci tensor vanishes only in vacuum. So if your objects are not in a vacuum, e.g. if there is a mass at the center of them (my example) or they are contained within a mass (Claudio's example), then it won't vanish and the volume won't be preserved. $\endgroup$
    – Eric Smith
    May 20 at 23:57
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    $\begingroup$ I accept it. Just took a while. There are more places to put the ball and more shapes of balls. I gave you +1. $\endgroup$
    – Felicia
    May 21 at 1:39
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    $\begingroup$ By my reading, Baez's argument assumes that the objects are nearby objects free-falling in vacuum. Neither of the given counterexamples satisfy this condition, which is why Baez's result doesn't hold for them. $\endgroup$ May 21 at 14:39
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If mass is conserved then the product of volume and density is conserved. Volume alone is not necessarily required to be conserved.

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Imagine an astronaut Albert, whom we shall refer to as ‘A’, falling freely in space, a little away above the Earth’s atmosphere. It is helpful to think ofA as being just at the moment of dropping towards the Earth’s surface, but it does not really matter what Albert’s velocity is; it is his acceleration, and the acceleration of neighbouring particles, that we are concerned with. A could be safely in orbit, and need not be falling towards the ground.

Imagine that there is a sphere of particles surrounding A, and initially at rest with respect to A. Now, in ordinary Newtonian terms, the various particles in this sphere will be accelerating towards the centre E of the Earth in various slightly different directions (because the direction to E will differ, slightly, for the different particles) and the magnitude of this acceleration will also vary (because the distance to E will vary). We shall be concerned with the relative accelerations, as compared with the acceleration of the astronaut A, since we are interested in what an inertial observer (in the Einsteinian sense)—in this case A—will observe to be happening to nearby inertial particles. The situation is illustrated in Fig. 17.8a. Those particles that are displaced horizontally from A will accelerate towards E in directions that are slightly inward relative to A’s acceleration, because of the finite distance to the Earth’s center, whereas those particles that are displaced vertically from A will accelerate slightly outward relative to A because the gravitational force falls of with increasing distance from E. Accordingly, the sphere of particles will become distorted. In fact, this distortion, for nearby particles, will take the sphere into an ellipsoid of revolution, a (prolate) ellipsoid, having its major axis (the symmetry axis) in the direction of the line AE. Moreover, the initial distortion of the sphere will be into an ellipsoid whose volume is equal to that of the sphere. ... It should be noted that this volume preserving effect only applies initially so.

Page-396, 397 Roger Penrose's Road to Reality.

This effect is also explained with detailed calculations in Tristan Needham's Visual Differential Geometry.

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  • $\begingroup$ I love Penrose's road! +1! BTW, do you know how a new big bang occurs in his cyclic model? Or should I ask as a new question? $\endgroup$
    – Felicia
    May 25 at 11:35
  • $\begingroup$ Nope .. way beyond me at this point xd @Felicia $\endgroup$ May 25 at 11:38

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