# Does curved spacetime change the volume of the space?

Mass (which can here be considered equivalent to energy) curves spacetime, so a body with mass makes the spacetime around it curved. But we live in 3 spatial dimensions, so this curving could only be visualized in a chart with 4 dimensions, and the living being will notice only the effect that this bend causes on matter and light, the gravity, or the curve will expand-contract also the space itself? and can be measured by the change in volume?

Let's take an analog on 2 dimensions:

The distortion on space is visible only by a 3d being (such as a human.) The 2d creatures can't see the distortion, as they are 2d creatures, but they can see that space is 'bigger' where the distortion occurred, because the total area of the squares (as viewed from the top--i.e., from a different coordinate in a third, orthogonal dimension--increases.

In the real world, the axis of distortion is trickier to define. The space needs to be expressed as cubes, not squares, and what changes with distortion is the volume of those cubes.

So, if you have two identical boxes, one with a very massive object, and another empty, the one with the massive object fits more things inside it?

• In actuality you would need a lot more dimensions than that to embed a general solution of Einstein's equations. Flat spacetime alone takes six, I believe, and in general it's not less than ten, if I remember correctly. The cosmological solution does change the volume (it makes it larger in an expanding universe) and strong gravity even reduces dimensionality from three to two. – CuriousOne Jun 28 '16 at 0:11
• Essentially, the question is about the definition of the volume element, $\sqrt{|\det g_{\mu\nu}|} d^4 x$, which is multiplied by the jacobian determinant compared to the flat spacetime volume element. – auxsvr Jun 28 '16 at 6:18

In a question like this you need to ask what does the volume change relative to. So it's a little bit ambiguous.

However, the answer to your question is "yes" in the following restricted sense.

Imagine having a "swarm" of test objects, with mass so small that their effect on the spacetime around them is negligible. Assume that they are in freefall, i.e. all following spacetime geodesics. The swarm has a shape with volume; let's assume it's a sphere, and that their geodesic paths take them through varying curvature spacetime. Let's assume also that our space-bees are signalling their queen so that she can work out her distance to her fellows at all times.

Then, in general, the queen will find that the distance to her fellows changes following the geodesic deviation equation (see also Misner, Thorne and Wheeler, "Gravitation" Chapter 1 for a great intuitive explanation). So she will perceive a change in volume of her swarm.

Indeed, there is a tensor object whose primary meaning is that of volume change. If one writes the co-ordinates of the queen's swarm in Riemann Normal Co-ordinates (see also Misner Thorne Wheeler section 11.6), whereby something's position is named by naming a direction vector (a tangent vector to spacetime) and a distance along that vector, then the volume element $\mathrm{d} V$, compared to the volume element $\mathrm{d} V_f$ one would calculate assuming flat spacetime, is defined by the Ricci curvature tensor:

$$\mathrm{d} V \approx \left(1-\frac{1}{6}R_{j\,k}\,x^j\,x^k+\mathscr{O}(x^3)\right)\mathrm{d} V_f$$

Indeed one can "decompose" the full curvature tensor into expressions involving the Ricci tensor and so-called Weyl tensor. The former measures how the volume of our swarm changes, the latter tells us in detail how the shape of the swarm changes as the swarm freefalls.

You may also be interested in Chapter 42 of Volume II, called Curved Space of the "Feynman Lectures on Physics"

I can answer some of it, and in such a way that it has invariant general relativistic meaning. However, not a general answer. You do have to, and can, treat curvature and some measures of volume invariantly.

There are two questions. 1)Does negative/positive curvatures have more volume, that some (in some sense) equivalent spacetime with no curvature? And 2)Does more energy density in a spacetime cause more positive or negative curvatures? I will cover both.

For cosmological spacetimes, i.e. homogeneous and isotopic, a positive curvature has finite volume (the simplest topological possibilty is that the space hypersurfaces (i.e. Volumes) are 3D spheres). The negative and 0 curvature spaces have infinite volume, and they are open and infinite. It's got some general application because those spacetimes are the ones with constant curvature.

If you added more matter (or more correctly if the matter density was higher than estimated) it would tend to make the curvature more positive. If less, more negative.

So one would say more matter less volume, less matter more volume.

More generally negative curvature tends to have its geodesics diverge, and so open up more space. See the ref at

https://amathew.wordpress.com/2013/01/04/volume-growth-and-negative-curvature-i/

Still, I have not seen a general treatment other than in fairly symmetric spacetimes, and there is some concern that the energy you have to use to 'lower' mass from infinity into some volume causes the insertion of negative (potential) energy, the energy needed to take it out to infinity. So intuitively one has to be careful

It is true, eg, in the following link,

that the volume of a space at a physical distance (calculated using the metric weighting the coordinate distances) in the cosmological spacetimes is higher for negative curvatures than for flat spaces, and both more than positive curvatures

The next question is whether matter or energy makes a spacetime (or the space sections) more positively curved or less. Even in our universe, if there was no cosmological constant, a matter plus radiation density less than a certain amount would make the universe open and the volume infinite. And empty spacetime can have positive, flat or negative curvatures, so having more or less matter is not clearly determinative.

For other geometries there are various quantities that can be used to measure curvature, where there are a number of curvature invariants. The Schwarzchild solution (for a spherically symmetric space) has a curvature (defined as in https://en.m.wikipedia.org/w/index.php?title=Kretschmann_invariant&redirect=no, the Kretschmann invariant), which increases as the square of the mass, and which is positive. Thus, the higher the mass the higher the curvature. But notice that in this case the higher the mass, the larger the equivalent Schwarzschild radius, and if it was a Black Hole the larger then its horizon and the area of the horizon.

There are other dependencies, which involve other parts of the stress energy tensor, including stress and momentum and pressure. Solving the Einstein equations in a box is not easy (or possible except numerically?). However, there are general results simply from the definition of the Riemann and Ricci Tensors that the normalized (by volume) rate of change of the rate of change of volume (i.e., double time derivative) is proportional to minus the sum of energy density and pressures. So if those increase generally, for these somewhat simple cases, the volume change will decrease as

$\ddot{V}$/V = -{energy density + pressure}

Within the Schwarzschild metric, the volume does change.

It is the rectangle formed by the radial dimension and time which is invariant: The dilating effect of the Schwarzschild metric

$$\mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2 ~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2 \theta~\mathrm d\varphi^2)$$

compared to the corresponding flat metric

$$\mathrm ds^2 = -c^2 ~\mathrm dt^2 + \mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2 \theta ~\mathrm d\varphi^2)$$

does impact only the radial direction $\mathrm dr$ and time $ct,$ the rest of the geometry ($\theta$ and $\varphi$) remains unchanged.

You can easily see that the factor by which $\mathrm dr$ is multiplied is the same as the factor by which $c~\mathrm dt$ is divided. The factor

$$\sqrt{1 - \frac{2GM}{c^2 r}}$$

is equal to the gravitational time dilation.

By consequence, the rectangle formed by the radial dimension and time conserves its surface in Schwarzschild geometry.

If you now consider the additional dimensions you see that the spacetime volume must have changed. If you add e.g. one dimension, obtaining a cylinder, the 3D-volume is $\pi r^2 t$, that means that a reduced radial dimension r is appearing squared, while the time dimension remains the same.

OK, I can not show the math, but anything you try to fit inside the box, will itself be curved/stretched/contracted along with the space. Therefore, you should not be able to fit more stuff inside one box as compared to the other of the same size. Even saying same size involves space and its curving. So, even your box will be curved as there is nothing that does not curve alongside the space. There will be only one difference between the two cases, and that is given in your question - that of mass and gravity.

Consider fitting more stuff into a box while inside intense gravitational field. Now move the full box to a less intense gravitational field. Is it going to explode? Is the stuff going to leak out?

Yes, curved spacetime does change the volume of space. When space is curved by mass it is stretched more in some dimensions than others. Picture a balloon being stretched or squeezed--the volume changes.

## protected by Qmechanic♦Jun 28 '16 at 7:18

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