# Does mass create space?

Imagine two cubes each with the side 1 light year. One of the cubes is empty of any matter. The other cube contains a massive star.

• Does the cube which contains the massive star have a larger volume than the empty cube? (Because the mass makes space curvy)

• If the star is replaced by a gas cloud of the same mass but which is evenly distributed inside the cube, does the volume of the cube change?

• Is it not enough with geometry to calculate volumes? Is knowledge of the mass (and its distribution?) also necessary?

• Does mass (concentrations) create new space volume, or does it just change the properties of the space around it?

• Your question seems to assume an absolute reference frame, which we know does not exist. Jul 4, 2014 at 14:52
• How would you make sure that the two cubes are identical? Would you measure their side with a laser when they are already in place? Or would you build them out of some more or less rigid structure with the same 'mold' and then move them to place? I believe the answer could be different in the two cases. A rigid structure and a laser beam behave differently in a gravitational field. See for example Ligo's arms Jun 14, 2022 at 6:12

This is a tough question to answer because the dimensions of your proverbial cube would be affected by the mass inside (as dimensions can only exist in space, so anything that affects space will also bend your cube), and mass does not bend space, it bends spacetime. You have to keep in mind that when talking about special and general relativity, you are talking about four-dimensional spacetime. This is what is curved by mass. For example, due to the curvature of spacetime due to Earth's gravity, you feel the same effects on its surface that you would if you were accelerating upwards at 9.8 m/s^2 through deep space (where we assume there is no gravity). This doesn't mean the physical space is curved - if you draw a straight line on a piece of paper in space and then travel down to Earth's surface, it will remain straight. Instead, spacetime is curved, which can affect objects' paths of motion, but not the objects themselves. Now to your questions:

• The volume inside of the cubes would be exactly the same, but two identical objects flying through each cube would follow two different paths, and if either looked at the other, it would see that the other's clock is moving at a different speed than its own due to time dilation (due to the curvature of spacetime towards mass, time moves more slowly closer to massive objects). We can see evidence of the effect of gravity on paths with gravitational lensing. Light travels in a straight line through spacetime, but when looking at the space around a star, you will actually see objects that should be directly behind the star and blocked from sight. This is because light rays that, in the absence of gravity, would pass next to the star and continue off at an angle are pulled towards the star by it's gravitational field (in reality they're just following the curvature of spacetime around the massive star), so they curve around the star, becoming visible to us, even though, in space alone, they should be blocked by the star.
• The distribution of the mass has no effect on the gravitational field around it (same with electromagnetic fields). If you look at any equation that has to do with gravity (gravitational potential energy, gravitational force, etc.), you will find a term for the total mass of the object, but unless you are within the bounds of the control volume containing the mass, the distribution does not matter. This is one of the reasons why black holes are so hard to study - we can tell how much mass is inside of them, but without the ability to see inside, we have no other way of detecting their properties.
• Like I said above, the volume is not affected because space is not warped by gravity, only spacetime. If you made a physical box (or we'll say a cube frame) around empty space and then moved it to a space that contained a star, it would remain the same size and shape (assuming that it was strong enough to withstand the force of gravity pulling it towards the massive star).
• Neither, it simply curves spacetime.
• Why is space not curved if time is curved? In a curved space the angles of a triangle don't add up to 180 degrees, and pi has a different value. Lightrays don't go straight. A black hole, as seen from far away, contains no space inside the horizon. There IS no inside. This affects the volume. Jun 15, 2022 at 21:35

Yes, there's more volume with more mass. (However, there's also "less time", therefore, it's complicated with spacetime)

Let's assume the Schwarzschild-metric with Schwarzschild-coordinates for simplicity. And let's assume a sphere, not cubes, for simplicity (spherical symmetry) The general spherically symmetric spacetime is $$ds^2=-Bc^2dt^2+Adr^2+r^2d\Omega.$$ The volume of a radius-$$R$$ sphere with the spherical symmetry of the Schwarzschild-metric is $$V=4\pi\int_0^Rr^2\sqrt{|A|}dr$$ where the $$r$$ in the equation for $$ds^2$$ is not the same $$r$$ as in the equation for $$V$$. The coordinate origin of the sphere $$V$$ is put at an arbitrary position within the influence of the Schwarzschild solution $$ds^2$$. Regarding the Schwarzschild-metric in Schwarzschild-coordinates, A is always larger than 1. Therefore, volume is clearly added by adding mass.

A is larger than 1 in both cases: using the outer solution or the inner solution of the Schwarzschild-metric.

However, the argumentation is not as simple if the center of gravity is inside the sphere, as there resides the singularly of the Schwarzschild-solution. Yet, on the one hand, the Schwarzschild-solution is a good approximation for spherical symmetric masses, and, on the other hand, one can assume continuous behavior of the equations. Therefore, the argumentation holds for central masses without singularity and calculating the sphere including the center of mass.

• Nice answer! Could you expand the answer by covering the case of a uniform mass distribution inside the sphere? Is it larger or smaller? Also, I guess you are doing the integral only in the outside of the event horizon, so the sphere is missing 4 pi/3 Rs^3 of space. Could this compensate the added space thanks to the A>1 factor? Jun 14, 2022 at 6:21
• Thank you! It's always larger with mass. See added argumentation in the answer. The inner sphere is not missing. The coordinate origin of the sphere is not the same as the center of mass of the Schwarzschild-metric Jun 14, 2022 at 20:57
• In my opinion, an interesting addendum to this question would be: Is there a linear dependency between mass and added volume? Unfortunately, there is not. Jun 15, 2022 at 15:28
• I provided the exact dependency in my answer. You are right, it is definitely not linear Jun 16, 2022 at 7:03

In my opinion yes, mass affects volume. Instead of a cube, imagine a pyramid with a 3 sided base. All edges have the same length defined by the time to reflect light back and forth between the apexs. The number of smaller such pyramids that could be fit inside would increase with mass because the path of light would be bent. Bent light means bent space, as well as bent spacetime. They are not independent. Per Einstein, the speed of light is not constant when mass is present and depends on the coordinates. This amounts to more space. I believe that it was Eddington who first tried to quantify how much space is added by a given amount of mass.

Is it not enough with geometry to calculate volumes? Is knowledge of the mass (and its distribution?) also necessary?

To answer this question. Mass distribution does make a difference. For example, if a star collapses into a black hole, the mass of the black hole is less than the original mass of the star due to some matter lost during the supernova. But now the star is compressed into a more compact space. So, now one can reach much more closer to its center(and experience more gravitational force) than one could have before, So, a BH is a deadlier object than normal stars.

• Why the downvote? Is there anything wrong with the answer? Jun 30, 2014 at 7:49
• I didn't downvote your answer, but I think that the physical removing of mass (by a supernova) is beside what I'm asking about. Maybe my question was unclear. Jul 2, 2014 at 6:23
• What I was trying to say is mass distribution shouldn't affect gravity. Jul 2, 2014 at 12:25

The issue seems debated, since the two most upvoted answers give opposite conclusions. I decided to carry out the full calculation, hoping to settle the matter.

Let's settle the scene: we built a giant hollow spherical structure, made of a very resistant and rigid material. We place the sphere in empty space and we measure its circumference with a yardstick and find it is equal to a certain value, let's call it $$2 \pi \mathcal{R}$$. Then we measure the radius and we find it is equal to $$\mathcal{R}$$. All good, we have proven that empty space is flat. Nice to know. So, in empty space, the metric is given by

$$ds^2 = - dt^2 + dr^2 + r^2 d\theta^2 + r^2 \sin^2(\theta)d\phi^2$$

and the volume of the sphere is

$$V = \int_0^\mathcal{R} dr \cdot r d\theta \cdot r \sin(\theta) d\phi = 4 \pi \int_0^\mathcal{R} r^2 dr = {4 \pi \over 3}\mathcal{R}^3$$

Now we fill the sphere uniformly with gas that has total mass $$M$$. We need not to lose our sleep in wondering how this gas can be kept at uniform density, despite its self gravitational pull, for this is irrelevant for this question: as you will see in a moment, pressure and temperature of the gas are not involved in the spacial part of the metric.

The general expression of the metric of a spherically symmetric matter distribution can be found in this stack exchange answer

$$ds^2 = - B(r) dt^2 + A(r) dr^2 + r^2 d\theta^2 + r^2 \sin(\theta)d\phi^2$$

where

$$A(r) = \left(1-{2G m(r)\over r}\right)^{-1}$$ and $$m(r) = 4\pi \int_0^r r'^2 \rho(r') dr'$$. In our case, $$\rho$$ is constant inside the sphere and zero outside, so that $$m(r) = M {r^3 \over \mathcal{R}^3}$$ when $$r<\mathcal{R}$$ and $$m(r) = M$$ when $$r\geq \mathcal{R}$$.

Inside the sphere, $$A_{in}(r) = \left(1-{2G M \over \mathcal{R}} {r^2 \over \mathcal{R}^2}\right)^{-1}$$ and outside the sphere $$A_{out}(r) = \left(1-{2G M \over r} \right)^{-1}$$, as required by the usual Schwarzschild metric. As already noted by BarrierRemoval, $$A>1$$, and this means that the volume inside the sphere is increased. But let's proceed slowly.

Now if we measure the circumference again, we will still obtain $$2 \pi \mathcal{R}$$, because the tangential part of the metric has not changed. Instead, if we went and measured the radius we would obtain

$$\int_0^\mathcal{R}\left(1-{2G M \over \mathcal{R}} {r^2 \over \mathcal{R}^2}\right)^{-1} dr> \mathcal{R}$$

The space is curved! And the volume is

$$V = 4\pi \int_0^\mathcal{R} r^2 \left(\sqrt{1-{2G M \over \mathcal{R}} {r^2 \over \mathcal{R}^2}}\right)^{-1} dr > {4 \pi \over 3} \mathcal{R}^3$$

The volume has increased. Now imagine we compress the gas (but keeping our rigid sphere as it is) until it becomes a small ball at the center of our sphere, with uniform density, mass $$M$$ and radius $$L$$. Let's pretend that this homogeneous ball of gas represents a star. We already know how to calculate the volume inside our spherical rigid structure:

$$V = 4\pi \int_0^L r^2 \sqrt{A(r)_{in}} dr + 4\pi \int_L^\mathcal{R} r^2 \sqrt{A(r)_{out}} dr$$

The integral is actually solvable analytically (check it in Mathematica), but I won't write the solution here in matjax, because it is quite long. But I have implemented it in python and I leave you the script

import numpy as np
import matplotlib.pyplot as plt

# rs=2GM is the Schwarzschild radius
rs = 1
# the radius of the rigid sphere
R = 5
# the radius of the gas ball
L = np.linspace(1.001,R,1000)

def V(rs,L,R):
# calculate the volume of the sphere
return 4*np.pi*(I1(L,L)-I1(0,L) + I2(R)-I2(L))

def I1(r,L):
# primitive of the internal integral
return 1/2*(L**(9/2)/rs**(3/2)*np.arcsin(np.sqrt(rs/L**3)*r) - L**3*r/rs * np.sqrt(1-rs*r**2/L**3))

def I2(r):
# primitive of the external integral
return 2*rs**3*np.sqrt(1-rs/r)* (5*np.arctanh(np.sqrt(1-rs/r))/(16*np.sqrt(1-rs/r)) - r**3/(48*rs**3) * (-15*(1-rs/r)**2 + 40 *(1-rs/r) - 33))

plt.plot(L,V2(rs, L, R)/(4*np.pi*R**3/3))

plt.xlabel("$$L$$")
plt.ylabel("$$V/V_0$$")
plt.grid()
plt.show()


The figure here shows how $${V \over 4\pi/3\mathcal{R}^3}$$ varies as a function of $$L$$, where the other parameters have been fixed at $$r_s=1$$ and $$\mathcal{R} = 5$$. If $$L \to r_s$$ the uniform gaseous sphere gets close to become a black hole and the volume increases. But I don't think it diverges, I believe it tends to a finite value, although I have not attempted to prove it.

• Interesting answer! Especially the second-to-last sentence opens my eyes. Your whole answer sheds light on the "compression" part of the question which I did not take into account in my own answer. - - - As you show, compression of the mass leads to volume gain. However, this volume gain is only inside the compressed mass. Regarding the outside (imagine a second, spherical shell around the first sphere), only the sum of the inner mass leads to volume gain. Jun 17, 2022 at 11:12
• Compression produces volume inside the compressed mass... That's really counterintuitive; reminds me of the anomaly of water. Furthermore, it seems not to be common knowledge, for example, it is erroneously written that "in a black hole, mass is compressed to zero volume" here: astronomy.stackexchange.com/questions/20340/… Jun 17, 2022 at 11:44
• @BarrierRemoval unfortunately I had made a mistake, I had forgot the square roots. Now I have corrected it, but I don't think it diverges anymore Jun 17, 2022 at 14:56
• Ah, thank you for the correction!! Good job! If there is a singularity, shouldn't $r$ become infinite and therefore the volume as well? Jun 17, 2022 at 16:46
• @BarrierRemoval I don't think it's so straightforward. The volume integral might converge even if the radius integral doesn't. Jun 18, 2022 at 7:54