Regarding to the Schwarzschild solution, is there more volume with more mass?

Lets take for example a space shuttle (spheric for simplicity) in an stationary orbit at the position $r_1$. The black hole around which it is orbitting has a mass $M$. Now, the mass is increased (not interesting, how). $M_2=2\cdot M_1$for example. The position $r_1$ is kept the same. Doesn't the volume of the spaceshuttle increase due to the mass increase and curvature increase?

According to a comment, the question might be similar to the question whether the volume of the space shuttle increases while moving nearer to the center of the black hole. The curvature there is stronger then outside.

As the metric is $ds^2 = -Bdt^2 + Adr^2 + \text{angular terms}$, and A is increasing while decreasing r, the volume should increase either. However, this seems to be dependent on the sign convention: If (+---) is used instead of (-+++), then the volume is decreasing, isn't it? I read somewhere, that Einstein used (+---) and now we are using (-+++) mostly and it is pure convention - but isn't the volume change important to classify?

Addendum: To make the question clearer I want to change it to the following thought experiment: Mankind finally managed to develop a material which completely isolates from gravity (please, don't ask me how). Real hoverboards (see "Back to the future" 2) are used all-around the planet and, at last, the following experiment is done regularly in the physics undergrads course: They got an empty cube of this gravity-isolating material. In the middle of the cube, there is a (non-isolating) sphere which can be filled with a material of high gravitational mass. The volume of the cube is measured with and without the mass in the middle (probably by simply pressing water into it). Of course, the undergrads need to be very precise with the volume measurement, carefully keeping temperature the same and everything. Do they measure more volume inside the cube if the middle-sphere is filled with mass?

For clarification: volume is positive here, mass is also positive here. The material of the cube is magical. It just stays where it is and isolates gravity. (Hm, ok, I see: "It stays where it is" regarding inside the cube is different from "it stays where it is" regarding the room with the students. So, to define even this: It stays where it is regarding the room with the students)

  • $\begingroup$ For me, it seems to be a real simple beginner question in general relativity. Very basic. I expected plenty of canonical answers. Why is the question downvoted? Is it to easy to answer? Not well defined? Not the actual question? I don't get it. I try to understand general relativity and this is one of the first questions I asked. $\endgroup$ Nov 26, 2021 at 5:13
  • $\begingroup$ Please do not let posts look like revision histories. In particular, instead of saying "I want to change this question to the following question", just change it. On the other hand, please be mindful to not change the question in ways that invalidates existing answers. $\endgroup$
    – ACuriousMind
    Nov 30, 2021 at 13:13
  • $\begingroup$ If your magical water is weightless and can fill all the way to the horizon, then you can put an unlimited amount of water in the box. $\endgroup$
    – safesphere
    Dec 3, 2021 at 17:12

4 Answers 4


I'm not sure this fully answers your question, but I hope this can help to shed some light on the issue.

Instead of a spherical spaceship I consider a rod in free fall towards a Schwarzschild black hole.

Black hole and rod

In the Schwarzschild coordinates, the rod occupies the space between $r=r_1$ and $r=r_2$. The length $L$ of the rod can be found by considering the amount of time it would take a light signal to move from one end of the rod to the other. We have

$$L = c \int_{t_1}^{t_2} \text{ d}t,$$ where we assume that the light moves from $(t_1,r_1)$ to $(t_2,r_2)$.

If the light moves in a purely radial direction, we can describe its path by the coordinate functions $t(\lambda)$ and $r(\lambda)$. The equation of motion $ds^2 =0$ then takes the form $$ g_{tt} \left(\frac{dt}{d\lambda}\right)^2 + g_{rr} \left(\frac{dr}{d\lambda}\right)^2 = 0,$$ which we can rewrite as $$\left(\frac{dt}{dr}\right)^2 = -\frac{g_{rr}}{g_{tt}}.$$

The length of the rod is then $$L = c \int_{r_1}^{r_2} \frac{dt}{dr} \text{ d}r = c \int_{r_1}^{r_2} \sqrt{-\frac{g_{rr}}{g_{tt}}} \text{ d}r,$$ where I have taken the positive square root because $r_2 > r_1$.

Notice that the length is independent of the signature of the metric, so whether you work with the (-+++) or (+---) metric is purely conventional and will not change the physics.

For the Schwarzschild metric, we obtain explicitly $$L = r_2 - r_1 + r_s \ln\left(\frac{r_2 - r_s}{r_1 - r_s}\right) > r_2 - r_1.$$

Now what happens if you magically, instantaneously increase the mass of the black hole? I think the length $L$ of the rod stays the same (I'm here assuming that the rod is infinitely stiff), but that it would now "appear shorter" to the distant observer - i.e. it would no longer occupy the entire space between $r_1$ and $r_2$.

Situation after the mass of the black hole has been increased.

Applying this reasoning to the case of a spaceship, the astronauts inside the ship should not feel its volume change when mass is added to the black hole. However, a distant observer might see that the spaceship now looks smaller than it did before.

In this analysis I have assumed that the rod/ship can be considered to be stationary for the duration of our experiment. I am not sure if all the same conclusions would hold if the rod/ship were in motion during the experiment.

  • $\begingroup$ Thank you fir your answer! However, I'm still looking for the volume. Isn't the ricci tensor the volume gain? Can someone show rhis canonically? $\endgroup$ Nov 27, 2021 at 13:55
  • $\begingroup$ Does this not depend on what we mean by length? The difference between coordinates does not in general provide a meaningful definition of length, consider for instance the distance between two points $(r,\phi_1)$ and $(r,\phi_2)$ in polar coordinates. On the other hand, the time taken by a photon to propagate from an observer A to a point B, and back again, provides a coordinate-independent measure of the distance between A and B (as measured by the observer A). $\endgroup$
    – Jakob KS
    Dec 3, 2021 at 11:45

@BarrierRemoval, you must be aware of what kind of curvature do you speak. Schwarzschild vacuum spacetime is indeed curved. Its Riemann curvature $R^{\rho}_{\sigma \mu \nu}$ is non-zero. On the other hand, its Ricci $R_{\mu \nu}$, and correspondingly $R$ curvatures are zero. Therefore, for a co-moving observer a spaceship does not change its volume. For reference see "Physical and Geometric Interpretations of the Riemann Tensor, Ricci Tensor, and Scalar Curvature", by Lee C. Loveridge https://arxiv.org/abs/gr-qc/0401099v1

  • $\begingroup$ That's indeed important, thank you for pointing this out! I updated the question. $\endgroup$ Nov 29, 2021 at 18:55
  • $\begingroup$ It's the Ricci-tensor with produces volume, isn't it? And the Weyl tensor is the difference between volumes, is that right? Therefore, the Ricci-tensor is zero outside the mass, but the Weyl-tensor is not. $\endgroup$ Nov 29, 2021 at 19:15
  • $\begingroup$ And, together, they build up the Riemann tensor? $\endgroup$ Nov 29, 2021 at 19:16

For a Schwarzschild solution, the Ricci tensor vanishes identically. As the space shuttle moves towards the black hole (or if there is some change in the black hole's mass due to matter falling in) the volume of the shuttle won't change, since volume changes are governed by the Ricci tensor. The Weyl tensor (i.e. the components of the Riemann tensor which are not on the Ricci tensor) does not vanish, but it leads only to shape deformations. Hence, while the shuttle will undergo spaghettification, its volume remains constant while the shape changes. A similar statement will hold for any object in a region where the stress-energy tensor vanishes. However, there is no way to determine what would happen to the cube: since it is speculative, so is any affirmation we can make regarding it. Gravity as we understand can't be isolated. If you are assuming it can be isolated, then we are no longer working with GR and there is no meaning in discussing what would happen from the point of view of GR.

As for the sign conventions, they are just conventions. If you change your preferred metric signature, you must also change the signs on remaining formulae so that the predictions are still the same. Nevertheless, this is meaningless for a Schwarzschild solution, since the Ricci tensor vanishes and there is never any change in volume of infalling objects.

  • $\begingroup$ Sorry for that, but, unfortunately, I suspect this to be wrong. The Ricci-tensor changes the volume and the Ricci-tensor is increased inside the mass. So, outside the mass, in my point of view, it's like lowering the deepest point of the Flamm's paraboloid. On the one hand (as you say), the steepness (spaghettification, shape change) is increased. But, on the other hand, a specific point of spacetime outside the mass is lowering it's position (which in my opinion belongs to the volume), too. Do you see this? $\endgroup$ Nov 30, 2021 at 8:51
  • $\begingroup$ @BarrierRemoval I didn't really understand your comment. Could you rephrase it? Notice that everywhere outside of the mass the Ricci tensor vanishes identically, and hence the shuttle will not have volume changes. In the case of a black hole, the Ricci tensor vanishes everywhere (except at the singularity, which is not a point of spacetime), so there is no volume change. $\endgroup$ Nov 30, 2021 at 9:10
  • $\begingroup$ I'm not absolutely sure (that's why I'm asking questions), but to me, the Ricci-tensor means volume gain at the position of the mass. OK, a black hole is a complicated thing - so, let's take something easier, just a sphere filled with mass (sun, neutron star). Then, inside the mass/sun, the ricci tensor is not zero. you add volume there. In infinity, the volume is still the same, the volume of the Minkowski space (probably, I get something wrong here). So in between it's basically as a potential difference. If you lift up potential at point 1 and are in between point 1 and 2, then... $\endgroup$ Nov 30, 2021 at 11:08
  • $\begingroup$ you experience to things: 1. it is steeper a your point in between and 2. you are lift up a bit. That's all I tried to say. I'm not absolutely sure wether I'm right or not. It is named "general relativity" - so probably, there doesn't exist anything like "lifting up" and everything is only relative. $\endgroup$ Nov 30, 2021 at 11:10
  • $\begingroup$ @BarrierRemoval for a localized mass, such as the sun, the Ricci tensor vanishes everywhere but at the star (recall Birkhoff's theorem: everywhere where $T_{ab} = 0$ and there is spherical symmetry, the solution is Schwarzschild). Hence, outside of the star, the change in volume is always zero. Once you are inside the star, things can get different and volume might change, but there is not change of volume at all outside of the star $\endgroup$ Nov 30, 2021 at 11:45

The volume of cube will change most probably. If there is matter or negative matter added into the center of the, special gravity isolated cube. The cube itself is made of exotic matter as it defies gravity in most probability, as normal matter can't defy gravity. In your question , it must be determined whether the spherical mass is

M , -M.

This type of a situation could also result in negative volume in the universe. The radius gets larger. That’s “how” it changes.

Oh, did you want to know “why” the radius of the Event Horizon changes when you add mass to a Black Hole? Well, then, why didn’t you say so?

The more mass a singularity (or any object) has, the greater its gravity. The greater its gravity, the farther away from the singularity is the point where light cannot escape the gravitational pull of the singularity. And, since the Event Horizon is defined as the maximum distance from a singularity where light cannot escape the gravitational pull of the singularity; that means the radius of the Event Horizon gets larger. The equation is, *R_schwarzschild=2GM/c^2 *
and if you understand equations you can see how R is directly proportional to M.

  • $\begingroup$ Ok, probably that was not clear enough. I pointed out that the students add positive mass M to the cube. $\endgroup$ Nov 30, 2021 at 7:33
  • $\begingroup$ Regarding the rest of your answer: It points in the right direction, but it is not really satisfying: You say that the event horizon of the black hole increases, ok. And, therefore, the volume inside the event horizon increases. However, that doesn't say anything about the volume of the space-time itself. You did not show that the event horizon isn't just moving outwards without changing the volume of the space-time itself. $\endgroup$ Nov 30, 2021 at 7:39
  • $\begingroup$ I was actually trying to tell u that the cube maybe is made of exotic matter as it defies gravity according to you and treated the the spherical mass as normal matter $\endgroup$ Nov 30, 2021 at 7:39
  • $\begingroup$ The volume of spacetime shall not change if I consider spacetime and matter differently $\endgroup$ Nov 30, 2021 at 7:41
  • $\begingroup$ ok, thank you for the clarification. I did not get this because you wrote "In your question, it must be determined whether the spherical mass is M, -M. This type of a situation could also result in negative volume in the universe" $\endgroup$ Nov 30, 2021 at 7:45

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