# Concept of negative mass in solving problems

I have searched a lot on the internet regarding negative mass. And in the meantime, I came across this question.

The question states:

Consider two spherical empty regions (C1 and C2) in an otherwise uniform and essentially infinite intergalactic gas cloud of density $\rho$ as shown in the figure. As a result of gravitational effects, describe the movement of the empty regions. One very obvious approach is to consider the spherical empty regions as a superposition of two spherical regions of densities $\rho$ and $-\rho$. From this one can easily deduce that the spherical regions will move towards each other. However, I am not satisfied with this because the theory of negative mass is not defined in reality. Also, what would happen if the density varied according to some law? Will the same approach work?

## 4 Answers

The method of superposition you suggested in your question is sound. Knowing that gravity obeys an inverse square law we describe the force (per test mass) at point $\vec{x}$ as

$$\vec{F}(\vec{x})=\int_V \frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x'$$

where $\vec{x}'$ describes positions within a massive volume $V$ with corresponding densities $\rho(\vec{x}')$ that contribute to the gravity force felt at $\vec{x}$ (and the negative ensures attraction). Now I define a cavity volume $\bar{V}$ that describes everywhere that isn't $V$ (loosely, $\bar{V}\equiv\infty \setminus V$). Then the force is equivalently:

$$\vec{F}(\vec{x})=\int_\infty \frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x' - \int_{\bar{V}}\frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x'$$

$\rho(\vec{x}')$ appears within the cavity, but its effects are perfectly cancelled by the first integral - you can choose whatever you want! In general, if the density varies this doesn't do you any good.

For the specific situation you've described, let $\rho(\vec{x}')=\rho$ within the cavity. Then there is translational symmetry at any point $\vec{x}$ in the medium and the first term cancels based on symmetry (a mass embedded in an infinite mass feels no net force). The remaining term is:

$$\vec{F}(\vec{x})=\int_{\bar{V}}\frac{(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x' = \int_{C_1}\frac{(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x' + \int_{C_2}\frac{(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x'$$

since $\bar{V}$ describes the two volume of the two cavities. This tells us that a mass at $\vec{x}$ is repelled by the cavities. Conceptually, you will see the mass between the cavities flowing upwards and downwards (imagine the field lines between two positively charged spheres). The results is that the cavities converge. (They will also deform; that's not terribly important here).

Main Points

1. The apparent "negative mass" attracting the cavities is due to a change in sign to describe mass moving away from the cavities (towards an area with more mass). None of the equations described the force experienced by the cavities because they don't actually interact with anything.
2. This approach that gives rise to mass-cavity repulsion is only effective in an infinite system with the right symmetries.
• Thanks for your answer.Can you please explain your first integral.I am used to only the basic vector form of Newton's Gravitation.What does '$d^3'$ signify. – Pink Dec 23 '16 at 7:03

Assume we have mass $m_1$ and $m_2$. Newton's second law of motion for the two masses is then $$m_1\vec a_1~=~\frac{Gm_1m_2\hat n_{12}}{r^2},~r~=~|\vec r_1~-~\vec r_2|$$ $$m_2\vec a_2~=~\frac{Gm_1m_2\hat n_{21}}{r^2},~\hat n_{21} = -\hat n_{12}.$$ Now assume $m_2~<~0$ and so $m_2~=~-|m_2|$. We now have the equations of motion $$\vec a_1~=~-\frac{G|m_2|\hat n_{12}}{r^2}$$ and $$\vec a_2~=~-\frac{Gm_1\hat n_{12}}{r^2}$$ It is clear that something curious is going to happen, the the accelerations of the two masses are in the same direction. if the magnitude of the two masses are equal the two will remain a constant distance from each other and accelerate off endlessly. This of course ignores other perturbing gravitational fields. If $m_1$ is very large and $|m_2|$ small the small negative mass will levitate or antigravitate away.

In the case of the magnitude of the two masses equal it is clear that in a funny sense this is "nothing" accelerating away. So there is no violation of conservation of momentum. However, this is a violation of the Hawking-Penrose weak energy condition. Also the quantum fields or quantum systems composing the negative mass will have no lower bound to its specturm. This could lead to pathologies. This is related to the problem of traversable wormholes, warp drives and other spacetimes that may not be permitted by quantum gravity once that is established.

As my previous answer was not understood and the wrong one was chosen as correct, I'll try to present a clearer answer.

The integrals in forky's answer are fine, but the bubbles repel!

Imagine that in stead of completely empty bubbles you have balls of styrofoam, with $\rho'= (1/10000)\rho$, or any other density, as small as you like. Now things get easier, the balls still have a practical density of almost 0 compared with the rest of the bulk, but now they also have a well defined gravitational and inertial mass.

Now if you do the integrals, you will see that the gravitational force on $C_1$ will be: $$F=\frac{\rho'V_{C_1}(\rho'-\rho)V_{C_2}\hat{R_{12}}}{R_{12}^2}=-9999\frac{\rho'^2V_{C_1}V_{C_2}\hat{R_{12}}}{R_{12}^2}=\rho'V_{C_1}a$$ $$a=-9999\frac{\rho'V_{C_2}\hat{R_{12}}}{R_{12}^2}=-\frac{9999}{10000}\frac{\rho V_{C_2}\hat{R_{12}}}{R_{12}^2}$$

The acceleration is in the direction of $-\vec{R}$, meaning the balls repel. As we take $\rho'\rightarrow 0$, the acceleration will approach $a\rightarrow-\frac{\rho V_{C_2}\hat{R_{12}}}{R_{12}^2}$, but is will still be in the "out" direction.

• One confusion.are you calculating the force on one cavity due to the other .if that is so I mean that both cavities have equal densities.so you should substitute $\rho'-\rho$ with $\rho'$ in your integral.please forgive me if I got you wrong. – Pink Dec 23 '16 at 8:57
• Using forky's notation, the new integral will read: $\vec{F}(\vec{x})=\int_\infty \frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x' - \int_{\bar{V}}\frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x'+\int_{\bar{V}}\frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho'(\vec{x}')d^3x'=\int_\infty \frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} \rho(\vec{x}')d^3x' - \int_{\bar{V}}\frac{-(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3} (\rho(\vec{x}')-\rho'(\vec{x}'))d^3x'$. And then you proceed like forky and are left with only the second integral. – milo Dec 23 '16 at 9:27
• Then you integrate it over the ball, using Gauss's theorem to get the total force on the ball that is written in the answer. – milo Dec 23 '16 at 9:30
• And $d^3x$ is just a short way of writing $dx dy dz$ or $dx_1dx_2dx_3$, and it generalizes well to other dimensions. – milo Dec 23 '16 at 9:35
• Yes perhaps forky missed out the negative sign,for attraction,in his last para.hence the particle between the two cavities will be attracted to the cavities.this will apparently lead to the two cavities coming towards each other .So it is attraction in this case too. – Pink Dec 23 '16 at 9:58

They would actually repel. Yes, the force will act towards the middle but as $F=ma$ and you're looking at negative masses, it will cause repulsion.

How can we see this without negative mass voodoo? Easy, just use test particles.

Put a test particle on the left side of $C_1$, there are 2 holes on the right and 0 from the left, the total gravitational force will be to the left!

Now put a particle to the right of $C_1$, the hole on its left is much closer, so it will be pulled to the right! The forces try to stretch the bubble! (counter-intuitive, but yes gravity like to clump matter together and make the vacuum bigger).

But notice that the force to the left is stronger than the one to the right!

Now if you imagine the the bubbles are contained in a thin hard shell of some small mass, you will see that $C_1$ will move to the left and $C_2$ to the right, without using negative mass. Of course, to really calculate the force on the shell one needs to perform an integral over all the points on its surface, but I don't believe you'd get a different result.

• This is incorrect, see forky40's answer for both quantitative and qualitative explanations. – Kyle Oman Dec 23 '16 at 3:56
• You are wrong, and forky did the right integrals but has failed explaining is qualitatively, and got to the right conclusions. I'll post another answer in hopes that you will understand. – milo Dec 23 '16 at 7:55
• *got to the wrong conclution – milo Dec 23 '16 at 7:56

## protected by Qmechanic♦Dec 23 '16 at 3:36

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