# Negative mass effect?

What's wrong with the following argument to "produce" negative mass?:

Consider two charged bodys of weight $$m,M$$ of distance $$r$$ with the electric charge $$Q$$ and $$-Q$$. Then the energy of this system is: $$E = (m+M)c^2-k\frac{Q^2}{r}-G \frac{mM}{r},$$ where $$k$$ is the Coulomb constant. Hence the whole mass of this system is: $$\hat{m} = m+M - \frac{k Q^2+GmM}{rc^2}.$$ If we set this whole mass to $$-0.001\text{kg}$$ we might solve for $$Q$$ and get: $$Q = \sqrt{\frac{-GmM+rc^2(0.001+m+M)}{k} }$$ Now for this to make sense we must have $$r > \frac{G m M }{c^2(0.001+m+M)}.$$ For instance if $$r = 0.01 \text{m}$$, and $$m=M=0.001\text{kg}$$, we get $$Q = 17.33\text{C} = C \cdot U$$. If we let $$U=12 \cdot 10^3 \text{V}$$, the we get for the capacity of the capacitor: $$C= \frac{Q}{U}=0.001444178 \text{F}$$. Is this practically achievable? (It would be nice if some experimental physicists would perform an experiment to see if this thought experiment is valid or not)

Edit: I read about the "positive mass theorem" and it seems like it is based at least on the "null energy condition" which is a conjecture, which is violated by the Casimir effect. In this article, which I do not understand fully, it is investigated what happens if the null energy condition is violated: https://arxiv.org/abs/gr-qc/0003025

It seems like there is no contradiction (if one assumes that the null energy condition can be violated, which is shown by the Casimir effect) to assume there exists negative mass:

Suppose two neutrons are located at distance $$r$$. Then the whole energy of this system is (by the Yukawa Potential):

$$E = \hat{m}c^2 = 2m_pc^2 - g^2 \frac{e^{-m c r / h}}{r}$$ where $$h = 6.58\cdot 10^{-16}$$ is the reduced Planck constant. If we let $$-k m_p = \hat{m} = 2m_p - g^2 \frac{e^{-m c r / h}}{rc^2}$$ where $$k=10^{27}$$ and $$m_p$$ is the mass of the neutron, hence $$\hat{m}=-1.67 \text{kg}$$ then solving for $$r$$ (for example with wolfram alpha), we get: $$r = \frac{h W(\frac{g^2m}{2chm_p+chkm_p})}{cm},$$ where $$W$$ is the Lambert W-function. If we set $$g=m=1$$, which I did, because I am not comfortable with the values of $$g$$ and $$m$$, we get: $$r = \frac{h W(\frac{1}{2chm_p+chkm_p})}{c} \equiv 4.14 \cdot 10^{-20} \text{m},$$ which is still greater then the Planck length.

So, is it possible to bring two neutrons this near?

Second edit: The thought experiment with the differently charged bodys might be thought of as charging a capacitor. It is shown here experimentally, that it is possible to reduce the mass of the capacitor when charging it: https://arxiv.org/abs/1502.06915. This effect is called the Biefeld-Brown effect. It would be interesting to calculate if this thought experiment can explain the Biefeld-Brown effect.

Update for those interested: It seems that this concept of the first thought experiment has already been explored by others: https://tu-dresden.de/ing/maschinenwesen/ilr/rfs/ressourcen/dateien/forschung/folder-2007-08-21-5231434330/ag_raumfahrtantriebe/JPC---Propellantless-Propulsion-with-Negative-Matter-Generated-by-Electric-Charges.pdf?lang=en The first thought experiment can be dated back to: http://aapt.scitation.org/doi/citedby/10.1119/1.11328

• There is a difference between gravitational repulsion and negative net force. – MaxW Jan 6 '18 at 16:44
• You forgot the units for $Q$. – freecharly Jan 6 '18 at 16:51
• What are the units for $Q$? – orgesleka Jan 6 '18 at 16:52
• This post (v19) is an inconsistent mixture of relativity and Newtonian physics. – Qmechanic Jan 6 '18 at 21:18
• @stackExchangeUser Coulombs. $1~{\rm C}\approx 6.2\cdot 10^{18}~\rm e$, where $\rm e$ is the charge of a proton. – Chris Jan 6 '18 at 21:38

If $E = (m+M)c^2+k\frac{Q^2}{r}-G \frac{mM}{r} < 0$, then $(m+M)c^2-G \frac{mM}{r}$ must also be less than zero. Thus we must have $r < \frac{GmM}{(m+M)c^2}$.

This is less than the Schwarzschild radius of a particle of mass $m+M$, and this gets to the heart of the matter: You are using Newtonian gravitation in a field where it does not apply.

• what would happen if the two bodies are charged with different signs? then $Q^2$ becomes $-Q^2$ and your argument is not directly applicable – orgesleka Jan 6 '18 at 18:52
• @stackExchangeUser The main point of the post still holds- all other issues aside, you're trying to apply Newtonian gravity in a domain where you need general relativity. – Chris Jan 6 '18 at 21:35
• No, If one changes the sign, then Newton gravitation can be applied, as the distance $r$ becomes larger. – orgesleka Jan 7 '18 at 17:48
• @stackExchangeUser You are still using Netwonian gravity in environments where it does not apply. If Q is great, then you have an electrically charged black hole, which has a different situation. You have to look at Reissner–Nordström metrics or Kerr–Newman's, depending on whether it is also rotating or not. In either case, you run into the fundamental issue that Netwonian gravity is only an approximation for real effects. It happens to be a very useful approximation in an amazingly large number of cases, but this is not one of those. – Cort Ammon Jan 7 '18 at 18:39
• Q does not have to be great m and M also. please look up the biefeld brown effect – orgesleka Jan 7 '18 at 18:42

You already have several good answers. In short, the error lies in (inconsistently) mixing Newtonian gravitation and relativistic effects.

That being said, one may wonder what general relativity has to say. In other words, we already know what the naïve picture predicts; now we want to know what the more correct one does.

In full generality, the answer is given by the Positive energy theorem. This theorem proves that, under some natural assumptions, the mass of any system is always positive (or null, for flat spacetime). In this sense, one would always have $M>0$ for physical situations. General relativity predicts that mass cannot be negative. Neat, right?

Let me illustrate this theorem through a couple of (relevant) examples. Your description corresponds to a two-body problem, for which no exact solution is known. It can nevertheless be approximated by a one body problem by taking one of the masses to be much larger than the other one, which we assume from now on. Under this circumstance, you can describe your charged system as a Reissner-Nordström black hole. This black hole is essentially the same as the Schwarzschild one, but with a non-zero electric charge. Its mass is given by $$M=M_0+\frac{Q^2}{16\pi GM_0}$$ where $M_0$ is the mass it would have for $Q=0$. Here we see that $M>0$ for any value of $Q$, in accordance with the theorem above.

A more general situation is given by the Kerr-Newman one, where we also have a non-zero angular momentum (the BH is rotating). Here the expression for $M$ is more complicated, but it can be shown that it is also positive (in fact, it is larger than the Reissner-Nordström one, which is itself larger than $M_0$; this is easy to understand: charge adds energy, and rotation adds even more energy).

Interestingly, it can also be shown that in both cases $Q$ cannot be too large, or otherwise the solution becomes pathological. In other words, even if $M$ is always positive, if you keep increasing $Q$ (or the angular momentum in the KN case), at some point the black-hole starts violating basic principles (naked singularities, which are no good). To some extent, this is the analogous situation to your "negative mass" result, but taking into account relativistic effects. For a very large value of $Q$ the model does also break down, but for different reasons.

• Thank you for your answer. Didn't know of the positive energy theorem. – orgesleka Jan 7 '18 at 6:26
• I read about the positive energy "theorem" and it seems like it is based on at least the null energy condition which is a conjecture. In : arxiv.org/pdf/gr-qc/0003025.pdf it is investigated what happens, if this conditition is broken. – orgesleka Jan 7 '18 at 8:26
• @stackExchangeUser Casimir corresponds to a quantum-mechanical situation. Here we consider classical gravitation only. If you introduce quantum-mechanical phenomenons, you are again inconsistently mixing different regimes. The full answer would require a quantum theory of gravity, which we for now lack. A good candidate, String Theory, does also predict (at least, as long as we have SUSY) that mass is always positive (the so-called BPS bound $M\ge|Z|$). The bosonic string does predict states with $m^2<0$, but the model is very unrealistic anyway. – AccidentalFourierTransform Jan 7 '18 at 10:47

Following your reasoning, you don't need any electric charge to obtain a negative energy $E$ and with $E=mc^2$ also a "negative mass" $\hat{m}$ if the distance $r$ between your masses becomes small enough: $$E = \hat{m} c^2=(m+M)c^2-G \frac{mM}{r}$$

• Yes, I know this, thank you for your answer. But the problem is, one will get unrealistic values for $r$, below planck length i think. – orgesleka Jan 6 '18 at 17:29
• @stackExchangeUser: Not Planck length (since there is no $\hbar$ in your calculations), but Schwarzschild radius – A.V.S. Jan 6 '18 at 17:46
• indeed, and if you solve for the radius in terms of the masses, you arrive at the Schwarzschild radius of the mass system – lurscher Jan 19 '18 at 14:43
• @lurscher: How do you derive this? I get: $r = G \frac{m M}{(m+M-\hat{m}) c^2}$ – orgesleka Jan 19 '18 at 16:42

It is shown by Martin Tajmar ( https://tu-dresden.de/ing/maschinenwesen/ilr/rfs/ressourcen/dateien/forschung/folder-2007-08-21-5231434330/ag_raumfahrtantriebe/JPC---Propellantless-Propulsion-with-Negative-Matter-Generated-by-Electric-Charges.pdf?lang=en , Figure 4 and related equations) that if the capacitor is symmetric, then the "self-energy" is always greater than the "interaction-energy". So in order to achieve negative mass, one needs an asymmetric capacitor. So to answer the question above: If the bodys are charged with $Q$ and $-Q$ then the mass of the system will decrease, but can not become negative. On the other hand if the two bodys are asymmetrically charged, then the mass of the system can theoretically become negative, as Martin Tajmar shows in his paper.

## protected by Qmechanic♦Jan 7 '18 at 0:09

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).