For a very large value of $Q$ the model does also break down, but for different reasons. To some extent, this is the analogous situation to your "negative mass" result, but taking into account relativistic effects. 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). Interestingly, it can also be shown that in both cases $Q$ cannot be too large, or otherwise the solution becomes pathological. 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). Here we see that $M>0$ for any value of $Q$, in accordance with the theorem above.Ī more general situation is given by the Kerr-Newman one, where we also have a non-zero angular momentum (the BH is rotating). Where $M_0$ is the mass it would have for $Q=0$. This black hole is essentially the same as the Schwarzschild one, but with a non-zero electric charge. Under this circumstance, you can describe your charged system as a Reissner-Nordström black hole. 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. Your description corresponds to a two-body problem, for which no exact solution is known. Let me illustrate this theorem through a couple of (relevant) examples. It is hypothesised that macroeconomic conditions in which the company operates would moderate the relationship that already exists, so that capital structure, asset growth and profitability are no longer considered by investors as a credible signal on the value of the company's future.The purpose of this study was to determine the effect of. General relativity predicts that mass cannot be negative. In this sense, one would always have $M>0$ for physical situations. This theorem proves that, under some natural assumptions, the mass of any system is always positive (or null, for flat spacetime).
#Negative moderation effect stack exchange full
In full generality, the answer is given by the Positive energy theorem. In other words, we already know what the naïve picture predicts now we want to know what the more correct one does. That being said, one may wonder what general relativity has to say. In short, the error lies in (inconsistently) mixing Newtonian gravitation and relativistic effects. The first thought experiment can be dated back to: It seems that this concept of the first thought experiment has already been explored by others: It would be interesting to calculate if this thought experiment can explain the Biefeld-Brown effect. This effect is called the Biefeld-Brown effect. It is shown here experimentally, that it is possible to reduce the mass of the capacitor when charging it: The thought experiment with the differently charged bodys might be thought of as charging a capacitor. So, is it possible to bring two neutrons this near? $$E = (m+M)c^2-k\frac,$$ which is still greater then the Planck length. What's wrong with the following argument to "produce" negative mass?:Ĭonsider two charged bodys of weight $m,M$ of distance $r$ with the electric charge $Q$ and $-Q$.
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