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So I was thinking about something for the past while

Consider a large spherical foam-ball with homogeneous density. Where a foam ball is defined as an object that can absorb matter with 0 friction (example: a gravitational well without an object inside). This is a purely theoretical construct.

If the foam-ball has a radius R and a charge Q. What charge must the foam-ball have, such that there is well defined sphere or horizon such that any object with a negative charge (even if it is the slightest bit) must travel faster than C in order to escape the field of the ball.

Aka what charge would turn this into a black-hole for all objects with opposite sign on their charge?

Once again I assume the foam-ball is a single particle, that doesn't repel itself... its just a very large homogenous 'thing' with charge.

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  • $\begingroup$ Can this approached using a similar style of math as the construction of a gravitational black hole, except mass is substituted with charge, etc... ? $\endgroup$ – frogeyedpeas May 28 '13 at 19:14
  • $\begingroup$ I think you should just use basic equations as $F_e = k*\frac{Q_1*Q_2}{r^2}$ and $a = \frac{F}{m}$. You then put e constant instead of $Q_2$. I'm only not sure how to deal with aceleration. $\endgroup$ – Tomáš Zato May 28 '13 at 19:24
  • $\begingroup$ @frogeyedpeas Given the flaws that have been pointed out, please consider changing the accepted answer to this question. $\endgroup$ – Emilio Pisanty Mar 30 '17 at 13:22
  • $\begingroup$ @EmilioPisanty yes, but i'll need some time to digest these new answers (I didn't set the bounty, someone else appears to have) $\endgroup$ – frogeyedpeas Mar 31 '17 at 1:39
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Such a singularity would not occur, if you have no lower bound on the negative charge. For gravity, the singularity occurs because gravitational potential energy and relativistic kinetic energy both depend on the mass of the smaller object, which allows it to divide out when you solve for escape velocity. However, in this case, only the electromagnetic potential energy, not the kinetic energy, depends on the negative charge. This means the negative charge never gets divided out, and hence you can arbitrarily decrease the magnitude of this charge to make the escape velocity as low as you want.

However, if you fix the negative charge at some value $q_2$, it will be unable to escape when its electromagnetic potential energy equals its rest energy (up to a sign). So just set $\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} = mc^2$ (again, up to the correct sign) and solve for $r$.

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  • $\begingroup$ do u think u might have a link or something that explains how the gravitational potential energy and relativistic kinetic energy 'divide out'? I'm curious if there is an analog of kinetic energy that applies to E&M. $\endgroup$ – frogeyedpeas May 28 '13 at 19:44
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    $\begingroup$ Yep, try here. It explains how to derive relativistic escape velocity in a gravitational field, and you should see $m_0$ divide out in the algebra. $\endgroup$ – Izzhov May 28 '13 at 19:46
  • $\begingroup$ Looks good! this is much clearer than I expected. $\endgroup$ – frogeyedpeas May 28 '13 at 19:55
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    $\begingroup$ This answer is right in that the answer depends on the charge of the escaping particle, but it oversimplifies the relativistic dynamics; in particular, it's perfectly possible to have a particle moving at close to the speed of light with many times more kinetic energy than its rest energy. $\endgroup$ – Emilio Pisanty Mar 23 '17 at 10:29
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To provide a consistent relativistic treatment of the problem, I will model the "foam-ball" by the Reissner-Nordström metric which represents the relativistic gravitational and electromagnetic field of a point of effective mass $M$ and charge $Q$. The reason why I am calling $M$ the effective mass is that even the pure electromagnetic field necessarily carries some energy and will thus look from afar as a gravitating mass of some mass $M_Q$. I will not discuss here what $M_Q$ should be and just leave $M$ as a free, in principle non-zero parameter. (The only influence on this analysis is that this allows for the existence of an event horizon.)

I will also use the term event horizon, which is the usual horizon known from general relativity from behind which no particle can escape, and the term "electromagnetic horizon", which would be a point from behind which a particle of certain charge is unable to escape.


The Reissner-Nordström metric (in geometric units) reads $$ds^2 = -\left(1 - \frac{2 M}{r} + \frac{Q^2}{r^2}\right) dt^2 + \frac{dr^2}{1 - 2M/r + Q^2/r^2} + r^2 d\Omega^2$$ This solution has an event horizon (horizon for any particle) at $1 - 2M/r + Q^2/r^2 = 0$ if $M^2>Q^2$.

The electromagnetic potential $A_\mu$ has only one non-zero component which is $A_t = -Q/r$. A charged test particle in this field obeys the Hamilton's equations generated by the Hamiltonian $$ H = \frac{1}{2} g^{\mu \nu} (\pi_\mu - e A_\mu)(\pi_\nu - e A_\nu)\,, $$ where $\pi_{\mu} = m u_\mu + eA_\mu$ is the canonical momentum conjugate to $x^\mu$. Both the field and the metric are static and we thus know that the total energy of the particle $\mathcal{E}\equiv-\pi_t$ will be an integral of motion. Particles at infinity have total energy larger than $m$, $\mathcal{E}>m$. I.e., a particle which has escaped the potential wells of the electromagnetic and gravitational fields will necessarily have total energy larger than just its rest energy.


Let us now investigate the condition $\mathcal{E}>m$ for a purely radially moving particle. For this particle we can use the four-velocity normalization $g^{\mu\nu}u_\mu u_\nu = -1$ to obtain $$ u_t = -\sqrt{1 - \frac{2M}{r} +\frac{Q^2}{r^2} + \dot{r}^2}\,, $$ where $\dot{r} = u^r = dr/d\tau$ can attain any value. (No speed-of-light violations, you can easily check that $dr/dt \to 1$ as $\dot{r} \to \infty$.)

If we now substitute this into $\mathcal{E}>m$, we easily obtain the necessary condition for escape $$ m\sqrt{1 - \frac{2M}{r} +\frac{Q^2}{r^2} + \dot{r}^2} +\frac{eQ}{r} >m $$ Note that if $e$ and $Q$ are of opposite sign, the $eQ/r$ term is negative and makes it "harder" to make $\mathcal{E}>1$. However, it is obvious that there will always be values of $\dot{r}$ for which $\mathcal{E}>m$. These values are $$ \dot{r}^2 > \frac{2(M - eQ/m)}{r} + \frac{Q^2(1 + e^2/m^2)}{r^2}\,. $$ It is important to remember that this is not a sufficient but just a necessary condition. However, it shows that it is always possible to set the velocity of the particle so that it acquires an "unbound" energy.

The two cases when $\mathcal{E}>m$ but the particle does not reach infinity are the following: First, the case when $\dot{r}<0$, because that corresponds to a radial infall and a geodesic ending in the central singularity never to reach infinity. Second, the case when $\dot{r}>0$ inside the horizon (if it exists), and the particle then escaping outside with $dt/d\tau<0$; this case is in fact the radial infall "played backwards". In any other case $\mathcal{E}>m$ is sufficient for the particle escape.


To conclude, the electromagnetic field does not create an "electromagnetic horizon" which would restrict particles of a certain charge from escaping. The only horizon which puts an ultimate bound on the escape of any particle is the event horizon. As long as we are outside of the event horizon, a particle of any charge can be endowed with enough kinetic energy to escape the potential well.

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  • $\begingroup$ Could you clarify how we can see from these equations that gravity does create an event horizon? Looking at the last inequality, it appears that increasing $M$ and increasing $Q$ could play similar roles. $\endgroup$ – Alon Navon Mar 23 '17 at 22:09
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    $\begingroup$ @AlonNavon The condition $\mathcal{E}>m$ does not tell you anything about the event horizon because it asks whether the particle is bound or not. (If we have chosen the right root so that it is travelling forward in time.) The event horizon is a point of no return, no matter what the energy of the particle is. I.e., from the horizon radius $r_H$ you are unable to get even to $r_H + \delta r$ let alone to infinity. In this sense the commonly used analogy of "escape velocity" from Newtonian mechanics is not correct; the event horizon of a black hole is much more badass. $\endgroup$ – Void Mar 23 '17 at 23:20
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Here is my attempt to the question:

We have a sphere of charge $Q$ and radius $R$. At a distance $r>R$ the potential $V(r)$ is given by

$$V(r)=\frac Q {4\pi\epsilon_0 r} $$

Now, total energy of the particle X of mass m, charge -q , velocity v is given by $$E=-qV(r)+\gamma m c^2 $$ where $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}$ and observe that when $v\rightarrow c$, $\gamma\rightarrow \infty$

Say escape velocity of X at a distance r is given by $v_e$ and $\gamma(v_e)=\gamma_e$.

From energy conservation if X escapes \begin{align} E(r)&=E(r= \infty)=mc^2 +K.E \geqslant mc^2 \\ \implies -qV(r)+\gamma_e m c^2&= mc^2 \end{align} This gives \begin{align} \mathrm{KE}_\mathrm{initial}&=(\gamma_e-1) m c^2=qV(r)\\ \implies \gamma_e&=\frac {qQ} {4\pi\epsilon_0 r m c^2} + 1\\ \implies v_e &=c\left(1-\frac {1} {\gamma_e^2}\right)^\frac {1} {2} \end{align} This shows if $v\geqslant v_e$ then $\gamma\geqslant \gamma_e$ and X escapes.

But in order to obtain sufficiently high $\gamma$, $v$ just need to be grater than $v_e$ and can also be less than $c$.This shows that if sphere has finite charge Q then for $v_e \leqslant v\leqslant c$, X escapes. Therefore X as well as any particle can escape from any $r>R$ if $v>v_e$ although it is negatively charged without even approaching the speed of light. Thus, the sphere is not a black hole for the negatively charged particles.

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  • $\begingroup$ There's several good answers with a variety of approaches, both simpler and more technical, but I think that this one strikes the best balance. $\endgroup$ – Emilio Pisanty Mar 30 '17 at 13:28
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I'm not sure whether the OP is assuming the ball has mass $M$ and are considering its gravitational effects, or if the OP's only considering electromagnetic effects. If it's the latter, then clearly no such radius exists, because at any finite radius a test particle has only a bounded potential energy, and its kinetic energy can be made arbitrarily large without its exceeding lightspeed.

An even easier way to see this is just to appeal to dimensional analysis; any such radius would need to equal $R$ times a function of dimensionless ratios of the remaining relevant quantities $Q$ and $c$, but no such dimensionless ratio exists. So if the radius existed, it would need to be independent of $Q$, and setting $Q = 0$ we clearly see that it can't exist.

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Aside from everything that was said by the others I would like to lay down the theoretical framework for a generalized solution (any speed, any mass, any charge, any distance, as long as the "balls" don't fall into a singularity).

There are 2 ways of looking at this problem.

The easiest one is to choose special relativity if the masses of the charges are relatively small, in which case we can neglect the gravitational effects. In such a situation we can use

$$ \frac{d(m_{0}U^\mu)}{d\tau} = - e (F'')^\mu_\nu U^\nu $$

with $$ (F'')^\mu_\nu = \Lambda^\mu_\alpha(-v_{rcv}) \Lambda_\nu^\beta(-v_{rcv}) (F')^\alpha_\beta $$ and $$ (F')^\mu_\nu = \Lambda^\mu_\alpha(v_{src}) \Lambda_\nu^\beta(v_{src}) F^\alpha_\beta $$ (for the rest of equations, what is which, how to combine them, see here)

We use this eq for each of foam-balls, and then we solve (using retarded positions) "orbits" at speeds close to the speed of light.
We vary $q_1$, $q_2$,draw graphs, deduce what happens.
Of course we need to define some border (surface) conditions, that are very important because they define what happens when the 2 balls collide.
Will they scatter ? Will they combine ? Will charge density combine to form peculiar new (kind of) matter ? Will they annihilate to create lots of EM waves or other kind of radiation ?
That is why it is very important to have the proper definition of what is under the exterior surface of the ball.
Simply assuming that is just a singularity under the exterior surface may disagree when it will came to compare with real life experiments.


The other way is to use general relativity.
There are two paths we could take here.
The simpler one is to assume that one of the ball has the charge and mass far far smaller than the other one : $m_1 >> m_2$ and $q_1 >> q_2$.

For such a case @Void provided here an answer in the framework of Reissner-Nordström metric, but I will try to answer from a bit different perspective the one of the theory of the bridges.

Einstein derived the metric in case of spherical symmetry for combined electrity and gravity a little bit different; he choose the sign of the energy tensor in such a way that by solving the field equations we obtain the metric $g_{\mu\nu}$ :

$$ ds^2 = (1 - \frac{2m}{r} - \frac{q^2}{2r^2}) dt^2 - \frac{1}{1 - \frac{2m}{r} - \frac{q^2}{2r^2}}dr^2 - r^2(d\theta^2 + \sin^2{\theta} d\phi^2) $$

So for such a metric the event horizon will be defined at $$\left(1 - \frac{2m}{r} - \frac{q^2}{2r^2} \right) = 0$$ This means that even without the help of mass we can get an event horizon.
Since we used the square of charge implies that it does not matter which sign has the charge.
For this using traditional black hole analysis we will came to the conclusion that anything that passes the event horizon will have no way to get out, no matter how close to the speed of light we go.

On the other hand Einstein suggested a change of variable that would help us get rid of the singularity of the event horizon.

The first step would be to choose $u^2 = r^2 - \frac{q^2}{2}$, set mass $m = 0$ and then apply it to the metric to obtain:

$$ ds^2 = - du^2 - \left(u^2 + \frac{q^2}{2} \right) \left(d\theta^2 + \sin^2{\theta} d\phi^2 \right) + \frac{2u^2}{2u^2+q^2} dt^2 $$

So as we can see if $u$ varies from $-\infty$ to $+\infty$ but $r$ will only have positive values between $\sqrt{\frac{q^2}{2}}$ and $+\infty$. Our smaller ball will move from one spacetime sheet to another.

The last way and the most complicated way but which will give answers for general cases, no matter how big/small, how many, how fast/slow the "balls" are.

We have $p$ singularities. We enclose each singularity denoted by $s$ in a closed surface.

$$ \int^s{(\Phi_{\mu k} + 2 \Lambda_{\mu k}) \cos{(x^k \cdot N)} dS } = 0$$

We assign to each singularity $s$ the position $\overset{s}{\xi}$ with $\xi^k(x^0)$ being actually a 3-vector.

A distance from $s$ singularity will be defines as: $$ \overset{s}{r}^2 = (x^1 - \overset{s}{\xi}^1)^2 + (x^2 - \overset{s}{\xi}^2)^2 + (x^3 - \overset{s}{\xi}^3)^2 $$

The generalized field equations are: $$ \Phi_{\mu \nu} + 2 \Lambda_{\mu \nu} = C_{\mu \nu} $$

where

$$ C_{mn} = - \sum^p_{s=1}{\left( \left(\frac{\overset{s}{C}_m}{\overset{s}{r}} \right)_{,n} + \left(\frac{\overset{s}{C}_n}{\overset{s}{r}} \right)_{,m} - \delta_{mn}\left(\frac{\overset{s}{C}_k}{\overset{s}{r}} \right)_{,k} \right)} $$

$$ C_{00} = - \sum^p_{s=1}{\left(\frac{\overset{s}{C}_k}{\overset{s}{r}} \right)_{,k}}$$

$$ C_{0n} = - \sum^p_{s=1}{\left(\frac{\overset{s}{C}_0}{\overset{s}{r}} \right)_{,n} + \left(\frac{\overset{s}{C}_n}{\overset{s}{r}} \right)_{,0} }$$

having $$ \overset{s}{C}_{m} = \frac{1}{4\pi} \int^s{ 2 \Lambda_{mn} \cos{(x^n \cdot N)} dS }$$ $$ \overset{s}{C}_{0} - \frac{1}{3}\overset{s}{C}_{k}\overset{s}{\dot{\xi^k}} = \frac{1}{4\pi} \int^s{ 2 \Lambda_{on} \cos{(x^n \cdot N)} dS }$$

We can now apply an approximation method such as one defined here and as stated here we can use this for charged particles in electromagnetic field.

We now want to find the distance for which the two singularities become inseparable, in the sens that none of them can get out of that distance.
We call that the event horizon(if there exists such a thing) $ r_H(x^0_j) = \sqrt{(\overset{i}{\eta^k} - \overset{s}{\xi^k})(\overset{i}{\eta^k} - \overset{s}{\xi^k})} $ such that $ r_H(x^0_c) \le r_H(x^0_j) $ for all $ x^0_c > x^0_j$ (here $x^0$ is obviously the time component).
We are therefore looking for a field that which is max of $|C_{\mu\nu}|$.
By either simplifying the assumptions or by using different methods of approximation suitable at each stage we could come to an answer - or we could setup a supercomputer and wait for a result.

There is one more note. To really generalise we need to consider the influence of all the other fields from all the energy in the universe ($10^{53}$ Kg), which as it turns out, will manifest itself at very small distances in the form of cosmic range Casimir effect or vacuum energy/entanglement, having values and influences in the range $ \Delta x \Delta p = h/2$. If this influences are small as compared to local fields we can rely on our result for having experimental observance, we can predict same thing as the experiments.

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I'll take it that the definition of a black hole's event horizon you're thinking of is the point that the escape velocity equals the speed of light.

This point has the property where the electric potential energy plus the kinetic energy of the object at the speed of light equals 0.

$0=E_{potential}+0.5mc^2$

The formula for the electric potential energy is related to the formula for the attraction. It is the integral of it from infinity to the radius.

$Attraction(r)=-\frac{(r+R-\left|r-R\right|)^3*Q*q}{32*\pi*\epsilon_0*R^3*r^2}$

$E_{potential}=-\frac{Q*q}{32*\pi*\epsilon_0*R^3}\int_r^\infty \frac{(r+R-\left|r-R\right|)^3}{r^2} dr=-\frac{Q*q}{4*\pi*\epsilon_0*R^3}(R^3*\int_\frac{r+R+\left|r-R\right|}{2}^\infty \frac{1}{r^2}dr+\int_\frac{r+R-\left|r-R\right|}{2}^R r dr)=\frac{Q*q}{4*\pi*\epsilon_0}\left(\frac{2}{r+R+\left|r-R\right|}+\frac{4R^2-(r+R-\left|r-R\right|)^2}{8R^3}\right)$

$0=mc^2+\frac{Q*q}{2*\pi*\epsilon_0}\left(\frac{2}{r+R+\left|r-R\right|}+\frac{4R^2-(r+R-\left|r-R\right|)^2}{8R^3}\right)$

We can use this equation to calculate the minimum mass needed to escape the foam ball and the maximum combined charge you can escape the ball with.

$m=-\frac{Q*q}{2*\pi*\epsilon_0*c^2}\left(\frac{2}{r+R+\left|r-R\right|}+\frac{4R^2-(r+R-\left|r-R\right|)^2}{8R^3}\right)$

$-Q*q=\frac{2*\pi*\epsilon_0*c^2*m}{\frac{2}{r+R+\left|r-R\right|}+\frac{4R^2-(r+R-\left|r-R\right|)^2}{8R^3}}$

If we set r's value to 0 then we can check if the foam ball is a black hole to the particle.

if $0\ge mc^2+\frac{3*Q*q}{4*R*\pi*\epsilon_0}$ then foam ball is a black hole to the particle.

If Q's magnitude is this or larger then the foam ball is a black hole to the particle.

$Q=-\frac{4*R*\pi*\epsilon_0*mc^2}{3*q}$

If the particle's mass-charge ratio's magnitude is this or lower then the foam ball is a black hole to the particle.

$\frac{m}{q}=-\frac{3Q}{4R*\pi*\epsilon_0*c^2}$

As you an see, there is no one charge which will make all charged objects view the foam ball as a black hole. If the object is very massive or has an extremely low charge, it could easily escape the foam ball. The foam ball will affect all objects with the same mass-charge ratio equally though. I hope this answers your question.

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  • $\begingroup$ The kinetic energy of a massive particle going at the speed of light is not $\frac12 mc^2$ (and the kinetic energy of a particle at close to the speed of light isn't $\frac12 mv^2$, either). As you may have heard, massive particles can't reach the speed of light. $\endgroup$ – Emilio Pisanty Mar 27 '17 at 8:06
  • $\begingroup$ I know that but if I acknowledge that relativistic fact then there ceases to be any possible values of Q and R where the foam ball is a black hole to a particle traveling at the speed of light since a particle going that fast has $\infty$ kinetic energy. $\endgroup$ – Laff70 Mar 27 '17 at 16:21
  • $\begingroup$ So... you're saying that if you do things correctly then your results go away? $\endgroup$ – Emilio Pisanty Mar 27 '17 at 16:23
  • $\begingroup$ I suppose. Maybe I should've acknowledged that in my post. It's the reason why I used that rather classical definition of a black hole's event horizon at the beginning of my post. The answer to this question is rather meaningless since there are no possible "electromagnetic" black holes. I figured I'd explain it with classical physics instead since that allows "electromagnetic" black holes to exist. $\endgroup$ – Laff70 Mar 27 '17 at 16:33
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    $\begingroup$ If the answer is that the premises of the question are impossible, I normally suggest to just say that and provide an appropriate proof of the fact, rather than set up yet another impossible premise, whose consequences are therefore meaningless. $\endgroup$ – Emilio Pisanty Mar 27 '17 at 17:04

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