# Does GR provide a maximum electric field limit?

Does GR provide a limit to the maximum electric field?

I've gotten conflicting information regarding this, and am quite confused. I will try to quote exactly when possible so as not to confuse things more with my paraphrasing.

The author of the Motion Mountain physics textbook claims in his book there is a limit, and clarifies on his site that ( http://www.motionmountain.net/wiki/index.php?title=Dislike_Page ):

"electromagnetic fields are limited in magnitude. Now, every electromagnetic field contains energy, and energy density is limited by general relativity: if energy density is too high, a black hole appears. The smallest possible black hole then leads to a field limit. If you deny an upper field limit, you deny general relativity. However, general relativity has been confirmed in every experiment so far."

This sounds very obvious and intuitive to me. However one of my physics TA's got very upset when I used this in a thought experiment when discussing some limits in physics. When I told him the textbook I got that from, he looked it up, and commented on the Motion Mountain wiki website his argument:

"I'd like to add something here. Suppose for a moment the energy density limit is correct, then if an object is sped up until it length contracts enough such that the energy density is greater than this limit, does it turn into a black hole? No, quoting John Baez "The answer is that a black hole does not form"* http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html *The claim that "if energy density is too high, a black hole appears" is an incorrect oversimplification. The issue is that in GR, gravity depends on more than just the energy density (component T^00 of the stress energy tensor). So we can't just look at the "relativistic mass" (E/c^2) to judge whether a black hole forms. It actually isn't even enough to look at the "invariant mass", for while the trace T of the stress energy tensor for a particle is just its invariant mass, for an electromagnetic field it is identically zero even though electromagnetic fields curve spacetime in GR. So none of these concepts of mass are sufficient when discussing gravity using GR (especially when considering electromagnetic fields), because gravity couples to the entire stress energy tensor. I hope this was helpful."

This makes much less sense to me, and I don't understand how energy density tending to infinity could EVER avoid being a black hole. No offense to my TA, but I'm skeptical as he's disagreeing with a textbook author. Plus, the author's response was that my TA is another Einstein denier, and not worth responding to.

So I'd like a third party's answer on this. Does GR provide a theoretical limit to the strength of an electric field? Is it best to just ignore my TA on this one?

• There is enough discussion on this that the given Mountain book needed to prove mathematically its assertion that EM fields alone can generate BHs. Does it actually claim to do this? – Roy Simpson Mar 19 '11 at 16:48
• GR has general coordinate invariance, you can always choose coordinate system to make the electric field at a point anything you like, as large and as small as you want it to be. You can make the components of it encode your birthday if you wish. To discuss physical consequences such as black hole formation you'd have to make general coordinate invariant statements. – user566 Mar 19 '11 at 19:03
• Is this not simply a statement about the hoop conjecture? If the experimental apparatus satisfies that criteria, then yes a black hole most likely forms. Details are important here though, which is why John Baez added a caveat. – Columbia Mar 19 '11 at 19:50
• Schiller, the author of Motion Mountain, describes this part of the book as an attempt to explain, at a level understandable to high school students, the ideas from T. Jacobson, "Thermodynamics of spacetime: the Einstein equation of state," PRL 75 (1995) 1260, arxiv.org/abs/gr-qc/9504004 . You can get yourself tied up in all kinds of knots on this material if you don't carefully consider the specific definitions. Because Schiller's treatment fluctuates wildly in its level of presentation, this can be difficult to do. If you really want to understand this, go to the Jacobson paper. – Ben Crowell Aug 14 '11 at 23:16
• Aren't we forgetting particle-antiparticle creation when the EM field energy density crosses the Schwinger pair creation threshold? – Rho Phi Jan 3 '18 at 13:45

Your TA is right that energy density alone does not trigger black hole formation. Consider a ball that's sitting still. Now speed up and look at the ball again. It will have gained (kinetic) energy. Relativistically, you can make the ball's energy density arbitrarily large by moving sufficiently near the speed of light. But the ball hasn't done anything in this process. It's you that has been changing speed. The notion of a black hole is not observer-dependent, so energy density alone cannot make a black hole form.

That said, there are senses in which electromagnetic fields can form black holes in general relativity. Two colliding electromagnetic plane waves (which are necessarily gravitational plane waves as well) can do this. The field strength for a single plane wave can be arbitrarily large, however. Any "limits" are highly dependent on the specifics of the configuration.

• Okay, I guess this makes sense. If there is an electromagnetic wave, I can boost such that this looks like it has more and more kinetic energy, but its still the same physical situation so no black hole forms. Is that the idea? How do I relate the kinetic energy to the electric field strength? – John Mar 19 '11 at 17:47
• Yes, exactly. The analog of energy density for an electromagnetic field $\frac{1}{2} (E^2+B^2)$. Note that the electric and magnetic fields are observer-dependent, so this density is not a scalar. – Stingray Mar 19 '11 at 18:00
• @Stingray: that density most definitely is a scalar--it's proportional to $F^{ab}F_{ab}$. – Jerry Schirmer Mar 19 '11 at 18:29
• $F^{ab} F_{ab}$ is proportional to $E^2-B^2$. – Stingray Mar 19 '11 at 18:32
• @Jerry, wrong. What you've written is Lagrangian, nothing to do with energy. – Marek Mar 19 '11 at 20:17

All the answers to this question are wrong. They are right that there is no limit to the magnitude of the electric field limit in classical GR, but they are wrong to claim that this is because electric fields by themselves can't make a black hole.

• There is no bound to the magnitude of the electric field in GR:

It is not enough to have a high energy density to make a black hole, you need an total energy greater than R in a sphere of radius R. If the sphere is arbitrarily small, the energy density can be arbitrarily big. For a specific example of a strong field without collapse, an extremal black hole solution in general relativity has a metric:

$$ds^2 = (1-Q/r)^2 dt^2 - {dr^2 \over \left(1-{Q/r}\right)^2 } + r^2 (d\theta^2 + \sin^2\theta \, d\phi^2)$$

And an electric field in the direction of r of magnitude $Q/r^2$. This electric field is maximum at $r=Q$ (at least outside the black hole), and the maximum value is 1/Q, which is as big as you like. To make a non singular solution with the same properties, just replace the black hole horizon with a slightly bigger sphere with the same mass and the same surface charge.

• An electric field which has a large enough magnitude, all by itself, in a large enough region of space, causes gravitational collapse.

The easiest way to see this is to consider a parallel plate capacitor of area $L^2$ and charge Q with no net momentum, and pull apart the plates to distance L, then you have order $|Q/L^2|^2 L^3$ amount of mass in a region of radius L, and when Q is bigger than L (which happens with small charge densities when L is large) you get a total energy at large distances which is too big to avoid collapse just by considering the total mass and zero momentum.

• There is nothing wrong with John Baez's examples of highly boosted noncollapsing stress, but they are irrelevant.

These are not like a static electic field because they have a large net mometum. The static electric field has zero field momentum.

The answer is: the Motion Mountain book is wrong, your TA and John Baez are correct

A couple answers here are discussing specific solutions in GR. I think there is a much easier and more general way to answer this.

In special relativity, we can choose to use a different coordinate system (we don't need to physically change our motion as some posters seem to claim), and by merely changing the coordinate system the electric field changes. In fact, the electric field can be made arbitrarily large this way. All that this relies on is the Lorentz symmetry in special relativity.

This is possible because the value of the electric field is just a coordinate system dependent component value of a tensor - the electromagnetic tensor.

Now the global Lorentz symmetry of SR doesn't survive when transitioning to GR, but there is still local Lorentz symmetry. This means we can still change our coordinate system locally, to have the electromagnetic field tensor change (at a point at least) the same way as before. So what the Motion Mountain book is claiming it can derive from GR, is actually violating GR. So we don't even need to look at any specific solutions in GR. The local Lorentz symmetry of GR is enough to show that the textbook is wrong.

As other posters have mentioned, the same goes with energy density. So the textbook author's starting comment of "energy density is limited by general relativity: if energy density is too high, a black hole appears" is wrong. In my opinion, if this is an example display of the textbook author's understanding of GR, it is probably best to avoid his books altogether.

Trust John Baez over your TA. Another thing that you need in addition to a large mass density in order to form a black hole is a matter distribution that behaves in a certain way--in particular, you can fill the whole universe with radiation in a certain Robertson-Walker spacetime, and the density of radiation, and thus the magnitude of the RMS E-field is just a parameter with no upper limit, and the Robertson-Walker spacetime contains no black hole for any value of its parameters.

I don't know about what Baez is talking about, but i assure you that speeding up an object (relative to my "funny" frame) definitely increases the relative kinetic energy of the object in that frame, but not all energy transforms the same;

kinetic energy transforms as the 0-component of a cuadrivector, while rest-energy ($mc^2$) transforms as a scalar under Lorentz transformations. Also $T^{00}$ transforms as a tensor component, because it is the integral density of a 0-component vector against a oriented 3-form defining flow over a 3D volume element. Once you integrate $T^{00}$ over a volume you get a scalar density. Gravitational contributions only come from energy tensors like densities or rest-energies integrated over volumes. A binding energy in an atom would also behave as a scalar so it couples directly to gravity. A high-speed electron has an unbounded kinetic-energy but doesn't not contribute to any tensor energy density in the same unbounded way

hope this clarifies the issue

• So is the energy in an electric field one of these scalars and therefore gravitate and thus limited in magnitude by GR? – John Mar 19 '11 at 16:24
• "Gravitational contributions only come from energy scalars like densities or rest-energies." -> what? You'll have to clarify this because gravity obviously couples to all of $T^{\mu\nu}$. You don't need rest mass to interact gravitationally. Just think about light. – Marek Mar 19 '11 at 16:26
• No. Gravitational contributions come from the entire stress energy tensor--there is nothing special about $T^{00}$ in $R_{ab}-\frac{1}{2}R g_{ab} = 8\pi G T_{ab}$. Having pressures definitely changes the way the gravitational field behaves--a FRW spacetime with radiation expands and collapses at a different rate than one with just dust, for example. – Jerry Schirmer Mar 19 '11 at 16:26
• This is not true at all. You can make $T^{00}$ almost anything you want by an appropriate change in coordinates. Additionally, gravity couples to a rank-2 tensor field, not a collection of scalars. – Stingray Mar 19 '11 at 16:28
• sorry i corrected the post – lurscher Mar 19 '11 at 16:32