# Energy conservation in a closed universe

1. Feb 16, 2010

### JustinLevy

Consider a 4 dimensional spacetime which is everywhere flat and is closed in along all three spatial dimensions. Since spacetime is everywhere flat we can use global inertial frames. We will not consider any gravitational interactions in this problem.

Now a valid vacuum solution to maxwell's equations is:
$$\vec{A}(x,y,z,t) = -E_0 t \hat{x}$$
and V=0, such that the magnetic field is zero, and the electric field is everywhere a constant.

Since the volume of this universe is finite, the energy contained in this field is constant. Now take a hydrogen atom and separate the proton and electron and let them accelerate away in this field.

Spacetime is flat so the usual maxwell's equations and lorentz force law are the same, which show that energy is everywhere locally conserved. If energy is everywhere locally conserved, and we don't have to worry about subtleties of GR, then shouldn't energy be globally conserved as well?

I've talked this over with a friend and we could only see two possibilities:
1] There is a subtle flaw in that the closed universe affects the local equations somehow (we couldn't find such a flaw, but maybe we are missing something).
2] The radiation of the accelerating charges (the radiation is also a vacuum solution), somehow collects to cancel the original constant electric field vacuum solution. But that would require the radiation to have no spatial dependence, which it of course does. So I can't see how that would work either.

Any ideas anyone?

2. Feb 16, 2010

### IttyBittyBit

I don't understand how your spacetime can be flat everywhere, yet still closed. Unless it is compact, in which case it would have 'solid' boundaries that could not be crossed, causing the acceleration of your charged particle to eventually cease.

3. Feb 16, 2010

### Physics Monkey

The simplest space which works is a 3-torus. Like the 2-torus, a 3-torus can support a completely flat metric despite the periodic boundary conditions. Now on to the question. I don't want to give everything away all at once because this problem is so amusing. I hope people don't mind if I begin by just throwing out some interesting directions to think in.

Let the background field configuration be a constant electric field $$E_0$$ on a 3-torus pointing in the x-direction, say. This configuration looks time independent, but it really isn't. How can we tell? The vector potential looks time dependent, but one might think this is a gauge artifact. However, if one requires gauge transformations to be periodic then actually the coefficient of the linear t and space independent piece in the gauge potential is physical (it's just the constant electric field).

In fact, there is a new gauge invariant observable available on the torus, namely the integral of the vector potential around the entire space. In the background I specified above, this observable is $$\int A \cdot dx = - E_0 L t$$ where I integrated along the x-cycle and L is the length of the torus in the x-direction. This observable is time dependent and effects the physics. For example, charged quantum particles taken around the torus in the x-direction will experience interference with a time dependent phase. The constant electric field will also cause pair production. You can even think about the amusing situation of "pair-producing" a pair of oppositely charged capacitor plates, taking them around the x-cycle of the torus, and then annihilating them. It's interesting to think about what this does to the field.

Still another point of view on this type of field configuration can be had by imagining an extra fictitious dimension in which you thread a solenoid "through" the x-cycle of the torus. The constant electric field can now be thought of as arising from a linearly increasing current in the imaginary solenoid. Indeed, the gauge invariant observable I introduced earlier can be thought of as the flux through this imaginary solenoid.

Now we can get more technical. Since energy conservation is associated with time translation invariance, should this non-translation invariant (in time) background conserve energy?

4. Feb 16, 2010

### bcrowell

Staff Emeritus
I don't think you need a closed universe in order to create this paradox.

Let's just say we take Minkowski space, with the usual topology. Put a parallel-plate capacitor in it. Put a point charge between the plates. The charge accelerates, gaining KE. The gain in KE is canceled by a change in the integrated energy density $U=(\text{const.})\int E^2dV$ of the electric field. The only reason the energy of the electric field decreases is because of the decrease in the energy of the part of the point charge's field that hangs *outside* the capacitor plates.

Now if you take the capacitor plates and make them very far apart, conservation of energy is still satisfied, but it's satisfied because of the decrease in the electric-field energy in very distant regions of space.

If we make the plates sufficiently far apart, then there isn't time for the effect to propagate to the exterior region. It appears as though conservation of energy will be violated for some amount of time.

So I think this is the same paradox, and it doesn't have much to do with relativity.

I think the answer has to be that the radiation from the accelerating charge has an electric field that partially cancels with the ambient electric field, so the integrated electric field energy decreases.

5. Feb 17, 2010

### pervect

Staff Emeritus
You don't seem to have considered that energy may not be (and probably isn't) conserved globally in the situation you describe.

It's interesting that your example doesn't involve curvature at all - the problems of global energy conservation in GR are well known. I think yours may be a particularly simple example of why it isn't.

You'd probably need to add in a cosmological constant to balance out the energy density rho in your electric field universe to keep it from collapsing for an exact GR formulation.

6. Feb 17, 2010

### bcrowell

Staff Emeritus
I think the radiation is the solution to the paradox. Why do you say that it requires the radiation to have no spatial dependence? This is really a purely non-relativistic paradox, and the resolution is just an E&M resolution. The charge accelerates, and therefore radiates. The radiation pattern includes a circulating magnetic field, which is qualitatively similar to the B field you'd get from a current-carrying wire. This time-varying B field induces an E field, and the induced E field is in the direction that tends to cancel the EMF that produced the current. That means that the induced E field partially cancels the ambient E field. This reduces the total value of $U_E=(\ldots)\int E^2 dv$ throughout the whole space. This decrease in UE is matched by the increase in the KE of the particle plus the increase in $U_B=(\ldots)\int B^2 dv$.

I don't think you need to go to cosmological time or distance scales in order to get this paradox, nor do you need the closed topology to get the paradox. In the parallel-plate capacitor case, you can get conservation of energy to a good approximation while ignoring radiation as long as the size, x, of the capacitor is small compared to ct, where t is the time the particle takes to fall from one plate to the other. All you need in order to get the paradox is $x \gg ct$, but x can still be many orders of magnitude smaller than cosmological distance scales. All you need in order to resolve the paradox is to include the effects of radiation.

One good way to see that the topology is irrelevant is to take a flat space with the same spatial topology as the Euclidean plane, but stick in an infinitely long string of charges, each separated from its neighbors by the same distance L (an infinite, linear necklace of charges). Now impose an electric field along the axis of the necklace. The view of an observer looking at the initial configuration is exactly the same as the view an observer would have in the universe with the wrap-around topology (where you get a hall of mirrors effect). The transverse equilibrium is unstable, but that's no big deal -- you can just surround the necklace with a plastic tube to constrain the charges from moving transversely. Once they start to accelerate, their radiation is what preserves conservation of energy. The only observable difference between the necklace-in-flat-space version and the single-charge-in-wrapped-around-space version is that in the latter, you will eventually see a different pattern of wave superposition for the radiation fields. But this happens for $t>L/c$, whereas the paradox of nonconservation of energy occurs immediately, unless you take radiation into account properly.

7. Feb 17, 2010

### JustinLevy

WOW! There seems to be consensus that energy is not conserved, but not agreement on why. I was not expecting this at all.

bcrowell,

Physics Monkey,
I see you are arguing that (paraphrasing) the lagrangian is written with the potentials as the degrees of freedom, and the vacuum potential breaks time invariance here, and therefore energy is not conserved. (I love your discussion of this in quantum mechancis, and would like to return to that later.)

I don't really agree with either line of reasoning there, since I thought we could show local energy conservation just using the fields and the lorentz force law. Maybe there is a subtle flaw in that this assumes something about the fields at infinity (maybe some of the vector calc manipulation?), but I don't know where. So let me explicitly write out my understanding of the local energy conservation in electrodynamics.

---------------------------------------

Local conservation of energy from Maxwells equations and the Lorentz force law (shortened, but following along the derivation from Griffith's book):

The work done by the fields on a charge q is:
(eq1) $$W= \vec{F} \cdot d\vec{l} = \vec{F} \cdot \vec{v} \ dt = q(\vec{E} + \vec{v} \times \vec{B}) \cdot \vec{v} \ dt = q\vec{E}\cdot\vec{v} \ dt$$
Unfortunately, I don't know how to do this with point charges, so generalizing to continuous distributions instead, we have:
the rate at which work is done all all charges in a volume V is
(eq2) $$\frac{dW}{dt} = \int_V \vec{E} \cdot \vec{J} \ d^3r$$
using Maxwell's equation to substitute in for J
(eq3) $$\frac{dW}{dt} = \int_V \vec{E} \cdot [\frac{1}{\mu_0}\vec{\nabla} \times \vec{B} - \epsilon_0 \frac{\partial}{\partial t} \vec{E}] d^3r$$
Noting the vector calc identity:
(eq4) $$\vec{\nabla} \cdot (\vec{E} \times \vec{B}) = \vec{B} \cdot (\vec{\nabla} \times \vec{E}) -\vec{E} \cdot (\vec{\nabla} \times \vec{B})$$
And using Maxwell's equations for the curl of E, we get
(eq5) $$\vec{E} \cdot (\vec{\nabla} \times \vec{B}) = - \vec{B} \cdot \frac{\partial}{\partial t} \vec{B} - \vec{\nabla} \cdot (\vec{E} \times \vec{B})$$
Putting this all together, we currently have:
(eq6) $$\frac{dW}{dt} = - \int_V \left[ \frac{1}{\mu_0}\vec{B} \cdot \frac{\partial \vec{B}}{\partial t} + \frac{1}{\mu_0}\vec{\nabla} \cdot (\vec{E} \times \vec{B}) + \epsilon_0 \vec{E} \cdot \frac{\partial \vec{E}}{\partial t} \right] d^3r$$
(eq7) $$\frac{dW}{dt} = - \int_V \left[ \frac{1}{2\mu_0} \frac{\partial}{\partial t} B^2 + \frac{\epsilon_0}{2} \frac{\partial}{\partial t} E^2 + \frac{1}{\mu_0}\vec{\nabla} \cdot (\vec{E} \times \vec{B}) \right] d^3r$$

Now, defining
(eq8) $$u_{em} = \frac{\epsilon_0}{2}E^2 + \frac{1}{2\mu_0}B^2$$
and
(eq9) $$\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}$$
we get rate work is done on the charge density by the fields in a volume V is:
(eq10) $$\frac{dW}{dt} = - \frac{\partial}{\partial t} \int_V u_{em} \ d^3r - \int_{V} \vec{\nabla}\cdot \vec{S} \ d^3r$$

Now of course the work done on the charge density will increase their mechanical energy (kinetic, potential, or whatever). So denoting the mechanical energy density as u_mech, we get:
(eq11) $$\frac{d}{dt} W = \frac{\partial}{\partial t} \int_V u_{mech} \ d^3r$$
thus
(eq12) $$\frac{\partial}{\partial t} \int_V (u_{mech} + u_{em}) \ d^3r = - \int_{V} \vec{\nabla}\cdot \vec{S} \ d^3r$$
Since this has to be true for any volume, this requires:
(eq13) $$\frac{\partial}{\partial t} (u_{mech} + u_{em}) = - \vec{\nabla}\cdot \vec{S}$$

Which is just a local conservation law like the familiar charge conservation law:
(eq14) $$\frac{\partial \rho}{\partial t} = - \vec{\nabla}\cdot \vec{J}$$
So quite generally, if we consider u_em to be the energy density of the fields, and S to be the energy "current" of the fields, then energy is locally conserved.

8. Feb 17, 2010

### JustinLevy

Okay, I needed to split my reply up.

I intended this discussion to focus on classical mechanics (but the QM examples are beautiful, thanks! I'd like to return to that later), so let's try to analyze this classically first.

Basically, because of the local conservation law being in terms of just the fields, and the "background" field is independent of time, it appears to me that energy should be conserved. So the time dependence of the potential is just a gauge issue (at least classically).

So your answer is essentially that just because we have local energy conservation doesn't mean I can demand global energy conservation. I agree if the metric is dynamic. I don't see how this local -> global problem can arise if the metric, and the "size" of universe, is static like it is here.

Are you saying this example shows the local -> global problem is much more general than I realized?

Look at it in fourier space. The particles would have to emit radiation with zero frequency dependence for it to cancel the background field.

That is probably an over simplification, but should get the point across.

The radiation from an accelerated particle has sin(theta) dependence (where theta is the angle from its direction of motion), right? With that symmetry, I don't see how the energy in the fields decreases, since:
$$(E+\delta)^2 + (E-\delta)^2 > 2E^2$$
Yes, the radiated field goes against the background field in some areas, but by symmetry goes with the fields elsewhere. Overall the radiation ADDS energy to the fields.

Without a closed universe, I still don't know how to do this with a point charge. But I can start with a ball of constant charge density gas in a huge capacitor. The energy in the field decreases since the gas also expands as it accelerates.

If it is a solid instead of a gas, I'm not sure how to do the problem. Imagine just two point charges travelling paralel... the magnetic field tries to pull them together. So the mechanical forces to hold an object together are non-trivial. And since Newton's third law doesn't hold in E-M (if you look at the force on charge 1 due to charge 2 and visa versa at the same time, the forces do not need to be equal and opposite), maybe the mechanical forces can be shown to do net negative work or something weird if the particle is accelerating.

EDIT:
Just realized a simple example of how the closed universe can mess with the LOCAL laws.
Consider a uniform charge gas everywhere. By symmetry the electric field is zero, but this violates:
$$\vec{\nabla}\cdot\vec{E} = \rho / \epsilon_0$$

So there may be subtle flaws in a lot of the vector manipulation?

9. Feb 17, 2010

### bcrowell

Staff Emeritus
I think this doesn't work because it's not a discrete Fourier spectrum, of the type that you could use to describe a periodic wave. The Fourier integral is $E=\int a_\omega dE_\omega$. Its interaction with the ambient field Eo is the energy $U_i=E_o \cdot \int a_\omega dE_\omega$. You're essentially arguing that $\int a_\omega dE_\omega$ has a vanishing average component along the axis of the ambient field, because all the field components oscillate symmetrically in both directions relative to that axis. I don't think that's true for a continuous Fourier spectrum. E.g., you can use a continuous Fourier spectrum to represent a delta function, which doesn't average to zero.

No, I don't think that's true. I don't know if you have access to Electricity and Magnetism, by Purcell, but he has some nice diagrams of this at p. 162 and 164 in the 1963 edition. The field of a charge moving at constant velocity has the kind of nice, symmetric field you're thinking of. But after that he shows the field of a charge that's recently been abruptly accelerated from rest. There is a spherical region surrounding the charge, where everything inside the sphere is the space that the effect has had time to propagate to. Within this sphere, the field has the kind of symmetry you're thinking of *but* -- the charge is not at the center of the sphere; it's close to the front. So inside the sphere there is a high-volume region behind the charge where the field is in the right direction to partially cancel your ambient field, and a smaller region in front of it where the fields add constructively.

10. Feb 17, 2010

### Physics Monkey

Haha, I actually think that energy is conserved, it just looks like it isn't. One interesting argument is as follows.

Start with a 3-torus with no electric field. Now imagine creating an electron positron pair by supplying some external energy. Take the positron around the x-cycle on the torus. After annihilating it with the electron and removing the initial energy one is left with an electric field in the x-direction. This is easiest to see if you imagine all the electric flux is collimated between the electron and positron. So charged particles wrapping the torus can change to total electric field around the x-cycle.

Now let's imagine that there is a big electric field, but not so big that it creates large accelerations and generates appreciable radiation. Let's also imagine that there is some overall neutral charge configuration in the system. For simplicity, focus on the motion of one positive charge in the configuration (imagine everybody else is far away). Assuming it starts from rest, this charge moves around the torus in the x-direction one time, and it gains an energy $$\Delta U_q = \frac{1}{2} m v^2 = q E_0 L$$ just using Newton and Lorentz. However, as I argued above, the winding motion of the charge must reduce the global electric field. How much is the reduction? Well, if a charge $$q$$ passes through a gaussian surface it reduces the integral $$\int E \cdot dA$$ by $$4 \pi q$$, so in the simplest approximation we have that $$\Delta E_0 = - 4 \pi q / A \ll E_0$$ where $$A$$ is the area in the y-z plane. The energy of the constant electric field is $$L A \frac{E_0^2}{8 \pi}$$, so a change in the size of $$E_0$$ yields a change in the energy of $$\Delta U_E = L A \frac{E_0}{4 \pi} \Delta E_0 = - q E_0 L$$. Thus in this approximation we find that $$\Delta U_q + \Delta U_E = 0$$ and energy is conserved! It's rather like a capacitor discharging, as bcrowell mentioned.

Clearly I've made a lot of approximations here. In particular, I've worked in some kind of quasi-static approximation. To be more complete one needs to check that radiation, magnetic fields, the motion of other charges, inhomogeneities, etc do not alter the results. Nevertheless, I think at the end of the day energy is conserved. One can say that this ultimately follows from the underlying time translation invariance of the system, despite the appearance of background that seems to violate time translation invariance. Agree?

11. Feb 17, 2010

### JustinLevy

Hmm...
I need to think about all this for a bit to let everything internalize. Too much seems counter-intuitive right now. In particular that the "boundary conditions" in some sense ruining the divergence law for electrodynamics. Because of that, I'm not sure what equations I'm even allowed to trust now!

Does anyone know how to do that local energy conservation proof with point charges? I can't figure out how to do the equivalent of the J -> B and E step when we have point charges.

12. Feb 17, 2010

### Physics Monkey

It's actually not possible to have a net charge density in a closed universe. Intuitively, one should ask where the field lines would go? In the case at hand, one way to prove this is to expand the electric field in periodic plane waves. The integral of the divergence of the these plane waves over the torus is always zero. Another way of saying it amounts to the statement that the integral of $$\nabla \cdot E$$ over a compact space is zero. See the discussion https://www.physicsforums.com/showthread.php?t=376980 for more information on the pathologies of a space filled with charge

13. Feb 18, 2010

### JustinLevy

But I can have the problem appear without a net charge density either.
Consider a "band" of uniform charge density going around the universe. You can also have a band of the opposite charge elsewhere so that there is no net charge density.

Again due to symmetry the electric field along the band is zero, thus violating the divergence condition. Somehow the boundary condition seems to be messing with the local laws!

14. Feb 18, 2010

### Physics Monkey

I don't think I understand exactly the setup you're imagining, but it seems like you don't have enough symmetry to conclude there is no field. It sounds like you're describing something like a capacitor, but maybe I misunderstood you? It is true that boundary conditions matter, but this doesn't conflict with the local laws. The boundary conditions can be seen to enter because of the freedom to add a solution of the homogeneous equation $$\nabla \cdot E = 0$$ to any solution of the equations with sources.

15. Feb 18, 2010

### atyy

If you are looking for a steady state solution with radiation, why not try sinusoidal motion, like an LC circuit?

16. Feb 18, 2010

### bcrowell

Staff Emeritus
This is incorrect. The local laws of physics (like Maxwell's equations) always have the same form, regardless of the curvature and topology of the space. To determine the solution of a differential equation, you need both the equation and the boundary conditions. In your example of the uniformly charged universe, the boundary conditions could be that the field vanishes at infinity, or the boundary conditions could be that the field has a certain nonzero value at infinity.

17. Feb 20, 2010

### pervect

Staff Emeritus

My intuition can imagine this field configuration, and it comes to the conclusion that there should be some background field left behind as you state. It's not a very formal argument, but I can see where you're coming from.

I think the safest answer is to say "I don't know" - but IMO you're asking the right question. If there is some explicit time translation invariance, one can define a global conserved energy using Noether's theorem. Otherwise, it's not clear that there is a globally conserved energy in this situation.

18. Feb 20, 2010

### JesseM

Well, the laws of electromagnetism are time translation invariant in an infinite minkowski spacetime right? And a closed flat spacetime should be exactly equivalent to an infinite flat spacetime where all particles and fields precisely repeat in a periodic way over space.