Energy Density of Electric Fields: Is it the Sum of All Electrons?

In summary, the energy density of an electric field is not simply the sum of all the individual electron fields surrounding it. This is due to the fact that the energy density goes as the square of the field, and the fields from different sources add as vectors. This leads to a singularity when considering a point charge, and classical electrodynamics breaks down. To solve this issue, a classical particle of finite extension is considered instead. The energy of an individual electron's field is still not fully understood, and its size and energy contribute to the overall mass/energy of the electron. Bringing together many electrons to create a more energetic electric field also involves new field energy from the work done to bring them together. There is no expected percentage for this
  • #1
Herbascious J
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I understand that the energy of an electric field arises from the work put into gathering the electrons together to create the field. Bringing electrons close together requires energy because they naturally want to repel. This potential energy is stored in the field itself and the field has an energy density. I am assuming that the energy field of an individual electron must also have a small energy density associated with it. My question is; is the energy density of an electric field actually just the 'sum' of all the little electric fields surrounding each individual electron? Or does the resultant electric field have more energy than the net sum of all the electrons field energies? Implying that the extra energy is from the work done to accumulate the electrons.
 
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  • #2
Herbascious J said:
is the energy density of an electric field actually just the 'sum' of all the little electric fields surrounding each individual electron?

No. The energy density goes as the square of the field.
 
  • #3
If there's more than one source, the energy density is proportional to the square of the sum of the fields from all the sources. For some locations, this is larger than the sum of the squares of the fields. For other locations, it's smaller. It depends on how the fields add, as vectors.
 
  • #4
You are asking a very difficult question here. In fact at this very point indeed classical electrodynamics breaks down. It asks nothing less then after the "self-energy" of the electron. In the classical picture usually one considers the electron to be a point particle. Let's consider one electron sitting at rest in the origin of the coordinate system. Its electric field is the Coulomb field (written in SI units),
$$\vec{E}(\vec{r})=-\frac{e \vec{r}}{4 \pi \epsilon_0 r^3}.$$
Now the very argument you bring in #1 shows that for continuous charge distributions at rest the total energy of the electrostatic field is given by
$$\mathcal{E}=\frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 r \vec{E}^2(\vec{r}),$$
i.e., the electric field's energy density is
$$u(\vec{r})=\frac{\epsilon_0}{2} \vec{E}^2(\vec{r}).$$
But now comes the trouble, if you want to apply this equation to the case of a pointlike electron! Plugging in the Coulomb field gives the energy density
$$u(\vec{r})=\frac{e^2}{32 \pi \epsilon_0 r^4}.$$
This is highly singular at the origin, i.e., the location where the point charge sits. To regularize it we have to integrate over all space taking out a little sphere of radius ##a##. Then using spherical coordinates, we get
$$\mathcal{E}_{a}=\frac{e^2}{8 \pi \epsilon_0} \int_a^{\infty} \mathrm{d} r \frac{1}{r^2} = \frac{e^2}{8 \pi \epsilon_0 a}.$$
The total energy in the Coulomb field of a point charge thus indeed is
$$\mathcal{E}=\lim_{a \rightarrow \infty} \mathcal{E}_a =\infty.$$
The regularized energy would be correct, if we'd consider the electron as a spherical shell of radius ##a## carrying the electron's charge. So the assumption of a mathematical point charge leads to trouble in classical electrodynamics, and that already for the most simple static case!

In fact, the entire issue becomes even worse when considering a point charge moving in an arbitrary electromagnetic field: Then it's accelerates and radiates off electromagnetic radiation. This carries off energy from the particle's kinetic energy, which means it should be damped. Trying to take this into account for a strict point charge leads to an equation (the Lorentz-Abraham-Dirac equation) which gives rise to completely unphysical motions of the electron: Rather than simply radiating off some energy as e.m. waves and being damped the equation also leads to the prediction that the electron self-accelerates (i.e., out of nothing even without external fields the electron accelerates!) and there are actions from the future too (runaway solutions and preacceleration).

The solution within classical physics is to consider a classical particle of finite extension, i.e., not a point charge but something like a little charged sphere. One has to take into account some stresses to keep the charges together too. The equation of motion is, however, now not only a differential equation in time as is usual in mechanics (basically Newton's ##\vec{F}=m \ddot{\vec{x}}##) but a socalled differential-difference equation. There's a vast literature on the subject. A very nice pedagogical discussion of the model of a charged spherical shell as a model for an electron of finite spatial extent is

https://aapt.scitation.org/doi/abs/10.1119/1.3269900
 
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  • #5
Thank you for the response. This is impressive. So there is a mathematical rationale which suggests that the diameter of an electron is not point-like and this is because the energy density of the field approaches infinity as the radius of the particle becomes zero?? Obviously that can't be correct (so my intuition tells me). It looks like I will be needing to do some self study. My questions now are, 1). Do we have a good idea of what the energy of an individual electron's field is? 2.) How does this imply an expected radius of the sphere compared to the size of other particles, the plank length and the pauli exclusion principle, etc.? 3.) How does this energy level (assuming a sphere structure) contribute to the overall mass/energy of the electron and what portion of it's total energy content. Taken all of this to be true, my original question restated would look like; 4.) If we have a good idea of what the energy of an individual electron's electric field is, and we then bring together many electrons to build up a more energetic electric field (like in a capacitor?) how much of this resultant field is from the individual particles' fields, and what proportion is new field energy supplied by the work done to bring the particles together? Is there an expected percentage? Am I comparing apples to oranges somehow? (I don't expect that there are direct answers to these questions, and that I will have to do some reading. I am doing the best I can to frame this problem, and I am prepared to have it restated and redirected in a way that is more correct and useful for learning.) Thank you!

Edit: I am also curious how an electrons' wavelength (as understood in quantum mechanics) would have any relation to, or comparable size to, the radius of the 'shell' we imagine an electron might be. (perhaps this is for another thread).
 
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  • #6
The resolution of the apparent paradox is, of course, quantum theory. There's no way to understand the microscopic behavior from the underlying fundamental microscopic dynamics in terms of classical physics. On the most fundamental level, as we understand it today, instead of using classical electrodynamics, we have to use quantum electrodynamics (QED), where both the particles (in the minimal version electrons and positrons) and fields (here the electromagnetic field) are described by quantum fields.

In quantum theory, particularly relativistic quantum theory, the idea of a point particle becomes obsolete, because you cannot localize a particle at a point more precisely than its de Broglie wavelength, ##\hbar/(mc)##. The reason is that if you try to localize the particle better than that you need a large uncertainty of momentum that the typical energy scales become ## \gtrsim 2 m c^2##, with ##m c^2 \simeq 511\; \text{keV}## the electron mass, at which energies you'd rather not localize the electron better but create electron-positron pairs.

On the other hand, QED also shows that the classical approximation works well for macroscopic systems, which are described by continuum mechanics with (usually smooth) charge-current distributions of finite extent.
 
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  • #7
vanhees71 said:
The resolution of the apparent paradox is, of course, quantum theory.

As a final inquiry. Is there a convenient or approximate 'radius of the electron particle sphere' so-to-speak that we use to describe the size of an electron physically that allows all the math and energy to make sense? Something related to quantum mechanics? I'm imagining there are some creative guesstimates that pronounce that we are not talking about a point like particle, but instead something that has boundaries to it's 'size'. I'm putting all this in quotes only because I realize at this scale size itself becomes difficult to define.
 
  • #8
On the most fundamental level an electron is described as a spin-1/2 Dirac spinor field. As far as we know today the electron is an elementary particle, i.e., it is not composed (in the sense of a bound state) of other entities. Thus there's no more and no less than the description as the quanzized spin-1/2 Dirac spinor field, coupling (on the tree level) to the electromagnetic and the weak gauge bosons as well as (via Yukawa couplings) to the Higgs field. That's all you can say about electrons within the standard model, but it's a lot! It describes in fact correctly everything we know about the electron and in high precision. There's in principle no need for a classical model of the electron which was so pupular around 1910-1920 when electron theory was a hot topic.

For practical purposes, however, a classical picture is still a good description. E.g., in accelerator physics usually electrons and other particles are treated as if they are classical particles or in the sense of classical fluid (plasma) physics. The latter description is less problematic than the point-particle approach. For literal point particles you have all these problems.

An analysis of the issue in terms of classical models as extended objects is pretty complicated. I like most a recent paper by Medina [1] using the (very artificial!) model of a charged Born rigid body to derive the equations of motion including self-consistently the radiation reactions. This shows that the point-particle limit is inconsistent for the fully consistent equation, but you can make an approximation to it, which finally leads to the Landau-Lifshitz equation of motion, which is an approximation of the old Abraham-Lorentz-Dirac equation, which is plagued with all the unphysical features like preacceleration and run-away solutions. All this is cured in the Landau-Lifshitz equation of motion. Similar results also come out of reinterpreting continuum-mechanical models (relativistic fluid dynamics) in the sense of particle motion of its constituents.

[1] https://arxiv.org/abs/physics/0508031
https://iopscience.iop.org/article/10.1088/0305-4470/39/14/021/meta

Also see

https://doi.org/10.1119/1.3269900
 

1. What is energy density of electric fields?

The energy density of electric fields is the amount of energy per unit volume that is contained in an electric field. It represents the potential energy that is stored in the electric field and is measured in joules per cubic meter (J/m^3).

2. Is energy density of electric fields the same as electric field strength?

No, energy density of electric fields and electric field strength are two different concepts. Electric field strength refers to the force per unit charge that a charged particle experiences in an electric field, while energy density of electric fields refers to the amount of energy stored in the electric field.

3. How is energy density of electric fields calculated?

Energy density of electric fields can be calculated using the formula U = 1/2 * ε0 * E^2, where U is the energy density, ε0 is the permittivity of free space, and E is the electric field strength.

4. Is energy density of electric fields the sum of all electrons?

No, energy density of electric fields is not the sum of all electrons. It is a property of the electric field itself and is not dependent on the number of electrons present. However, the energy density can be affected by the presence of charged particles in the electric field.

5. How does energy density of electric fields affect the behavior of charged particles?

The energy density of electric fields plays a crucial role in the behavior of charged particles. It determines the amount of potential energy that is available for the charged particles to move and interact with each other. The higher the energy density, the more force the charged particles will experience and the more they will be affected by the electric field.

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