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In classical (c-number) theory, it depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory or Lorentz microscopic theory with single EM field, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian/Hamiltonian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can, because potentials $\varphi, \mathbf A$ can have arbitrarily negative value.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

But it is not usual because usually the interaction energy is much lower than sum of rest energies of the particles $mc^2$. Total energy of the system being negative would mean negative mass, a very strange concept.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \int \sum_i \sum_k' \mathbf E_i \cdot \mathbf E_k~ d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear.

In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative. Hypothetically, the particles can be so close that magnitude of $W$ is higher than total rest mass of the particles and then total energy of the system is negative. But such a system has very strange behaviour (negative mass) and AFAIK hasn't been observed.

In classical (c-number) theory, it depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory or Lorentz microscopic theory with single EM field, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian/Hamiltonian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can, because potentials $\varphi, \mathbf A$ can have arbitrarily negative value.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

But it is not usual for total energy to be negative because usually the EM interaction energy is much lower than sum of rest energies of the particles $mc^2$. Total energy of the system being negative would mean negative mass, a very strange concept.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \int \sum_i \sum_k' \mathbf E_i \cdot \mathbf E_k~ d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear.

In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative. Hypothetically, the particles can be so close that magnitude of $W$ is higher than total rest mass of the particles and then total energy of the system is negative. But such a system has very strange behaviour (negative mass) and AFAIK hasn't been observed.

In classical (c-number) theory, it depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory or Lorentz microscopic theory with single EM field, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian/Hamiltonian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can, but it is not usual because usually the interaction energy is much lower than rest energy $mc^2$ of particles. Total energy of the system being negative would be negative mass, a very strange concept.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \sum_i \sum_k' \int \mathbf E_i \cdot \mathbf E_k d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear.

In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative. Hypothetically, the particles can be so close that magnitude of $W$ is higher than total rest mass of the particles and then total energy of the system is negative. But such a system has very strange behaviour (negative mass) and AFAIK hasn't been observed.

In classical (c-number) theory, it depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory or Lorentz microscopic theory with single EM field, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian/Hamiltonian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can, because potentials $\varphi, \mathbf A$ can have arbitrarily negative value.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

But it is not usual because usually the interaction energy is much lower than sum of rest energies of the particles $mc^2$. Total energy of the system being negative would mean negative mass, a very strange concept.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \int \sum_i \sum_k' \mathbf E_i \cdot \mathbf E_k~ d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear.

In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative. Hypothetically, the particles can be so close that magnitude of $W$ is higher than total rest mass of the particles and then total energy of the system is negative. But such a system has very strange behaviour (negative mass) and AFAIK hasn't been observed.

It depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \sum_i \sum_k' \int \mathbf E_i \cdot \mathbf E_k d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear. In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative.

In classical (c-number) theory, it depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory or Lorentz microscopic theory with single EM field, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian/Hamiltonian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can, but it is not usual because usually the interaction energy is much lower than rest energy $mc^2$ of particles. Total energy of the system being negative would be negative mass, a very strange concept.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \sum_i \sum_k' \int \mathbf E_i \cdot \mathbf E_k d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear. 

In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative. Hypothetically, the particles can be so close that magnitude of $W$ is higher than total rest mass of the particles and then total energy of the system is negative. But such a system has very strange behaviour (negative mass) and AFAIK hasn't been observed.