What is relationship between electromagnetic mass and rest mass?

Question

Is there a direct equation which compares rest mass $m_°$ and electromagnetic mass $m_{em}$?

Nothing on web I found.

$m_{em} = \frac{4 E_{em}}{3c^2}$

4/3 problem

The final solution of the problem was found by Valery Morozov (2011).[36] He gave consideration to movement of an imponderable charged sphere. It turned out that a flux of nonelectromagnetic energy exists in the sphere body. This flux has an impulse exactly equal to 1/3 of the sphere electromagnetic impulse regardless of a sphere internal structure or a material, it is made of. The problem was solved without attraction of any additional hypotheses. In this model, sphere tensions are not connected with its mass, so Poincare hypothesis can resolve the paradox 4/3 in no way

Does it mean if $m_{em}$ is the em mass than total mass would be $m_{°} = \frac{4 m_{em}}{3}$

Answer 1

For the case of a charged hollow sphere the relationship is:

$m_{em}=\frac{4}{3}m_0$

because

$m_{em}=\frac{4}{3}E_{em}/c^2$

The electromagnetic mass depends on the shape you assume for the charged object. In the case above it is assumed the object is a charged, hollow sphere. In general the electromagnetic mass for a charged object producing electric and magnetic fields $E$ and $B$ is:

$m_{em}=\frac{\int{\epsilon_0E \times B}d^3x}{v}$

where the integral is carried out over all space and $v$ is the magnitude of the object's velocity.

To get to a comparison to rest mass, the assumption usually made is that relativistic energy of the object is its electromagnetic energy, then:

$m_0=E/c^2=E_{em}/c^2$

so

$m_{em}/m_0=(c^2/E_{em})\frac{\int{\epsilon_0E \times B}d^3x}{v}$

where the electromagnetic energy is calculated in the usual way from the fields of the object. In general the 4/3 factor is not applicable.

Answer 2

If we integrate the energy and momentum for the field created by a moving charged sphere shell moving with velocity v, the relationship is indeed $\mathbf{p}=\frac{4}{3}m_{em}\mathbf{v}$.

However, this result does not consider the effect of the charge-field interaction, that is, the J·E term, that happens in the shell.

The integral for this term is zero, but the energy is transferred from the field to the charge in the "front" of the sphere and it is transferred from the charge to the field in the back (I call "front" to the part of the sphere aligned with v).

The fact that the field is losing energy in the front and gaining it in the back is equivalent to a displacement of the mass-energy that can be computed as a momentum whose value is:

$$\mathbf{P_{int}}=\frac{1}{c^2}\int (\mathbf{x}-\mathbf{X_0})(-\boldsymbol{J}\cdot \boldsymbol{E})d\boldsymbol{x} $$

When you integrate this term over the sphere it gives you a value of $-\frac{1}{3}m_{em}\mathbf{v}$ that when added to the momentum obtained by Poynting vector integration, leads to the correct result.