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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 $m_{°} = \frac{4 m_{em}}{3}$

Qmechanic
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Anubhav Goel
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    @Omry Electromagnetic mass was initially a concept of classical mechanics, denoting as to how much the electromagnetic field, or the self-energy, is contributing to the mass of charged particles. It was first derived by J. J. Thomson in 1881 and was for some time also considered as a dynamical explanation of inertial mass per se. Today, the relation of mass, momentum,velocity and all forms of energy, including electromagnetic energy, is analyzed on the basis of Albert Einstein's special relativity and mass–energy equivalence. As to the cause of mass of elementary particles, the Higgs mechanism – Anubhav Goel May 22 '16 at 14:31
  • in the framework of the relativistic Standard Model is currently used. In addition, some problems concerning the electromagnetic mass and self-energy of charged particles are still studied. – Anubhav Goel May 22 '16 at 14:32
  • When you apply Special Relativity, the 4/3 term vanishes. This paper from 1962 explains it. http://www.philsoc.org/1962Spring/1526transcript.html – Peter R May 23 '16 at 14:10
  • @PeterR Do I have to read it all? It will then take me some days to progress further. Your paper is of 1962 and my citation in question above is of 2011 – Anubhav Goel May 23 '16 at 15:02
  • The Wikipedia article cited above also states that the mass is resolved when the Relativistic effects are taken into consideration, The 4/3 issue is not new. it goes back to over 100 year and the discrepency was resolved as explained in the above citings. Sometimes, text books incorrectly incude the 4/3 term. – Peter R May 23 '16 at 15:38
  • @PeterR They solved using relativity by increasing $E_{em}$ , I think. So, actual question is as it is. – Anubhav Goel May 23 '16 at 15:48
  • Possible duplicate: http://physics.stackexchange.com/q/80856/2451 – Qmechanic May 26 '16 at 14:35

2 Answers2

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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.

czechmea
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  • You mean , $m_{°} = \frac{4 m_{em}}{3}$ is correct? – Anubhav Goel May 26 '16 at 14:33
  • I asked for relation between electromagnetic mass and rest mass and not EM mass and EM energy. – Anubhav Goel May 26 '16 at 14:34
  • I've given the general formula for the electromagnetic mass, you just need to compare this to the rest mass of the particle... – czechmea May 26 '16 at 14:50
  • I've added to my answer to make it clearer. – czechmea May 26 '16 at 14:56
  • In "$m_0=E/c^2=E_{em}/c^2$ " , how did you considered Energy of rest mass is equal to electromagnetic mass? – Anubhav Goel May 29 '16 at 09:16
  • If I put , $E_{em}/c^2 = m_o$ in first equation , my answer and your answer are opposite. Which is wrong one? – Anubhav Goel May 29 '16 at 09:20
  • In $m_{em}=\frac{\int{\epsilon_0E \times B}d^3x}{v}$ if I put v= 0, I get $m_{em}= ∞$ . It's ambiguous. – Anubhav Goel May 29 '16 at 09:27
  • We know from relativity that $m_0=E^2/c^2$ (using your notation). So, in order to be consistent with relativity, since we have calculated the energy to be $E_{em}$ we have to set that $=E$. Note also that, strictly speaking, the energy we're using is the electrostatic energy (since we aren't considering the magnetic field in the energy calculation). – czechmea May 29 '16 at 11:47
  • The correct expression is $m_0=3m_{em}/4$. (For the case of a charged spherical shell) – czechmea May 29 '16 at 11:48
  • You won't get an infinity if you carry out the integration over the fields of the object. For example, for the case of a charged spherical shell the integral is also proportional to $v$ so the $v$'s cancel. – czechmea May 29 '16 at 11:50
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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.

Sergio Prats
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