# Composition of energy of electron? [closed]

Mass energy of electron is 0.510 998 9461MeV/ c2

1) Does it include energy of electron due to electric field too?

2) If yes, how much?

I am more interested in second part.

Nevertheless, you could use the above reasoning to claim that a charged electron actually has a lower energy than an uncharged one would. Now there's an unexpected result :-)

Can anyone tell how much exact less energy it will have?

Or is it just, we can't because we don't have uncharged electron.

## closed as unclear what you're asking by ACuriousMind♦, John Rennie, user36790, AccidentalFourierTransform, CuriousOneMay 19 '16 at 2:32

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• @AnubhavGoel The relativistic mass is an obsolete concept, mass/inertia is invariant under Lorentz transformations – Oswald May 18 '16 at 11:20
• The charge is an invariant quantity. Then how could it emit more electric field? The mass varying equation has no physical significance. The best way to see it as the relativistic momentum. For a relativistically moving charged particle, it's the momentum that varies relativistically:$p=mv_0=mv/\gamma$. That variation is solely due to the difference between proper velocity and ordinary velocity – UKH May 18 '16 at 12:33
• @ACuriousMind and others: whats not clear? – Anubhav Goel May 20 '16 at 5:13
• @JohnRennie: Is it clear now? – Anubhav Goel May 25 '16 at 10:08
• @MAFIA36790 : What is unclear? – Anubhav Goel Jun 2 '16 at 7:52

There is the concept of the self energy of the electron that can be described by its electric field. The model, which goes back to classical physics, can be described as a spherically distributed charges (that equal the charge of the electron) at infinity and the work required to bring the charges to the radius of the electron. The calculation would show the total energy of the electron due to its own charge and ensuing electric field. The standard model describes the electron as a point charge. If you do this calculation to a zero point, the electron would have infinite mass, however we know what it should be. The calculated radius however is much larger that the upper limit of the electron radius as measured in high energy probe experiments into subatomic particles. The electron should have much more mass due to its own self energy of its own electric field. This is a big puzzle and renormalization techniques are used to arrive at the mass of the electron. Feynman did a lot of work in this area and spends quite some time describing it in his Nobel Prize speech.

• Has some result been founded now? – Anubhav Goel May 18 '16 at 14:08
• Modern quantum field theory has some answers. There is the concept of the polarization of the vacuum surrounding the electron due to the filed of virtual particle pairs that surround the electron. This tends to smear out the charge to a larger radius. Uncertainty principle at work. – Peter R May 18 '16 at 14:13
• can you please give a link for *renormalization techniques*😃 – Anubhav Goel May 18 '16 at 16:25
• This Wikipedia article explains the issues fairly well, especially for the electron, although the math behind it may be somewhat advancced for you en.wikipedia.org/wiki/Renormalization – Peter R May 18 '16 at 17:47
• *" The electron should have much more mass due to its own self energy of its own electric field. This is a big puzzle and renormalization techniques are used to arrive at the mass of the electron. " * Can you please tell how much more mass does it have than self energy of electric field? – Anubhav Goel Jul 4 '16 at 15:30

To begin with, I can't think of any reason why an increase in the mass of an electrically charged object should affect the magnitude of the electric field it produces. You can easily check that $E=kq/r^2$ doesn't depend on $m$. Moving charges do produce magnetic fields, but this is a different matter (and the magnitude of the magnetic field has nothing to do with the mass of the object, neither).

On the other hand, as it was noted in the comments, mass is a Lorentz scalar, then a relativistic invariant quantity (i. e., it is the same on any reference frame). In fact, the Minkowski norm (length) of the 4-momentum is $p^{\mu} p_{\mu}= -m^2c^2$. Only the components of a 4-vector are modified when subject to Lorentz transformations, never its norm.

Edit: To sum up, as far as I know charge/electric field doesn't affect to the mass/energy of any charged object, is it moving or at rest. By "electric energy" you might be referring to the electric potential energy or to the electric field density, you can see that both concepts are unrelated with mass.

• That was my first question does energy of electron include electric field energy as well. If electric energy comes from mass energy , increase in mass can cause increase in charge and hence electric field. – Anubhav Goel May 18 '16 at 12:20
• Where did you get that "electric energy" (I don't exactly know what are you referring to) comes from mass? There's no increase in mass, that's what I just argued. – David Herrero Martí May 18 '16 at 12:25
• That's what I asked does above energy I mentioned in question includes energy in electric field around charge. – Anubhav Goel May 18 '16 at 12:28
• @AnubhavGoel Does the edit clarify my answer/satisfy your doubts? – David Herrero Martí May 18 '16 at 12:37
• Nop! 😭 It would be better if you can answer my first question – Anubhav Goel May 18 '16 at 13:31

2) Does it include energy of electron due to electric field too?

We do not know. If the electron was composed of charged elementary parts, its apparent inertial mass would be greater than sum of inertial masses of the parts. This is because such parts will act on each other with electromagnetic forces and in accelerated motion, sum of these internal forces is not zero, but can be expressed as $-m_{em}\mathbf a$, where $m_{em}$ is a positive number. This force can be put next to the expression $m\mathbf a$ in the equation of motion to get the approximate equation

$$(m+m_{em})\mathbf a = \mathbf F_{ext}.$$

Hence $m_{em}$ is called electromagnetic mass(for slow motions of charged sphere, this was derived by Lorentz). The apparent mass $m+m_{em}$ will be greater than sum of individual masses $m$.

If the electron was a point and therefore not composed of parts, there would be no such effect, for there are no internal forces due to any parts.

It is necessary that at least part of apparent mass is non-electromagnetic; for electron made of parts, non-electromagnetic forces need to be present to keep the parts of electron together and these forces probably have some similar effect on the apparent mass; for point electron, there are no mutual electromagnetic forces at all, so they cannot result in any change in apparent mass.

• Please see the reference to the self energy of the electron, which is an elementry particle with no other stucture. – Peter R May 18 '16 at 19:17
• Which one? What is the point? – Ján Lalinský May 18 '16 at 19:31
• I don't understand the comment. Are you saying that the electron does not interact with its own electric field? – Peter R May 18 '16 at 19:41
• No, I was asking which reference did you mean and why do you suggest it. Still, to answer your question, in case of point electron in vacuum, I do not think it makes sense to talk about self-interaction, the common rationale for it (Larmor formula for radiated energy, limiting radius of extended particle to 0) is misguided. Self-interaction only makes sense for extended objects. For point-particle, it can also make sense if it is meant to be effective description of interaction with actual material medium. But not if the particle is alone in vacuum. – Ján Lalinský May 18 '16 at 20:27
• How do you explain electromagnetic radiation from accelerating electrons in space? – Peter R May 18 '16 at 21:52