I Self Energy?

[Meir Achuz]

TL;DR

This shows thata point charge in classical electromagnetism has no 'self energy'.

This PDF shows that a point charge in classical electromagnetism has no 'self energy'.

 

[renormalize]

This PDF shows that a point charge in classical electromagnetism has no 'self energy'.

In your attachment you assert:

This is certainly true for a point "test charge" with infinitesimal $q\,$, which by assumption does not contribute to the potential $\phi$. But can you cite a textbook reference that justifies and agrees with your last two sentences above for a point charge in the case where $q$ is finite?

 

[Meir Achuz]

"Chapter2 Electrostatics so,ifyouhavesetthereferencepointatinfinity, W=QV(r)."
You added the word "infinitesimal".
"W= 1/ 8π\\epsilon_0 sum n i=1 n j=i qiqj/rij (2.41) (wemuststillavoidi=j,ofcourse)."

 

[renormalize]

"Chapter2 Electrostatics so,ifyouhavesetthereferencepointatinfinity, W=QV(r)."
You added the word "infinitesimal".
"W= 1/ 8π\\epsilon_0 sum n i=1 n j=i qiqj/rij (2.41) (wemuststillavoidi=j,ofcourse)."

Can you please clarify the citation by providing the author, title and page number of the text where this quotation appears? Thanks!

 

[Meir Achuz]

I thought you would recognize Griffiths. Actually, the shoe is on the other foot. Can you cite a textbook reference that justifies and agrees for an electron in the case where its charge is infinitesimal??

 

[renormalize]

Actually, the shoe is on the other foot. Can you cite a textbook reference that justifies and agrees for an electron in the case where its charge is infinitesimal??

Absolutely. From Jackson, Classical Electrodynamics 2nd ed. pg. 28 (emphasis added by me):

So by definition, the electrostatic field $\vec{E}$ in eq.(1.1) ignores the effect of the charge $q$ on itself, but only in the limit $q\rightarrow 0$ (i.e., infinitesimal charge). This same restriction holds for the relation between the work and the potential $U$:$$\vec{E}=-\nabla U\Rightarrow\vec{F}=-q\nabla U\Rightarrow\text{work}\equiv\int\vec{F}\cdot d\vec{r}=-q\int\nabla U\cdot d\vec{r}=-q\,U+\text{const.}$$On the other hand, for $q$ a finite charge, $U$ will contain a contribution from the charge's own (self) $\vec{E}$-field, which is indeed divergent if that charge is concentrated at a point.

 

[Meir Achuz]

Don't ignore the footnote, which I have attached. Jackson points out that infinitesimal charge is mathematical and NOT PHYSICALLY POSSIBLE, The point here is that to define E, even a small charge can polarize the source of E, making the force equation bilinear. It is completely different for energy where Jackson knows the charge cannot be infinitesimal. (See Section I.1. See also pages 40,41, which is more like Griffiths. Jackson use i<j

 

[renormalize]

See also pages 40,41, which is more like Griffiths. Jackson use i<j

What Jackson says there regarding discrete charge distributions is (my emphasis added):

So the self-energy terms are there in principle but are dropped by convention. This is allowed because they are (infinite) constants that simply set the zero of the energy $W$ and are thus ignorable.
That situation changes when we consider continuous charge distributions:

because the integrand in (1.52) includes the points $x=x^{\prime}$; i.e., self-energy is inevitably included in the double integral that defines $W$.
Feynman comments on how this affects the energy of a point charge in his lecture https://www.feynmanlectures.caltech.edu/II_08.html:

So in classical electrodynamics it seems to be unavoidable that infinite self-energy must accompany any truly point charge.

 

[Meir Achuz]

This is getting silly. Neither of us will be convinced, but you keep putting up half truths, half wrong.
I will try again. You neglect what you don't want. Jackson's equations 1.47, 1.48, 1.49, 1..50 clearly show there is no i=j term. For some reason, he says it in words for Eq,. (.51). You, not he, adds, "So the self-energy terms are there in principle but are dropped by convention." His equations clearly show that it is not there by absence, not "convention". You made up your "convention". When you quote someone don't make up what they didn't say.
My original post was limited to discrete charges. Do you find anything wrong there? I will respond to the question of going from discrete to integrals in another post.

Reference: https://www.physicsforums.com/threads/self-energy.1083048/

 

[A.T.]

Jackson's equations 1.47, 1.48, 1.49, 1..50 clearly show there is no i=j term. For some reason, he says it in words for Eq,. (.51).

Not for "some reason", but simply because in equation 1.51 (as it's written) there is a i=j term, so he needs to specify that it should be ignored. Whether you call that a "convention" or not doesn't really matter.

 

[renormalize]

My original post was limited to discrete charges.

Yet the very first sentence of your original post was:

TL;DR: This shows thata point charge in classical electromagnetism has no 'self energy'.

which contradicts the Feynman lecture I quoted and is clearly wrong.
Are you espousing a personal theory?