Why Does Feynman Use a Single Electron Charge for Dipole Calculations?

[Seidhee]

Hello,

I am reading the volume 2 of the Feynman's Lectures on Physics, and something is bothering me when he calculates the dipole moment of a single atom induced by an extern field :

https://books.google.co.uk/books?id=uaQfAQAAQBAJ&pg=SA11-PA2&lpg=SA11-PA2&dq=feynman+dipole+single+atom&source=bl&ots=6nmZCqHDMk&sig=cd6wOGCKh9E9227ZX-a8KKhTqlM&hl=fr&sa=X&ei=NuI8VbHdI8v9UOSvgOAD&ved=0CDEQ6AEwAQ#v=onepage&q=feynman%20dipole%20single%20atom&f=false

Indeed, he states that : " p = q_ex "

But why ? I would use in general : " p = Zq_ex " where Z is the number of electrons in the atom.

x is the displacement of the center of charges of the electrons, and thus x is also the displacement of each electron.

Could you explain his reasoning ? It is not the first time he uses a single electron charge instead of Z in his calculations, and I do not understand.

Thanks.

PS : And q_e is obviously equal to the charge of one electron and note equal to Z*(charge), because he uses then e² = q²_e/4piε₀ (we can also see it in the mechanical equation of motion).

Moreover, when he the does his calculations for the Helium atom, he keeps using q_e = charge of one electron and does not use a new value for Z. And that is exactly what I find weird: he gets the good results event without having taken Z = 2.

 

[Seidhee]

Nobody has an idea ? :)

 

[Omega0]

Hello,
I am reading the volume 2 of the Feynman's Lectures on Physics, and something is bothering me when he calculates the dipole moment of a single atom induced by an extern field :
https://books.google.co.uk/books?id=uaQfAQAAQBAJ&pg=SA11-PA2&lpg=SA11-PA2&dq=feynman+dipole+single+atom&source=bl&ots=6nmZCqHDMk&sig=cd6wOGCKh9E9227ZX-a8KKhTqlM&hl=fr&sa=X&ei=NuI8VbHdI8v9UOSvgOAD&ved=0CDEQ6AEwAQ#v=onepage&q=feynman%20dipole%20single%20atom&f=false
Indeed, he states that : " p = q_ex "
But why ? I would use in general : " p = Zq_ex " where Z is the number of electrons in the atom.
x is the displacement of the center of charges of the electrons, and thus x is also the displacement of each electron.
.

I think that you have an idea about an atom as follows: There is a bunch of protons in the kernel and there is another bunch of electrons around it. If you apply an electric field the electrons move to the "plus" of the field. Reality is much more complex. Atoms are subjects to the quantum mechanics and this means basically the following: Say you have Hydrogen. The electron feels fine in its states but it is alone. If you apply an electric field the electron and the proton will be influenced and the electron will move a bit from its mean radial distance to the core. Nevertheless, the electron is without any shelter, it will feel the force strongly, so you will have a certain dipole moment.
Now let's take Helium. Regarding your critics you expect now both electrons moving to one side and having twice the dipole moment - but you didn't take into account that they feel pretty well in their state and that they have mighty repulsive forces against each other. The more they move around a lot. Helium is normally not impressed by an electric field. Otherwise it would be chemically very active but the contrary is valid. See: Electronegativity in chemistry.

 

[Seidhee]

I would accept the justification by quantum mechanics, but then it seems very awkward to try to explain this result by classical electromagnetism like Feynman did, throwing arbitrarily the factor Z in order to get the good result.

Moreover, even one is reasoning with the repulsion of the electrons (and the more they move around), one must not forget that the nucleus is compact and thus still feels the force ZqE.
Whatever, if the only way to get the good result is thanks to quantum mechanics, I think Feynman should have made a note or something (like he did when he calculated the energy levels of the hydrogen atoms), but he did not.

 

[blue_leaf77]

Atoms and molecules are quantum objects. According to the postulate of QM, any observable/variable such as displacement cannot be determined with perfect accuracy, there is always nonzero chance that one measurement result will differ from another measurement. In this regard, one does not normally say "displacement" in its actual meaning, as a result of the uncertainty in QM it's more practical to talk about the mean value (or expectation value) of an observable, in this case displacement.

Unfortunately I couldn't get access on that book so I couldn't really say what the author referred to. But as my statement above says, x in that expression would very likely mean the expectation value taken w.r.t the single/multi particle state.

 

[Seidhee]

Yep, I know quantum mechanics. Thanks for your answer anyway. My conclusion is then that Feynman tried to "prove" a quantum result using classical properties, by using the good factors to get the good numerical results.

 

[blue_leaf77]

And by the way one should be careful when trying to calculate the classical correspondence of the electric dipole moment of an atom. Not all cases of electric dipole moment computed using QM result in a linear dependency on the displacement. For the nondegenerate and parity-definite states, such as the ground state of H atom, the induced dipole moment is quadratic in displacement.

 

[Omega0]

I like that book - but it is really basic in a "theoretical sense". The depth of insight into the practical parts of electrodynamics is much above what you've been told in say 2 years in the university but from a theroetical point of view it is lightyears away from before getting your master degree. Little examples: Not mentioning differential forms (sure, due to the publishing date), only cartesian coordinates, no insight into PDEs (search for boundary conditions!) and so on. For a deep classical view of electrodynamics (means: no differential forms) I still love J.D. Jackson. He's been going so very much above the Feynman but isn't such an easy read... I love it.