Paradox regarding energy of dipole orientation

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gabbagabbahey
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Re: "paradox" regarding energy of dipole orientation

Sam,
I thought you could work out problems with that involve M in two ways:

Method 1:
Given M and the free current:
- Calculate the bound current
- Calculate the magnetic field using these current densities
- Calculate whatever you need using the magnetic field now

method 2:
Given M and the free current:
- Calculate H
- Calculate whatever you need using B and H now

I thought these two methods (use the bound current, or use H) were equivalent. In particular, when I had to take an EM course, I remember them requiring us to do many problems using both methods to apparently drive home that either way worked. So shouldn't I get the same energy either way?

What you brought up seems to clearly answer no. It seems obvious now, but I will need to read more to jive these two thoughts in my head. This however seems to raise another issue since using the relations already derived, and given that H is defined as [itex]H= \frac{1}{\mu_0} B - M[/itex] we arrive at the relation between B_external and M inside the sphere as [itex]H= \frac{1}{\mu_0} B_{ext} + (\frac{2}{3} M) - M = \frac{1}{\mu_0} B_{ext} - (\frac{1}{3} M)[/itex]. This seems to raise another "magnitude of the energy" issue.

Thanks for the input. I think you've sketched together enough that I can see all the answers are in your suggestions and examples there ... I just need to sort through this some more. At least now it is abundantly clear how the magnetized and non-magnetized situations arrive at different signs.
My previous post was in error (I hadn't fully read all of the posts to see that you were in fact having a very different issue:redface:); Sam's resolution of the "paradox" is correct. For a detailed discussion of it, see R.H. Young, Am. J. Phys 66, 1043 (1998) --- In which Young answers Griffiths' question on why the Hamiltonian is [itex]-\vec{\mu}\cdot\vec{B}[/itex]--- and references cited therein.
 
Meir Achuz
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Re: "paradox" regarding energy of dipole orientation

This the whole point. [itex] F =\nabla (\mu \cdot B) [/itex] is the correct force (with the right sign)
on the dipole when the Lorentz force is applied on a circular current !

The author assumes a mysterious sign flip (stemming from elementary
particle physics). There are however no sign flips. The sign is always like this.
There are no mysterious sign flips in the force.
Read it again.
The force on a magnetic dipole is always [itex] F =\nabla (\mu \cdot B) [/itex].
But the energy on a current loop where the current is kept fixed is given by
[itex]U=\mu \cdot B[/itex], with the force [tex]F=\nabla U[/tex]. The energy of a permanent magnetic dipole in a magnetic field is given by [itex]U=-\mu \cdot B[/itex], with the force [tex]F=-\nabla U[/tex].
The force [itex] F =\nabla (\mu \cdot B) [/itex] for a permanent magnet cannot be derived from the Lorentz force because a permanent magnet is not due to current loops but to the electron's magnetic moment, which is unrelated to current loops.
 
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Hans de Vries
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Re: "paradox" regarding energy of dipole orientation

Read it again.
The force on a magnetic dipole is always [itex] F =\nabla (\mu \cdot B) [/itex].
But the energy on a current loop where the current is kept fixed is given by
[itex]U=\mu \cdot B[/itex], with the force [tex]F=\nabla U[/tex].

The force is per definition [tex]F=-\nabla U[/tex] and there is no sign change which would
violate the conservation of energy. A static magnetic field is conservative and
can only transform one kind of kinetic energy into another, in this case rotational
kinetic energy into translational kinitic energy as described in post #75.


The energy of a permanent magnetic dipole in a magnetic field is given by [itex]U=-\mu \cdot B[/itex], with the force [tex]F=-\nabla U[/tex].
The force [itex] F =\nabla (\mu \cdot B) [/itex] for a permanent magnet cannot be derived from the Lorentz force because a permanent magnet is not due to current loops but to the electron's magnetic moment, which is unrelated to current loops.

According to the Dirac equation there IS a current loop associated with the electron's
magnetic moment. See for instance: Sakurai, Advanced Quantum Mechanics eq(3.205)
and further in the section on the Gordon Decomposition.

A paper which discusses this in the context see:
http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf" [Broken]
Abstract said:
According to the prevailing belief, the spin of the electron or some other
particle is a mysterious internal angular momentum for which no concrete
physical picture is available, and for which there is no classical analog.
However, on the basis of an old calculation by Belinfante [Physica 6 887
(1939)], it can be shown that the spin may be regarded as an angular
momentum generated by a circulating flow of energy in the wave field
of the electron. Likewise, the magnetic moment may be regarded as
generated by a circulating flow of charge in the wave field. This provides
an intuitivelyl appealing picture and establishes that neither the spin nor
the magnetic moment are “internal” — they are not associated with the
internal structure of the electron, but rather with the structure of the field.
Furthermore, a comparison between calculations of angular momentum in
the Dirac and electromagnetic fields shows that the spin of the electrons
is entirely analogous to the angular momentum carried by a classical
circularly polarized wave.
see equation (22)


Regards, Hans
 
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Hans de Vries
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Re: "paradox" regarding energy of dipole orientation

According to the Dirac equation there IS a current loop associated with the electron's
magnetic moment. See for instance: Sakurai, Advanced Quantum Mechanics eq(3.205)
and further in the section on the Gordon Decomposition.

A paper which discusses this in the context see: http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf" [Broken]

Regards, Hans

For an intuitive picture see the introductory chapter of my book (section 1.9)
http://physics-quest.org/Book_Chapter_EM_basic.pdf

Or the chapter on the Gordon decomposition itself (section 18.2 and 18.3)
http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf


Regards, Hans
(P.S. Some work in progress here regarding the decomposition of the axial current I see now)
 
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Meir Achuz
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Re: "paradox" regarding energy of dipole orientation

The force is per definition [tex]F=-\nabla U[/tex] and there is no sign change which would
violate the conservation of energy.
I am a bit surprised that you say that so late in this thread. The thread started with the fairly simple demonstration that [tex]U=+\mu\cdot{\bf B}[/tex] for a loop with a constant current in a B field. This required [tex]{\bf F}=+\nabla U[/tex] to get the correct form that
[tex]{\bf F}=+\nabla(\mu\cdot{\bf B})[/tex]. F also equals [tex]\nabla U[/tex] for the force of one capacitor plate on the other, when the capacitor is kept at constant voltage. This is shown in most EM texts. In every case, [tex]{\bf F}=+\nabla(\mu\cdot{\bf B})[/tex], but the energy can have either sign depending on whether or not a constant voltage or constant current source is providing energy to keep the voltage or current constant.

For an electron, [tex]U=-\mu\cdot{\bf B}[/tex] because there is no external source feeding energy to the electron. This is independent of any speculation about the source of the moment.
 
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Hans de Vries
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Re: "paradox" regarding energy of dipole orientation

I am a bit surprised that you say that so late in this thread. The thread started with the fairly simple demonstration that [tex]U=+\mu\cdot{\bf B}[/tex] for a loop with a constant current in a B field. This required [tex]{\bf F}=+\nabla U[/tex] to get the correct form that [tex]{\bf F}=+\nabla(\mu\cdot{\bf B})[/tex]. F also equals [tex]\nabla U[/tex] for the force of one capacitor plate on the other, when the capacitor is kept at constant voltage. This is shown in most EM texts. In every case, [tex]{\bf F}=+\nabla(\mu\cdot{\bf B})[/tex], but the energy can have either sign depending on whether or not a constant voltage or constant current source is providing energy to keep the voltage or current constant.

For an electron, [tex]U=-\mu\cdot{\bf B}[/tex] because there is no external source feeding energy to the electron. This is independent of any speculation about the source of the moment.

This is all correct Meir,
My original response was to Justin Levi's attack on section 4 of this paper:

http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3421v3.pdf [Broken]

Which discusses the magnetic moment of elementary particles. Which have, as you
say, [tex]U=-\mu\cdot{\bf B}[/tex] because they are passive.

As stated before, I READ THAT PAPER. Please everyone stop bringing up that paper. It is not useful for this situation. Here's what they say about the dipole problem:

"In fact, [ [itex] F =\nabla (\mu \cdot B) [/itex] ] should be considered a separate force law for permanent magnetic dipoles on a par with the Lorentz force on charge or the j×B force on currents, since it cannot be derived from those force laws."

Their answer is a non-answer. Their "solution" to the dipole problem is to postulate a new force. If you want to take that as an answer, fine

From "Here's what they say about the dipole problem" I concluded that the
author says that Justin's "paradox": [tex]U=+\mu\cdot{\bf B}[/tex] was explained by telling that
"elementary particles obey different laws".

Re-reading section 4 shows that the author correctly states that [tex]U=-\mu\cdot{\bf B}[/tex]
So the author of the paper is correct here, besides the other independent
discussion about the nature of the magnetic moment of elementary particles.


Regards, Hans.
 
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