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The Wang paper "An experimental proposal to test the physical effect of the vector potential" proposes an experiment to decide between two interpretations of the Aharonov-Bohm effect:

“the interpretation of electromagnetic potentials”

or

“the interpretation of interaction energy”

The Konopinski paper "What the electromagnetic vector potential describes", takes the interaction energy interpretation and shows the motional scalar potential of the curl-free vector potential:

$$\vec{v} · \vec{A}$$

has the derived electromotive ($\vec{E}_m$) field:

$$\vec{E}_m = - \nabla(\vec{v} · \vec{A})$$

With force $\vec{F}$ measurable by a test charge q moving with velocity $\vec{v}$:

$$\vec{F} = q ( - \nabla(\vec{v} · \vec{A}) ) $$

The Lorentz force:

$$q \left( -\nabla\phi - \frac{\partial \vec A}{\partial t} + \vec{v} \times (\nabla \times \vec{A}) \right)$$

does not contain this term in its electromotive field. Adding it:

$$q \left( -\nabla\phi - \frac{\partial \vec A}{\partial t} + \vec{v} \times (\nabla \times \vec{A}) - \nabla(\vec{v} \cdot \vec{A})\right)$$

is, by the vector identity of the total derivative:

$$q \left( -\nabla\phi - \frac{D \vec A}{D t} \right)$$

Hence, the Wang paper proposes an experiment that can determine whether

$$\vec{E}_m = -\nabla\phi - \frac{D \vec A}{D t}$$

is true.

This conclusion appears to result from main stream physics but flies in the face of the Lorentz force accepted since the time of Maxwell thence Einstein's 1905 paper "On the Electrodynamics of Moving Bodies".

Apocryphal, non-mainstream apologetics for the $ - \nabla(\vec{v} \cdot \vec{A})$ term maintain it has escaped classical notice because it integrates to 0 around any circuit, and should appear only in open electrodynamics as occur in plasmas and antennas.

It is hard to believe there is a contradiction this long-standing between the recognized classical Lorentz force:

$$\vec{F} = q \left( -\nabla\phi - \frac{\partial \vec A}{\partial t} + \vec{v} \times (\nabla \times \vec{A}) \right)$$ and quantum physics implied by the interaction energy interpretation of the Aharonov-Bohm effect: $$\vec{F} = q \left(-\nabla\phi - \frac{D \vec A}{D t}) = q ( -\nabla\phi - \frac{\partial \vec A}{\partial t} + \vec{v} \times (\nabla\times \vec{A}) - \nabla(\vec{v} \cdot \vec{A})\right)$$ Why is the conclusion reached above, $\vec{E}_m = -\nabla\phi - \frac{D \vec A}{D t}$ incorrect?

This question is not the same as asking, "Why is the conclusion... in conflict with mainstream theory." precisely because it is mainstream theory, apparently, that leads to this conclusion from its empirical grounding in the ABe if one adopts the interaction energy interpretation.

There is either a problem with the interaction energy interpretation of the ABe, or there is a problem with the steps taken in reaching the above conclusion from the interaction energy interpretation of the ABe. Which is it?

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  • $\begingroup$ 1. Please use MathJax to typeset formulae. 2. You ask for an "error in the above derivation", but there is no derivation, just a bunch of formulae whose relation to each other and what you say in the preceding/following text remain unclear. 3. Wang does not argue what you claim he argues in the linked paper, he just refers to other of his papers, where the "argument" are a few sentences that handwavingly talk about the "uncertainty principle" forbidding "measurements without back-reaction". $\endgroup$ – ACuriousMind Mar 27 '16 at 20:34
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    $\begingroup$ I made the required edits and I simplified the question to avoid Wang's assertion about the uncertainty principle. $\endgroup$ – James Bowery Mar 27 '16 at 21:57
  • $\begingroup$ possibly relevant: physics.stackexchange.com/q/56926 $\endgroup$ – Rococo Mar 28 '16 at 18:15
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    $\begingroup$ Notational comment to the question (v16): It seems that two unrelated quantities are denoted with the same symbol $\vec{E}_m$. $\endgroup$ – Qmechanic Apr 5 '16 at 13:39
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If you write $\frac{D \vec A}{D t}$, this is the derivative w.r.t. the specific velocity of a specific test charge. That doesn't make sense as the "electric field" since this value depends on the velocity of the test charge, but the electric field is by definition the force on a stationary test charge divided by the magnitude of the test charge.

Even worse, the $\nabla(v\cdot A)$ term is not gauge invariant. Under $A\mapsto A+\nabla \chi$, we have (for a constant velocity) \begin{align} \nabla(v\cdot A) & \mapsto \nabla (v\cdot (A+\nabla \chi)) = v\times (\nabla \times A) + (v\cdot \nabla) (A+\nabla \chi) \\ & = \nabla(v\cdot A) + (v\cdot \nabla)\nabla\chi \end{align} and the second summand does not vanish.

The very definition of the scalar and vector potential of electromagnetism is that they are the functions $\phi,A$ fulfilling \begin{align} E & = -\nabla \phi - \partial_t A \\ B & = \nabla\times A \end{align} which exist locally as a purely mathematical consequence of $E$ and $B$ fulfilling Maxwell's equations. No amount of physical reasoning can ever change these relations because they are the definition of the potentials.

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  • $\begingroup$ In the Lorentz force equation, the subexpression $ ( -∇Φ - ∂\vec{A}/∂t + \vec{v} × (∇ × \vec{A}) )$ contains a velocity, yet it does not become a force until multiplied by a charge, q. The physical dimension of that subexpression matches what you call an "electric field" so I don't understand what you would call it, unless you would qualify it, as do some authors, as a "motional electric field". Would it be acceptable to you for me to replace all such references to "electric field" with "motional electric field" even if they contained non-motional terms? $\endgroup$ – James Bowery Mar 28 '16 at 2:53
  • $\begingroup$ Your observation of the lack of gauge invariance is simply an elaboration of my observation of the contradiction, which is the basis of my question. It is not an answer to my question. It begs my question. $\endgroup$ – James Bowery Mar 28 '16 at 2:59
  • $\begingroup$ The "definition" of the vector potential you cite is under-constrained. In that sense it is not a definition but a constraint. $\endgroup$ – James Bowery Mar 28 '16 at 3:03
  • $\begingroup$ @JamesBowery: The $\vec v \times (\nabla \times A)$ is the magnetic force per unit charge. The magnetic field $B$ is by definition the field that gives the force on a test charge moving with velocity $v$ when taking the cross product with it and multiplying by the charge. Having units of some quantity does not mean that you actually are that quantity (just think of torque and energy: Same unit, different things). Since the equations of motion of electromagnetism are Maxwell's equations in which only $E$ and $B$ appear, making any further constraint on the potentials is wholly unphysical. $\endgroup$ – ACuriousMind Mar 28 '16 at 11:32
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    $\begingroup$ @JamesBowery: Indeed, the Aharonov-Bohm effect is invisible in classical electrodynamics. That is because the "phase difference" it is about makes no sense at all in classical physics - the electron beams are not wave-like in classical mechanics, so it doesn't make sense to ask about their phase difference. Maxwell's equations of motion as well as the Lorentz force law are only equations in the fields, and they are the complete description of what happens in classical electrodynamics. $\endgroup$ – ACuriousMind Mar 28 '16 at 13:35
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I) In this answer we will consider the microscopic description of classical E&M only. The Lorentz force reads

$$ \tag{1} {\bf F}~:=~q({\bf E}+{\bf v}\times {\bf B})~=~\frac{\mathrm d}{\mathrm dt}\frac{\partial U}{\partial {\bf v}}- \frac{\partial U}{\partial {\bf r}}~=~-q\frac{\mathrm d{\bf A}}{\mathrm dt} - \frac{\partial U}{\partial {\bf r}}, $$

where $$ \tag{2} U~:=~ q(\phi - {\bf v} \cdot {\bf A})~=~ q\left(\phi - \sum_{i=1}^3v^i A_i\right)$$

is the corresponding velocity-dependent potential, see e.g., Herbert Goldstein, Classical Mechanics, Chapter 1, for details. Here we use the Minkowski sign convention $(-,+,+,+)$.

The potential (2) is Lorentz invariant, and it changes with a total time derivative under gauge transformation. The total time derivatives is

$$ \tag{3} \frac{\mathrm d}{\mathrm dt}~=~\frac{\partial }{\partial t}+{\bf v}\cdot\frac{\partial }{\partial {\bf r}}+{\bf a}\cdot\frac{\partial }{\partial {\bf v}}+\ldots, $$

where ellipsis "$\ldots$" denotes possible higher time-derivative terms.

II) Newton's 2nd law

$$\tag{4} \frac{\mathrm d{\bf p}_{\rm kin}}{\mathrm dt} ~=~F, \qquad {\bf p}_{\rm kin}~:=~ m_0 \gamma {\bf v}, \qquad \gamma~:=~\frac{1}{\sqrt{1-\left(\frac{{\bf v}}{c}\right)^2}} ,$$

for a relativistic charged point particle can be written as

$$ \tag{5}\frac{\mathrm d{\bf p}_{\rm can}}{\mathrm dt}~=~ - \frac{\partial U}{\partial {\bf r}}.$$

The momentum

$$ \tag{6}{\bf p}_{\rm can}~:=~\frac{\partial L}{\partial{\bf v}}~=~{\bf p}_{\rm kin} +q{\bf A}, \qquad L~:=~-\frac{m_0c^2}{\gamma}-U, $$

is the conjugate/canonical momentum ${\bf p}_{\rm can}$, which is different from the kinematic/mechanical momentum ${\bf p}_{\rm kin}$, cf. e.g. my Phys.SE answer here. It is instructive to compare eqs. (5) and (6) with OP's Phys.SE post here. The relativistic Lagrangian (6) is mentioned in this Phys.SE answer.

III) It seems that OP is pondering what would happen if we remove the force term

$$ \tag{7} {\bf F}_m ~:=~ q \frac{\partial {\bf v} \cdot {\bf A}}{\partial {\bf r}}~=~ q\sum_{i=1}^3v^i\frac{\partial A_i}{\partial {\bf r}} $$

from the term

$$ \tag{8} - \frac{\partial U}{\partial {\bf r}}~=~{\bf F}_m- q\frac{\partial \phi}{\partial {\bf r}}$$

on the right-hand sides of eqs. (1) and (5)?

Answer: Besides breaking gauge symmetry, as correctly observed in ACuriousMind's answer, it would break Lorentz covariance. Also one may prove that the force term (7) has no velocity-dependent potential via methods similar to this Phys.SE post. Hence the modified theory would be without a stationary action principle.

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  • $\begingroup$ This is a further elaboration of my question's premise: That the conclusion is in conflict with mainstream physics. Reiterating the end of my OP: This question is not the same as asking, "Why is the conclusion... in conflict with mainstream theory."... There is either a problem with the interaction energy interpretation of the ABe, or there is a problem with the steps taken in reaching the above conclusion from the interaction energy interpretation of the ABe. $\endgroup$ – James Bowery Apr 5 '16 at 13:26

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