Vector field equality Curl Proof of Moving Magnet & Conductor Problem

In summary, the moving magnet and conductor problem is a famous and intriguing scenario in early 20th century electromagnetics that has been cited by Einstein in his seminal 1905 special relativity paper. It involves two frames of reference, the magnet's frame and the ring conductor's frame, in which there are two vector fields, (v × B) and E', respectively. Classical electromagnetics shows that the curls of both vector fields are mathematically equivalent, leading to the statement that both vector fields themselves are equivalent. However, the steps to derive this equivalency are not readily available in current research on the mathematics of this problem. One approach could be to consider the electromagnetic four-potential for the system, but it is not
  • #1
tade
702
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magnet4.gif

The moving magnet and conductor problem is an intriguing early 20th century electromagnetics scenario famously cited by Einstein in his seminal 1905 special relativity paper.

In the magnet's frame, there's the vector field (v × B), the velocity of the ring conductor crossed with the B-field of the magnet.

In the ring conductor's frame, there's the induced electric field E'.Using classical electromagnetics, it can be shown that the curls of both vector fields are mathematically equivalent. It is then usually stated that both vector fields themselves are mathematically equivalent.

I've been researching documents on the mathematics of the moving magnet and conductor problem, and I haven't been able to find any mathematical steps showing how we can arrive at both fields being equivalent if their curls are equivalent.

Multiple different vector fields can produce the same single curl field, and vice versa, a single curl field can be indicative of multiple different vector fields.
So I would like to know how to derive the equivalency of the moving magnet and conductor vector fields.
 
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  • #2
One good method would be to consider (4) vector potential for the system. I feel certain this is done in detail in a textbook but I don't know off the top of my head where. Someone else will
 
  • #3
hutchphd said:
One good method would be to consider (4) vector potential for the system. I feel certain this is done in detail in a textbook but I don't know off the top of my head where. Someone else will
is it the electromagnetic four-potential?

because this is supposed to be an early 20th century scenario pre-dating special relativity
 
  • #4
If you wish to ask specific questions about particular papers or complete derivations perhaps I can help. Otherwise this seems too nebulous for my help. Sorry.
 
  • #5
hutchphd said:
If you wish to ask specific questions about particular papers or complete derivations perhaps I can help. Otherwise this seems too nebulous for my help. Sorry.

Sure, let's make it very clear:

From another thread:
Orodruin said:
the Maxwell-Faraday equation
$$
\nabla \times \vec E' = - \frac{\partial \vec B'}{\partial t}
$$
leading to the given relation. The behaviour of ##\vec B'## is given in the equation before, but that equation seems incomplete. Approximately (for small velocities),
$$
\vec B'(\vec r', t) = \vec B(\vec r'+\vec v t).
$$
This leads to
$$
\partial_t \vec B' = \partial_t \vec B(\vec r+ \vec v t) = (\vec v \cdot \nabla)\vec B.
$$
For the left-hand side with ##\vec E' = \vec v \times \vec B## you would have
$$
\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,
$$
because the magnetic field is divergence free. Thus ##\vec E' = \vec v \times \vec B## solves the Maxwell-Faraday equation.

If indeed ##\vec E' = \vec v \times \vec B## then of course ##\nabla \times \vec E' = \nabla \times (\vec v \times \vec B)##

But so far the steps have only shown ##\nabla \times \vec E' = \nabla \times (\vec v \times \vec B)##

I'm trying to show that it really does lead to ##\vec E' = \vec v \times \vec B##
 
  • #6
tade said:
so far the steps have only shown ##\nabla \times \vec{E}^\prime \rightarrow \nabla \times \left( \vec{v} \times \vec{B} \right)##

No, what @Orodruin showed in what you quoted was that if you assume ##\vec{E}^\prime = \vec{v} \times \vec{B}##, then you can solve the Maxwell-Faraday equation.
 
  • #7
PeterDonis said:
No, what @Orodruin showed in what you quoted was that if you assume ##\vec{E}^\prime = \vec{v} \times \vec{B}##, then you can solve the Maxwell-Faraday equation.
yes, I didn't disagree with that. so, do you know how to show that it is indeed ##\vec{E}^\prime = \vec{v} \times \vec{B}##, instead of just assuming
 
  • #8
tade said:
do you know how to show that it is indeed ##\vec{E}^\prime = \vec{v} \times \vec{B}##, instead of just assuming

You're not "just assuming". You assume a hypothesis: that ##\vec{E}^\prime = \vec{v} \times \vec{B}## is a solution. You prove that hypothesis by showing that ##\vec{E}^\prime = \vec{v} \times \vec{B}## is, indeed, a solution. That's it: you're done. What else would you need to show?
 
  • #9
PeterDonis said:
You're not "just assuming". You assume a hypothesis: that ##\vec{E}^\prime = \vec{v} \times \vec{B}## is a solution. You prove that hypothesis by showing that ##\vec{E}^\prime = \vec{v} \times \vec{B}## is, indeed, a solution. That's it: you're done. What else would you need to show?

Well, I believe ye old moving magnet and conductor case-study takes ##\vec{E}^\prime = \vec{v} \times \vec{B}## as the solution, instead of just a solution, then it would be just one of a number of different possible solutions
 
  • #10
tade said:
it would be just one of a number of different possible solutions

How many possible solutions can the Maxwell-Faraday equation have?
 
  • #11
PeterDonis said:
How many possible solutions can the Maxwell-Faraday equation have?
Another possible solution might be ##\vec{E}^\prime = \vec{v} \times \vec{B} + \vec{F_1}##
where ##\vec{F_1}## is a curl-free vector field
 
  • #12
tade said:
Another possible solution might be ##\vec{E}^\prime = \vec{v} \times \vec{B} + \vec{F_1}##
where ##\vec{F_1}## is a curl-free vector field

Please show your work: plug that into the equation and show what you get.
 
  • #13
PeterDonis said:
Please show your work: plug that into the equation and show what you get.
we take the curl of the LHS and the curl of the RHS.

The curl operator is distributive, and the curl of F1 is zero, so we end up with ##\nabla \times \vec E' = \nabla \times (\vec v \times \vec B)##
 
  • #14
For the moving magnet problem you don't even need explicitly relativity but just Maxwell's equations. The reason is that as long as you don't make any approximations (e.g., by assuming that the moving charges making up a current follow non-relativistic kinematics, as is done in many textbooks on electromagnetism, and this is often justified but still leads to confusion when it comes to the question of "electrodynamics of moving bodies" as Einstein's famous paper is titled).

All you need here is that Faraday's Law in differential form simply is (in Heaviside Lorentz units)
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
Applying Stokes's integral theorem leads to
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_{A} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Now the crucial point is that you want to bring the time-derivative on the right-hand side out of the integral. In the case of the moving magnet on the right-hand side you want to integrate along the moving wire making up the loop and a corresponding surface whose boundary this loop is on the right-hand side. That means the surface and/or boundary are time-dependent and you must take this time-dependence into account when taking the time-derivative out of the integral. The corresponding correction adds a line integral along the boundary, which can be brought to the other side of the equation. Using also Gauss's Law for the magnetic fields, ##\vec{\nabla} \cdot \vec{B}=0## leads to the general form of Faraday's Law in integral form:
$$\int_{\partial A} \mathrm{d}\vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=-\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
For a derivation, see the Wikipedia:

https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof

This is valid in any reference frame. In the frame, where the magnet is at rest, you have ##\vec{E}=0## and the entire electromotive force (acting on the electrons in the wire as a force in the usual thing) is the ##\vec{v} \times \vec{B}## term. The corresponding Lorentz force acts on the electrons in the wire and make them move relative to the wire.

In the frame, where the wire is at rest there's by definition ##\vec{v}=0##, but of course there must also be some electromotive force making the electrons moving relative to the wire also in this frame. The only conclusion must be, since also in this frame Maxwell's equations hold true due to the principle of relativity according to which the natural laws are the same in all inertial frames, that there's an electric field, and indeed as Einstein has derived in his paper according to the transformation laws of the electromagnetic field components a moving magnet leads to both electric and magnetic field components, i.e., in this reference frame there's an electric (vortex) field leading to the electromotive force making the electrons move in the wire loop. Since these Lorentz transformations are a one-to-one mapping there's never a contradiction between what each of the observers observe, and there's also one and only one cause of the current in the wire: the electromotive force due to the electromagnetic field, which is one field, i.e., you have to consider both the electric and magnetic "parts" of the field as one entity, and as in this case, where you have a magnetostatic field only in the reference frame where the magnet is at rest, in the other reference frame, where the magnet is moving, you have both electric and magnetic field components, i.e., the split of the one and only electromagnetic field in electric and magnetic components is observer dependent, but the Lorentz transformations of these field components tell you how to relate the components with respect to one inertial reference frame to another, and these transformations are such that the Maxwell equations hold in unaltered form for the field components in both frames.
 
  • #15
tade said:
Sure, let's make it very clear:
From another thread:
If indeed ##\vec E' = \vec v \times \vec B## then of course ##\nabla \times \vec E' = \nabla \times (\vec v \times \vec B)##

But so far the steps have only shown ##\nabla \times \vec E' = \nabla \times (\vec v \times \vec B)##
I'm trying to show that it really does lead to ##\vec E' = \vec v \times \vec B##
I don't know if your math is relevant but I do know that a charge in a conductor with motion relative to a B field experiences either a force qE (observer stationary wrt the conductor) or q(v x B} if the observer is stationary wrt the B field. Same force! Both represent the force on a charge and there are no other forces (excepting negligible gravity).

Maxwell's equations do not directly address moving media; thus the Lorentz force q(v x B) is additionally needed, so
## \nabla \times \mathbf E \rightarrow \nabla \times \mathbf E + \nabla \times (\mathbf v \times \mathbf B) ##.

Later, special relativity showed that the B field itself is really only an E field also, but in classical physics the Lorentz force is distinguished from the Coulomb force.
 
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  • #16
rude man said:
I don't know if your math is relevant but I do know that a charge in a conductor with motion relative to a B field experiences either a force qE (observer stationary wrt the conductor) or q(v x B} if the observer is stationary wrt the B field. Same force! Both represent the force on a charge and there are no other forces (excepting negligible gravity).

Maxwell's equations do not directly address moving media; thus the Lorentz force q(v x B) is additionally needed, so
## \nabla \times \mathbf E \rightarrow \nabla \times \mathbf E + \nabla (\mathbf v \times \mathbf B) ##.

Later, special relativity showed that the B field itself is really only an E field also, but in classical physics the Lorentz force is distinguished from the Coulomb force.

well, without the math its kinda moot
 
  • #17
sorry, but I'm still unsure about how we show the proof that ##\vec{E}^\prime = \vec{v} \times \vec{B}##

vanhees71 said:
For the moving magnet problem you don't even need explicitly relativity but just Maxwell's equations. The reason is that as long as you don't make any approximations (e.g., by assuming that the moving charges making up a current follow non-relativistic kinematics, as is done in many textbooks on electromagnetism, and this is often justified but still leads to confusion when it comes to the question of "electrodynamics of moving bodies" as Einstein's famous paper is titled).

All you need here is that Faraday's Law in differential form simply is (in Heaviside Lorentz units)
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
Applying Stokes's integral theorem leads to
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_{A} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Now the crucial point is that you want to bring the time-derivative on the right-hand side out of the integral. In the case of the moving magnet on the right-hand side you want to integrate along the moving wire making up the loop and a corresponding surface whose boundary this loop is on the right-hand side. That means the surface and/or boundary are time-dependent and you must take this time-dependence into account when taking the time-derivative out of the integral. The corresponding correction adds a line integral along the boundary, which can be brought to the other side of the equation. Using also Gauss's Law for the magnetic fields, ##\vec{\nabla} \cdot \vec{B}=0## leads to the general form of Faraday's Law in integral form:
$$\int_{\partial A} \mathrm{d}\vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=-\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
For a derivation, see the Wikipedia:

https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof

This is valid in any reference frame. In the frame, where the magnet is at rest, you have ##\vec{E}=0## and the entire electromotive force (acting on the electrons in the wire as a force in the usual thing) is the ##\vec{v} \times \vec{B}## term. The corresponding Lorentz force acts on the electrons in the wire and make them move relative to the wire.

In the frame, where the wire is at rest there's by definition ##\vec{v}=0##, but of course there must also be some electromotive force making the electrons moving relative to the wire also in this frame. The only conclusion must be, since also in this frame Maxwell's equations hold true due to the principle of relativity according to which the natural laws are the same in all inertial frames, that there's an electric field, and indeed as Einstein has derived in his paper according to the transformation laws of the electromagnetic field components a moving magnet leads to both electric and magnetic field components, i.e., in this reference frame there's an electric (vortex) field leading to the electromotive force making the electrons move in the wire loop. Since these Lorentz transformations are a one-to-one mapping there's never a contradiction between what each of the observers observe, and there's also one and only one cause of the current in the wire: the electromotive force due to the electromagnetic field, which is one field, i.e., you have to consider both the electric and magnetic "parts" of the field as one entity, and as in this case, where you have a magnetostatic field only in the reference frame where the magnet is at rest, in the other reference frame, where the magnet is moving, you have both electric and magnetic field components, i.e., the split of the one and only electromagnetic field in electric and magnetic components is observer dependent, but the Lorentz transformations of these field components tell you how to relate the components with respect to one inertial reference frame to another, and these transformations are such that the Maxwell equations hold in unaltered form for the field components in both frames.
 
  • #18
rude man said:
Maxwell's equations do not directly address moving media; thus the Lorentz force q(v x B) is additionally needed, so
## \nabla \times \mathbf E \rightarrow \nabla \times \mathbf E + \nabla \times (\mathbf v \times \mathbf B) ##.

Later, special relativity showed that the B field itself is really only an E field also, but in classical physics the Lorentz force is distinguished from the Coulomb force.
Maxwell's equations perfectly address moving media, because they are in fact a realtivistic classical field theory. The only thing which has not been taken properly into account was the necessity of using the correct relativistic constitutive relations for macroscopic electrodynamics. The first consistent scheme has been proposed by Minkowski, but it's indeed not a trivial issue.
 
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  • #19
tade said:
sorry, but I'm still unsure about how we show the proof that ##\vec{E}^\prime = \vec{v} \times \vec{B}##
I guess you mean the transformation properties of the electromagnetic field components under Lorentz boosts.

It is most convenient to use the electromagnetic potentials. In 3D formulation you solve the homogeneous Maxwell equations identically by introducing the scalar and vector potentials. I use Heaviside-Lorentz units, because relativistic electrodynamics within the SI system is spoiling the beauty of the subject:
$$\vec{E}=-\vec{\nabla} \Phi-\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
To derive the Lorentz-transformation properties, it's better to use the 4D notation in Minkowski space, and the potentials can be put together to build a four-vector field
$$(A^{\mu})=\begin{pmatrix} \Phi \\ \vec{A} \end{pmatrix},$$
i.e., ##A^0=\Phi## is the time component.

A glance at the ##\vec{B}##-field tells you that the electromagnetic field components in four-tensor notation should be given by the antisymmetric Faraday tensor,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
Indeed
$$F_{0 \nu}=-F_{\nu 0}=\partial_0 A_{\nu} - \partial_{\nu} A_0=-\frac{1}{c} A^{\nu} - \partial_{\nu} A^0=E^{\nu}$$
for ##\nu \in \{1,2,3 \}## and
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}=-\partial_{\mu} A^{\nu} + \partial_{\nu} A^{\mu} = \epsilon^{\nu \mu \rho} B^{\rho},$$
for ##\mu \nu \in \{1,2,3 \}## and ##\epsilon^{\mu \nu \rho}## the usual 3D Levi-Civita symbol.

Not the transformation property of a vector field like our four-vector potential is
$$\bar{A}^{\mu}(\bar{x})={\Lambda^{\mu}}_{\nu} A^{\nu}(\Lambda^{-1} \bar{x}).$$
Take ##\Lambda## to be a boost in ##x^1## direction. Then
$$\bar{A}^{\mu}=\begin{pmatrix}\gamma (A^0-\beta A^1) \\ \gamma (-\beta A^0+A^1) \\ A^2 \\ A^3 \end{pmatrix}.$$
Here ##\beta=v/c## and ##\gamma=1/\sqrt{1-\beta^2}##.

Then you use
$$\bar{\partial}_{\mu} = \frac{\partial}{\partial \bar{x}^{\mu}} = \frac{\partial x^{\nu}}{\partial \bar{x}^{\mu}} \partial_{\nu}$$
and evaluate ##\bar{F}_{\mu \nu}##, which is a bit of work but straight-forward algebra. Then you can read off the field components in 3D notation as:
$$\begin{pmatrix} \bar{E}^1 \\ \bar{E}^2 \\ \bar{E}^3 \end{pmatrix}=\begin{pmatrix} E^1 \\ \gamma (E^2-\beta B^3) \\ \gamma (E^3+\beta B^2) \end{pmatrix}$$
and
$$\begin{pmatrix} \bar{B}^1 \\ \bar{B}^2 \\ \bar{B}^3 \end{pmatrix} = \begin{pmatrix}B^1 \\ \gamma(B^2+\beta E^3) \\ \gamma (B^3-\beta E^2) \end{pmatrix}.$$
A more convenient notation is
$$\bar{\vec{E}}_{\parallel}=\vec{E}_{\parallel}, \quad \bar{\vec{E}}_{\perp} = \gamma (\vec{E}_{\perp} + \vec{\beta} \times \vec{B})$$
and
$$\bar{\vec{B}}_{\parallel}=\vec{B}_{\parallel}, \quad \bar{\vec{B}}_{\perp} = \gamma (\vec{B}_{\perp} - \vec{\beta} \times \vec{E}),$$
where for any vector ##\vec{E}##
$$\vec{E}_{\parallel}=\frac{\vec{\beta}}{|\vec{\beta}|^2} \vec{\beta} \cdot \vec{E}, \quad \vec{E}_{\perp}=\vec{E}-\vec{E}_{\parallel}.$$

If you now have a situation, where in the reference frame ##\Sigma## you have only a magnetic field, i.e., ##\vec{E}=0##, then
$$\bar{\vec{E}}=\gamma \vec{\beta} \times \vec{B}, \quad \bar{\vec{B}}=\gamma \vec{B}.$$
For ##\beta \ll 1## you can neglect the Lorentz factor ##\gamma=1+\mathcal{O}(\beta^2)##.

For more details on the relativistic formulation of electrodynamics, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
 
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  • #20
vanhees71 said:
I guess you mean the transformation properties of the electromagnetic field components under Lorentz boosts.

but weren't they able to show it pre-relativity?
 
  • #21
Indeed there was an attempt by H. Hertz (the discoverer of electromagnetic waves) to formulate an electrodynamics of moving media within Newonian physics. If I rememeber right, he came close to the transformation laws for the field components you get when using the non-relativistic limit of the full Lorentz transformations, i.e., neglecting all effects of order ##\mathcal{O}(\beta^2)## and higher. What results is the socalled "aether theory". It is correct to order ##\mathcal{O}(\beta)## but not at higher orders. That's why you need experiments accurate to corrections of the order ##\mathcal{O}(\beta^2)## to distinguish between aether theory and special relativity like, e.g., the Michelson-Morley experiment.
 
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  • #22
vanhees71 said:
Indeed there was an attempt by H. Hertz (the discoverer of electromagnetic waves) to formulate an electrodynamics of moving media within Newonian physics. If I rememeber right, he came close to the transformation laws for the field components you get when using the non-relativistic limit of the full Lorentz transformations, i.e., neglecting all effects of order ##\mathcal{O}(\beta^2)## and higher. What results is the socalled "aether theory". It is correct to order ##\mathcal{O}(\beta)## but not at higher orders. That's why you need experiments accurate to corrections of the order ##\mathcal{O}(\beta^2)## to distinguish between aether theory and special relativity like, e.g., the Michelson-Morley experiment.

sorry, I'm awfully interested in this subject, if it isn't too much trouble could you show me how Hertz's method leads to ##\vec{E}^\prime = \vec{v} \times \vec{B}##?
 
  • #23
I can't tell you, how Hertz derived his equations. The paper (in German of course in Wiedemanns Annalen) is very hard to read for us today, because as usual in his time, he wrote everything out in components. I'm not aware of any textbook which gives Hertz's theory in modern notation.

If you want to understand the physics, it's much better to learn the right thing from the very beginning, i.e., in terms of (special) relativity. On of the best treatments of classical electrodynamics in the "relativity-first approach" is Vol. 2 of Landau and Lifshitz (which also contains a very lucid treatment of general relativity in its second part).
 
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  • #24
vanhees71 said:
I can't tell you, how Hertz derived his equations. The paper (in German of course in Wiedemanns Annalen) is very hard to read for us today, because as usual in his time, he wrote everything out in components. I'm not aware of any textbook which gives Hertz's theory in modern notation.
i see, do you have a link to the paper?

or maybe you can just output whatever you know to get E'=v×B haha
 
  • #26
  • #27
tade said:
thanks, though I'm not sure where Hertz's work is in that pdf, and sorry, i meant whatever you know of the Hertzian method lol
@vanhees71 sorry, still grappling with this physics problem
 
  • #28
Maybe I didn't get the question then. Since Einstein's paper of 1905 and particularly its mathematical clarification by Minkowski there's no problem with moving magnets and conductors anymore. So what precisely is your problem with this example?
 
  • #29
vanhees71 said:
Maybe I didn't get the question then.

So what precisely is your problem with this example?

from #5 and onwards, I exchanged messages with Pete, though he didn't provide an answer to my question

so I'm looking for a way to show that if the curls of the fields are equal, the fields are equal, without resorting to SR.

SR wasn't required to show that the curls are equal.
 
  • #30
Of course, it's not true what you want to prove, and it's very important in the connection with electrodynamics, because it's basically what's behind gauge invariance, which is the most important concept of electrodynamics to begin with. So let's think about the electromagnetic potentials carefully again:

Observable is the electromagnetic field. In the usual 3D formalism it's described by the electric and magnetic fields ##\vec{E}## and ##\vec{B}##. There are two sets of Maxwell equations, the homogeneous ones, which are constraints to the fields, not involving the sources ##\rho## (charge density) and ##\vec{j}## (current density). For sake of convenience I use Heaviside-Lorentz units, in which ##\vec{E}## and ##\vec{B}## have the same dimensions as it should be from a relativistic point of view:
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
These constraints can be fulfilled identically by introduction of the potentials, which however are not unique, as we shall see soon.

Indeed, Helmholtz's fundamental theorem of vector calculus says the from ##\vec{\nabla} \cdot \vec{B}=0## it follows that there exists a vector potential such that
$$\vec{B}=\vec{\nabla} \times \vec{A},$$
and it's immediately clear that ##\vec{A}## is not uniquely defined by ##\vec{B}##, because any other vector potential
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi$$
with an arbitrary scalar field ##\chi## leads to the same ##\vec{B}##, because
$$\vec{\nabla} \times \vec{A}' = \vec{\nabla} \times \vec{A} + \vec{\nabla} \times \vec{\nabla} \chi=\vec{\nabla} \times \vec{A}=\vec{B}.$$
Now we use ##\vec{B}=\vec{\nabla} \times \vec{A}## in the other homogeneous Maxwell equation (Faraday's Law of induction):
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \vec{\nabla} \times \partial_t \vec{A}=0$$
or
$$\vec{\nabla} \times \left (\vec{E} + \frac{1}{c} \partial_t \vec{A} \right)=0.$$
This implies that the expression in the bracket is derivable from a scalar potential ##\Phi##,
$$\vec{E}+\frac{1}{c} \partial_t \vec{A}=-\vec{\nabla} \Phi$$
or
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi.$$
Of course, for a given ##\vec{E}## the scalar potential ##\Phi## depends on the choice of ##\vec{A}##. Since ##\vec{E}## is observable and thus must be uniquely defined, ##\Phi## must change when changing ##\vec{A}## using the gauge field ##\chi##:
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}' - \vec{\nabla} \Phi'.$$
Now
$$\vec{E}=-\frac{1}{c} \partial_t (\vec{A}-\vec{\nabla} \chi) - \vec{\nabla} \Phi' = -\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} \left (\Phi'- \frac{1}{c} \partial_t \chi \right),$$
which implies that the gauge transformation for the scalar potential is
$$\Phi=\Phi'-\frac{1}{c} \partial_t \chi \; \Rightarrow \; \Phi'=\Phi+\frac{1}{c} \partial_t \chi.$$
Due to this gauge invariance of the entire formalism under gauge transformations of the potentials
$$\Phi'=\Phi+\frac{1}{c} \partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
the potentials ##\Phi## and ##\vec{A}## are not uniquely determined but only up to such a gauge transformation.

To see how this is consistent with the inhomogeneous Maxwell equations,
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho,$$
we note that this equations imply the equation of continuity, i.e., the local form of charge conservation. Indeed taking the divergence of the first equation (the Ampere-Maxwell Law) and using the second equation (Gauss's Law for the electric field) leads to
$$-\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{E}=-\frac{1}{c} \partial_t \rho = \frac{1}{c} \vec{\nabla} \cdot \vec{j} \; \Rightarrow \; \partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$
Now plugging in the potentials in the inhomogeneous equations leads to
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A}) + \frac{1}{c^2} \partial_t^2 \vec{A} + \frac{1}{c} \vec{\nabla} \partial_t \Phi=\frac{1}{c} \vec{j}, \quad \frac{1}{c} \partial_t \vec{\nabla} \cdot{\vec{A}} + \Delta \Phi=-\rho.$$
Since now the potentials are only determined up to a gauge transformation, i.e., up to an arbitrary scalar field ##\chi## we can impose one constraint, a socalled gauge-fixing condition. A very convenient choice is the Lorenz gauge condition,
$$\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0.$$
Plugging this into the above equations leads to
$$\left (\frac{1}{c^2} \partial_t^2 - \Delta \right) \vec{A}=\frac{1}{c} \vec{j}, \quad \left (\frac{1}{c^2} \partial_t^2 - \Delta \Phi \right)=\rho.$$
In the first equation we have used the identity
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}$$
and the Lorenz gauge condition. In the 2nd equation we just used the gauge condition. This procedure shows that the Lorenz gauge condition decouples the equations for the scalar and vector potentials in 4 wave equations for each of these four field degrees of freedom, ##\Phi## and the three Cartesian components of ##\vec{A}##.

Now it's easy to see that the Lorenz gauge condition is consistent with these wave equations thanks to the continuity equation for ##\rho## and ##\vec{j}##. This shows that the Maxwell equations are gauge invariant when formulated in terms of the potentials.

One can show this equivalence also for other gauge fixing conditions, e.g., the sometimes also convenient Coulomb gauge, which reads
$$\vec{\nabla} \cdot \vec{A}=0.$$
One can show again that thanks to the continuity equation for ##\rho## and ##\vec{j}## the resulting equations for ##\Phi## and ##\vec{A}## are compatible with this gauge condition either. One can also show that the Lorenz-gauge potentials are related with the Coulomb-gauge potentials by a gauge transformation:

https://arxiv.org/abs/physics/0204034
 
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  • #31
vanhees71 said:
Of course, it's not true what you want to prove

The moving magnet and conductor problem hinges upon the notion that if the curls of the fields are equal, the fields are equal.

Einstein thought, they have equal magnitudes but totally different origins, how silly, they are both the exact same phenomena of a relative motion.

So its not correct then?@Delta2
 
Last edited:
  • #32
tade said:
So its not correct then?

What are you asking? Are you asking if the observed results of the experiment with the moving magnet and the conductor are different depending on which frame we use to analyze the experiment? The answer to that is obvious: no, they aren't.

If your question is something else, it would be a good idea to restate it, since this thread has been dormant for a while.
 
  • #33
PeterDonis said:
What are you asking? Are you asking if the observed results of the experiment with the moving magnet and the conductor are different depending on which frame we use to analyze the experiment? The answer to that is obvious: no, they aren't.

If your question is something else, it would be a good idea to restate it, since this thread has been dormant for a while.
vanhees71 said, "Of course, it's not true what you want to prove"And these are the relevant posts:
tade said:
Another possible solution might be ##\vec{E}^\prime = \vec{v} \times \vec{B} + \vec{F_1}##
where ##\vec{F_1}## is a curl-free vector field
tade said:
but weren't they able to show it pre-relativity?

to be able to show, pre-relativity, that not only are the curls equal, the original fields are equal as well
 
  • #34
tade said:
vanhees71 said, "Of course, it's not true what you want to prove"

I'm not sure what he was referring to, because I'm not sure anyone in the previous discussion was entirely clear about what the issue was. That's why I asked you to restate the issue.

tade said:
not only are the curls equal, the original fields are equal as well

If you mean, to show that ##\vec{E}^\prime = \vec{v} \times \vec{B}## (with the prime on the ##\vec{E}^\prime## to indicate that it is the field in a different frame), try looking at the Lorentz transformation equations for the fields.
 
  • #35
PeterDonis said:
I'm not sure what he was referring to, because I'm not sure anyone in the previous discussion was entirely clear about what the issue was. That's why I asked you to restate the issue.
i see, i thought you were asking for his sake
PeterDonis said:
If you mean, to show that ##\vec{E}^\prime = \vec{v} \times \vec{B}## (with the prime on the ##\vec{E}^\prime## to indicate that it is the field in a different frame), try looking at the Lorentz transformation equations for the fields.
but, as mentioned, I'd like to figure out how they proved it pre-relativity

as it was cited by Einstein in his 1905 seminal SR paper
 

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