What's the derivation in a moving magnet & conductor problem?

[Orodruin]

As in, in this specific case of the moving magnet, ε represents t?

No. It is just a number. Again, the relation to $t$ comes from the chain rule.

@Orodruin @vanhees71 Is it correct to say that the x-component of $$(\vec{v} \cdot \vec{\nabla}) \vec{B}$$ reads $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?

Yes.

 

[vanhees71]

Sorry, not familiar with Ricci calculus.

@Orodruin @vanhees71 Is it correct to say that the x-component of $$(\vec{v} \cdot \vec{\nabla}) \vec{B}$$ reads $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?

Yes, that's the x-component of what I wrote from the very beginning. Sorry, I was not aware that you didn't know Ricci calculus. In the usual matrix-vector notation you have
$$(\vec{v} \cdot \vec{\nabla} \vec{B})=\begin{pmatrix}
v_x \partial_x B_x + v_y \partial_y B_x + v_z \partial_z B_x \\
v_x \partial_x B_y+ v_y \partial_y B_y + v_z \partial_z B_y \\
v_x \partial_x B_z + v_y \partial_y B_z + v_z \partial_z B_z
\end{pmatrix}.$$

 

[tade]

$$\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,$$
because the magnetic field is divergence free. Thus $\vec E' = \vec v \times \vec B$ solves the Maxwell-Faraday equation.

by the way, the divergence of E should be zero right?

 

[Orodruin]

by the way, the divergence of E should be zero right?

Yes, this follows directly from $\vec B$ being divergence free.

 

[tade]

Yes, this follows directly from $\vec B$ being divergence free.

I was thinking about the divergence of v × B. It's a triple product, so it is equal to -v ⋅ (∇ × B).

If the divergence of E is zero, then so must the divergence of v × B.

But the curl of B is not always zero, as it is proportional to the current density.

 

[Orodruin]

In vacuum the curl is zero. If there is not vacuum, the divergence of E will not be zero.

 

[tade]

In vacuum the curl is zero. If there is not vacuum, the divergence of E will not be zero.

If there's a current density but no net charge density, will the divergence of E be zero?

 

[Orodruin]

If there's a current density but no net charge density, will the divergence of E be zero?

No, but in the case described here there is no net charge density in the unprimed frame so since there is a current in the unprimed frame there will be a net charge density in the primed frame.

 

[tade]

No, but in the case described here there is no net charge density in the unprimed frame so since there is a current in the unprimed frame there will be a net charge density in the primed frame.

hmm, Einstein used this moving magnet and conductor problem as an introduction to his original paper on special relativity.

He argues that the electric and magnetic force explanations from different frames imply a preferred rest frame even though the effect is clearly one of relative motion.

So it seems like they managed to equate it mathematically without resorting to relativity.

 

[vanhees71]

I must say, I lost track of what's the issue discussed here, and to say it friendly, the Wikipedia in this case is pretty incomplete (Pauli comes to my mind, who was much less friendly in such cases).

First of all, although it's not explicit in the usual notation of the local Maxwell's equations, they are relativistic equations, and they can be written covariantly, i.e., you can use the Maxwell equations to calculate what happens in this famous case using any frame of reference (it's what Einstein discusses in the introduction in his famous "electrodynamics of moving bodies" paper of 1905 as the motivation for his reanalysis of the space-time description necessary to make these Maxwell equations obeying the special principle of relativity).

That said, I think it's not a very simple problem though, because you need the full Maxwell equations, and it's not easy to find a formulation, where you can really do the calculation analytically to the end, but you can do it at least formally, i.e., just keeping the fields, currents etc. in general terms. It's also good to start by thinking about what can be concretely measured.

I'd say, we take the conductor as a conducting ring and having put a volt meter in series (comoving with the ring). Supposed the motion is slow enough such that the voltmeter can follow the changes of the fields, what's measured is the electromotive force along the ring. I'm using Heaviside-Lorentz units (which are much more convenient for relativistic arguments than the SI).

(a) Moving-ring frame

In this case we have a given magnetostatic field $\vec{B}(\vec{x})$ obeying the corresponding static Maxwell equations, $\vec{\nabla} \cdot \vec{B}=0$, $\vec{\nabla} \times \vec{H}=\vec{j}=0$, $\vec{H}=\mu \vec{B}$, but that's not relevant for our question here.

Now we have the moving ring, described by $C:\vec{x}=\vec{r}(t,\lambda)=\vec{r}_0(\lambda)+\vec{v} t$, where $\lambda$ is the parameter for the curve. Also let $\vec{v}(t,\lambda)=\partial_t \vec{r}(t,\lambda)=\vec{v}=\text{const}$. Further let $A$ be an arbitrary surface with $C=\partial A$. Now we use Faraday's Law in integral form to get the electromotive force. We need the magnetic flux through the moving surface with the ring as its boundary:
$$\Phi(t)=\int_{A} \mathrm{d} \vec{x} \cdot \vec{B}.$$
The time derivative is not 0 although the magnetic field is static, because $A$ is moving, i.e., we get for the electromotive force
$$\mathcal{E}=-\frac{1}{c} \dot{\Phi}(t) = \frac{1}{c} \int_{\partial A} \mathrm{d} \vec{x} (\vec{E}+\vec{v} \times \vec{B}) =\frac{1}{c} \int_{C} \mathrm{d} \vec{x} \vec{v} \times \vec{B}.$$
That's the reading of the volt meter when calculated in this frame of reference.

(b) Moving-magnet frame

That's slightly more complicated ;-)). We have to use the Lorentz transformation of the fields first. In this frame the em. field becomes time dependent and has both electric and magnetic components. As is derived in any complete textbook of electromagnetics (most elegantly using the four-tensor formalism and writing the em. field components in terms of the antisymmetric Faraday four-tensor $F_{\mu \nu}$),
$$\vec{E}_{\parallel}'(t',\vec{x}')=\vec{E}_{\parallel}(t,\vec{x})=0, \quad \vec{E}_{\perp}'(t',\vec{x}') = \gamma [\vec{E}_{\perp}(t,\vec{x}) +\vec{\beta} \times \vec{B}(t,\vec{x}),$$
$$\vec{B}_{\parallel}'(t',\vec{x}')=\vec{B}_{\parallel}(t,\vec{x})=\vec{B}_{\parallel}(\vec{x}), \quad \vec{B}_{\perp}'(t',\vec{x}')=\gamma[\vec{B}_{\perp}(t,\vec{x})-\vec{\beta} \times \vec{E}(t,\vec{x})]=\gamma \vec{B}_{\perp}(\vec{x}).$$
Now
$$\vec{x}=\gamma(\vec{x}'+\vec{v} t') \qquad (*),$$
and since the area and its boundary don't move, we simply have
$$\mathcal{E}'=\int_{\partial A} \mathrm{d} \vec{x}' \cdot \vec{E}' = \int_{\partial A} \mathrm{d} \vec{x}' \cdot \gamma [\vec{\beta} \times \vec{B}(\vec{x})] = \frac{1}{c} \int_{\partial A} \frac{\mathrm{d} \vec{x}}{\gamma} \cdot (\gamma \vec{v} \times \vec{B})=\mathcal{E},$$
as it should be. In the prelast step we have used (*) to transform the line element $\mathrm{d} \vec{x}'=\mathrm{d} \vec{x}/\gamma$. Note that this is the correct transformation since the integral is to be taken at constant coordinate time, $t'$!

 

[tade]

I must say, I lost track of what's the issue discussed here, and to say it friendly, the Wikipedia in this case is pretty incomplete

I was asking Oroduin how physicists had established the mathematics of the moving magnet and conductor problem before relativity.

 

[tade]

$$
\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,
$$
because the magnetic field is divergence free. Thus $\vec E' = \vec v \times \vec B$ solves the Maxwell-Faraday equation.

I was thinking of the magnetic field at the center of a Helmholtz coil.

When we increase the size of the coil, but keep the magnetic flux density at the center the same, $\nabla \vec B$ gets smaller.

This should make $\vec E'$ smaller. But it has no effect on $\vec v \times \vec B$. Is this correct?

 

[vanhees71]

I was asking Oroduin how physicists had established the mathematics of the moving magnet and conductor problem before relativity.

Before relativity, Heinrich Hertz, among others, tackled the problem and came pretty close to the modern relativistic view, but it's pretty difficult to understand compared to the very clear solution of all the problems concerning "electrodynamics of moving bodies" within the completed (special) theory of relativity, which is due to Minkowski.

 

[tade]

Before relativity, Heinrich Hertz, among others, tackled the problem and came pretty close to the modern relativistic view

thanks, do you know what their solution to the scenario mentioned in #42 was?