# I Struggling with Moving Conductor and Magnet Problem

1. Jun 3, 2016

2. Jun 4, 2016

### greswd

Because the magnetic flux is dependent on the longitudinal component of the B-field, and the Lorentz force is dependent on the transverse component, I'm wondering how they can be shown to be equivalent.

3. Oct 4, 2016

4. Oct 4, 2016

### vanhees71

First of all in components you have
$$\vec{v} \cdot \vec{\nabla} =v^j \partial_j$$
So applied to a vector field it's
$$[(\vec{v} \cdot \vec{\nabla}) \vec{B}]^k = v^j \partial_j B^k,$$
which is different from
$$[\vec{v} (\vec{\nabla} \cdot \vec{B})]^k=v^k \partial_j B^j,$$
which vanishes because of $\vec{\nabla} \cdot \vec{B}=0$.

The derivation on Wikipedia concerning the Galilei transformation of em. fields is, naturally, pretty mediocre. The correct derivation would be to take limits $c \rightarrow \infty$ of the Lorentz transformations. What they seem to do is the following: For the spacetime coordinates you have
$$t \rightarrow t'=t, \quad \vec{x} \rightarrow \vec{x}'=\vec{x}-\vec{v} t.$$
Then one simply assumes that the magnetic field is a scalar field under Galileo transformations, i.e.,
$$\vec{B}'(t',\vec{x}')=\vec{B}(t,\vec{x})=\vec{B}(t',\vec{x}'+\vec{v} t').$$
Next they assume that Faraday's Law holds in the $(t',\vec{x}')$ reference frame, i.e., (in SI units)
$$\vec{\nabla}' \times \vec{E}'=-\partial_{t'} \vec{B}'.$$
Now according to the above transformation law you have via the chain rule (for a time-independent $\vec{B}$ field in the unprimed frame)
$$\partial_{t'} \vec{B}'(t',\vec{x}')=(\vec{v} \cdot \vec{\nabla}) \vec{B}(t',\vec{x}'+\vec{v} t').$$
Now you have
$$[\vec{\nabla} \times (\vec{B} \times \vec{v})]_k=\epsilon_{kij} \partial_i \epsilon_{jlm} B_l v_m = (\delta_{kl}\delta_{im}-\delta_{km} \delta_{il}) v_m \partial_i B_l =v_m \partial_m \partial_i B_k-v_k \partial_i B_i = [(\vec{v} \cdot \vec{\nabla}) \vec{B}]_k.$$
So you can write Faraday's Law of induction as
$$\vec{\nabla}' \times \vec{E}'=-\vec{\nabla}' \times (\vec{B}' \times \vec{v}).$$
Now just ignoring a possible gradient term they conclude
$$\vec{E}'=-\vec{B} \times \vec{v}=\vec{v} \times \vec{B}.$$
As you see, it's not really convincing since there are many ad-hoc assumptions and inaccuracies going into the "derivation".

As we know nowadays, the true problem is that Maxwell's theory is a relativistic field theory and thus not compatible with the Galileo transformations. The solution of these problems is special relativity, and indeed, as cited in the Wikipedia article, that's how Einstein came to discover it as a solution of the inconsistency between Galileo spacetime structure and Maxwell electromagnetics.

5. Oct 4, 2016

### greswd

Sorry to ask again, but, what is the meaning of the notation: (v ⋅ ∇) B ?

Can you explain it in terms of breaking things down into the x,y and z components?

Like how curl is explained in this Wikipedia article: https://en.wikipedia.org/wiki/Curl_(mathematics)#Usage

6. Oct 4, 2016

### Ibix

Both v and the gradient are vectors (or a vector and a co-vector if you want to go the whole differential geometric hog - if that means nothing to you, don't worry about it for now). You take the dot product in the obvious way: $$\vec v\cdot\nabla=v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}$$That's just a scalar, so you apply it to the vector in the same way you multiply any scalar and vector:$$(\vec v\cdot\nabla)\vec B=\left(\begin{array}{c} v_x\frac{\partial B_x}{\partial x}+v_y\frac{\partial B_x}{\partial y}+v_z\frac{\partial B_x}{\partial z}\\ v_x\frac{\partial B_y}{\partial x}+v_y\frac{\partial B_y}{\partial y}+v_z\frac{\partial B_y}{\partial z}\\ v_x\frac{\partial B_z}{\partial x}+v_y\frac{\partial B_z}{\partial y}+v_z\frac{\partial B_z}{\partial z} \end{array}\right)$$This is what the first two equations you quoted from @vanhees71 say ($\vec{v} \cdot \vec{\nabla} =v^j \partial_j$ and $[(\vec{v} \cdot \vec{\nabla}) \vec{B}]^k = v^j \partial_j B^k$), but he's put them in a more compact form using upper and lower index notation and the Einstein summation convention.

It's different from $\vec v(\nabla\cdot B)$. Again, $\nabla\cdot B$ is just a dot product, which gives a scalar which you multiply by the vector v. I'll leave you to figure out the components if you want.

Last edited: Oct 4, 2016
7. Oct 24, 2016

### greswd

thanks.

It's interesting that they ignore the gradient term.

We can re-write the problem with an electromagnet that obeys the Biot-Savart Law, and apply Faraday's Law in Integral Form in the conductor frame, and integrate the Lorentz force Law across the conductor in the magnet frame.

This time there would be no need to ignore any terms. Would we get the same result?

8. Oct 28, 2016

### greswd

Do you know how they prove that ∂B'/∂t = (v ⋅ ∇) B?