# Struggling with Moving Conductor and Magnet Problem

• greswd
In summary, the Moving Conductor and Magnet Problem deals with the transformation of fields and their components under Galilean transformations, which are not compatible with Maxwell's theory. The notation (v ⋅ ∇) B represents the operation of taking the dot product of the velocity vector with the gradient operator, resulting in a scalar operator. This is different from the notation v ⋅ (∇B), which results in a vector operator. The derivation of this expression is not very convincing and the true solution to the problem lies in special relativity.
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.

greswd said:
What is the meaning of the notation: (v ⋅ ∇) B ?
Is it the same as v ⋅ (∇B) ?

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.

vanhees71 said:
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.
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

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

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Ibix said:
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:$$(\vec v\cdot\nabla)\vec B=\left(\begin{array}{c} v_x\frac{\partial B_x}{\partial x}+v_y\frac{\partial B_x}{\partial y}+v_z\frac{\partial B_x}{\partial z}\\ v_x\frac{\partial B_y}{\partial x}+v_y\frac{\partial B_y}{\partial y}+v_z\frac{\partial B_y}{\partial z}\\ v_x\frac{\partial B_z}{\partial x}+v_y\frac{\partial B_z}{\partial y}+v_z\frac{\partial B_z}{\partial z} \end{array}\right)$$

thanks.

vanhees71 said:
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".

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?

vanhees71 said:
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##.

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

## 1. What is the "Struggling with Moving Conductor and Magnet Problem"?

The "Struggling with Moving Conductor and Magnet Problem" is a physics problem that involves a conductor (such as a wire) moving through a magnetic field created by a magnet. The movement of the conductor causes the magnetic field to change, which in turn induces an electric current in the conductor.

## 2. Why is this problem important?

This problem is important because it helps us understand the relationship between electricity and magnetism, also known as electromagnetism. It is also a fundamental concept in many everyday devices, such as generators, motors, and transformers.

## 3. How do you solve the "Struggling with Moving Conductor and Magnet Problem"?

To solve this problem, you will need to apply the principles of Faraday's law of induction, which states that the induced electromotive force (EMF) in a closed circuit is equal to the rate of change of the magnetic flux through the circuit. You will also need to use the right-hand rule to determine the direction of the induced current.

## 4. What are some common mistakes when solving this problem?

Some common mistakes when solving this problem include not considering the direction of the induced current, not using the correct equation (Faraday's law), and not understanding the concept of magnetic flux. It is also important to pay attention to the units of measurement and make sure they are consistent throughout the calculations.

## 5. How can I practice and improve my skills in solving this problem?

You can practice and improve your skills in solving this problem by working through similar examples and exercises, watching online tutorials or lectures, and seeking help from a teacher or tutor if needed. You can also try to apply the concepts to real-life situations to better understand the practical applications of electromagnetism.