In the Wikipedia page of the moving magnet and conductor problem, it asserts "This results in: E' = v x B", but does not elaborate why.
What's the full derivation?
What's the full derivation?
What is being referred to is 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.
The "better" way I think is to use the Lorentz transformation properties of the electromagnetic field.
I would like the derivation so I can understand how the two formulas give the same numerical results.A derivation of the Lorentz force would be somewhat circular: the expression for the LF is derived from observations. That's the way it is ...
If you find this view unsatisfactory, I can understand. But digging deeper doesn't really change this situation, I think.
As Oroduin mentioned, it is: $$\partial_t \vec B(\vec r+ \vec v t)=(\vec v \cdot \nabla)\vec B$$ with some intermediate steps, and I'd like to figure out the detailsLorentz force is ##\vec F_L = q (\vec v\times \vec B) ## and with ##\vec F = q \vec E## you are back at the 'result' expression.
oh no, which part is wrong? and what are the steps of the correct derivation? thanksI'm a bit worried about the fact that the OP is mislead by using utterly wrong transformation properties of the em. field. Here Lorentz transformations rather than Galilei transformations have to be applied. BTW it's the very problem Einstein used in his famous 1905 paper on special relativity as a motivation for the reformulation of the space-time model!
The transformation rules are being given in the low-velocity approximation, not in its full relativistically invariant glory.oh no, which part is wrong?
and what are the steps of the correct derivation?
The "better" way I think is to use the Lorentz transformation properties of the electromagnetic field.
I see, I guess that's not "utterly wrong" though.The transformation rules are being given in the low-velocity approximation, not in its full relativistically invariant glory.
Thanks, though sorry, could you explain the meaning of this notation: $$(\vec v \cdot \nabla)\vec B$$
And also the intermediate steps required to go from: $$\partial_t \vec B(\vec r+ \vec v t)$$ to: $$(\vec v \cdot \nabla)\vec B.$$
Hope its not too much of a hassle with a lot of TeX
@vanhees71 i'm good with the low-velocity approximations, keeping it simple :)oh no, which part is wrong? and what are the steps of the correct derivation? thanksI'm a bit worried about the fact that the OP is mislead by using utterly wrong transformation properties of the em. field. Here Lorentz transformations rather than Galilei transformations have to be applied. BTW it's the very problem Einstein used in his famous 1905 paper on special relativity as a motivation for the reformulation of the space-time model!
I do not think it needs to be didactically wrong as long as one is clear about being in the low velocity limit. To the contrary, keeping only the leading order terms in a small parameter expansion is an important tool in phenomenology.It's utterly wrong not only in a physical sense but also in a didactical. Classical electrodynamics is among the most difficult subjects in the undergraduate curriculum. It's not necessary to make it even more complicated by using oldfashioned concepts which have been solved more than 110 years ago by the development of special relativity, finalized by Minkowski in 1908.
I didn't get where the problem is. The operator is self-explaining by notation. Maybe it helps to write it down in the concrete Ricci calculus for Carstesian coordinates ##x^j##:
$$(\vec{v} \cdot \vec{\nabla}) B^j=v^k \partial_k B^j.$$
Also, the only thing necessary to reach that expression is the chain rule for derivatives.
sorry, I’m not clear about what the operator $$(\vec v \cdot \nabla)$$ meansThanks, though sorry, could you explain the meaning of this notation: $$(\vec v \cdot \nabla)\vec B$$
And also the intermediate steps required to go from: $$\partial_t \vec B(\vec r+ \vec v t)$$ to: $$(\vec v \cdot \nabla)\vec B.$$
Hope its not too much of a hassle with a lot of TeX
I’m not familiar with Ricci calculus sadlyIt is a directional derivative in the direction of ##\vec v##. It is written explicitly how it is expressed in post #18. I am sorry, but I do not see how it can be any clearer than that ...
@vanhees71I’m not familiar with Ricci calculus sadly
ε is the quantity of time right?Then just look at it as a directional derivative
$$
(\vec v \cdot \nabla) f(\vec x) = \lim_{\epsilon\to 0}\left(\frac{f(\vec x + \epsilon \vec v) - f(\vec x)}{\epsilon}\right),
$$
where ##f## is any field (scalar, vector, tensor, etc).