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tade
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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?
Thanks, though sorry, could you explain the meaning of this notation: $$(\vec v \cdot \nabla)\vec B$$Orodruin said: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.
BvU said: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 detailsBvU said:Lorentz 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.
vanhees71 said:I'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.tade said:oh no, which part is wrong?
tade said:and what are the steps of the correct derivation?
Orodruin said: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.Back to my previous question, as the low-velocity approximation is fine for me, I'd like to know the notation's meaning and the intermediate steps, thanksOrodruin said:The transformation rules are being given in the low-velocity approximation, not in its full relativistically invariant glory.
tade said: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
tade said:oh no, which part is wrong? and what are the steps of the correct derivation? thanksvanhees71 said:I'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.vanhees71 said: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.
vanhees71 said: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##:
$$(\vec{v} \cdot \vec{\nabla}) B^j=v^k \partial_k B^j.$$
Orodruin said:Also, the only thing necessary to reach that expression is the chain rule for derivatives.
tade said: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
I’m not familiar with Ricci calculus sadlyOrodruin said:It 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 ...
@vanhees71tade said:I’m not familiar with Ricci calculus sadly
Orodruin said: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).
No. This is just the standard definition of a derivative, it has nothing to do with any physical quantity. As I already said, you will also need the chain rule for derivatives.tade said:ε is the quantity of time right?
As in, in this specific case of the moving magnet, ε represents t?Orodruin said:No. This is just the standard definition of a derivative, it has nothing to do with any physical quantity. As I already said, you will also need the chain rule for derivatives.
vanhees71 said:Again, in Cartesian components it reads
$$\vec{v} \cdot \vec{\nabla}) \vec{B} = \vec{e}_j v^k \partial_k B^j.$$
yes I believe that is correct.tade said: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 $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?
No. It is just a number. Again, the relation to ##t## comes from the chain rule.tade said:As in, in this specific case of the moving magnet, ε represents t?
Yes.tade said:@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 havetade said: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 $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?
Orodruin said:$$\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,$$
because the magnetic field is divergence free. Thus ##\vec E' = \vec v \times \vec B## solves the Maxwell-Faraday equation.
Yes, this follows directly from ##\vec B## being divergence free.tade said:by the way, the divergence of E should be zero right?
I was thinking about the divergence of v × B. It's a triple product, so it is equal to -v ⋅ (∇ × B).Orodruin said:Yes, this follows directly from ##\vec B## being divergence free.