# Tag Info

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The Lie derivative has a geometrical meaning: it measures the change of a tensor field (including scalar function, vector field and one-form), along the flow of another vector field. For example, the Lie derivative of the metric tensor along a Killing vector is zero (this defines the Killing vector equation). It means that a tensor (for example the metric) ...

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We have the frame $\{e_\mu\}_{\mu=0,\dotsc,3}$ in terms of which the velocity vector is $v=v^\mu e_\mu$. There are a few properties of the affine connection which I would like to summarize: $$\nabla_{fX}Y=f\nabla_XY$$ $$\nabla_X(fY)=f\nabla_XY+X(f)Y$$ $$\nabla_{e_\mu}e_\nu=\Gamma^\lambda{}_{\mu\nu}e_\lambda$$ Using this, let's get to work. We have ...

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In a sense, parallel transport, covariant derivative and connection are all synonym for you can recover one from the other. So given a manifold one usually starts by giving one notion (e.g. how a vector field is transported parallel to itself along a family of curves) and then, if needed, the other objects are derived. In physics, when dealing with a ...

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$\nabla^2\vec A = (\nabla \cdot \nabla)\vec A$, not $\nabla(\nabla \cdot \vec A)$. The former is usually written $\nabla^2$, whereas the latter is usually written out explicitly.

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There is nothing wrong with having a divergence-less vector potential, as you can always arrange for this situation (cf. Coulomb gauge) without changing the fields, which are the physically relevant objects. Observe that the curl of a radial field is always zero.

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In light of the fact, documented in the paper by Carron, N. J., that the simplicity of the system in question allows for a gauge invariant formulation of Maxwell's Laws (not requiring Lorentz invariance -- since everything is motionless), the answers thus far given are inadequate due to their appeal to the arbitrary choice of gauge. It is for this reason ...

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Good terminology question. Let's work in some differentiable manifold $M$, our transformation is a smooth map $T: M \to N$. In the case of a rotation $M = N$. Our $\phi$ is a smooth function $\phi: M \to \mathbb{R}$. In classical field theory the fact that $\phi$ maps to $\mathbb{R}$ is often expressed by the statement "$\phi$ is a scalar field". Now ...

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As usual when dealing with transformations one has to be careful whether they are active or passive. If I understand your question you are implying a passive transformation, which is a mere change of coordinates. In this case all you are doing is changing the way you assign a scalar value to a "vector" of coordinates. Therefore $\phi(\mathbf x) = ... 0 You are correct that Aharanov and Bohm's effect means that the potential$A^\mu$cannot be blithely disregarded. However, that doesn't mean it's physically meaningful or well defined in the way you seem to imagine. It's still a gauge variant quantity meaning it can be locally set to zero by a gauge transformation. Its physical significance (independently of ... 0 The diagram for A resembles that of a field due to a dipole that varies inversely as r cube which is not the case here though directions of A are OK. However the situation here is that of a current sheet of a cylinder that has been deformed so that the ends meet. There is no B outside. By redefining A all over the space you end up with zero A outside and ... 1 There are two kinds of momentum. One kind is simply a frame dependant portion of a larger tensor. It is exemplified in the total stress-energy tensor, which is a symmetric rank two tensor that is divergence free. The divergence free part means that is is conserved locally in in the sense that the momentum (or energy) in a region at one time, is equal ... 6 Comments to the question (v3): I) The notions of vectors, tensors, scalars, etc, depend on contexts in physics, cf. e.g. this and this Phys.SE posts and links therein. II) In OP's context, these notions refer to representations$\rho$of the Lie group$SO(3)$[and the corresponding Lie algebra$so(3)$] of 3D rotations, cf. e.g. Ref. 1. Let$\mathrm{i}L_k$, ... 2 Nice question. This is connected to gauge invariance. The interaction term in the Lagrangian (interaction between charge and field) is not gauge invariant. Thus, a curl-free vector potential, which corresponds to zero magnetic field, appears to have a non-zero momentum. Nevertheless the equation of motion is not changed in a new gauge, i.e., you will get ... 3 The Poisson bracket you wrote only works for position (which is not a vector in general, as coordinates do not transform as vectors or any other tensor quantity). For general vectors the correct Poisson bracket is $$\{L_i,A_m\} = - (\mathbf p\times\nabla_{\mathbf p} + \mathbf r\times\nabla)_i A_m,$$ which reduces to the relation you wrote if you take$A_m = ...

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UPDATE Thank you for posting back your thought experiment. My calculations apply again. I think I found the source of your confusion. Your analysis is missing the displacement current. You seem to think that the linear momentum of the electron could only go back to the current in the torus: Hence there is at least one scenario in which the electron can ...

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It sounds like you want an electron, a statically charged torus, and then to consider the two cases of the torus with and without a current. You also bring up the canonical momentum conjugate to position: $\vec{p}+e\vec{A}$, which evolves according to: $$... 0 The important point is that both t^a and v^b are vectoror fields defined (at least) on C. The idea is that if t^a\nabla_av^b=0 and you take v^b at a particular point p on C then v^b at all other points on C is the parallel transported vector from p. Another way of thinking about it is that the covariant derivative in a given direction is ... 1 I think the expression you are trying to write is$$c \left( r \cdot\hat r + \cos \theta \cdot \hat \theta + z \cdot\hat k \right)$$As you say this will not represent a physical quantity since since \cos \theta is dimensionless and r, k have dimensions of length. The expression$$c \left( r \cdot\hat r + z \cdot\hat k \right) would be an ...

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You know of course how to calculate a vectorial product. Then, let's calculate the two vectorial products with which you have a problem, according to the formula that you indicated for $\nabla _T$: \$ \ (I) \ -j \omega \mu \nabla _T \times \hat z H_z = \begin {bmatrix} {\hat {\vec x}} & {\hat {\vec y}} & {\hat {\vec z}} \\ {\frac {∂}{∂x}} & ...

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