# Tag Info

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I'm going to vote to close, for lack of a clear question statement. For the record: The first expression looks like a scrambled version of the solution via Green's function for the vector potential in the Coulomb gauge. (Among other things, there's a missing $e^{-i \omega t}$, and $\beta=\omega/c$ should not be allowed to escape the integral.) The ...

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If the particle is moving along some constrained curve then there must be some external force, $\mathbf{F_e}$, acting to constrain it to the curve. So the net force on the particle is: $$\mathbf{F_{net}} = \mathbf{F} - \mathbf{F_e}$$ and the work done on the particle is: $$W_{net} = \int (\mathbf{F} - \mathbf{F_e}).d\mathbf{s} = \int ... 0 If we consider Newton's laws, then$$m\ddot{ \vec x}=\sum_i \vec F_i,\tag1$$where \vec F_i is force from ith source. When you calculate work of force \vec F_i, you have$$W_i=\int_C \vec F_i d\vec s.\tag2$$Actual motion, whatever you set it to, results from the sum of all the forces acting on the partice, and the work of any given force is given ... 2 Here's a derivative-free explanation. For readers who are doing E&M at the college level, the other answers posted here are more comprehensive, but since the OP has stated a high-school knowledge with little math and physics knowledge, here's the primer: A vector is a quantity that, in order to be fully measured and described, needs to include both a ... 1 The two maxwell equations using divergence are$$ div \vec{D} = \rho \\ div \vec{B} = 0 $$at least in differential form. In integral form they are maybe more clearer for you. They are$$ \iint_{\partial V} \vec{D} \ d \vec{A} = \iiint_{V} \rho \ dV = Q(V) \\ \iint_{\partial V} \vec{B} \ d \vec{A} = 0$$The first equation just means the electrical flux D ... 1 Taking your (lack of) knowledge about differential geometry into account, this might be too hard to follow, but here it goes anyway: Let u_1,\dots,u_n be some tangent vectors with base point p and \omega the volume form, ie V = \omega_p(u_1,\dots,u_n) is the (possibly negative) volume of the parallelepiped spanned by these vectors. In case of three ... 4 Divergence is an operation that maps a vector field \vec D(x,y,z) to a scalar field {\rm div}\,\vec D(x,y,z). How do you calculate {\rm div}\,\vec D(x,y,z)? Either you follow the definitions using derivatives, which you can't if you don't know what a derivative is. Or you imagine the following: the vector field \vec D(x,y,z) tells you about the ... 3 In general, the statement that \nabla_\mu V^\mu transforms as a scalar does not quite fix the transformation properties of V^\mu. Rather, the most general such transformation would be$$V^\mu \mapsto V'^\mu + C^\mu,$$where V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu is the ordinary vector transformation law, and C^\mu is any quantity ... 1 When you write V^\mu you mean that V is a vector. Next \nabla_\mu V^\mu is called divergence of a vector. Finally answering your question, a vector field with constant zero divergence is called incompressible or solenoidal – in this case, no net flow can occur across any closed surface (according to the Gauss law). If it is a constant, but not zero, ... 3 1) One way to show that a map constitutes a 1-1 correspondence is by showing that it has an inverse. This is what is done here: elements of T_eG are tangent vectors to G at the unit element, and one-parameter subgroups of G are smooth homomorphisms \mathbb R\to G. Note that this is a notion more general than a closed 1-dimensional subgroup, for which ... 2 I ended up spending an hour or so wading through the papers, so here's my main conculsion from that: No, I don't think the critique of the papers is wrong; Nor do I think that the basic algebraic arguments in the three papers addressed by the critique are wrong either. The authors describe a particular "decomposition" of the electromagnetic fields into a ... 0 I think the answer to this is basically yes. You do have to be careful, because in general, you can't interpret inner products in relativity as measures of whether something is "orthogonal" to something else in the Euclidean sense. E.g., a lightlike vector has a zero inner product with itself. This is because the metric isn't the Euclidean metric. However, ... 1 To address the first question you can consider$$ \nabla_av^b := v^b ,_a$$since$$ \nabla : \Gamma(E) \rightarrow \Gamma(E\otimes T^*M)$$where E is any section(e.g. the vector field in question) and contracting it with the tangent vector field of the curve you get$$t^av^b,_a = 0 $$this is similar to contracting a vector field with a dual vector$$ ...

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You are on the right lines. I think the answer to the question has to be, yes, this is a valid form for the vector potential, but only if ... [insert your conditions here]. The conditions come from the analysis you have shown and from taking the divergence of the potential and equating to zero (the Coulomb gauge).

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To add to Qmechanic's Answer and TwoBs's Answer and answer "....what the heck is h then? Any arbitrary function?": $h$ is pretty much arbitrary. It is wontedly taken to be at least differentiability class $C^1$ (all first derivatives continuous) so that the Lie bracket of vector fields is defined as in Qmechanic's answer. You need to assume it is of class ...

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Think of an infinitesimal Diff as of a translation where the the shift is space dependent, $x^\mu\rightarrow x^\mu+\epsilon^\mu(x)$. Now, you get that the generators are $L_\epsilon=\epsilon^\nu(x)\partial_\nu$ since $L_\epsilon x^\mu=\epsilon^\mu(x)$. They form an infinite space since $\epsilon^\mu(x)$ is a function that can be expanded in infinitely many ...

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Formally speaking, given a (differentiable, finite dimensional) manifold $M$, then the (infinite dimensional) Lie group of (globally defined) diffeomorphisms (with composition $\circ$ as group structure) has the set $\Gamma(TM)$ of (globally defined, differentiable) vector fields as corresponding Lie algebra. This (infinite dimensional) Lie algebra ...

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