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## Hot answers tagged vector-fields

19

Killing fields are one of the most important concepts in general relativity both in its classical as well as quantum versions. Classically, one thing we are always interested in is the world-line/trajectory of a free-falling observer in curved space-times. These world-lines are described as geodesics and satisfy the equation $$\frac{d^2 x^\mu}{d\tau^2} + \... 16 There is a sort of analog called gravitomagnetism (or gravitoelectromagnetism), but it is not discussed that often because it applies only in a special case. It is an approximation of general relativity (i.e. the Einstein Field Equations) in the case where: The weak field limit applies. The correct reference frame is chosen (it's not entirely clear to me ... 16 The fact is that, in the general case$$ \vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t}; $$(signs depend on conventions used) where \vec{A} is called vector potential. You can consult for example Wikipedia. Let us consider homogeneous Maxwell equations:$$ \begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + \frac{\...

13

Actually, the electric and magnetic fields from one combined tensor called the electromagnetic field tensor. This is a rank-2 tensor and takes the form* $$F^{\mu\nu}=\left(\begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{array}\right)$$ It ...

12

I think https://en.wikipedia.org/wiki/Killing_vector_field answers your question pretty good: "Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. More simply, the flow generates a symmetry, in the sense that moving each point on an object the same distance in ...

12

We have also the same notions of derivation, curl, etc... for functions that are less regular. When you write Maxwell's equations, you are writing a system of partial differential equations. To investigate them, you have to specify the type of solution you look for (in the language of PDEs: classic, mild, weak...) and the functional space you set your ...

12

Let me try this more clearly than the other answers, which aren't wrong. You ask: So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition? There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable ...

11

With suitable boundary conditions, the decomposition is unique. Without them, it's not. Suppose that $(\phi,{\bf G})$ and $(\phi',{\bf G}')$ are two different decompositions for the same function. Then $$\nabla(\phi-\phi')+\nabla\times({\bf G}-{\bf G}')=0.$$ Take the divergence of both sides to find that $$\nabla^2(\phi-\phi')=0. ... 11 I was always told that to find whether or not a field is conservative, see if the curl is zero. This is almost always true, but not always true. I have now been told that just because the curl is zero does not necessarily mean it is conservative. Correct! To illustrate what's going on, let's do an example. Conside the following vector field:$$\vec{...

11

On spherical coordinates, the gradient of a general function $V$ is: $$\nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi$$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...

10

As long as the field can be Fourier transformed, $$\tilde{\mathbf F}(\mathbf k) = \frac1{(2\pi)^{3/2}} \iiint e^{-i\mathbf k\cdot\mathbf r} \mathbf F(\mathbf r) d^3\mathbf r,$$ we can separate $\tilde{\mathbf F}$ into the longitudinal and traverse parts $$\tilde{\mathbf F}(\mathbf k) = \tilde{\mathbf F}_\parallel(\mathbf k) + \tilde{\mathbf F}_\perp(\... 10 In general, if \xi^\mu is a Killing vector field on a spacetime, and if u^\mu is a tangent field along a geodesic in that spacetime, then \xi_\mu u^\mu is a conserved quantity along the geodesic. (See for example Wald's GR proposition C.3.1). To illustrate the physical significance of this, consider a particle moving in 2-dimensional Minkowski ... 10 Curl can be equated with the closed line integral in the limit that the encircled area \Delta S goes to zero. However, we would have to do this in three components because curl is a vector.$$ (\nabla \times \vec{v})_x = \lim_{\Delta S \rightarrow 0} \frac{1}{\Delta S} \oint \vec{v}\cdot d\vec{l} $$in the yz plane and so on. But what does it mean? ... 9 No, the statement is false even in the electric case. At the very beginning, the acceleration is \vec a \sim \vec E so they have the same direction at t=0: the tangents agree. However, as soon as the particle reaches some nonzero velocity \vec v \neq 0, its acceleration is still \vec a\sim \vec E, in the direction of the field lines, however its ... 8 In terms of classical general relativity: Einstein's equations$$ G_{ab} = 8\pi T_{ab} $$can be formulated, in local coordinates, as a system of second order partial differential equations for the metric unknown g_{ab}. The matter field equations further generate some family of partial differential equations. Given a continuous symmetry (as guaranteed ... 8 The answer to your question depends on the context, but the basic unifying theme distinguishing different kinds of fields (like vector fields, scalar fields, etc.) is how these fields transform when they are acted on by Lie Groups (and or Lie Algebras) which falls under the mathematical subject of representation theory of Lie groups and Lie algebras. Here ... 8 In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ... 7 The group of isometries of a given connected smooth (semi) Riemannian manifold is always a Lie group. However, a Lie group can include subgroups of discrete isometries that, barring the identity, cannot be represented by continuous isometries and thus they have no Killing vectors associated with them. (Actually, only some elements of the connected component ... 7 Nice question. One way to think about it is that given a metric g, the statement \mathcal L_Xg = 0 says something about the metric, whereas \nabla_Xg = 0 says something about the connection. Now what \mathcal L_Xg = 0 says, is that the flow of X, where defined, is an isometry for the metric, while \nabla_Xg = 0 says that \nabla transports a ... 7 Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function f(x) to a field, which is a function of many variables f(x, y, z), we suddenly have ... 7 Imagine water flowing steadily in a stream, steadily enough that the surface of the water never changes shape. The water forms a three-dimensional manifold M. Watching how the water moves over a period of t seconds gives a diffeomorphism \phi^t \colon M \to M. If the current is carrying a diatom down the stream, and you see it at the point p, you ... 7 When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form \mathbf{E}=-\nabla V. 6 Pretty sure the question is about \frac{\hat{r}}{r^2}, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution. 6 You don't need a vector field on the sphere - you just need vectors. Vectors don't have any intrinsic location, just a direction and a magnitude. The polarization of light is independent of the propagation direction of the light. Let's examine this with a simple experiment: Consider an ideal plane-wave laser beam, beam 1, propagating in the z-direction ... 6 The most obvious example from physics is the Maxwell equation$$ \nabla\cdot \mathbf{E} = 4\pi\,\rho $$which simply states that the electric field \mathbf{E} "comes out of" any charged particle (where there is a finite charge density \rho), and does not have any source at places where \rho is zero. 6 It is just$$\partial F_x/\partial x + \partial F_y/\partial y + \partial F_z/\partial z $$and measures whether the field is a source or sink at a given place. A basic introduction is here: http://en.wikipedia.org/wiki/Divergence and the most important relationship that gives the divergence an "intuitively comprehensible" meaning is Gauss' theorem ... 6 To answer this question, you need to have a full geometric understanding of the Maxwell equations and what they represent. Maxwell's equations are a garden-variety system of PDEs. In STA notation, it's simply$$\nabla F = -J We take it for granted that $F$ is a bivector, and thus has 6 components, and that $J$ is a vector, and thus has 4 components. ...

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 $\mathrm{i}L_k$, ...

6

If the field is not stationary, curl of $\vec{E}$ does not vanish. So generally you cannot identify electromagnetic field with the curl-free part of the decomposition. However, you can indeed introduce a complex vector combination of electric and magnetic field, in a certain system of units it is $\vec{E}+i\vec{H}$. This is the so-called Riemann-Silberstein ...

6

It depends on how the quantity in question transforms. Almost always, densities in the form of "stuff per unit volume" and generally the "stuff" (like a charge) is a scalar (a number of things - number of elementary charges), but the volume it is contained in is observer dependent, owing to the Lorentz contraction. Therefore the density is observer-dependent....

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