# Tag Info

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 ...

11

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 ...

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 $$\tilde{\mathbf F}(\mathbf k) = \tilde{\mathbf F}_\parallel(\mathbf k) + \tilde{\mathbf ... 10 With suitable boundary conditions, the decomposition is unique. Without them, it's not. Suppose that (\phi,{\bf G}) and (\phi',{\bf G}') are two different decompositions for the same function. Then$$ \nabla(\phi-\phi')+\nabla\times({\bf G}-{\bf G}')=0. $$Take the divergence of both sides to find that$$ \nabla^2(\phi-\phi')=0. ...

9

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

In general, if $\xi^\mu$ is a Killing vector field on a spacetime, and if $u^\mu$ is a tangent field along a geodesic, then it is true that $\xi_\mu u^\mu$ is a conserved quantity along the geodesic. (See for example Wald's GR proposition C.3.1). To illustrate the physical significance of this, let's take, for example, a particle moving in $2$-dimensional ...

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 ...

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

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 ...

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 ...

5

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.

5

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 ...

5

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 ...

4

Yes, your friend is right. Within electrostatics, an electric field $\vec{E}$ should be curl-free $\vec{\nabla} \times\vec{E}= \vec{0}$. The drawn electric field lines looks like the electric field is of the form $$E_x=E_x(y), \qquad E_y=0, \qquad E_z=0,$$ cf. the rule that to depict the magnitude $|\vec{E}|$, a selection of field lines is drawn such ...

4

This is more like a maths question to me. This is just an identity, which is true and facilitates the calculation and it is valid for any vector field. The proof, using Einstein summation convention would be something like: $$(\nabla \times \vec u )\times \vec u = \epsilon_{ijk}(\nabla \times u)_j u_k = \\ \epsilon_{ijk}\epsilon_{jlm}\partial_l (u_m) u_k ... 4 Edit edit: as has been pointed out, I was incorrect to say \partial_t = \partial_{t'} and so on. Serves me right for trying to look at it by inspection instead of being rigorous. Nevertheless, I do think cylindrical coordinates simplifies the problem somewhat. Recall the cylindrical line element:$$ds^2 = -dt^2 + dr^2 + r^2 \, d\phi^2 + dz^2$$Now, ... 4 I'm going to address the important concepts at play here in three dimensions. The issue here is to get straight the distinction between any function \mathbf v:\mathbb R^3\to\mathbb R^3, which we'll call a vector field, and an object that in addition to being a vector field in this sense, transforms in some prescribed way. To mathematically ... 4 Heat flux is a vector beceause it has a magnitude and a direction. Furthermore it has these properties in every point in space, which makes it a vector field. You can think of an analogy with the mass flux in a medium with inhomogenius density; diffusion will tend to equalize the denisty everywhere, so there will be specific motion of mass at every point ... 4 Edit: Note: I have posted another proof of this in another question, here. Those who prefer coordinates may find it slightly more palatable. I gather from your comments that you can do this if you have \mathcal{L}_X\text{Ric} = 0. Thus I will outline a somewhat more general result, assuming a certain identity connecting Killing vectors and Riemann ... 4 This is not a complete answer, but fills in some of the missing pieces Trimok asked about in the comments to Stan's answer: Note that I did not verify that Stan's proof actually works. Re 1)$$ \begin{align} \nabla_Z \nabla_Y X &= Z^\lambda(Y^\mu X^\nu_{;\mu})_{;\lambda} \partial_\nu \\&= Z^\lambda(Y^\mu_{;\lambda} X^\nu_{;\mu} + Y^\mu ...

4

This is just a consequence of Helmholtz's theorem - Any vector field can be written as: $$\mathbf{F}=-\boldsymbol{\nabla}\Phi+\boldsymbol{\nabla}\times\mathbf{A}$$ Now take the curl of both sides - since the curl of $\mathbf{F}$ is zero by assumption, and the curl of a divergence is also zero (write it out), we must have ...

4

Yes, you can have such "isolated" isometries. Consider the real line $\mathbb R$ and the inversion mapping $x\to -x$. This isometry does not arise from a killing vector because it's not "continuously connected to the identity."

4

Let $\lambda$ be an affine parameter of the integral curves of $\xi^{\nu}$ then you question translates as $$\frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = \xi^\mu \nabla_\mu(\xi_{\nu}\xi^{\nu}) = (\xi^\mu \nabla_\mu\xi_{\nu})\xi^{\nu} + \xi_{\nu}(\xi^\mu \nabla_\mu\xi^{\nu})$$ if the connection is Levi-Civita (i.e metric compatible) $$... 4 This is more an extended comment to address the comments to Kyle's answer For example if there was a time component to the electric and magnetic field In a relativistic context, the electric and magnetic field components are not components of separate, related vector fields but, rather, are components of a 2nd rank tensor field; the electric and ... 4 To answer this question, you need to have a full geometric understanding of the Maxwell equations and what they represent. Maxwell's equations are a garden-variety system of PDEs. In STA notation, it's simply$$\nabla F = -J$$We take it for granted that F is a bivector, and thus has 6 components, and that J is a vector, and thus has 4 components. ... 4 You have two options here: 1) Forget you ever had x,y,z coordinate system and plug H_{i'} into determinant. 2) Compute curl in x,y,z coordinates and see how it looks in x',y',z'. You can easily check that these two give different results. The reason is, curl operator really maps vectors not to vectors but to antisymmetric tensors. Antisymmetric ... 4 Gradient is covariant! Why? The components of a vector contravariant because they transform in the inverse (i.e. contra) way of the vector basis. It is customary to denote these components with an upper index. So, if your coordinates are called q's, they are denoted q^i. Therefore, the gradient (or a derivative if you prefer) is$$\partial_i = ...

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