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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 ...


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

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 ...


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 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 ...


9

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: ...


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

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

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

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

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

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

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 ...


6

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

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

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 = ...


5

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

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

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

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

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

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

There is a gravitational analogue of the magnetic field. See gravitoelectromagnetism and frame dragging on Wikipedia.


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 ...


4

Indeed the answer is not zero but $-4\pi\delta(r)$ (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at $r=0$ we must be careful. At any other points is easy to calculate it. It is ...


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 ...



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