# Tag Info

0

If the metric is Riemannian (positive) your conjecture (a maximally symmetric spacetime is a constant curvature spacetimes) is a known theorem: Theorem 3.1 in Transformation Groups in Differential Geometry by S. Kobayashi. From the proof, it seems to me that the result should hold in the Lorenzian case too, but without a closer scrutiny I am not completely ...

0

So actually I just computed it by hand from the very beginning. Starting from the expression of the field $A_i(\mathbf{x},t)=\sum_s\int\frac{\text{d}^3\mathbf{k}}{\sqrt{(2\pi)^3 2|\mathbf{k}|}}\left[a_s(\mathbf{k})\epsilon_i(\mathbf{k},s)e^{i(|\mathbf{k}|t-\mathbf{k\cdot x})}+\text{H.c.}\right]$ where $\epsilon_i(\mathbf{k},s),\,s=1,2,\, i=1,2$ are ...

-1

The search term you want is "gauge transformation," and if you read up on that (google comes up with lots of good hits), you'll find lots of different ways of thinking about this problem. But it seems what has you confused is this: I think you're assuming that if two functions have the same divergence, then they must be the same. But think about what that ...

2

First, the physical thing we care about is $\vec B$. So we can do anything to $\vec A$ we like as long as we get the same $\vec B$. That is, we can do anything that doesn't change the curl of $\vec A$. Now, suppose that $\vec \nabla\cdot\vec A = f$. Here's where Purcell neglects to stress what he means by "analogue of $\vec E$ in electrostatics" - the curl ...

4

Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation $$\frac{dx}{ds}= X(x(s))\tag{1}$$ with a generic initial condition $z$ at $s=0$ in the manifold $M$ (Minkowski spacetime in your example). $X$ is your vector field on ...

1

There are many conditions for identifying that a vector field $\vec v$ is conservative or not: $\nabla \times \vec v =0$ A conservative field vector is essentially irrotational. $\oint_c \vec v \cdot d\vec r =0$ Work done by a conservative vector field about any closed path $C$ is $0$. $\vec v=\nabla \phi$ A conservative vector field can always ...

2

Yes. 1. option: Another (equivalent) criterion: If the work done on a arbitrarily chosen closed path is zero, then the field is conservative. I.e: $$\oint \vec{F} \cdot \mathrm{d}\vec{r} = 0$$ means $\vec{F}$ is conservative. 2. option: More general definition of curl $$\left(\mathrm{curl} \ \vec{F} \right)\cdot \vec{n} = \lim_{A \rightarrow 0} ... 0 We remove an overall constant for simplicity. Let us use cylindrical coordinates (\rho,\phi,z), where$$\tag{1} x ~=~\rho \cos\phi, \qquad y ~=~\rho \sin\phi .$$Also assume the standard metric$$\tag{2} ds^2~=~\mathrm{d}x\odot \mathrm{d}x +\mathrm{d}y\odot \mathrm{d}y +\mathrm{d}z\odot \mathrm{d}z ~=~\mathrm{d}\rho\odot \mathrm{d}\rho ...

2

A physical system in GR is never isolated, in general, as it interacts with the curved metric, i.e., the gravitational background. (However a notion of isolated system can be given in the particular case of an asymptotically flat spacetime as discussed in auxsvr's answer.) Apparently this fact prevents the existence of conserved quantities because the ...

0

If the metric is asymptotically flat, it is straightforward to assign meaning to a quantity, such that it resembles the energy we know from special relativity. In particular, for the Kerr metric we may regard $(\partial_t)^a$ as a vector representing the stationary observer at infinity, where space-time is Minkowski, and the rest appears to said observer as ...

0

I doubt that your expression is correct. Your original equation is of the form \begin{equation*} \partial _{\mathbf{x}}\cdot \mathbf{E(x})=\rho (\mathbf{x}) \end{equation*} where $\rho (\mathbf{x})$ vanishes away from the $x_{3}$-axis. You can write \begin{equation*} \mathbf{E(x})=\mathbf{E}_{1}\mathbf{(x})+\mathbf{E}_{2}\mathbf{(x})=\partial ...

2

$\vec{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}$, and $\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$; the coordinate dependence is encoded in the $\vec{r}$. These expressions are in Cartesian components, but if you ever need to calculate this in curvlinear coordinates, the logic would be the same. If $\phi$ is not specified, you can probably assume ...

0

$B$ actually behaves as you explain, however there's a problem with $H$. You say: "However, the field lines must also curl around and meet the top surface of the magnet, where $H$ will therefore need to point in the negative z direction." Why the field lines need to meet the top surface of the magnet? Outside the magnet $B$ and $H$ are proportional so ...

0

The answer to my question is simpler than I suspected. It is fairly easy to describe the movements of both soldiers mathematically. The first soldier's spear is being transported along $X_1 = [y^2+z^2, -x^2, 0]$. The second soldier's spear along $X_2 = [y^2+z^2, 0, -x^2]$. Both are valid parallel transports relative to the tangent vector field $V=[0, -z, ... 0$\vec{v}=\vec{w}\times \vec{r}\nabla \times \vec{v}=\nabla \times (\vec{w}\times \vec{r}) =\vec{w}(\nabla .\vec{r})-(\vec{w}.\nabla)\vec{r}\nabla .\vec{r} = 3$and$(\vec{w}.\nabla)\vec{r}=w.(\nabla\vec{r})=w$therefore$\nabla \times \vec{v}=2w$or see:$\$\nabla \times \left( {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} ...

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