18

You don't. Take a set of short permanent magnets. Chain them together. Make a knot out of the chain, and connect the ends. Or form two chains. Make them into linked closed loops.


17

The problematic line in your reasoning is in assuming that $\nabla \times \vec F=0$ implies that $\vec F$ is conservative, i.e. that $\vec F = \nabla \phi$ for some $\phi$. If this were true, then it would imply that $\oint \vec F \cdot \mathrm d\vec r = 0$ via the fundamental theorem of vector calculus, but alas it is not. We are only permitted to make the ...


8

Based on OP's elaboration in the comments on their background, I will try to make this answer pedagogical and self-contained, erring ont he side of simplicity in the beginning; however, since OP mentioned some familiarity with topological manifolds and vector bundles, the concluding section will become more sophisticated. Before beginning, however, I'll do ...


7

A magnetic field line is a line that is tangent to the magnetic field vector at every point along the line. The question asks if two magnetic field lines can be linked, if the magnetic field is consistent with Maxwell's equations in a vacuum. The answer is yes, which I'll prove by constructing an example. The example In terms of the complex-valued field $\...


6

The core idea in this problem is that integrating the curl over the region inside the unit circle is not a valid operation, because the curl is not defined at the origin.* You can integrate over a region that looks like the unit circle, but has a wedge cut out of it so that the boundary goes most of the way along the unit circle, comes in along an almost-...


6

First of all we fix the general setup. Let $M$ be the space of phases, I assume that all structures I am henceforth describing are $C^\infty$ (a $C^2$ Hamiltonian would be pretty sufficient for many issues actually, $C^3$ for the validity of Liouville theorem). The Hamiltonian flow $\Phi$ is defined as follow. $$I_a \ni t \mapsto \Phi_t(a) \in M$$ is the ...


6

No, it is quite easy to have a situation where $A$ and $B$ are spacelike separated, as are $B$ and $C$, but $A$ and $B$ are timelike separated. Consider an example with $A=(t_{0},0,0,0)$ and $B=(t_{0},d,0,0)$; clearly the separation between these two is spacelike. However, any point $C=(t_{1},0,0,0)$ in the direct future (or past) of $A$ will be spacelike ...


6

QFTs are definitely sensitive to the topology of spacetime. As a matter of principle, you can of course measure this topology. You can even do it classically: get on a rocket and keep going; if the universe turns out to be a torus, you'll eventually come back home (and fly arbitrarily close to Henri Poincaré, infinitely many times). In practice, of course, ...


6

There is no physical singularity. $r{=}0$ isn't part of the manifold. This isn't just nitpicking, because the behavior you get in the $r{\to}0$ limit doesn't really fit with the idea of the singularity as a point at the center. In the Schwarzschild case, two geodesics that approach $r{=}0$ at the same Schwarzschild $t$ but different $θ,\phi$ end up out of ...


6

The misunderstanding many have is that topology just the study of topological spaces. It is really also about continuous functions between two topological spaces. If one has an infinite lattice model, where the Hilbert spaces is more-or-less $\ell^2(\mathbb{Z}^d)$, and if the Hamiltonian is periodic (a huge limitation) then one has momentum space that is ...


5

A relatively recent review by Jean-Pierre Luminet discusses this subject at some length. Broadly, it can be summarized as follows: The requirements of homogeneity and (local) isotropy do not necessarily mean that the spatial slices of the cosmos are diffeomorphic to $S^3$, $\mathbb{R}^3$, or $H^3$; these are merely the simplest examples of spaces with ...


5

I mean how do we know that the topology is the euclidean topology even in very very very small (or infinitesimal) intervals of time. We don't. Spacetime could very well be discrete, or maybe our entire conception of space and time is ultimately wrong and needs to be revised. The reason we treat spacetime as continuous is because no experiment has ever ...


4

In general the representation of the time-reversal operator depends on the system that you consider. Let us start with some basic remarks on time-reversal symmetry prior to the spinor question that you are asking. Time-reversal means to go backwards in time, $t \rightarrow -t$. When we want to know the time-evolution of a quantum-mechanical state, we need to ...


4

As the three-form $(g^{-1} dg)^3$ is proportional to the Haar volume form on ${\rm SU}(2)$, and there is a tacit pull-back of this to $S^3$, the quantity $Q/24 \pi^2$ will be the integer winding number (Brouwer degree) of the map $g: S^3 \to {\rm SU}(2)\equiv S^3$.


3

Yes, it's possible that the uniformity that we observe is only local. The Lemaître–Tolman–Bondi family of exact solutions to GR generalizes the FLRW and Schwarzschild geometries, and can describe, for example, a huge uniform FLRW region with a slight positive curvature (small enough to be consistent with current data) surrounded by an infinite Schwarzschild ...


3

If the universe had a positive average curvature then it would be finite. The converse is not correct - a universe with zero or positive average curvature could be either finite or infinite, depending on its overall topology. Note that even a finite universe could still be so large that the limited size of the observable universe might mean we can only ...


3

The metric and the Riemann tensor are local constructs, which means that they completely specify the geometry of your manifold. These quantities don't say anything about the topology of the manifold, which is a global (in contrast to geometric/local) construct. Thus, you can in principle change the topology of a space while keeping $R_{\alpha\beta\mu\nu}=0$. ...


3

Think of a cone without the tip, e.g. , $$ds^2~=~-dt^2+ dr^2 +r^2d\theta^2, $$ $$ t~\in~\mathbb{R}, \qquad r~\in~\mathbb{R}_+, \qquad \theta~\in~[0,\theta_0],$$ where the angle $\theta=\theta_0$ is identified with the angle $\theta=0$. The Riemann curvature tensor vanishes everywhere, and this spacetime is locally isometric to the Minkowski spacetime in 2+1D....


3

An example that covers the whole spacetime with a finite number of charts is Kruskal-Szekeres coordinates. The only reason you need multiple charts for these coordinates is that the angular coordinates describe a two-sphere, and you can't cover a two-sphere with a single chart. Does the central singularity have to be excluded? In the standard black hole ...


3

I did not originally answer this question because I felt someone else might be able to offer a more authoritative answer on the matter. But in the intervening time I have become more confident in the answer I am about to give, though if anyone sees any issues with it, please feel free to let me know. Also, sorry if this is a little long, but I wanted to try ...


3

When you say topological hole, the implication is that you're talking about a spacetime that can be made by taking some other spacetime and removing some points. For example, I can take Minkowski space, define $r$ as the distance from the $t$ axis, and remove all points with $r\le a$. Relativists are generally not interested in this kind of spacetime, and ...


3

Ref. 1 is likely referring to the integral quantization condition $$\frac{1}{2\pi\hbar}\omega~\in~H^2(M,\mathbb{Z})\tag{1} $$ in geometric quantization of a symplectic manifold $(M,\omega)$, aka. classical phase space, cf. Ref. 2. The condition (1) is related to the Bohr-Sommerfeld quantization condition, cf. e.g. this Phys.SE post. Note that the remainder ...


3

It could make sense for a nematic liquid crystal, where $\uparrow$ and $\downarrow$ correspond to the same state. Otherwise the configuration is discontinuous and does not have a winding number. Where did you get this particular configuration?


3

Side note: As already pointed out in the comments, curvature of spacetime in general relativity is intrinsic, i.e. does not require extra spatial dimensions. As described in the Wikipedia article Shape of the universe, there are multiple aspects which describe the geometry of the universe, e.g. Boundedness (whether the universe is finite or infinite) Flat (...


2

One approach is to consider the Noether charge associated to this symmetry. In the absence of sources, it turns out that the Noether current associate to the electric-magnetic duality symmetry takes the form \begin{align} J^\mu=(h,\vec{s}), \qquad h=E\cdot C-B\cdot A, \quad \vec s=E\times A+B\times C \end{align} where $F=dA, \star F=dC$. The physical meaning ...


2

The source $S^1$ is, as is usual in the homotopical classification of defects, coming from considering a loop around a defect line (in this case a cosmic string). If no such nontrivial loops would exist, then you can always deform the string to a "pointlike" object, i.e. it is not a stringy (one-dimensional) defect.


2

Considering that, in the Schwarzshild interior, the time and space directions are flipped, a 2D analogy that might be more instructive is to imagine a globe, where your latitude coordinate is time, and your longitude coordinate is space. Start at the equator at $t=0$. You're free to travel around the circle that is your spatial part of the world, but you ...


2

You need to compare the topological data of these two manifolds. In this particular example they have different homotopy group, i.e. in the second example you do not need to cross the throat to get into the other side but in the first one you do need cross the throat.


2

As @benrg mentioned in his comment, there are papers on exactly this topic, e.g., Carlos Hoyos, Nilanjan Sircar and Jacob Sonnenschein, New knotted solutions of Maxwell’s equations. Hridesh Kedia, Iwo Bialynicki-Birula, Daniel Peralta-Salas, and William T.M. Irvine, Tying knots in light fields. There, the electromagnetic field is constructed using the ...


2

Yes a 2d object has an edge. You can draw a square in 2d and it will have 4 edges. A circle in 2d has one edge.


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