# Tag Info

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In general, boundary conditons must be adapted to the real situation. Zero boundary conditions are just for the sake of simplicity. But they are realistic only when the field is really zero for some definite reason. If the boundary is at infinity, zero boundary conditions means that everything of interest happens in a finite domain and cannot be noticed ...

7

Different boundary conditions represent different models of cooling. The first one states that you have a constant temperature at the boundary.This can be considered as a model of an ideal cooler in a good contact having infinitely large thermal conductivity The second one states that we have a constant heat flux at the boundary. If the flux is equal zero, ...

6

It is a very delicate matter to decide when solutions are unphysical or not. A classic example is Dirac's discovery of anti-particles: he found them as negative energy eigenstates for a relativistic Hamiltonian. A less insightful theorist might have discarded the negative energy solutions as unphysical, although we now know that those solutions mean a great ...

5

I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature. Example: To be concrete: an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given, while a boundary value problem could be to ask ...

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In the Lorentzian case: I am not aware of anyone studying it, and don't know explicit counterexamples off-hand. But I have doubts on the uniqueness. With the Lorentzian case, the nature candidate to draw comparisons with is the wave equation. And we see that on something as simple as the unit square $[0,1]\times [0,1]$, the wave equation with vanishing ...

5

There is a standard book which contains everything about electrostatics, the Laplace/Poisson equation and boundary conditions: Classical Electrodynamics by J. D. Jackson. Get the book from the library of your choice, read all chapters labeled "Electrostatics", and you will find the answers to all your questions (if you are simulating this, you need to know ...

5

Topological degeneracy is only defined in the thermodynamic limit on a closed manifold. The ground state degeneracy of a finite-sized system or on an open manifold is not "topological", and can not be called topological degeneracy. Considering your examples. (1) The ground state degeneracy is ill-defined with open boundary condition. Because there might be ...

4

A naive guess would be that there isn't any real difference. The theoretical logic behind my guess is that at low-energies FQH states are described by 2+1D Chern-Simons theories, which are topological gauge theories. Although the bulk does not have any local degrees of freedom, the boundary does. This is because in the presence of a boundary $\partial M$ ...

4

The most difficult part is to actually get a set of consistent boundary conditions in the first place - this requires a combination of educated guessing, physical insights, prior experience with related problems, detailed calculations and trial-and-error. In short, it is a bit of an art. However, once you have a set of boundary conditions (as in your case ...

4

The second condition is saying that there is no discontinuity in the slope of the rope at the junction. In other words, there is no "kink" in the rope. Imagine if this assumption were to fail in the following way: $$\frac{\partial D_1}{\partial x}(0,t) = -1, \qquad \frac{\partial D_2}{\partial x}(0,t) = 1$$ Then near the origin, the rope would look ...

4

The Fourier transform of $y\left(x,t\right)$ from the time to the frequency domain is given by $Y\left(x,\omega\right)=\int_{-\infty}^{\infty}y\left(x,t\right)e^{i\omega t}dt$ and satisfies the differential equation: $$EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial ... 3 Dear James, there is no reason why \psi should be periodic. First, if you have a problem, imagine that \psi are just auxiliary variables but the true ones are the bilinears \psi_i \psi_j which are still periodic. Only things such as the world sheet stress-energy tensor T_{++} and T_{--} have to be periodic and they are because they're bilinear in ... 3 Good questions; I'm sure a lot of people are confused on this stuff (as I was the first time I used Jackson). Essentially your confusion boils down to being careful to consider the following fact: The Green's function for a particular boundary value problem depends on the boundary conditions. In particular, let's say you have a Dirichlet boundary value ... 3 You cannot have a total vorticity with periodic boundary conditions, since if you take a path around all of your vortices, it will have a non-zero circulation. But you have periodic bc, so you can continuously deform that path to a point, and a point has zero circulation. Mirror images are not quite the same as in electrostatics. We want periodic boundary ... 2 However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric. I guess you mean that putting another set of image charges on both sides of the conductor plane so that the system is symmetric ... 2 This is a very complex problem to solve so you will probably want to start with some simplifying assumptions such as N=2 to make it more tractable. You will be looking for a minimum energy solution where the energy is a combination of the gravitational potentials and the energy in the surface tension. Depending on parameters there may be some meta stable ... 2 You may think this way: take a perfect infinite crystal where Bloch theorem perfectly work and add potential which makes real crystal finite. Next question you may ask how this potential is "seen" by quasiparticles which have been obtained from infinite crystal consideration. This procedure is perfectly self-consistent and is applicable in all cases. Also, ... 2 If you were to stir an ideal no-slip liquid, your spoon would get stuck if it is to touch the bottom. You can try taking 2 glass plates, dipping them in water, and moving them relatively to each other, then get them touching together and try moving again, the friction increases and for ideal no-slip liquid the force required to move surfaces at given speed ... 2 The usual Fourier series formula states that a function g(x) with period 2\pi can be expressed as$$ g(x) = \sum_{k\in {\mathbb Z}} c_k \, e^{ikx}$$where c_n (the coefficients of the Fourier series) are given by$$c_k = \frac{1}{2 \pi} \int_{0}^{2 \pi} g(x) \, e^{-i x \, k}dx$$If we have f(x) with period L, we can relate with the above by ... 2 I) Let us just consider 1 dimension for simplicity. (The generalization to higher dimensions is straightforward). Then the volume factor V is just a length factor L. II) The standard Fourier series formulas can be derived from (12.1.7) and (12.1.6) by taken the length L to be L=2\pi. Then (12.1.7) and (12.1.6) become the standard Fourier ... 2 Since this has just been asked again, let me attempt an intuitive explanation. The real explanation is of course to match \vec{E} and \vec{B} at the interface and the direction of the reflected wave drops out, but this isn't especially intuitive. Let's calculate the ratio E_r/E_i as a function of the ratio n_t/n_i, and let's start with the ... 2 The usual derivation of the Euler-Lagrange equations forces us to assume that both the initial and final conditions are fixed. However, when one actually derives the equations, he sees that there are differential equations in time so from the knowledge of the initial state, including the velocities (or whatever derivatives are needed to specify the initial ... 2 The parity operator commutes with the Hamiltonian because of the symmetry in your potential. This says that all eigenstates of the Hamiltonian are eigenstates of the parity operator. Therefore, the only possible eigenstate solutions to the system are ones with even or odd parity. This fact will allow you to simplify the process of applying the boundary ... 2 The Hilbert space for a particle in a box has no associated momentum operator (as a momentum operator implies that the state space is invariant under space translations). So your attempt do define one leads to artifacts.  In general, the Hamiltonian must be a densely defined operator on the physical Hilbert space. But a square well potential is too ... 2 Its a method called co-ordinate grid transformations that is used to transform an arbitrarily shaped geometry into a square; in the computational domain. Grid transformations work as parametric transformations do, in co-ordinate geometry. what these transformations basically do is, they map the complex geometry (viz. a curve) into a simpler geometry (a line; ... 2 The formal structure shows, that the matrix in brackets is of same dimension as the matrix \xi^j_q. Therefore b_q^j is just a proportionality constant. The coatings/media of the optical structure are labeled by index j.$$\xi^j_q=b_q^j(R^{-1}V_q^j)\qquad \text{(22)} And then what is the meaning of the $\bf b^{j+1}$? The set of equations (22) is ...

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