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Hot answers tagged boundary-conditions

11

This is a general property of waves. If you have waves reflecting off a clamped point (like waves running on a string that you pinch hard at one point), the waves get phase inverted. The reason is the principle of superposition and the condition that the amplitude at the clamped point is zero. The sum of the reflected and transmitted wave must be the ...

9

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 ...

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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, ...

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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 ...

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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 ...

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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

The domain of the problem is the entire real line, not $[0,L]$. Otherwise the potential would not be specified for $x>L, x<0$. Thus, calculate the total energy of any wavefunction that does not vanish at $x = 0,L$, you will find it to be infinite. Therefore for all eigenstates that do not have infinite energy, i.e. the entire spectrum, the wavefunction ...

4

The charge is distributed uniformly on a spherical surface, but that is a function of the high degree of symmetry on the situation. In general the charge tends to accumulate most strongly near pointy bits and most weakly in depressions in the surface. There are several way to understand this. My favorite is not necessarily the most helpful for a beginner, ...

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 ...

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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: $$EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial ... 4 The most difficult part is to actually get a set of consistent boundary conditions in the first place - this requires a combination of educated guessing, physical insights, prior experience with related problems, detailed calculations and trial-and-error. In short, it is a bit of an art. However, once you have a set of boundary conditions (as in your case ... 4 A naive guess would be that there isn't any real difference. The theoretical logic behind my guess is that at low-energies FQH states are described by 2+1D Chern-Simons theories, which are topological gauge theories. Although the bulk does not have any local degrees of freedom, the boundary does. This is because in the presence of a boundary \partial M ... 4 By mirroring V(x) about x = 0, i.e., by setting V(-x) = V(x), the wavefunction can be taken to be even or odd. The even solution satisfies the Neumann boundary condition since the derivative of an even function is odd and thus zero at x = 0. 3 Dear James, there is no reason why \psi should be periodic. First, if you have a problem, imagine that \psi are just auxiliary variables but the true ones are the bilinears \psi_i \psi_j which are still periodic. Only things such as the world sheet stress-energy tensor T_{++} and T_{--} have to be periodic and they are because they're bilinear in ... 3 You cannot have a total vorticity with periodic boundary conditions, since if you take a path around all of your vortices, it will have a non-zero circulation. But you have periodic bc, so you can continuously deform that path to a point, and a point has zero circulation. Mirror images are not quite the same as in electrostatics. We want periodic boundary ... 3 One may view both (i) the infinite wall$$\tag{1} V(x)~=~\left\{\begin{array}{ccc}\infty & \text{for} & x>0, \\ 0 & \text{for} & x\leq 0, \end{array} \right. $$and (ii) the delta function potential$$\tag{2} V(x)~=~A\delta(x),$$as an appropriate limit of a finite barrier wall \tag{3} V(x) ~=~ V_0 ... 3 Good questions; I'm sure a lot of people are confused on this stuff (as I was the first time I used Jackson). Essentially your confusion boils down to being careful to consider the following fact: The Green's function for a particular boundary value problem depends on the boundary conditions. In particular, let's say you have a Dirichlet boundary value ... 3 Its a method called co-ordinate grid transformations that is used to transform an arbitrarily shaped geometry into a square; in the computational domain. Grid transformations work as parametric transformations do, in co-ordinate geometry. what these transformations basically do is, they map the complex geometry (viz. a curve) into a simpler geometry (a line; ... 3 You should not imagine a virtual photon as an individual object wandering from one charged particle to another. This picture is simply inapplicable. Unfortunately, Feynman diagrams mislead people to imagine such things. Actually, Feynman diagrams are good for calculation and bad for imagination. Feynman diagrams have been introduced to help physicists to ... 3 A j^\mu_a current is conserved if it satisfies the equations$$ \partial_\mu j^\mu_a = 0 \implies \partial_t j^0_a + \partial_i j^i_a = 0 ~~~~~~~~~~~...... (1) $$The first equation in your question is simply a definition. We can define an object called "charge" as$$ Q_a = \int d^{d-1} x j_a^0 $$Conservation of the current then implies that the charge ... 2 In the first method (the method of the Green function) you have removed the Laplacian operator from both sides of the equation. That is not a warranted operation in general case, for the potential V and the integral of \rho over the domain considered may differ by some non-vanishing function u(x,y,z) that obeys$$ \Delta u = 0. $$Only in special case ... 2 The 3D case is different than the 1D case, so don't feel too committed to the analogy. However, in the 1D case, you get only sine functions as solutions if you place a "hard wall" (an infinite potential barrier) at x=0. That is basically what is happening here: the origin is a sort of "hard" boundary for the radial equation since "negative radius" makes no ... 2 First of all, normalization is not a condition to fix the constant. Schodinger equation is a linear equation because the solutions form a linear space - Hilbert space. A physical state is an equivalent class of the vectors in Hilbert space, where two vectors belong to the same class if they differ by only a nonzero constant complex number. And it's clear ... 2 The formal structure shows, that the matrix in brackets is of same dimension as the matrix \xi^j_q. Therefore b_q^j is just a proportionality constant. The coatings/media of the optical structure are labeled by index j.$$\xi^j_q=b_q^j(R^{-1}V_q^j)\qquad \text{(22)}$$And then what is the meaning of the \bf b^{j+1}? The set of equations (22) is ... 2 When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate x must match that at coordinate x+L, so we have:$$\Psi(x)=\Psi(x+L) In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be ...

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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 ...

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