# Tag Info

## New answers tagged boundary-conditions

0

As long as $\mathbf{B}$ is a continuous (once-differentiable) function, when you look at small enough sizes, $\mathbf{B}$ has a Taylor series, the first term of which is a constant. As you let the loop size shrink, only the constant term matters. But then $\int_S \mathbf{B}\cdot d\mathbf{a}\rightarrow \mathbf{B}\int_S d\mathbf{a}\rightarrow 0$ since the area ...

0

The walls do not really matter that much here. I would say that the only points that matter to understand that quote are: The confining potential is conservative so that $K+U = constant$ The confining potential is virtually flat $U = 0$ almost everywhere inside the box (say 99% of the box) except very close to the walls These two conditions are enough to ...

0

Here's an image of the 'particle in a box' model: At the endpoints, you can see that there is an infinite potential outside the boundary of the box and zero potential inside. The unrealistic assumption in this model is that every time the particle reaches the boundary, an infinite force repels it to keep it in the box.

3

123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means ...

-1

The real answer is: it doesn't really. Or rather: we can still extract some physics out of it! Lets derive $$S[\phi] = \int d^d x \mathcal{L}[\phi,\partial\phi]\\ \delta S[\phi] = \int d^dx \delta \mathcal{L} = \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta ... 4 The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does ... 10 Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ... 3 This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ... 0 If it wouldn't be its tangential component would exert a force on the charges and they would move. The condition would then not be static. After some movement and redistribution of charges (when there would be no force on charges) the condition will again become static thus making the field only normal to the surface 1 Here's a simple explanation based on the dipolar nature of the medium: For most of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an isotropic medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor ... 1 The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the plane of incidence. Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave. Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the ... 1 There is a simple answer: Symmetry. Suppose the material is isotropic, and consider the initial condition of the p-polarized case, with a p-polarized light wave about to hit the surface. In this case, reflecting in the plane that contains the incident and scattered wave vectors leaves both the (vector) electric field and (pseudovector) magnetic field ... 2 As far as I know, and I'm only an undergraduate student, the boundary conditions in Schrodinger equations are there to hold some special subspace of the Hilbert space of the system or the Hilbert space as a whole. Bound states, for example, form a subspace on the Hilbert space. The boundary condition to that is that \psi\sim e^{-r} at infinity, for every ... 3 In the thermodynamic limit (linear size of the system L to infinity), boundary conditions don't really matter, and most physical observables will be the same for all boundary conditions. The use of periodic boundary conditions is mostly for practical reasons, in particular, translation symmetry is conserved, which really helps. One could in principle do ... 4 The choice of boundary conditions fixes the domain and thus a self-adjoint extension of your Schroedinger operator$$-\frac{\hbar^2}{2m} \Delta + V\:. In turn, this choice determines the spectrum of that operator, i.e., the eigenvalues $E$. (There is no analogy with initial conditions here.) No circumstances! Boundary conditions are uncorrelated with ...

7

In the examples you've given, the boundary conditions simply say "don't have infinite energy" and "don't be non-normalizable". These don't really have a physical interpretation. Moreover, no boundary conditions ever have an interpretation like "initial position and velocity" because the time-independent Schrodinger equation describes stationary states. ...

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