New answers tagged

0

When you compute a path integral, you must fix a start time and an end time, say $t_i$ and $t_f$. If you want to compute the amplitude to go from $q_i$ at time $t_i$ to $q_f$ at $t_f$, you sum over all paths where $q(t_i) = q_i$ and $q(t_f) = q_f$. If you are doing the phase space path integral, then you must also sum over all "momentum paths" as ...


2

I believe I see the misunderstanding. You are thinking of a ball, not a sphere. The closed 3-ball is the set of points $(x,y,z)$ such that $x^2+y^2+z^2\leq R^2$, like a basketball. The 2-sphere is the set of points $(x,y,z)$ such that $x^2+y^2+z^2=R^2$, like the surface of a basketball. A closed $n$-ball is compact and has a boundary, namely the $(n-1)$-...


0

So it is said that, as the areas $\delta A_1$ and $\delta A_2$ shrink together, the total charge remains finite. But why would the two areas not shrinking together cause the total charge to become infinite? This isn't what they're saying. The misreading you've done here is analogous to the following: "As $x \to 0$ in $(\cos x)/x$, the numerator remains ...


3

This is because of a mathematical property of variational calculus that I propose to call 'Jacob's lemma', after the mathematician who first pointed it out. Presumably this mathematical property has been rediscovered independently several times. The 'Jacob' is 'Jacob Bernoulli', brother of Johann Bernoulli. To put Jacob's lemma in context: some history of ...


0

The Euler-Lagrange equation establish the equivalence between the boundary value problem and the initial value problem. Here’s one way to think about it: Suppose we only start only knowing about the principle of least action. That is, if you know $x(t_i)$ and $x(t_f)$, as the initial and final positions of the particle, then you can figure out $x(t)$, the ...


1

Unlike what OP seems to suggest (v3), Newton's 2nd law and Euler-Lagrange (EL) equations are strictly speaking just differential equations (DEs) without conditions. Rather the context provides the appropriate conditions, such as, e.g., initial conditions (ICs) or boundary conditions (BCs). Together with the DEs, they constitute an initial value problem (IVP)...


-1

Correct me if I'm missing something @HeyDosa, but for the Laplace equation, $\nabla^2 \phi =0$, $\phi$ has the uniqueness property that if it is specified for the boundary of the region(volume) where you want to find it, then it is uniquely determined. Your region of interest is (in cylindrical coordinates), $S = [0,a]\times [0,2\pi]\times [-\infty,0] \cup [...


2

I'm using my own symbols, for readability and writability but it'll be clear what's what, I hope. Steady state wave equation in cylindrical coordinates: $$\nabla^2u(r,\varphi,z)=0$$ $$\frac1r \partial_r (ru_r)+\frac{1}{r^2}u_{\varphi \varphi}+u_{zz}=0$$ $$\frac1r u_r+u_{rr}+\frac{1}{r^2}u_{\varphi \varphi}+u_{zz}=0$$ The domain for $R$, $[0,+a]$, for $\Phi$...


0

Magnetization produces a bound current on the boundary, $$\mathbf{K}_b=\mathbf{M}\times\mathbf{\hat n}.$$ This current produces a B-field perpendicular to both the surface normal and the current, as you can easily confirm using Ampere's Law. Since there is no normal component of the B-field due to the magnetization, the normal component of the total B-field ...


0

The normal component of the B-field at an interface is always continuous, since $$\oint {\bf B}\cdot d{\bf S} = 0$$ and so $$B_{1\perp} = B_{2\perp}\ .$$ If there are no (non-induced) surface currents however, the component of the H-field parallel to the interface is the same on either side, which means the parallel component of the B-field will change ...


1

This form (1) is just an idealization for a matter-energy distribution which is very thin (in space), in such a way that the integrated quantities (energy, momentum flow, internal forces) are supposedly finite. This leads (through Einstein equations) to constraints on how "bad" the spacetime geometry can "vary" through boundaries (3d ...


0

The momentum transfer $q$ has to be an even integral multiple of $\pi/L$. There is no need for it be in the allowed momenum lattice.


1

These are very good questions. Refs. 1 & 2 are not entirely consistent on these issues. Let us analyze the situation. In general a Hamiltonian version of the stationary action principle is of form $$ S_H[z]~=~\int_{t_i}^{t_f}\! dt~L_H(z,\dot{z},t),\tag{1}$$ where the $2n$-dimensional phase space has (not necessary canonical) coordinates $(z^1,\ldots,z^{...


0

I found two sources which discuss microscopic explanations for the no-slip condition. The first$^1$ is from 1973: It has been argued that the no-slip boundary condition, applicable when a viscous fluid flows over a solid surface, may be an inevitable consequence of the fact that all such surfaces are, in practice, rough on a microscopic scale : the energy ...


0

I would say this happens because energy can't be destroyed, remember that for all waves amplitude is proportional to energy. So if the wave travelling through the more dense medium reflects with the same amplitude, and on top of that creates another wave in the less dense medium, then that would just create energy out of thin air. P.S. Perhaps this question ...


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