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2

A few comments: First - you have a sign error in your code: solt = NDSolve[{ rc D[u[x, y, t], t] - k (D[u[x, y, t], x, x] + D[u[x, y, t], y, y]) + q[t] == <<<<<<<<<< NeumannValue[omega[t], True], DirichletCondition[u[x, y, t] == minT, t == 0]}, u, {x, 0, maxX}, {y, 0, maxY}, {t, 0, time}, Method -> ...


2

I second of course 'user115350's comment about $Q$, it should really be something like $Q(t)=Q_0\sin\Big(\frac{2\pi}{\tau}t\Big)$, where $\tau$ is the period of the heating cycle. I would also recommend a small transformation like $T \to u$, where: $u=T-T_{min}$ because it makes the initial condition homogeneous: $$u(x,y,0)=0$$ Let's have a look at the ...


2

A few comments: unit of Q will be $\frac{W}{m^3}$ $Q=sin(...)$ only gives you maximum heat source of 1 $\frac{W}{m^3}$. It seems too small to make significant change. Could you try $Q=1000*sin(...)$ My understanding Dirichlet is boundary condition but not for initial condition. I'm not sure if it is the case in your code library. You can check.


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I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason: In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$) tend to zero. Then we arrive at the result that the ...


2

Gauß law for electrodynamics states that $$\nabla \vec{E}= \frac{\rho}{\varepsilon_0} $$ at any point and no matter by what the electric field $\vec{E}$ is produced. $\vec{E}$ here stands for the total electric field at a point produced by all sources that are present, not just any selected few. So the whole derivation is valid for any surface and any ...


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The previous answers and comments inspired to me what follow: maybe this is the simplest (and so the best) way to see why $\frac{d}{dt} \int_S \mathbf{B} \cdot d \mathbf{a}$ vanishes when the loop contracts indefinitely. I can simply consider a small surface that doesn't vary in time, but it is essential take into account the possibility that the magnetic ...


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You're right that a function can be "small" at a point but have a "large" derivative at that point. But maybe the confusion is that you're imagining the surface $S$ shrinking in time, so that it's only "small" at one instant. But the surface doesn't shrink in time - you're taking the limit where it's "small" at all times. And if a function is always small,...


1

Your proposed path has a VERY LARGE action. As @tparker pointed out, you have to minimize the path subject to the constraint that the average velocity doesn't change. Now, the action is quadratic in velocity. A little fiddling around with the math should convince you that to minimize the integral of $v^2$ subject to the constraint that the average velocity ...


2

You're missing that Dirichlet boundary conditions $$ x(t_i)~=~x_i \quad\text{and} \quad x(t_f)~=~x_i $$ are implicitly implied. The stationary action principle is not well-posed without boundary conditions.


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You have to minimize the integral subject the the constraint that the initial and final positions $x(t_i)$ and $x(t_f)$ are held fixed. In particular, $\Delta x = \int_{t_i}^{t_f} v(t)\, dt$ is held fixed. If the particle slowed down than sped up as you suggested, the action would be less, but it wouldn't have a high enough average speed to cover the full $...


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This question seems to have been posted for quite a while... I don't know if you ve already found the answer.. Anyway, there are at least two methods to derive it (as far as I know). One popular method is the ``equation of motion'' method which formulates the contour Dyson equation first and then uses Langreth rules to obtain similar recursive equations for ...


2

I will answer with an example on how eigen functions of momentum do not exist in a Hilbert space in general. When they do not, we say the momentum operator is not self-adjoint. Consider a one dimensional infinite potential well. Lets us place the walls at $x=0$ and $x=L$. For $x\ge L$ and $x\le 0$, the potential $V(x)=\infty$ and therefore we put the ...


3

The boundary conditions determine whether an operator is Hermitian or not. Once you know your operator is Hermitian, you have the results on the spectrum. Without boundary conditions the momentum operator need not be Hermitian, hence its spectrum can have non-real values (here I am assuming that by Hermitian you actually mean self-adjoint). As an example, ...


0

The reason for your conflicting results has to do with the subtleties of hermiticity on finite intervals. Look carefully at the formal steps in the derivation of the Ehrenfest theorem: $$ \frac{d}{dt} \langle \psi(t) | x |\psi(t)\rangle = \langle \frac{d\psi}{dt} | x |\psi\rangle + \langle \psi | x |\frac{d\psi}{dt}\rangle = \frac{i}{\hbar} \left[ \langle H\...


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It doesn't make sense to apply periodic boundary conditions to the infinite square well, since the solutions in each well will be independent from each other (it is not possible for a particle to tunnel between different wells). Electrons will be strictly confined within the infinite square well of width $L$. When you consider the finite well, there is a ...



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