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3

For any given electric field $\mathbf{E}(\mathbf{r},t)$, and assuming no constraints on the magnetic field, there is always a set of charges and currents (both possibly time-dependent) that produce that electric field. What's more, while the charges required are uniquely determined, the currents are not. Here's the construction: Let $\rho(\mathbf{r},t) = ...


3

Maxwell's equations only give a unique electric field subject to a set of boundary conditions and an initial condition for the field.


4

First of all, the tangential component of ${\bf E}$ (i.e. the component parallel to the interface) is always continuous across interfaces, so $E_z$ must be continuous at $r = R$, which fixes $C = 1/R$. There is no physical configuration of sources that could produce any other value for $C$. ${\bf E}$ does not depend on time, so ${\bf \nabla \times E} = -\...


6

The problem is that you calculated the curl wrong; you missed a delta function arising from the discontinuity. We can write $E_z$ as $$ E_z = \ln(Cr) \Theta(R-r)$$ and, using that $\Theta'(x) = \delta(x)$, we get that $$\nabla \times \mathbf{E} = -\frac{1}{r} \Theta(R-r) \hat{\phi} - \ln(CR)\delta(r-R)\hat{\phi}$$ so when giving $\nabla \cdot \mathbf{E}$...


1

Even with a delta function potential, continuity of the wave function is still required. (Please see comment from ACuriousMind below on this). The derivative of the wavefunction is obviously not continuous, however. You can find the discontinuity by integration about the delta function from +s to -s, where s is a small parameter. You then let s go to 0 ...


1

Within distribution theory, a mathematically rigorous formulation of OP's eq. (2) is $$ \lim_{z\to 0^+} z^{\Delta-d}K_{\Delta}(z,x)~=~\delta^d(x), \tag{A}$$ where $$ K_{\Delta}(z,x)~:=~ \frac{\Gamma(\Delta)}{\pi^{\frac{d}{2}} \Gamma(\Delta\!-\!\frac{d}{2})} \left( \frac{z}{z^2+x^2}\right)^{\Delta},\qquad x~\in~\mathbb{R}^d, \qquad \Delta >\frac{d}{2},\...


1

In order to show that some function is the Delta distribution, you have to show that It is zero except for where the argument of the function vanishes. It integrates to 1 when integrated over the full coordinate range. We can see these two properties explicitly. If $x-x'\not=0$, then $$ \lim_{z\to0}K(z,x,x')=\lim_{z\to0}\frac{z^\Delta}{(x-x')^{2\Delta}...


3

As a general rule, whenever you are setting sudden discontinuities (as in your example with the Heaviside function) it is not a surprise that these may reflect in discontinuous distributions, or derivatives, or infinities here or there. Do keep in mind that it is just a result of the mathematical simplifications that we are introducing (although ...


0

You shouldn't expect the occupation of the ground state to be one (or zero). That would imply that, for $T>0$, the ground state is ALWAYS (never) occupied. That isn't necessarily true, as you can see from your formula. What is true is that at ZERO temperature states with $E<\mu$ are always occupied, while states with $E>\mu$ are never occupied. At $...


1

But in reality, when calculating the $N(\epsilon)$, we just divide the whole energy range into small steps with equal size. Then count the number of electrons whose energy lies in each small steps. After further scaling we can get the $N(\epsilon)$. I just want to know why is this holding. You are right. But then, if we want to know the number of electrons ...



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