New answers tagged

0

In my answer to the physics stack question Discrete Cavity Radiation I answer your questions 2 and 3. The summary is: In classical EM a cavity with infinite conductivity walls only supports modes of certain frequencies because $E_{Transverse}=0$ and $B_{Perpendicular}=0$ at infinite conductivity walls. Real walls have some resistivity and thus can be ...


2

Yes, $$\nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0,$$ cf. e.g. my Phys.SE answer here. OP ponders if the covariant derivative $\nabla_{\mu} T^{\mu\nu}$ of the SEM tensor (3.54) could produce contributions proportional to the derivative of the Dirac delta distribution $\delta(\ell)$? Well, let's check. First of all, it is enough to consider the partial ...


1

Surface tension can be physically interpreted as the variations in the free energy arising at the interface/surface. Since the surface is normally comprised of multiple phases or components, the surface tension is, hence, a property of liquid and the medium at it's interface. For a water droplet in contact with rigid solid and air, the definition of contact ...


2

Some remarks first: no, the electric field cannot be considered approximately constant along the "horizontal" sides. Rather, we're applying the mean-value theorem for integrals. After that, yes, we have to shrink the length of the curve to $0$ as well. For the rest of the discussion, let us fix a point $p$ on the surface, and consider a "...


1

I found it finally using this (complicated) paper: http://verl.npre.illinois.edu/Documents/J-08-01.pdf As I said in comment we need to apply BC and we find a matrix equality to zero with coefficients, which gives a zero determinant equation, able to find the $\lambda_n$ eigenvalues.


2

The surface tension of a liquid is a basic property of the liquid alone- but since solids also possess surface energy, the nature of the solid upon which the liquid is sitting will affect the contact angle that the droplet of liquid assumes after it has equilibrated with the solid surface. Think of the edge angle of a droplet on a solid surface in the ...


0

The physical meaning for that, I'm not sure if this is something you already know, there is no potential difference at the same surface. There's potential difference between one surface and another, but at the same surface the potential difference is zero, and this is what we call an equipotential surface. The definition of an equipotential surface is a ...


2

Gauss's theorem is another name for the divergence theorem, which says that if you have some region $M$ with boundary $\partial M$ then $$\int_M \mathrm{div}(\mathbf F) \mathrm dV = \oint_{\partial M} \mathbf F \cdot \mathrm d\mathbf S$$ Let $M$ be a ball of radius R centered at the origin. We would then have that $$\int_{|\mathbf r|<R} \mathrm{div}(\...


3

It seems the core of OP's question is the following. Question: When we derived Snell's law for a single interface from Fermat's principle, we held 2 points fixed. How can we then use repeatedly Snell's law for a double interface, since we are not allowed to hold 3 points fixed during the variation? Answer: Although it is true that we're only allowed to ...


0

Well, after consulting my professor, it seems it was a Neumann boundary condition with zero flux at both boundaries ($\phi(0,t) = \phi(L,t) = 0$)


4

$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|}...


0

If there are boundary conditions $$ \phi_0(t) \equiv U(0,t),\\ \phi_1(t) \equiv U(L,t), $$ and the initial condition $$ f(x) \equiv U(x,0), $$ then you have what is called time-dependent boundary conditions, see e.g. these lecture notes. The problem statement in your question represents a special case, where only the values of $f(x)$ at two points are ...


1

This is a case where using hardcore Fourier series in terms of sines and cosines would be less confusing than starting with a more general Fourier integral. Another approach would be to use the separation of variables to find the eigenmodes of this equation (there is relation between the mode number $k$ and the corresponding frequency). The delta-function ...


Top 50 recent answers are included