Recent content by lautaaf

Oh, sorry, it seems i missed that. You can also compute the components of ##R^{a}{}_{bcd}## in a orthonormal frame, via the connection one-forms and Cartan's structure equations.

In a 2 dimensional (semi)-Riemannian manifold there is only ##\mathbf{one}## linearly independent component of the Riemann tensor (I think in Wald's General Relativity there's a proof of this). That should make it easier (##\simeq## shorter)!

The problem we are looking to solve it is: $$ \Delta \Phi = - \frac {\rho}{\epsilon_0} \qquad \qquad \text{ in } \Omega_{out}=\{\vec x \in \mathbb R^{3} : \|\vec x \| > R\} \quad (1) \\ \Phi \big|_{r=R} =0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \: \: \quad \quad(2) \\...

Think the first relation as a definition for an operator \frac{d}{dt}=\sum_k \frac{\partial}{\partial q_k}\dot{q}_k+\frac{\partial }{\partial t} The second equation follows immdiately from applying ##\frac{d}{dt}## to ##\frac{d\mathbf{r}_i}{dq_j}## and the last one from applying...

There's another (somewhat shorter) way, in which you always work in the non-coordinate frames. Noting that $$ \nabla_{\hat e_{a}} \hat \theta^{b} = - \Gamma^b{}_{ac} \hat \theta^{c}$$ and replacing this relation into the metric compatibility $$\nabla_{X}g=0 \Rightarrow X^{a} [\nabla_{\hat e_{a}}...

Exactly. Moreover, you can prove that, if we define ##\delta U = \frac{<U>-U}{U}## then the standard deviation of ##\delta U## (the relative fluctuations) decay as: $$ <(\delta U)^{2}> \sim \frac{1}{N}$$

In order to find the density matrix, you have to minimize the Gibb's entropy: $$S= -k_B Tr(\hat \rho ln \hat \rho)$$ With the constraints: $$Tr(\hat \rho)=1$$ and $$ <U>=Tr(\hat \rho \hat H) =E= constant$$ i.e, you let your system explore all the microstates (which include microstates with...

The ##\underline{\text{expected value}}## of the internal energy ##<U>## is the (macroscopical) quantity that depends of ##N##,##T##.

Contract both sides of (3) with ##g^{\mu\nu}## and you will get a condition for ##R##

I should have been clearer. When I said "well defined" i meant to say piecewise well defined, in the sense that the limit in the definition of the derivative exists, but is different depending on whether z<0 or z>0.

(Yes it would work in the case of a spherical conductor. (Check Jackson's Classical Electrodynamics 3rd edition, p 58)) Perhaps if I state all the equations it will become clearer. The solution of the problem is: $$ \Phi(r,\phi, z) \left\{ \begin{array}{l} \Phi_1=\frac{1}{4 \pi \epsilon_0 }...

The discontinuity of ##\vec E## (and hence the discontinuity of ##\frac{\partial \Phi}{\partial n}##) ##\underline{\text{at the boundary}}## is crucial in order to have a non-vanishing surfarce charge. Note that, if ##\frac{\partial \Phi}{\partial n}## were continuous, ##\sigma## would be...

The ##\underline{derivative}## of the potential in the normal direction (which is proportional to the normal component of the electric field) is discontinuous at ##z=0## You can check from the solution that i posted earlier that ##\Phi_2(z=0) = \Phi_1(z=0)## (i'm being inconsistent with the...

No, the electric field is discontinuous at ##z=0##, but the potential is not. The solution (in cylindrical coordinates) is $$ \Phi_1(r,\phi, z) = \frac{1}{4 \pi \epsilon_0 } \bigg ( \frac{q}{\sqrt{r^2 + (z-d)^2}} - \frac{q}{\sqrt{r^2 + (z+d)^2}} \bigg ) \quad \quad z>0$$ $$\Phi_2 = 0 \quad...

Let ##\Omega## be the region ##z>0##. We want to solve the following boundary value problem(s): $$\Delta \Phi_1 = - \rho \quad \text{ in } \Omega$$ $$\Phi_1(z=0)=0$$ where ##\rho=-q \delta(z-d)## And let ##\Omega^{'}## be the region ##z<0## $$\Delta \Phi_2 = 0 \quad \text{ in }...