Recent content by blenx

Thanks for Mute's answer. But I think there are two flaws in the definition of the contour delta function. First, this definition is only valid for the analytic function, namely it cannot apply to function such as f(z)=Re(z). Second, the way it selects the marked point differs from the usual...

This method is only valid in the case when u is real.

It is easy to integrate the delta function of real variable. But when the argument of the delta function is complex, I get stuck. For example, how to calculate the integral below, where u is a complex constant: \int_{ - \infty }^{ + \infty } {f\left( x \right)\delta \left( {ux}...

The equations I wrote are general, as long as the charge/current density is understood as the bound charge/current density in media. Of course one can use the polarization and magnetization to replace them, but that dose not change the number of unkonwn quantities. From the equations in Coulmb...

But if we express the Maxwell equation with potential in Coulmb gauge, [/tex] \begin{gathered} {\nabla ^2}\varphi = - \rho /{\varepsilon _0}\quad ,\quad {{\boldsymbol{E}}_{\text{L}}} = - \nabla \varphi \\ \square {{\boldsymbol{A}}_{\text{T}}} = {\mu...

It is no doubt that light is a transverse wave in vaccum. But is it also holds true for the case when light is in a medium?

Thanks for your reminding. The data I used in the previous post may be improper.

It is approximately the product of the current and the displacement between the centers of the positive and negetive charges, viz. {\boldsymbol{\dot d}} \sim I\Delta {\boldsymbol{l}}.

In classical electrodynamics, the volume integral of the current density is the time derivative of the electric dipole moment of the system: \int {{\boldsymbol{J}}{{\text{d}}^3}V} = \sum\limits_i^{} {\int {{{\boldsymbol{J}}_i}{{\text{d}}^3}V} } = \sum\limits_i^{} {{q_i}{{\boldsymbol{v}}_i}}...

For a closed electromagnetic system, there is a similar relationship between the energy density and the energy flux density which you can think as the energy current density: \dfrac{{\partial u}}{{\partial t}} + \nabla \cdot{\boldsymbol{S}} = 0

Here is the equation I don't know how to solve: \begin{aligned} \left( {\frac{{{{\rm{d}}^2}}}{{{\rm{d}}{t^2}}} + \beta _1^2} \right){u_1} = {g_1}u_2^{}{u_3} \\ \left( {\frac{{{{\rm{d}}^2}}}{{{\rm{d}}{t^2}}} + \beta _2^2} \right){u_2} = {g_2}u_1^{}{u_3} \\ \left(...

\Delta t = \frac{{3\pi {\varepsilon _0}\hbar {c^3}}}{{{\omega ^3}{{\left| {\left\langle {f\left| {{\boldsymbol{\hat d}}} \right|i} \right\rangle } \right|}^2}}} \approx \frac{{3\pi {\varepsilon _0}\hbar {c^3}}}{{{\omega ^3}{e^2}{a^2}}} = \frac{{3{c^2}}}{{4\alpha {\omega ^3}{a^2}}}=...

Because {\boldsymbol{\hat d}} = e{\boldsymbol{\hat x}}, the expression of the radiant power actually has contained the alpha constant. What confuse me very much is that if you evaluate the Δt, you will find it quite large, which means that we have to wait for a very long time before the dipole...

Does somebody know the answer?

Suppose there is an electic dipole that starts to oscillate with frequency ω at t=0, then how long does it take the electric dipole to emit a photon? We know that the radiant power of such electric dipole calculated from quantum physics is P\left( \omega \right) = \frac{{{\omega ^4}}}{{3\pi...