Recent content by hiyok

Now I agree with you. I thought your were referring to the part due to ##V## only. Putting all pieces together, we agree.

Dear DrDu, I'm much grateful for your comments which have led me to a good understanding of my problem. As this understanding is significant in improving one of my recent manuscripts, which I'll submit for publication, I'm thinking formally acknowledging you or adding you as a coauthor. So, if...

The z-component is obtained as ##E_z = -\partial_z V(\vec{x},t) - c^{-1}\partial_tA_z(\vec{x},t)##, where ##A_z## is the z-component of the vector potential. The final expressions are a little complicated, because now the vector potential depends on the current density, which further depends on...

That is exactly what I did. I have worked out the vector potential as well. The final result was shown in the figure attached in previous post.

The retarded potential I gave is a solution to this equation: ##(\partial^2_{\vec{x}}-\partial^2_t/c^2) V(\vec{x},t)= -\rho_0e^{i(\vec{q}\vec{r}-\omega t)}\delta(z)##.

This is what looks like:

I also found that, if both scalar and vector potentials are included, the plasmon dispersion is determined by an even simple equation: ##\left(2\pi n_0 e^2/m\right)^2 |q^2-\omega^2/c^2| = \omega^4##. Obviously, the dispersion becomes linear for small ##q##.

I found the electric potential to be ##V(\vec{r},z) = \frac{\rho_0}{2}\kappa^{-1}e^{i(\vec{q}\vec{r}-\omega t)}e^{-\kappa|z|}##, where ##\vec{r}## denotes the in-plane components. The electric field strength in the x-direction is ##E_x =...

Thank you very much for your insightful advice !

Thanks for your suggestion. But that kind of dispersion had been experimentally verified: see e.g. Phys. Rev. Lett. 36: 145(1976).

Here is a reference: http://rmf.smf.mx/pdf/rmf/39/4/39_4_640.pdf

Now plasma represents collective wave-like motions of charged particles. In 3D, their frequency is well known to be almost a constant, ##\omega^{3D}_p \approx \sqrt{4\pi ne^2/m}## with n=charge density, m=particle mass. However, in 2D, one can show that it becomes ##\omega^{2D}_p \sim...

Thanks for response. 1. Yes, I mean \lim_{k\rightarrow 0}(\omega+ik). 2. As pointed out by D H, it is x^l, x raised to the power of l. 3. Initially, I tried to do it by this formula, \frac{1}{\omega+i0_+-x^m} = \mathcal{P}\left(\frac{1}{\omega-x^m}\right)-i\pi \delta(\omega - x^m), with...

hi, I have difficulty in figuring out the following integral: I(l,m;z) = \int^1_0 dx~\frac{x^l}{z - x^m} , where l and m are integers, while z = \omega + i0_+ is a complex number that is infinitely close to the real axis. What is interesting to me is when \omega is close to zero, so...

Thanks for your response. I have worked it out. It amounts to zero by symmetry.