- #1

nykon

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I have got a question about the induced scalar potential. I will present the problem from beginning.

Lets say we have a Poisson's equation in form:

[itex]\epsilon \nabla^2 \phi = -4\pi \varrho(r,t)[/itex]

where [itex]\epsilon[/itex] is the dielectric constant. By use of the Fourier transform:

[itex]f(r,t) = \int \frac{d^3q}{(2\pi)^3} \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} e^{i(qr-\omega t)}f(q, \omega)[/itex],

(where [itex]q[/itex] is the momentum, and [itex]\omega[/itex] is the energy) one can write:

[itex]\phi_{q, \omega}= \frac{4\pi \varrho_{q, \omega}}{q^2 \epsilon}[/itex]

where [itex]\varrho_{q, \omega} = 2\pi Z \delta(\omega - q v)[/itex], v is the velocity of the incident ion.

Now if we assume that the incident ion is moving through the electron gas we can write the induced scalar potential in form:

[itex]\phi^{induced}_{q, \omega}= \frac{8\pi^2 Z}{q^2} \delta(\omega - q v)(\frac{1}{\epsilon} - 1)[/itex]

Now the energy loss per unit time W is:

[itex]W = - Z v E^{induced}, \qquad E^{induced} = -\nabla \phi^{induced}[/itex]

the final result is:

[itex]W = \int \frac{d^3q}{(2\pi)^3} \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} 2 \omega Z Im[-\phi^{induced}][/itex],

My question is: why I need the imaginary part in the final result of the induced scalar potential? The result is taken from the "dynamic screening of ions in condensed matter" written by Echenique, Flores and Ritschie. I just do not understand the last formula. I will be gratefull for any tip or advise.

nykon