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In Chapter 13.3 (Density Effect in Collisional Energy Loss) of Classical Electrodynamics by Jackson, it says:

In calculating the energy loss to an electron in an atom at impact parameter $b$, we evaluate: \begin{equation} \Delta E = -e \int_{-\infty}^{\infty} \vec{v}\cdot\vec{E}\,dt = 2 e \hspace{0.1cm} \Re \int_0^{\infty} i \omega \vec{x}\cdot\vec{E}^{*}(\omega)\, d\omega \end{equation} where $\vec{x}(\omega)$ is the Fourier transform in time of the electron's coordinates and $\vec{E}(\omega)$ is the Fourier transform in time of the eletromagnetic fields at a perpendicular distance $b$ from the path of the particle moving along the $x$ axis.

Could someone explain me how we could derive this expression?

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2 Answers 2

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I think you can take the Fourier transformation of $x$ and $E$ and just put it.

Full solution: $$\vec{x}(t) =\frac{1}{\sqrt{2\pi}}\int d\omega \exp(-i\omega t) \vec{x}(\omega)$$ So, $$\vec{v}(t) =\frac{1}{\sqrt{2\pi}}\int d\omega\ (-i\omega)\ \exp(-i\omega t) \vec{x}(\omega)$$ Also, $$\vec{E}(t) =\frac{1}{\sqrt{2\pi}}\int d\omega \exp(-i\omega t) \vec{E}(\omega)$$ So your equation 1 becomes, $$-e\int dt \vec{v}\cdot \vec{E} = -e \int dt\left(\frac{1}{\sqrt{2\pi}}\int d\omega\ (-i\omega)\ \exp(-i\omega t) \vec{x}(\omega)\right)\left(\frac{1}{\sqrt{2\pi}}\int d\omega' \exp(-i\omega' t) \vec{E}(\omega')\right)$$ Rearranging the term and using the property of $\frac{1}{2\pi}\int dt \exp(i\omega +i\omega')t = \delta_{\omega,-\omega'}$ $$= e\int d\omega\ i\omega\ \vec{x}(\omega)\ \vec{E}(-\omega) $$ Now use the property of Fourier transformation $E(-\omega) = E^*(\omega)$ $$= e\int_{-\infty}^{\infty} d\omega\ i\omega\ \vec{x}(\omega)\ \vec{E}^*(\omega) $$ $$= e\int_{0}^{\infty} d\omega\ i\omega\ \vec{x}(\omega)\ \vec{E}^*(\omega) + e\int_{0}^{\infty} d\omega\ -i\omega\ \vec{x}(-\omega)\ \vec{E}^*(-\omega) $$ Again using the same property, $$= e\int_{0}^{\infty} d\omega\ i\omega\ \vec{x}(\omega)\ \vec{E}^*(\omega) + e\left(\int_{0}^{\infty} d\omega\ i\omega\ \vec{x}(\omega)\ \vec{E}^*(\omega)\right)^* $$ $$= 2e\ Re \int_{0}^{\infty} d\omega\ i\omega\ \vec{x}(\omega)\ \vec{E}^*(\omega)$$

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Note that the Fourier transform of $\vec{v}$ and $\vec{E}$ at the required point on the trajectory can be written as $$ \vec{v} = \frac{d}{dt} \vec{x}(t) =\frac{d}{dt} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty d\omega \ \vec{x}(\omega) e^{-i\omega t} = - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty d\omega \ i\omega \vec{x}(\omega) e^{-i\omega t} $$ and $$ \vec{E}(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty d\omega' \vec{E}(\omega) e^{-i\omega t} = \frac{1}{\sqrt{2\pi}} \int_{\infty}^\infty d\omega' \vec{E}^*(\omega) e^{i\omega t}. $$ In the second equality on the second line, we have used the fact that $\vec{E}(t)$ is real. Thus, $$ \int_{-\infty}^\infty dt \vec{v} \cdot \vec{E}(t) = -\frac{1}{2\pi} \int_{-\infty}^\infty dt \int_{-\infty}^\infty d\omega \int_{-\infty}^\infty d\omega' e^{-i(\omega - \omega') t} i\omega \vec{x}(\omega) \cdot \vec{E}^*(\omega').$$ Using the property $\int_{-\infty}^\infty e^{-ikx} dk = 2\pi\delta(x)$, we find that $$ \int_{-\infty}^\infty dt \vec{v} \cdot \vec{E}(t) = -\int_{-\infty}^\infty d\omega \int_{-\infty}^\infty d\omega' \delta(\omega - \omega') i\omega \vec{x}(\omega) \cdot \vec{E}^*(\omega') $$ $$ = -\int_{-\infty}^\infty d\omega \ i\omega \vec{x}(\omega) \cdot \vec{E}^*(\omega) .$$ The left hand side is real, so we may replace the right hand side by its real part via $$ -\int_{-\infty}^\infty d\omega \ i\omega \vec{x}(\omega) \cdot \vec{E}^*(\omega) = -2 \Re \int_0^\infty d\omega \ i\omega \vec{x} \cdot \vec{E}^*(\omega) . $$ This equality is true since $\vec{x}^*(\omega) = \vec{x}(-\omega)$ and similarly for $\vec{E}$. Thus, $$ \int_{-\infty}^\infty dt \vec{v} \cdot \vec{E}(t) = -2 \Re \int_0^\infty d\omega \ i\omega \vec{x} \cdot \vec{E}^*(\omega) $$ as required.

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