Absorption/Extinction formula of nanoparticles

I know the absorption/extinction equations in nanoparticle physics should be: $$Q_{abs}=\frac{1}{2}\mathbf{Re}\int \mathbf{J}_{tot}\cdot\mathbf{E}_{tot}^\ast dV=\frac{\omega}{2}\mathbf{Im}(\epsilon)\int|\mathbf{E}_{tot}|^2dV$$ also, for the extinction, it reads: $$Q_{ext}=\frac{1}{2}\mathbf{Re}\int \mathbf{J}_{tot}\cdot\mathbf{E}_0^\ast dV$$

But I see in some papers people use the following equations: $$Q_{abs}=\frac{\omega}{2}\mathbf{Im}(\mathbf{d}\cdot\mathbf{E}_{inside}^\ast)$$ where $\mathbf{d}$ is the total dipole moment of nanoparticles. Also, for the extinction, it reads: $$Q_{ext}=\frac{\omega}{2}\mathbf{Im}(\mathbf{d}\cdot\mathbf{E}_0^\ast)$$

I failed to derive the two equations. Can anyone give some help? Or, some reference papers would be also very helpful. Thanks a lot for the help.

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I think the difference between two is that one of them is a spectrum and, in one case the field depends on time and in the other one it depends on frequency. It is only guess, I can be wrong. –  freude May 6 '13 at 8:58
No, both the two equations are dependent on frequency, and are the time-averaged results. –  Hui Zhang May 6 '13 at 17:20
What is V? How large is it? –  freude May 7 '13 at 7:09
V is the total volume of object, and $\int\cdots dV$ is a volume integral. The size of object is usually assumed to be much smaller than the wavelength of incident light. –  Hui Zhang May 10 '13 at 15:00

This mistery has an easy answear: in absence of external currents the total current is the current induced by the electric field, which is just the derivative of the nanoparticle polarization. This holds in general in absence of external current:

$$\mathbf{J}_{pol} = \frac{\partial\mathbf{P}}{\partial t}$$

Now in the case of a nanoparticle (or a dielectric/metallic sphere), it is known that the response to an external field is dipole-like. The total electric dipole $\mathbf{d}$ is given by:

$$\mathbf{d} = \int \mathbf{P}\, d\mathbf{r}$$

Now, since the total field inside the sphere $\mathbf{E}_{inside}$ and the exciting field $\mathbf{E}_0$ are both uniform (Attention: the total field outside the sphere is not uniform!), you can write: \begin{align} Q_{abs} &= \frac{1}{2}\mathbf{Re}\int \mathbf{J}_{pol}\cdot \mathbf{E}_{inside}^* = \frac{1}{2}\mathbf{Re}\left\{ \mathbf{E}_{inside}^*\cdot \int \mathbf{J}_{pol}\, d\mathbf{r} \right\}\\ &=\frac{1}{2}\mathbf{Re}\left\{ \mathbf{E}_{inside}^*\cdot \left(\frac{\partial}{\partial t}\int \mathbf{P}_{pol}\, d\mathbf{r} \right)\right\} = \frac{1}{2}\mathbf{Re}\left\{\left(\frac{\partial}{\partial t}\mathbf{d}\right)\cdot\mathbf{E}_{inside}^*\right\} \end{align}

If the external source is time harmonic, then all the time dependences can be assumed to be of the type $e^{-i\omega t}$, thus

$$\partial_t \mathbf{d} = -i\omega \mathbf{d}$$

Substituting into the above equation, you get:

$$Q_{abs} = \frac{1}{2}\mathbf{Re}\{-i\omega \mathbf{d}\cdot\mathbf{E}^*_{inside}\} = \frac{\omega}{2}\mathbf{Im}\{\mathbf{d}\cdot\mathbf{E}^*_{inside}\}$$

and similarly for the $Q_{ext}$.

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Very helpful, thank you:) –  Hui Zhang Nov 13 '13 at 18:23
You're welcome! –  Mattia Nov 13 '13 at 18:56