# Power absorbed from driving field by induced dipole

I don't understand the first equal sign in equation (4) of the paper Optical Dipole Traps for Neutral Atoms by Grimm and Weidemüller. Translated to a question here, the situation is:

Placed in an oscillating electric field $$\mathbf{E}$$, a dipole moment $$\mathbf{p}$$ is induced in an atom.
$$\mathbf{E}(\mathbf{r},t)=\hat{\mathbf{e}} \tilde{E}(\mathbf{r}) \exp (-i \omega t)+c.c.$$
$$\mathbf{p}(\mathbf{r}, t)=\hat{\mathbf{e}} \tilde{p}(\mathbf{r}) \exp (-i \omega t)+c.c.$$
$$\tilde{p}=\alpha(\omega) \tilde{E}$$
with $$\hat{\mathbf{e}}$$ the polarization vector, $$c.c.$$ the complex conjugate, and $$\alpha$$ the complex polarizability.

With $$U=\int_\limits{a}^b\mathbf{F}d\mathbf{r}$$, $$\mathbf{F}=(\mathbf{p}\mathbf{\nabla})\mathbf{E}$$, and $$\nabla(\mathbf{p}\mathbf{E})=\mathbf{p}\times(\nabla\times\mathbf{E})+\mathbf{E}\times(\nabla\times\mathbf{p})+(\mathbf{p}\mathbf{\nabla})\mathbf{E}+(\mathbf{E}\mathbf{\nabla})\mathbf{p}$$, I can understand

the potential $$U_{\mathrm{dip}}=-\frac{1}{2}\langle\mathbf{p} \mathbf{E}\rangle=-\frac{1}{2 \varepsilon_0 c} \operatorname{Re}(\alpha) I$$ with $$I=2 \varepsilon_0 c|\tilde{E}|^2$$
and the force $$\mathbf{F}_{\mathrm{dip}}(\mathbf{r})=-\nabla U_{\mathrm{dip}}(\mathbf{r})=\frac{1}{2 \varepsilon_0 c} \operatorname{Re}(\alpha) \nabla I(\mathbf{r})$$ in the paper,

but not the "power absorbed by the oscillator from the driving field" $$P_{\mathrm{abs}}=\langle\dot{\mathbf{p}} \mathbf{E}\rangle$$.

Where does it come from? Which first principles lead to it? It is not the same as $$\frac{d}{dt}U$$, if I calculated correctly.

Sorry if this is a dumb question.

It is based on the net work of external electric field acting on the charges forming the dipole.

The power of external electric forces (work per unit time) as function of time is, from definition,

$$P(t) = \sum_k q_k \mathbf E(\mathbf r_k,t)\cdot \mathbf v_k.$$

Expressing the velocities as $$\mathbf v_k = \frac{d\mathbf r_k}{dt}$$, we get $$P = \sum_k q_k \frac{d\mathbf r_k}{dt} \cdot \mathbf E(\mathbf r_k,t).$$ Now, we make an approximation that electric field everywhere is the same field $$\mathbf E(t)$$ - this means that the dipole is small enough so the external field can be approximated as uniform throughout the dipole extent. Then we get

$$P = \frac{d}{dt}\bigg( \sum_k q_k \mathbf r_k \bigg ) \cdot \mathbf E(t).$$

The expression in the brackets is electric moment of the system, sometimes called dipole moment, and often denoted as $$\mathbf p$$.

So we have the expression for work of external electric forces per unit time:

$$P(t) = \dot{\mathbf p} \cdot \mathbf E(t).$$

Now we can do time averaging to get time average power $$\bar{P}$$ that does not depend on time, or statistical averaging to get $$P_{average}(t)$$, which may or may not vary with time, depending on which particles we average over.

• @PedroHenriqueCookCunha see my answer above. Aug 10, 2023 at 22:00