Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In this paper, there's the following sentence:

...and the factor 1/2 takes into account that the dipole moment is an induced, not a permanent one.

Without any further explanation. I looked through Griffiths' electrodynamics to see if this was a standard sort of thing, but couldn't find anything. I was thinking it might be because the field of the dipole itself opposes the inducing field, but that doesn't quite seem right for some reason.

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

The force on a dipole placed in an electrical field is given by $\mathbf{F} = (\mathbf{p}\cdot \nabla)\mathbf{E}$ (see, e.g., Griffiths, 3rd edition, eq. 4.5). Recall that, $$ \nabla(\mathbf{p}\cdot\mathbf{E}) = \mathbf{p}\times (\nabla\times \mathbf{E}) + \mathbf{E}\times(\nabla\times \mathbf{p})+(\mathbf{p}\cdot\nabla)\mathbf{E} + (\mathbf{E}\cdot\nabla)\mathbf{p} $$ Assume $\nabla\times \mathbf{E} = 0$ (I'll justify this at the end, trust me for now). If the dipole moment is a permanent one, $\mathbf{p} = \mathrm{const}.$, the second and fourth terms above are zero, and the expression for the force can be rewritten, $$ \mathbf{F}=\nabla(\mathbf{p}\cdot\mathbf{E})\quad\Rightarrow\quad U=-\int_{r_a}^{r_b} \mathbf{F}\cdot d\mathbf{r} = -\mathbf{p}\cdot \mathbf{E} |_{r_a}^{r_b} $$ However, if $\mathbf{p}$ is not a constant, but rather $\mathbf{p} = \alpha\mathbf{E}$, where $\alpha$ is the polarizability, the fourth term is not zero, and $$\begin{split} \nabla(\mathbf{p}\cdot\mathbf{E}) &= 2\alpha\mathbf{E}\times (\nabla\times \mathbf{E}) + 2\alpha(\mathbf{E}\cdot\nabla)\mathbf{E}\\ &= 0+2(\mathbf{p}\cdot\nabla)\mathbf{E} \end{split} $$ Therefore, $$ U=-\int_{r_a}^{r_b} \mathbf{F}\cdot d\mathbf{r} = -\int_{r_a}^{r_b}\frac{1}{2}\nabla(\mathbf{p}\cdot\mathbf{E})=-\frac{1}{2} \mathbf{p}\cdot\mathbf{E} |_{r_a}^{r_b} $$ The only outstanding issue is justifying the assumption $\nabla\times\mathbf{E}=0$. From the relevant Maxwell equation, $\nabla\times\mathbf{E}= - \frac{\partial \mathbf{B}}{\partial t}$. If your fields are static, $\frac{\partial \mathbf{B}}{\partial t} = 0$, and we're done.

In an optical trap, the application discussed in the paper above, the field is not static and we have to be a bit more careful. An optical trap is arranged by counterpropagating two identical laser beams. Assuming beam fronts are approximately planar, $$ \mathbf{E} = \mathbf{E_1} + \mathbf{E_2}\\ \mathbf{B} = \mathbf{B_1} + \mathbf{B_2} = (\frac{1}{c}\mathbf{\hat{k}}\times\mathbf{E_1}) + (\frac{1}{c}(\mathbf{-\hat{k}})\times\mathbf{E_2}) $$ If the beams are arranged so that they're in phase ($\mathbf{E_1} = \mathbf{E_2}$), we have $\mathbf{B} = 0$ at all times and so $\nabla\times\mathbf{E}=0$.

That's the math, then, but what's the intuition? To first order, there are two contributions to any change in the quantity $-\mathbf{p}\cdot \mathbf{E}$: the change in $\mathbf{p}$ at constant $\mathbf{E}$ and the change in $\mathbf{E}$ at constant $\mathbf{p}$. But there is actually no force opposing the first of these changes: strictly speaking, the energy of the dipole should just be the integral of the second of them. For a permanent dipole, the first change is zero, so we get away with writing the energy as $-\mathbf{p}\cdot \mathbf{E}$. But for an induced dipole, this is no longer the case. Linear polarizability gave us a factor of $1/2$, but more general relations between $\mathbf{p}$ and $\mathbf{E}$ may give you more complicated answers.

share|improve this answer
add comment

Because the black area is half the box below.

enter image description here

To explain: move the dipole from an area of no field to an area of field strength E. As you do, there's a force proportional to the dipole moment and to the gradient of E. For a fixed dipole, this force depends only on the gradient (horizontal dashed line). But for an induced dipole, the dipole moment depends on E and grows linearly as you move from zero field to full strength, so on average it is only half as strong during that movement (solid diagonal line).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.