Finding electric dipole moment

Question

For a ideal electric dipole we have

$$\rho (\mathrm{\mathbf{x}})=-\mathrm{\mathbf{p_0}}\cdot\nabla\delta^{(3)}(\mathrm{\mathbf{x}}-\mathrm{\mathbf{x_0}})$$

The electric dipole moment formula is given as

$$\mathrm{\mathbf{p}}=\int_{\mathcal{V}}\mathrm{\mathbf{x}}\,\rho(\mathrm{\mathbf{x}})\,\mathrm{d^3x}$$

How do I use the above formula to find the electric dipole moment?

I was thinking of using the identity

$$\nabla\cdot(\psi\mathrm{\mathbf{A}})=\psi\nabla\cdot\mathrm{\mathbf{A}}+\mathrm{\mathbf{A}}\cdot\nabla\psi$$

However, I am not sure how to incoporate $\mathrm{\mathbf{p_0}}$, which is constant, into the formula.

Answer 1

I'll show it in one dimension, in case that's easier to understand. In this case,

$$\rho(x) = -p_0 \frac{\text{d}}{\text{d}x}\delta(x).$$

The dipole moment is

\begin{aligned} p &= \int_{-\infty}^\infty\text{d}x\,\, x \,\rho(x) = -\int_{-\infty}^\infty\text{d}x\,\, p_0 \,x\frac{\text{d}}{\text{d}x}\delta(x)\\ \end{aligned}

Now, we can integrate the above equation by parts, to show that $$p = - p_0 x \delta(x) \Bigg|_{-\infty}^\infty +\int_{-\infty}^\infty \delta(x) \frac{\text{d}}{\text{d}x}(p_0 x)$$

The first term is zero, because of the Dirac Delta, and the second term is just $p_0$, so you have $$p = p_0.$$

It should be straightforward to generalise this to higher dimensions.

Note: Another formula you could use (which we have effectively proved here) is that $$\int_{-\infty}^\infty \text{d}x\,\, f(x) \delta'(x) = -f'(0).$$ Setting $f(x) = -p_0 x$ gives you the same result.

Answer 2

You want to show that the first and second equations are consistent (if $\mathbf{x}=\mathbf{0}$). In particular, you want to show that $\mathbf{p}=\mathbf{p_0}$.

Begin with the second equation, which is the definition of the dipole moment $\mathbf{p}$.

Next use the multipole expansion of the electric potential:

$$V(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\left( \frac{q}{|\mathbf{x}|}+\frac{\mathbf{p}\cdot\mathbf{x}}{|\mathbf{x}|^3}+...\right)$$

For a dipole, all the terms are zero except the second.

Using Poisson's equation:

$$\nabla^2 \, V(\mathbf{x}) \left(= \nabla^2 \frac{\mathbf{p} \cdot \mathbf{x}}{4\pi\epsilon_0|\mathbf{x}|^3}\right) = -\frac{\rho(\mathbf{x})}{\epsilon_0}$$

it is possible to calculate the Laplacian and rearrange the resulting equation to show (see, for example, What is the charge density of a pure electric dipole?) that:

$$\rho(\mathbf{x}) = -\mathbf{p} \cdot \nabla\delta^{(3)}(\mathbf{x})$$

So you have that $\mathbf{p} = \mathbf{p_0}$. This is a trivial result because $\mathbf{p}$ was used to derive the charge density, but the derivation makes it clear that the two expressions you have are consistent.

An alternative approach is to set $\mathbf{p_0} = \mathbf{p}$ and $\mathbf{x} = \mathbf{0}$ in the first equation, then substitute into the second. I am quite unsure about the mathematics here, but since nobody else has provided an answer I will try:

$$ \mathbf{p} = \int_\mathcal{V} \mathbf{x} \left(-\mathbf{p} \cdot \nabla \delta^{(3)} (\mathbf{x})\right) \mathrm{d}^3x = \int_\mathcal{V} \left(-\mathbf{p} \cdot \nabla \delta^{(3)} (\mathbf{x})\right) \mathbf{x} \,\,\mathrm{d}^3x$$

Using the result that $\nabla\delta^{(3)} = -\delta^{(3)}\nabla$:

$$ \mathbf{p} = \int_\mathcal{V} \left( \mathbf{p} \cdot \delta^{(3)} (\mathbf{x}) \nabla \right) \mathbf{x} \,\,\mathrm{d}^3x \\ = \int_\mathcal{V} \mathbf{p} \cdot \delta^{(3)}(\mathbf{x})\nabla x \,\hat{\mathbf{i}} \, + \mathbf{p} \cdot \delta^{(3)}(\mathbf{x}) \nabla y \,\hat{\mathbf{j}} \, + \mathbf{p} \cdot \delta^{(3)}(\mathbf{x}) \nabla z \,\hat{\mathbf{k}} \, \,\,\mathrm{d}^3x \\= \int_\mathcal{V} \left(\mathbf{p} \cdot \delta^{(3)}(\mathbf{x}) \, \hat{\mathbf{i}}\right)\hat{\mathbf{i}} \, + \left( \mathbf{p} \cdot \delta^{(3)}(\mathbf{x}) \, \hat{\mathbf{j}} \right) \hat{\mathbf{j}} \, + \left( \mathbf{p} \cdot \delta^{(3)}(\mathbf{x}) \,\hat{\mathbf{k}} \right)\hat{\mathbf{k}} \, \,\,\mathrm{d}^3x \\ \\ = \int_\mathcal{V}\delta^{(3)}(\mathbf{x}) \left(p_x \hat{\mathbf{i}} + p_y \hat{\mathbf{j}}+ p_z\hat{\mathbf{k}}\right) \mathrm{d}^3x = \mathbf{p}$$.