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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.

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  • $\begingroup$ I think it is unclear what you are trying to do here. The second equation is the definition of the electric dipole moment (if the dipole is at the origin). The first equation is the charge density, which is derived from the electric dipole using Poisson's equation ($\mathbf{p}$ and $\mathbf{p_0}$ are the same thing in this context, and $\mathbf{x_0} = 0$). See for example physics.stackexchange.com/questions/384485/…. $\endgroup$
    – Tony
    Jan 16, 2021 at 10:44
  • $\begingroup$ I am trying to show that $\mathrm{\mathbf{p}}$ is $\mathrm{\mathbf{p_0}}$. $\endgroup$
    – user161157
    Jan 16, 2021 at 10:51
  • $\begingroup$ You might find it helpful to take a close look at the other answer I linked in my comment. In a way, what you want to do is to solve your second equation to get $\rho$, and then check that $\rho$ is equal to your first equation. The linked answer does something similar: it takes a function of your second equation (the second term in the multipole expansion, since all the others are zero for a pure dipole) and then uses Poisson's equation to solve for the charge. That expression is just the one you want. $\endgroup$
    – Tony
    Jan 16, 2021 at 11:00
  • $\begingroup$ Yes, that is one method to show what I wanted. However, I want to show that $\mathrm{\mathbf{p}}=\mathrm{\mathbf{p_0}}$ through direct integration. $\endgroup$
    – user161157
    Jan 16, 2021 at 11:15
  • $\begingroup$ I have attempted to show the result with direct integration. I am not at all certain about it because taking the gradient of the delta function is unpleasant, but it's a start. $\endgroup$
    – Tony
    Jan 16, 2021 at 12:30

2 Answers 2

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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.

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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 $\nabla\delta^{(3)} = -\delta^{(3)}\nabla$:

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

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