# Definition of electric polarisation and the potential due to a polarised body

I've two questions, the second one depends on the first.

$$\mathbf{1}$$

How exactly is polarisation defined? Griffiths says

$$\mathbf{P} \equiv$$ dipole moment per unit volume

How exactly do we go about calculating it?

For example if I need to find the value of $$\mathbf{P}$$ at some point do we take a small volume around that point enclosing few hundred/thousand atoms, add up the dipole moments and divide by the volume? And similarly repeat the process to find the value of Polarisation everywhere?

$$\mathbf{2}$$

Suppose I have to find the potential due to a polarised body far away from it. I can find it by adding up the individual contribution of each dipole. Since the field point is far away I can safely assume that the potential of each dipole can be written as $$V_{\mathrm{dip}}(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^{2}}$$

All I've to do is to add up the contributions of each individual dipole.

However an alternate equation is presented which too gives us the potential and is as $$V(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \int_{\mathcal{V}} \frac{\mathbf{P}\left(\mathbf{r}^{\prime}\right) \cdot \hat{r}}{r^{2}} d \tau^{\prime}$$

How can one justify that the second equation is correct and gives us the value of potential?

MORE DETAIL: Dear Urb said that :

"Instead of doing a sum $$\sum_{i=1}^N\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathbf{p}_i \cdot \hat{\mathbf{r}}}{r^{2}}$$ over all $$N$$ dipoles inside the body, we just chop the body into little pieces of volume $$d\tau'$$, assign to each piece a dipole moment $$\mathbf P(\mathbf r')d\tau'$$ and integrate over the entire body".

But we can't chop the body into $$d\tau'$$ elements and use the integral of $$\mathbf P(\mathbf r')d\tau'$$ . Because $$d\tau'$$ is infinitesimal.

We know that $$P$$ was an average over a small but not infinitesimal volume element and if we use $$V(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \int_{\mathcal{V}} \frac{\mathbf{P}\left(\mathbf{r}^{\prime}\right) \cdot \hat{r}}{r^{2}} d \tau^{\prime}$$

we have implicitly assumed $$P$$ to be an average over an infinitesimal volume element which isn't how we initially defined it.

• It may be that your question is primarily about the mathematical task of formulating a continuum approximation which smooths over the small-scale features of some structure, and justifying that the differential and integral calculus, applied to such a continuum, will give results of useful accuracy (i.e. which match sufficiently well what would be obtained if we did the averaging another way). Commented Jun 24, 2021 at 9:53

1. You are right regarding how to calculate $$\mathbf P$$. You should think of $$\mathbf P(\mathbf r)$$ as an average of the dipole moment around $$\mathbf r$$. A good definition is: $$\mathbf P$$ is something which integrated over a certain volume gives us the total dipole moment within that volume.

This is analogous to electric charge and charge density $$\rho(\mathbf r)$$. You could ask, what is $$\rho(\mathbf r)$$? Well, it's something that integrated over a volume gives us the total charge within that volume.

To calculate $$\mathbf P$$ at a point $$\mathbf r$$, choose a small volume around $$\mathbf r$$, sum all the dipoles inside and divide by the volume. Your assumption about a hundred/thousand molecules seems about right. If you choose a volume so small that there are only, say, three or four molecules, then it is not enough to make an average (remember that $$\mathbf P$$ is an average quantity). On the other hand, if you choose a volume very big, the orientations of the dipoles may be different at different points inside, so that making an average is no longer useful.

2. To answer your second question: yes, in principle, we could "just" sum all the contributions due to all the dipoles and we'll be done. But this would be a mess. You are missing the point: the purpose of introducing an average quantity $$\mathbf P$$ which varies smoothly over the body is to simplify your life. There are so many molecules in the body, that a discrete description is not useful, we need to treat the body as a continuous distribution of dipoles. Instead of doing a sum $$\sum_{i=1}^N\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathbf{p}_i \cdot \hat{\mathbf{r}}}{r^{2}}$$ over all $$N$$ dipoles inside the body, we just chop the body into little pieces of volume $$d\tau'$$, assign to each piece a dipole moment $$\mathbf P(\mathbf r')d\tau'$$ and integrate over the entire body.

Again, this is analogous to calculating an electric potential doing a sum over the charges $$V_{\mathbf E}(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\sum_{i=1}^N\frac{q_i}{|\mathbf r-\mathbf r_i|}$$ versus doing the integral $$V_{\mathbf E}(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\mathbf r')}{|\mathbf r-\mathbf r_i|}d\tau'$$ where in this case one assigns to each volume element $$d\tau'$$ a charge $$\rho(\mathbf r')d\tau'$$. When there are a few charges, summing is doable, when there are many, it can be a nightmare.

• Thank you, please see the added detail in my question. Commented Apr 25, 2021 at 6:27
• It doesn't matter how we define $\mathbf P(\mathbf r)$. If to define $\mathbf P(\mathbf r)$ we need a small but not infinitesimal volume around $\mathbf r$, then so be it. But once we have $\mathbf P$ defined everywhere, it doesn't matter how we do the integration. With this definition, two points that are infinitesimally close to each other will have the same $\mathbf P$, since the small volumes we need to define $\mathbf P$ at each of the points will overlap. So as we integrate, many elements $d\tau'$ close to each other will have the same $\mathbf P$, that doesn't mean we can't integrate.
– Urb
Commented May 3, 2021 at 10:16

A general remark, which is too long for a comment:
Landau's book on the subject is appropriately called Electrodynamics of continuous media - important thing here is that we are dealing with macroscopic quantities, i.e., the quantities averaged over a "macroscopically small volume", so that they vary smoothly in space (as opposed to actual atomic potentials that vary wildly on the microscopic scale). How one calculates a dielectric response from its microscopic structure of a media is an interesting and important subject, but this is not the point here.

Note that similar (but less unexpected) approximation is done in the elasticity theory, where the objects are considered as continuous, neglecting their atomic structure. Thus, instead of dealing with the intermolecular potentials we get something simple, such as Hooke's law. Indeed, the Hooke's law is the elastic equivalent of the relations such as $$\mathbf{D}=\varepsilon \mathbf{E}, \mathbf{B}=\mu \mathbf{H}$$

Remark: I fully agree with the answer by @Urb.