# Is dipole moment a vector?

Isn't it a misconception to assume that dipole moments are similar to vectors in their behavior? My reasons are as follows:

Let us take an example: Methane ($\ce{CH4}$) Clearly, the hydrogen atoms pushing their electrons into the more electronegative carbon only enforce the effects of each other. But, we assume for no particular reason that dipole moments must behave like vectors. We then go on to add the vectors and finally decide that the net dipole moment is $0$. Why?

• They are vector quantities. – Pritt says Reinstate Monica May 17 '17 at 7:03
• We don't assume or deduce anything here. We define dipole moments as vectors, hence they are vectors, as good as vectors can be. – Ivan Neretin May 17 '17 at 7:07
• @Themysteryoflife seems more interested in philosophy than physics. He may be interested to know that while methane does not have a net dipole moment, it might have a net quadrupole moment. en.m.wikipedia.org/wiki/Quadrupole?wprov=sfla1 – electronpusher May 17 '17 at 7:33
• Imagine applying every argument you've made here, @ThemysteryOflife, to the concept of distance as a vector quantity. – Zhe May 17 '17 at 15:04
• Dipoles have a magnitude and direction, therefore they are vector quantities. – gsurfer04 May 17 '17 at 23:00

we assume for no particular reason that dipole moments must be behaving like vectors

Ah, but there is a reason. Consider the interaction of a molecule with the scalar potential

$$E_{\text{int}} = \int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}$$

where the integral over all space is turned into an expansion:

$$\int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = Q^{}\phi^{} - \sum_{n=1}^{\infty} \frac{1}{n!} \mathbf{Q}_{j_{1}j_{2}...j_{n-1}}^{[n]}\cdot\mathbf{E}_{j_{1}...j_{n-1}}^{[n-1]}.$$

Now to identify some terms. $\phi^{} = 1$, so $Q^{}\phi^{} = E_0$, the interaction-free or field-free energy. $\mathbf{E}^{[n]}$ is the internal electric field or electrostatic potential generated by the atoms to the $n$th total order, and the part that would be the derivative term in a Taylor expansion is the set of electric multipole moments:

$$\mathbf{Q}_{j_{1}...j_{n-1}}^{[n]} = \int r_{j_1} r_{j_2} ... r_{j_n} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.$$

The first-order term is the (electric) dipole moment:

$$\mathbf{Q}_{i}^{} = \int r_{i} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = \mu_i.$$

The second-order term is the (electric) quadrupole moment:

$$\mathbf{Q}_{ij}^{} = \int r_{i} r_{j} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.$$

Convert the continuous electric dipole integral back into a finite sum by discretizing all space into atomic positions and representing the interaction of the electrostatic potential at zero strength with the atomic density as the atomic charge:

$$\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a}.$$

The set of atomic coordinates is nicely represented by each atom as a vector pointing from the Cartesian origin (0,0,0) to the atom $a$'s position $\mathbf{r} = \vec{r} = (r_x,r_y,r_z) = (a_x,a_y,a_z)$ or however you want to notate it, leading to

$$\vec{\mu} = \sum_{a}^{N_{\text{atoms}}} \vec{r}_{a}q_{a},$$

which can be seen as a vector of length 3 from the reduction $r_{ia} \rightarrow \vec{r}_{a}$. The quadrupole moment can be seen as a matrix describing the non-uniformity of the internal electric field, or its gradient.

Dipole moment is a concept introduced by man ryt? So why are they vector quantities? Or better: How did man know that they are vector quantities? How was that deduced?

The dipole moment is a concept "introduced" by us in only two ways that come to mind. The first is that the electric field interaction is truncated to first order in the summation, which is mathematically inexact, even if higher-order moments happen to be zero. Nature includes this interaction to infinite order. Related to the first point, take a molecular sample and apply a single, uniform, finite electric field to it. The above answer deals with a system's permanent or internal moments, but one can also consider the dependence of the energy on an external electric field

$$E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}E^{(n)}(\mathbf{a})\cdot(\mathbf{F}-\mathbf{a})^{n},$$

where the electric field is $\mathbf{F} = \vec{F} = (F_x,F_y,F_z)$ and $\mathbf{a}$ is the expansion point. In practice, we always expand around $\mathbf{a} = \mathbf{0}$, so it is a Maclaurin series:

$$E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}\mathbf{E}^{(n)}(\mathbf{0})\cdot\mathbf{F}^{n}.$$

Assume that $E(\mathbf{F})$ is well-approximated by the first 3 terms of this expansion ($0 \leq n \leq 2$) and truncate it:

$$E(\mathbf{F}) \approx E^{(0)}(\mathbf{0}) + \mathbf{E}^{(1)}(\mathbf{0})\cdot\mathbf{F} + \frac{1}{2}\mathbf{E}^{(2)}(\mathbf{0})\cdot\mathbf{F}^{2}$$

I've been rather sloppy, but this is another way of writing the first derivation, except we are using external applied fields $\{\mathbf{F}\}$ rather than the internal fields $\{\mathbf{E}\}$. For a single applied electric field, only up to the first term $\mathbf{E}^{(1)}(\mathbf{0})$ survives, which is the induced electric dipole moment. I bring this up because assuming the applied field is strong enough that the induced dipole is stronger than the permanent dipole, the total electron distribution will most likely move uniformly or linearly with the field direction, so it makes physical sense to consider just the dipole (up to a point). (Unrelated, $-\mathbf{E}^{(2)}$ is the molecular polarizability $\alpha$.)

I would also say the above shows that we don't define the dipole moment as a vector, it appears naturally that way from mathematics. I also wouldn't call these "thought inventions".

So yes, the dipole moment is represented as a vector, the quadrupole moment is represented as a matrix, the octupole moment is represented as a 3rd-order tensor, etc.

• Can you tell me what all i would need to learn in order to understand your answer and all that maths ?? – Tushar Aug 14 '20 at 8:43