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It is known that interaction energy = $-\vec{p}.\vec{E}$ where $\vec{p}$ is dipole moment and $\vec{E}$ is the electric field.

I have to calculate the interaction energy of a system whose dipole moment and potential are the only available data.

My assumption is since the negative gradient of potential is electric field, can we calculate the dipolar component of the potential as an alternative to electric field, i.e.

$$\begin{align} \phi_{dx}&=\phi_x(x-x_c) \\ \phi_{dy}&=\phi_y(y-y_c) \\ \phi_{dz}&=\phi_z(z-z_c) \end{align}$$ where $x$, $y$, $z$ are the positions of atoms and $x_c$, $y_c$, $z_c$ is the geometrical centre, and then calculate the dot product with the dipole moment $(p_x,p_y,p_z)$.

Does it sounds meaningful?? Kindly provide some suggestions in solving my problem.

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  • $\begingroup$ I've edited your question to typeset the equations using MathJax. Apologies if I've misunderstood any of them. $\endgroup$
    – John Rennie
    Commented Apr 8, 2014 at 9:34

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What you're calling the "dipolar component of the potential" is not actually that object. For something to be a 'component' of the potential it also needs to be a scalar; in that sense, the sum of those three terms, $$\Delta \phi=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})$$ could indeed be called the 'dipolar component', and it will work in any region where there are no charges other than test charges. You chould note in particular that this object is intimately related to the electric field, as $$\Delta\phi=-\nabla \mathbf{E}\cdot(\mathbf{r}-\mathbf{r}_\text{c}).$$ Because of this, using the dipolar component will work as long as you evaluate it at the dipole separation.

On the other hand, if you take the vector $$\mathbf{\Delta\phi}=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})\mathbf{e}_j$$ and then dot that with the atoms' positions, then you will get an interaction energy $$ U =\sum_a\sum_j\frac{\partial \phi}{\partial x_j}(x_{a,j}-x_{\text c,j})p_j =\frac2q\sum_j\frac{\partial \phi}{\partial x_j}p_j^2 $$ which is manifestly wrong.

So, bottom line: no.

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  • $\begingroup$ Thanks for the info. I am considering only the scalar dipolar component of the potential and I calculate the potential by keeping all charges zero except test charge. But, I want to know is: can the dot product of dipole and this dipolar component of potential be considered to be equivalent to the dot product of -p.E. I intend to hypothesize that if this angle<90 then interaction energies are favorable and vice versa. Am I correct in doing so? $\endgroup$
    – user44141
    Commented Apr 8, 2014 at 20:16
  • $\begingroup$ No, it is incorrect. There is no 'dot product' between the dipole (which is a vector) and (the dipolar component of) the potential, which is a scalar. You should use the electric field as I said. $\endgroup$
    – Emilio Pisanty
    Commented Apr 8, 2014 at 21:11
  • $\begingroup$ Sorry I shouldn't argue but since I come from bioscience background, I have another (possibly stupid) thought. Dot product of p.E is magnitude of p * magnitude of E and cos(theta). So, if I substitute magnitude of E by magnitude of my dipolar component of potential don't you think the hypothesis will still work?? I just want the theta angle to give me favorable interaction energies. $\endgroup$
    – user44141
    Commented Apr 9, 2014 at 14:39
  • $\begingroup$ No, it will not work. The dimensions do not check out: your proposal would have dimensions of [potential/distance]x[distance]x[dipole moment]=[potential x charge x distance], and it needs to have dimensions of [potential x charge]. Always do a dimensional analysis and correct your methods if the dimensions are incorrect. $\endgroup$
    – Emilio Pisanty
    Commented Apr 9, 2014 at 15:28
  • $\begingroup$ Yes I agree with you. But sorry I couldn't follow your dimension notations. What I meant is as follows: Instead of the actual equation as interaction energy=dipole moment.electric field, i.e. [dipole moment]x[potential/distance]=[charge x distance]x[potential/distance]=[potential x charge], my proposal is interaction energy=dipole moment.dipolar potential i.e. [charge x distance]x[potential x distance] which is obviously not true. And I vote your answer. But I have a last question. $\endgroup$
    – user44141
    Commented Apr 9, 2014 at 17:23
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The electric field is given by $$ E=-\nabla V-\frac{\partial A}{\partial t}, $$ where $V$ is the electric potential and $A$ is the vector potential. So you can get the electric field from the potential. Or if you have measurements of a large enough set of points that are close together compared to the scales over which the field varies you might be able to calculate a good approximation to the field.

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