June 9th: Dipoles and More Dipoles…

    Potential of a pair of oppositely charged particles on the plane

    WLOG, suppose the particles lie on the x-axis and their center of mass is on the origin. Let d be the distance from each particle to the origin and Q to be the magnitude of each of their charge.

    Then, the potential at a point (x, y) is given by:

    u(x, y) = Q ln(\frac{1}{\sqrt{(x-d)^2 + y^2}} - Q ln(\frac{1}{\sqrt{(x+d)^2 + y^2}}.

    For a fixed C \in \mathbb{R}:

    Q ln(\frac{1}{\sqrt{(x-d)^2 + y^2}} - Q ln(\frac{1}{\sqrt{(x+d)^2 + y^2}} = C

    \Leftrightarrow Q ln(\frac{\sqrt{(x+d)^2 + y^2}}{\sqrt{(x-d)^2 + y^2}}) = C

    \Leftrightarrow \frac{\sqrt{(x+d)^2 + y^2}}{\sqrt{(x-d)^2 + y^2}} = C_1

    \Leftrightarrow \sqrt{(x+d)^2 + y^2} = C_1 \sqrt{(x-d)^2 + y^2}

    \Leftrightarrow (x+d)^2 + y^2 = C_2 ((x-d)^2 + y^2)

    \Leftrightarrow (1-C_2)x^2 + 2(1-C_2)xd + (1-C_2)y^2 = (C_2 - 1)y^2, which is an equation of a circle.

    Key: the points whose distances to 2 other points has a fixed logarithm ratio form a circle. I’ve given an algebraic proof above. Is there a more intuitive way to understand this?

    From Arnold’s PDE

    Geometric interpretation of potential of double layer:

    This means that the potential at P corresponding to the double layer dS is approximately the angle at P subtended by the curve dS.

    Now, for a segment of the curve \gamma, the angle subtended at point P is given by:

    \int_\gamma \frac{x}{x^2 + y^2} dS (where dS is the infinitesimal approximation of \gamma at each point by its tangent line).

    Hence, if we view the angle as a function with respect to the point P, it is harmonic (because x and y are the two components of the curve with respect to P).

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