I have a 3 dimensional spherical topology, and I draw a curve onto the sphere labelled by $\vec{n}(\vec{r},t)$. The area bounded by the curve is termed the "Wess Zumino Action" (Hence my motivation to ask this site rather than math).

Now suppose I introduce an infinitesmal change to this curve, labelled $\delta \vec{n}$, then the variation in the area is given by:

\begin{equation} \delta S = \delta \vec{n} \cdot (\vec{n} \times \partial _0 \vec{n}) \end{equation}

I tried to reason it out, but I'm not quite able to get this expression. Let's say the curve is drawn along the equator, with a small variation in the z direction. $\partial _0 \vec{n}$ should point in the direction of the curve, so $\vec{n} \times \partial _0 \vec{n}$ should give a vector that points parallel the z axis, but then the dot product doesn't really give me any intuition on why this should be the change in the area. Could someone give me a derivation, and also point toward some resources that discuss similar intuition and proofs?

I'm currently looking at Eduardo Fradkin's Condensed Matter Physics but it doesn't have more than a highly condensed treatment of topology. This is taken from eqn 7.28 and 7.42 from Eduardo Fradkin's "Field Theories of Condensed Matter physics", in the chapter titled "sigma models and topological terms"

Please let me know of any other clarifications if this isn't enough information.

  • $\begingroup$ Is this from a reference? $\endgroup$ – Qmechanic Apr 16 '19 at 5:07
  • $\begingroup$ the derivative you write $\partial_0 \vec{n}$ is done with respect to which variable? $\endgroup$ – AoZora Apr 16 '19 at 6:41
  • $\begingroup$ @Qmechanic yes this is from a reference, eqn 7.28 and 7.42 from Eduardo Fradkin's "Field Theories of Condensed Matter physics", in the chapter titled "sigma models and topological terms" $\endgroup$ – pyroscepter Apr 16 '19 at 16:41
  • $\begingroup$ @france95 The derivative is with respect to the time, where varying the time traces out the curve on the sphere through the radial vector $\vec{n}(r,t)$ on the surface of the sphere. $\endgroup$ – pyroscepter Apr 16 '19 at 16:43
  • 2
    $\begingroup$ if you are un the surface of a sphere, the cross product just gives you something perpendicular to $\partial_0 n$, on the surface of the sphere (and proportional to the radius $|n|$). Then you do the scalar product with $\delta n$ that gives the area of the parallelogram, which is the change of area. The cross product can be seen as "best direction to move in order to change the area" $\endgroup$ – patta Apr 16 '19 at 16:57

I don't have your reference but I think I figured it out. Just in case it isn't clear from the question, this is a 2D sphere embedded in 3D Euclidean space. $n(x^0)$ is a unit vector that traces out a closed curve on the sphere as $x^0$ is varied. We can add a coordinate and extend this to a unit vector $n(x^0,x^1)$ that will cover the entire entire area enclosed by the curve as $x^0,x^1$ are varied.

The area enclosed could be written $$S=\int dx^0 dx^1 (\partial_0 n\times \partial_1 n)\cdot B$$ where $B$ is the magnetic field of a monopole at the center of the sphere. If it is normalized appropriately so the flux gives the area, this is just, $$B= n.$$ The reason for calling it a `magnetic field' is we can write $B=dA$, at least in the coordinate patch covering this closed curve, so by Stoke's theorem we can instead write, $$S=\int dx^0 A\cdot \partial_0 n$$ Note that $A$ varies on the sphere so it is a function of $n$. For what comes next I will use index notation and Latin indices refer to the Euclidean space $n$ lives in, not the $x^0,x^1$ coordinates. $$\delta S = \int dx^0 \left(\delta A^a\partial_0 n^a-\partial_0A^a\delta n^a\right)=\int dx^0 \left(\partial_b A^a \delta n^b\partial_0 n^a-\partial_bA^a \partial_0 n^b\delta n^a\right)=\int dx^0 \delta n^a\left(\partial_a A^b -\partial_b A^a \right)\partial_0 n^b$$

Now what's in the paranthesis is just the curl of $A$, which is $B=n$. Specifically, in terms of indices it is $\epsilon_{abc}n^c$ (where $\epsilon$ is the Levi-Civita symbol).

So going back to vector notation I have $$\delta S = \int dx^0 \delta n\cdot \left(\partial_0 n \times n\right)$$

  • $\begingroup$ Absolutely perfect! This is exactly what I was looking for. Thanks for being ultra clear with your answer. $\endgroup$ – pyroscepter Apr 16 '19 at 20:05
  • $\begingroup$ Incredible, how I had the exact same doubt when reading the exact same book and could find exactly that on the internet. $\endgroup$ – PedroDM Apr 26 '20 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.