Tension field

Remarks

If $M$ is compact without boundary and $u_s$ is a smooth family of maps with $u_0 = u$ and variation field $V = \partial_s u_s |_{s=0}$, then the energy satisfies $$\frac{d}{ds}\Big|_{s=0} E(u_s) = -\int_M h\big(\tau(u), V\big) \, dV_g.$$ So $-\tau(u)$ is the gradient of $E$, and $u$ is a harmonic map exactly when $\tau(u) = 0$.

Example: let $M = \R$ and $\gamma \colon \R \to N$ a curve; then $\tau(\gamma) = \nabla_{\dot\gamma} \dot\gamma$, the covariant acceleration, so $\tau(\gamma) = 0$ is the geodesic equation. For $\gamma(t) = (\cos t, \sin t, 0)$ into the unit sphere $S^2$, a great circle, $\tau(\gamma) = 0$.

Non-example: the curve $\gamma(t) = (\cos t, \sin t)$ into $\R^2$ with the Euclidean metric has $\tau(\gamma) = \ddot\gamma = (-\cos t, -\sin t) \ne 0$, so it is not harmonic. The same parametrized circle, seen as a map into the unit circle $S^1$, has tension $0$, so the tension field depends on the target. For $N = \R^n$ with the Euclidean metric the Christoffel symbols vanish and $\tau(u) = (\Delta_g u^1, \dots, \Delta_g u^n)$.