Let $u = (u_1,u_2,u_3)^T$ be a velocity vector.

Then, I know $\nabla u$ is a matrix whose $(i,j)$ entry is $\partial_{x_j} u_i$,

In the momentum equation of fluid dynamics, we have $u \cdot \nabla u$, which I think should be $u^T\nabla u$ in analogy with the usual dot product of two vectors, but this does not give the correct terms. Why is there notational inconsistency?

  • $\begingroup$ Do you know the correct terms? Also I might interpret $u\cdot \nabla u$ as $u_1\partial_1 u + u_2\partial_2 u + u_3\partial_3 u$. $\endgroup$
    – jgon
    Jan 19, 2018 at 1:58
  • $\begingroup$ Also what you have right now doesn't produce a vector, it produces a covector. $\endgroup$
    – jgon
    Jan 19, 2018 at 2:00
  • $\begingroup$ @jgon $u$ is a 3 by 1 vector and $\nabla u$ is a 3 by 3 matrix, so $u^T\nabla u$ is a 1 by 3 vector. This is again inconsistent as I want to get 3 by 1 vector $\endgroup$
    – nan
    Jan 19, 2018 at 2:02
  • $\begingroup$ Sorry, yeah my interpretation is that they mean $(\nabla u)u$. $\endgroup$
    – jgon
    Jan 19, 2018 at 5:01
  • $\begingroup$ For a coordinate system with covariant basis vectors $\vec g_i$, corresponding velocity components $\vec v^i$, you get $\vec v\cdot\nabla\vec v=\left(v^iv^j_{,I}+v^iv^k\Gamma^j_{ik}\right)\vec g_j$ where $v^j_{,I}$ represents partial differentiation of $v^j$with respect to the $i$-th basis variable and $\Gamma^j_{ik}$ are the Christoffel symbols of the second kind. Of course in an orthogonal coordinate system this doesn't look that bad. $\endgroup$
    – David
    Jan 23, 2018 at 22:33


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