Explain why the convective derivative of the position vector is the velocity.

Question

Can someone please explain why the convective derivative of the position is the velocity. It's not very obvious to me from the definition.

Thanks

Answer 1

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I assume by convective derivative you mean this: $$ \frac D{Dt} = \frac\partial{\partial t} + \vec v\cdot\nabla $$ where $\vec v = \vec v(x)$ is the velocity field and $\vec x$ is the position vector. The key fact is that $$ (\vec a\cdot\nabla)f(\vec x) = f(\vec a) $$ for any linear function $f$ and vector $\vec a$. This is particularly intuitive when $f(\vec x) = \vec x$ is the identity function: $$ (\vec a\cdot\nabla)\vec x = \vec a, $$ which is saying that the change of $\vec x$ as $\vec x$ varies in the direction $\vec a$ is just $\vec a$. Now in the convective derivative, the velocity $\vec v$ is not being differentiated; it's just being used like an ordinary vector. Hence $$ \frac{D\vec x}{Dt} = \frac{\partial\vec x}{\partial t} + (\vec v\cdot\nabla)\vec x = \frac{\partial\vec x}{\partial t} + \vec v. $$ We're almost there, but we still have that time derivative, which we want to be zero. There are two possible interpretations of $\vec x$ here:

1. $\vec x = \vec x(t)$ is the path of a particle as time varies.
2. $\vec x = (x, y, z)$ where $x, y, z, t$ are considered independent variables.

In interpretation (1) we need $\vec x(t) = \text{const}$, but in interpretation (2) we always have $\partial\vec x/\partial t = 0$ whence $$ \frac{D\vec x}{Dt} = \vec v. $$