Why is $(\mathbf{v} \cdot \nabla)\mathbf{v} = (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2)$?

Question

The Convective Derivative or Material Derivative is usually written as $\frac{D}{Dt}=\frac{\partial}{\partial t} + \mathbf{v} \cdot \nabla$. According to MathWorld, this equation, multiplied with ${\bf{v}}$ equals: $$ \frac{D \mathbf{v}}{Dt} = \frac{\partial \mathbf{v}}{\partial t} + (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2) $$

Clearly, it must hold that;

$$ (\mathbf{v} \cdot \nabla)\mathbf{v} = (\nabla \times \mathbf{v}) \times \mathbf{v} + \nabla (\frac{1}{2} \mathbf{v}^2) $$

However, I do not spot why this is true. What is the (trivial) identity that I am missing?

Answer 1

We can write expand the components using Levi-Civita symbol (using the convention of summing repeated indices):

$$ \left[(\nabla \times \mathbf{v})\right]_i = \epsilon_{ijk} \left(\partial_j v_k\right). $$

$$ \left[(\nabla \times \mathbf{v}) \times \mathbf{v}\right]_{l} = \epsilon_{lim} \left[\nabla \times \mathbf{v} \right]_{i} v_m= \epsilon_{lim} \epsilon_{ijk} v_m \left(\partial_j v_k\right) = -\epsilon_{ilm} \epsilon_{ijk} v_m \left(\partial_j v_k\right) $$

We also have this identity between Levi-Civita and Kronecker delta:

$$ \epsilon_{ilm} \epsilon_{ijk} = \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj}. $$

By plugging this identity in the second equation:

$$ \begin{align} \left[(\nabla \times \mathbf{v}) \times \mathbf{v}\right]_{l} &= - \delta_{lj} \delta_{mk} v_m \left(\partial_j v_k\right) + \delta_{lk} \delta_{mj} v_m \left(\partial_j v_k\right)\\ &= - v_k \partial_l v_k + v_j \partial_j v_l \\ &= - \frac{1}{2} \partial_l (v_k v_k) + (v_j \partial_j) v_l \\ &= - \left[\nabla (\frac{1}{2} \mathbf{v}^2)\right]_l + \left[(\mathbf{v} \cdot \nabla)\mathbf{v}\right]_l \end{align} $$

If you are not familiar with Levi-Civita symbol, expand $(\nabla \times \mathbf{v}) \times \mathbf{v}$ for its $x$ component. It will be similar for other components.

Answer 2

Here is another approach using the BAC-CAB identity and the definition of tensor product. I remind you of the BAC-CAB identity

$$\begin{align} A \times (B \times C) &= B(A \cdot C)-C(A \cdot B) \\ &= (B \otimes A) \cdot C - A \cdot (B \otimes C) \end{align}$$

So we can write

$$\begin{align} (\nabla \times \mathbf{v}) \times \mathbf{v} &= -\mathbf{v} \times (\nabla \times \mathbf{v}) \\ &= - [(\nabla \otimes {\bf{v}}) \cdot {\bf{v}} - {\bf{v}} \cdot (\nabla \otimes {\bf{v}})] \\ &\equiv - [(\nabla {\bf{v}}) \cdot {\bf{v}} - {\bf{v}} \cdot (\nabla {\bf{v}})] \\ &=-\nabla {\bf{v}} \cdot {\bf{v}} + {\bf{v}} \cdot \nabla {\bf{v}} \\ &=-\frac{1}{2}\nabla ({\bf{v}} \cdot {\bf{v}}) + {\bf{v}} \cdot \nabla {\bf{v}} \end{align}$$

or equivalently we get

$${\bf{v}} \cdot \nabla {\bf{v}}=(\nabla \times \mathbf{v}) \times \mathbf{v}+\frac{1}{2}\nabla ({\bf{v}} \cdot {\bf{v}})$$

Answer 3

$$\mathbf{v}\cdot(\nabla\wedge\mathbf{v}) = (\mathbf{v}\cdot\nabla)\mathbf{v} - \dot{\nabla}(\mathbf{v}\cdot\dot{\mathbf{v}})$$ This is a special case of a general identity in geometric algebra, namely $$\mathbf{a}\cdot(\mathbf{b}\wedge\mathbf{c}) = (\mathbf{a}\cdot\mathbf{b})\mathbf{c} - (\mathbf{a}\cdot\mathbf{c})\mathbf{b}$$

The over dots tell you what expression the $\nabla$ operator is operating on, so in $\dot{\nabla}(\mathbf{v}\cdot\dot{\mathbf{v}})$, the left $\mathbf{v}$ is being held constant. At any rate, the product rule gives us $$\nabla\mathbf{v}^2 = \nabla(\mathbf{v}\cdot\mathbf{v}) = \dot{\nabla}(\dot{\mathbf{v}}\cdot\mathbf{v}) + \dot{\nabla}(\mathbf{v}\cdot\dot{\mathbf{v}})$$ but commutativity of the dot product just says that that's $2\dot\nabla(\mathbf{v}\cdot\dot{\mathbf{v}})$ so we get $$\frac{1}{2}\nabla\mathbf{v}^2 = \dot\nabla(\mathbf{v}\cdot\dot{\mathbf{v}})$$ this gives in the original equation, slightly rearranged $$(\mathbf{v}\cdot\nabla)\mathbf{v} = \mathbf{v}\cdot(\nabla\wedge\mathbf{v}) + \frac{1}{2}\nabla\mathbf{v}^2$$

$\mathbf{v}\cdot(\nabla\wedge\mathbf{v})$ is the geometric algebra way of writing $(\nabla\times\mathbf{v})\times\mathbf{v}$ in the 3D case (though all the equations I've used are true in all dimensions). $\mathbf{a}\times\mathbf{b} = (\mathbf{a}\wedge\mathbf{b})I$ and $\mathbf{a}\cdot(\mathbf{b}I) =(\mathbf{a}\wedge\mathbf{b})I$ let you show that (where $I$ is the pseudoscalar).

If you aren't familiar with geometric algebra, then I highly recommend learning it. If you are doing vector calculus and worrying about material derivatives, you have more than enough prerequisites to learn geometric algebra. It is something that could be taught at high school or early undergrad, and it dramatically simplifies and unifies many formal systems used in physics.