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I'm trying to verify this identity using index notation: $$ \vec{V} \times \left(\nabla \times \vec{V}\right) = \frac{1}{2} \nabla\left(\vec{V}\cdot \vec{V}\right)-\left(\nabla\cdot\vec{V}\right)\vec{V}$$ Solving the LHS: $$v_i \hat{e_i} \times (\partial_j\hat{e_j} \times v_k \hat{e_k}) = v_i\hat{e_i} \times \varepsilon{_{jkm}} v_{k,j} \hat{e_m} = \varepsilon{_{qim}} \varepsilon{_{jkm}} v_i v_{k,j} \hat{e_q} $$

$$(\delta_{qj} \delta_{ik} - \delta_{qk} \delta_{ji}) v_i v_{k,j} \hat{e_q} = (v_kv_{k,q} - v_{q,j} v_j) \hat{e_q} $$

Which can be expressed as:

$$ (v_kv_{k,q} - v_{q,j} v_j) \hat{e_q} = \left(v_k \frac{\partial v_k}{\partial x_q} - \frac{\partial v_q}{\partial x_j} v_j\right) \hat{e_q}$$

Now, my question is about the way differentials can be "manipulated" so we can prove the identity. Is it possible to write the term $v_k \frac{\partial v_k}{\partial x_q}$ as $ \frac{\partial (v_k)^{2}}{\partial x_q}$ and then perform a derivation so we can end up with something similar to $ \frac{1}{2} \nabla(\vec{V}\cdot \vec{V})$? How could i express $\frac{\partial v_q}{\partial x_j} v_j$ as $(\nabla\cdot\vec{V})\vec{V}$ knowing that i need the subindexes $q$ and $j$ equal so i can write that part as the divergence of $\vec{V}$.

Sine of the Time
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J. J. Gonzalez
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1 Answers1

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I'll write out a full answer and hopefully that clarifies what manipulations and operations are required. \begin{align*} \mathbf V \times (\boldsymbol \nabla \times \mathbf V) &= \varepsilon_{ijk} V_i \left[ \boldsymbol \nabla \times \mathbf V\right]_j \mathbf e_k \\ &= \varepsilon_{ijk} V_i \varepsilon_{jpl} \partial_p V_l \mathbf e_k \\ &= \left( \delta_{kp} \delta_{il} - \delta_{kl} \delta_{ip}\right) V_i\partial_p V_l \mathbf e_k \\ &= V_i \frac{\partial V_i}{\partial x_k} \mathbf e_k - \frac{\partial V_i}{\partial x_i} V_k \mathbf e_k \\ &= \frac 12 \partial_k V_i V_i \mathbf e_k - (\boldsymbol \nabla \boldsymbol \cdot \mathbf V)V_k \mathbf e_k \tag{$*$} \\ &= \frac 12 \boldsymbol \nabla (\mathbf V \boldsymbol \cdot \mathbf V) - (\boldsymbol \nabla \boldsymbol \cdot \mathbf V)\mathbf V, \end{align*} where $(*)$ follows since \begin{alignat*}{2} \partial_k V_i V_i \mathbf e_k &= \frac{\partial V_i}{\partial x_k}V_i \mathbf e_k + V_i \frac{\partial V_i}{\partial x_k} \mathbf e_k && \quad\text{by the product rule} \\ &= 2 V_i \frac{\partial V_i}{\partial x_k} \mathbf e_k. \end{alignat*} Thus the result is obtained.

kipf
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