What does the symbol nabla indicate? - Mathematics Stack Exchange First up, this question differs from the other ones on this site as I would like to know the isolated meaning of nabla if that makes sense Meanwhile, other questions might ask what it means in rel
Vector triple product with $\nabla$ operator - Physics Stack Exchange $$\nabla_a \left ( \mathbf {a} \cdot \mathbf {a} \right) = \frac {1} {2}\nabla (\mathbf {a}^2) $$ Which establishes the correctness of your first equation The key takeaway as the other answers mention, is indeed not to try and apply regular vector identities to the $\nabla$ vector operator
vectors - Proof of $\nabla\times (\nabla\times \mathbf f)=\nabla . . . To give an example, in the derivation of the wave equation from maxwell's equations, the following identity is used: $$ \nabla\times (\nabla\times \mathbf f)=\nabla (\nabla\cdot \mathbf f)-\nabla^2\mathbf f \label {1} $$ I can prove it by direct calculation, but that would be very boring and mechanical
what does $ (A\cdot\nabla)B$ mean? - Mathematics Stack Exchange You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
Is $\nabla$ a vector? - Mathematics Stack Exchange Cross product: $\nabla \times (Vector)=Vector$ From the above equation of cross product we can say that $\nabla$ is a vector (specifically vector operator) However, a vector generally has magnitude and an associated direction
Gradient of a dot product - Mathematics Stack Exchange They are basically the same For the first identity, you could refer to my proof using Levi-Civita notation here And for the second, you should know that $\nabla a=\left (\frac {\partial a_j} {\partial x_i}\right)=\left (\frac {\partial a_i} {\partial x_j}\right)^T$ is a matrix and dot product is exactly matrix multiplication So the proof is $$ (\nabla a)\cdot b+ (\nabla b)\cdot a=\left
multivariable calculus - Proof for the curl of a curl of a vector field . . . $$ = ( \nabla ( \nabla \cdot \vec {A} ) - \nabla^2 \vec {A} )_k $$ which proves the identity This approach probably doesn't seem to simple to you if you aren't familiar with the tools I am using, but honestly this approach has really grown on me as the easiest way to get the result quickly