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  • What is the meaning of $A. \\nabla - Mathematics Stack Exchange
    If $\mathbf A = \pmatrix{a_x\\a_y\\a_z}$, then $$(\mathbf A\cdot\nabla)\phi = a_x\frac{\partial}{\partial x}\phi + a_y\frac{\partial}{\partial y}\phi + a_z\frac{\partial}{\partial z}\phi$$ Basically, you treat $\nabla$ as a vector of derivatives and do vector algebra, except that you are careful not to move terms across the derivative
  • Is there a well defined difference between $\\nabla$ and $D$?
    The gradient $\nabla f$ is a vector field (a tangent vector at each point) characterized by the property
  • homework and exercises - Vector triple product with $\nabla$ operator . . .
    The vector calculus identity of the cross product of a curl holds: $$\mathbf{v} \times \left( \nabla \times \mathbf{a} \right) = \nabla_a \left( \mathbf{v} \cdot \mathbf{a} \right) - \mathbf{v} \cdot \nabla \mathbf{a} $$ where the Feynman subscript notation $\nabla_\mathbf{a}$ is used, which means the subscripted gradient operates only on the
  • matrices - Meaning of $\nabla \cdot \mathbf{A}$ for matrix $\mathbf{A . . .
    What is $\nabla \cdot \mathbf{A}$ when $\mathbf{A} \in \mathbb{R}^{m \times n}$ is a matrix, and where is there a consise definition of this notation? The Euler equations on Wikipedia contain terms on the form $\nabla \cdot (\mathbf{u} \otimes \mathbf{u} - w\mathbf{I})$ where $\mathbf{I}$ is the identity matrix Other material on the Euler
  • what does $(A\\cdot\\nabla)B$ mean? - Mathematics Stack Exchange
    Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
  • Del. $\partial, \delta, \nabla - Mathematics Stack Exchange
    $\nabla$: Called Nabla or del This has four different uses, which can be easily distinguished while reading out loud, but it gets confusing when the first and last uses (grad and covariant derivative) get mixed up with $\partial$ and $\delta$ Gradient grad: $\vec{\nabla}\phi$ (phi is a scalar) Read as "nabla phi", or "del phi"
  • Proving $d^\\nabla( d^\\nabla \\omega) = F^\\nabla \\wedge \\omega$
    What I said above answers your question But here are some extra good-to-know things In Riemannian geometry, it is often the case that people do not go higher than second derivatives, because ‘everything else is encoded in the curvature’
  • How is $\\nabla (u\\cdot A) =u\\cdot \\nabla A+ u\\times (\\nabla . . .
    $$\vec{a} \cdot (\nabla \vec{b}) \neq (\vec{a}\cdot \nabla) \vec{b}$$ The answer that I linked derived a formula involving $ (\vec{a}\cdot \nabla) \vec{b}$ This is why they got an extra cross product term I derived a formula involving $ \vec{a} \cdot (\nabla \vec{b}) $, which is why I didn't get the cross product term
  • Why does $\\nabla \\to ik$ when you Fourier transform?
    Re: comments to the accepted answer There is a very natural interpretation: For a linear problem the Fourier transform is the same as a plane wave Ansatz





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