
What does the symbol nabla indicate? - Mathematics Stack …
Mar 27, 2018 · 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 …
Vector triple product with $\nabla$ operator - Physics Stack …
Oct 2, 2024 · Since the del operator, $\boldsymbol {\nabla}$, appears in the triple product, the normal $\text {BAC-CAB}$ vectorial rule for triple products cannot be applied directly.
differential geometry - Coordinate free definition of $\nabla
Jul 21, 2018 · The problem is that it only address the use of $\nabla$ in its function in taking the gradient. It is a satisfactory coordinate independent definition of $\nabla f$ but it isn't a …
multivariable calculus - Del. $\partial, \delta, \nabla $: Correct ...
$\nabla$: 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 …
Difference between the $\\nabla\\cdot a$ and $a\\cdot\\nabla$
Feb 9, 2019 · you shall take the proper orientation for the vectors (column/row) and the proper convention on how the operator $\nabla$ will apply to vectors on the left or on the right.
Is $\nabla$ a vector? - Mathematics Stack Exchange
Nov 19, 2014 · 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 …
differentiation - Directional derivative $ (\mathbf {A}\cdot\nabla ...
Mar 24, 2024 · $\nabla\cdot\mathbf {A}$ is the divergence of the vector field $\mathbf {A}$, which gives a scalar field filed at each point that represents the rate of expansion of the vector field.
vectors - Proof of $\nabla\times (\nabla\times \mathbf f)=\nabla ...
Oct 17, 2019 · To give an example, in the derivation of the wave equation from maxwell's equations, the following identity is used: $$ \nabla\times (\nabla\times \mathbf f)=\nabla …
Where does the relation $\nabla^2 (1/r)=-4\pi\delta^3 ( {\bf r ...
It is often quoted in physics textbooks for finding the electric potential using Green's function that $$\nabla ^2 \left (\frac {1} {r}\right)=-4\pi\delta^3 ( {\bf r}),$$ or more generally $$\nabl...
$\nabla$, $\cdot \nabla$, $\nabla \cdot$, $\nabla^2$ - What do …
Feb 19, 2023 · Finally, there's a $\nabla\cdot$ operator which seems to be the sum of the components of the first derivatives. So in the absense of an explanation, I'm somewhat …