What does the symbol nabla indicate? - Mathematics Stack Exchange
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 what it means …
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: $$ \nabla\times (\nabla\times \mathbf f)=\nabla (\nabla\cdot \mathbf f) …
Del. $\\partial, \\delta, \\nabla $: Correct enunciation
$\nabla$: Called Nabla or del. This has four different uses, which can be easily distinguished while reading out loud, but it gets confusing when the first and last uses (grad and covariant derivative) get …
Nabla and its rules - Mathematics Stack Exchange
Feb 25, 2021 · "Nabla" is a symbolic "vector differential operator". It can be written, symbolically, $\nabla= \frac {\partial} {\partial x}\vec {i}+ \frac {\partial} {\partial Y ...
A rigorous proof that $\nabla \cdot E = \frac {\rho} {\epsilon_0}$
Mar 26, 2021 · Edit: I think a key insight I was missing is that proving (1) is equivalent to showing that $$ \tag {2} \text {if} \quad \varphi (x) = -\frac {1} {4 \pi \epsilon_0} \int_ {\mathbb R^3} \rho (y) \| x - y \|^ { …
Is $\\nabla$ a vector? - Mathematics Stack Exchange
The following passage has been extracted from the book "Mathematical methods for Physicists": A key idea of the present chapter is that a quantity that is properly called a vector must have the
$\nabla$, $\cdot \nabla$, $\nabla \cdot$, $\nabla^2$ - What do they …
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 confused as to how the …
What does the equation $s=-\nabla \cdot (\rho \nabla u)$ mean?
Jan 18, 2023 · My understanding I think the equation $s=-\nabla \cdot (\rho \nabla u)$ holds in distribution sense. However, I'm not familiar with PDEs and not sure what it means exactly.
what does $ (A\cdot\nabla)B$ mean? - Mathematics Stack Exchange
Not really. $A\cdot\nabla$ is not quite what you say. There's only one cdot in $ (A\cdot\nabla)B$.
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...