PlanetPhysics/Position Vector

In the space $\mathbb {R} ^{3}$, the vector ${\vec {r}}\;:=\;(x,\,y,\,z)\;=\;x{\vec {i}}+y{\vec {j}}+z{\vec {k}}$ directed from the origin to a point \,$(x,\,y,\,z)$\, is the position vector of this point.\, When the point is variable, ${\vec {r}}$ represents a vector field and its length $r\;:=\;{\sqrt {x^{2}+y^{2}+z^{2}}}$ a scalar field.

The simple formulae

• $\nabla \!\cdot {\vec {r}}=3$
• $\nabla \!\times \!{\vec {r}}={\vec {0}}$
• $\nabla r={\frac {\vec {r}}{r}}={\vec {r}}^{0}$
• $\nabla {\frac {1}{r}}=-{\frac {\vec {r}}{r^{3}}}=-{\frac {{\vec {r}}^{0}}{r^{2}}}$
• $\nabla ^{2}{\frac {1}{r}}=0$

are valid, where ${\vec {r}}^{0}$ is the unit vector having the direction of ${\vec {r}}$.

If\, Failed to parse (syntax error): {\displaystyle \vec} \, is a constant vector,\, ${\vec {U}}\!\!:\mathbb {R} ^{3}\to \mathbb {R} ^{3}$\, a vector function and\, $f\!\!:\mathbb {R} \to \mathbb {R}$\, is a twice differentiable function, then the formulae

• Failed to parse (syntax error): {\displaystyle \nabla(\vec\cdot\!\vec{r}) = \vec}
• $\nabla \cdot ({\vec {\times }}{\vec {r}})=0$
• $({\vec {U}}\!\cdot \!\nabla ){\vec {r}}={\vec {U}}$
• $({\vec {U}}\!\times \!\nabla )\!\cdot \!{\vec {r}}=0$
• $({\vec {U}}\!\times \!\nabla )\!\times \!{\vec {r}}=-2{\vec {U}}$
• $\nabla f(r)=f'(r)\,{\vec {r}}^{0}$
• $\nabla ^{2}f(r)=f''(r)\!+{\frac {2}{r}}f'(r)$

hold.