# Gradient and Divergence in Orthonormal Curvilinear Coordinates

Gradient and Divergence in Orthonormal Curvilinear Coordinates Swapnil Sunil Jain Aug 7, 2006

Gradient and Divergence in Orthonormal Curvilinear Coordinates

## Gradient in Curvilinear Coordinates

In rectangular coordinates (where $f=f(x,y,z)$), an infinitesimal   length vector $d\vec{l}$ is given by

 $\displaystyle d\vec{l}=dx\hat{x}+dy\hat{y}+dz\hat{z}$

the gradient is given by

 $\displaystyle\nabla=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{% \partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$
 $\displaystyle df=\nabla f\circ d\vec{l}=\frac{\partial f}{\partial x}dx+\frac{% \partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$

Similarly in orthonormal curvilinear coordinates ( where $f=f(q_{1},q_{2},q_{3})$), the infinitesimal length vector is given by11See my article Unit Vectors in Curvilinear Coordinates for an insight into this expression.

 $\displaystyle d\vec{l}=h_{1}dq_{1}\hat{q}_{1}+h_{2}dq_{2}\hat{q}_{2}+h_{3}dq_{% 3}\hat{q}_{3}$

where

 $\displaystyle h_{i}=\sqrt{\sum_{k}\Big{(}\frac{\partial\vec{x}_{k}}{\partial q% _{i}}\Big{)}^{2}}\mbox{ and }\hat{q}_{i}=\frac{1}{h_{i}}\Big{(}\frac{\partial% \vec{x}_{k}}{\partial q_{i}}\Big{)}\mbox{ for }i\in{1,2,3}$

So if

 $\displaystyle\nabla=\alpha\frac{\partial}{\partial q_{1}}\hat{q}_{1}+\beta% \frac{\partial}{\partial q_{2}}\hat{q}_{2}+\gamma\frac{\partial}{\partial q_{3% }}\hat{q}_{3}$

then since we know that

 $\displaystyle df=\frac{\partial f}{\partial q_{1}}dq_{1}+\frac{\partial f}{% \partial q_{2}}dq_{2}+\frac{\partial f}{\partial q_{3}}dq_{3}$

and

 $\displaystyle df=\nabla F\circ d\vec{l}=\alpha h_{1}\frac{\partial f}{\partial q% _{1}}dq_{1}+\beta h_{2}\frac{\partial f}{\partial q_{2}}dq_{2}+\gamma h_{3}% \frac{\partial f}{\partial q_{3}}dq_{3}$

this implies that

 $\displaystyle\alpha=\frac{1}{h_{i}};\beta=\frac{1}{h_{2}};\gamma=\frac{1}{h_{3}}$

Hence,

 $\displaystyle\nabla$ $\displaystyle=$ $\displaystyle\frac{1}{h_{1}}\frac{\partial}{\partial q_{1}}\hat{q}_{1}+\frac{1% }{h_{2}}\frac{\partial}{\partial q_{2}}\hat{q}_{2}+\frac{1}{h_{3}}\frac{% \partial}{\partial q_{3}}\hat{q}_{3}$ $\displaystyle=$ $\displaystyle\sum_{i}\frac{1}{h_{i}}\frac{\partial}{\partial q_{i}}\hat{q}_{i}$

## Divergence in Curvilinear Coordinates

In the previous section  we concluded that in curvilinear coordinates, the gradient operator $\nabla$ is given by

 $\displaystyle\nabla=\sum_{i}\frac{1}{h_{i}}\frac{\partial}{\partial q_{i}}\hat% {q}_{i}$

Then for $\vec{F}=F_{1}\hat{q}_{1}+F_{2}\hat{q}_{2}+F_{3}\hat{q}_{3}$, the divergence of $\vec{F}$ is given by

 $\displaystyle\nabla\circ\vec{F}=\Big{(}\sum_{i}\frac{1}{h_{i}}\frac{\partial}{% \partial q_{i}}\hat{q}_{i}\Big{)}\circ\vec{F}$

which is not equal to

 $\displaystyle\Big{(}\sum_{i}\frac{1}{h_{i}}\frac{\partial}{\partial q_{i}}\hat% {q}_{i}\Big{)}\circ\vec{F}\neq\sum_{i}\frac{1}{h_{i}}\frac{\partial F_{i}}{% \partial q_{i}}$

as one would think! The real expression can be derived the following way,

 $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\frac{\partial}{\partial q_% {i}}\hat{q}_{i}\Big{)}\circ\vec{F}\Big{]}$ $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\Big% {(}\frac{\partial\vec{F}}{\partial q_{i}}\Big{)}\Big{]}$ $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\Big% {(}\frac{\partial}{\partial q_{i}}\Big{(}\sum_{j}F_{j}\hat{q}_{j}\Big{)}\Big{)% }\Big{]}$
 $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\Big% {(}\sum_{j}\frac{\partial}{\partial q_{i}}\Big{(}F_{j}\hat{q}_{j}\Big{)}\Big{)% }\Big{]}$ $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\Big% {(}\sum_{j}\hat{q}_{j}\frac{\partial F_{j}}{\partial q_{i}}+F_{j}\frac{% \partial\hat{q}_{j}}{\partial q_{i}}\Big{)}\Big{]}$
 $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\Big% {(}\sum_{j}\hat{q}_{j}\frac{\partial F_{j}}{\partial q_{i}}+\sum_{j}F_{j}\frac% {\partial\hat{q}_{j}}{\partial q_{i}}\Big{)}\Big{]}$ $\displaystyle=\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\sum% _{j}\hat{q}_{j}\frac{\partial F_{j}}{\partial q_{i}}+\Big{(}\frac{1}{h_{i}}% \hat{q}_{i}\Big{)}\circ\sum_{j}F_{j}\frac{\partial\hat{q}_{j}}{\partial q_{i}}% \Big{]}$
 $\displaystyle=\underbrace{\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big% {)}\circ\sum_{j}\hat{q}_{j}\frac{\partial F_{j}}{\partial q_{i}}\Big{]}}_{% \mbox{call it A}}+\underbrace{\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}% \Big{)}\circ\sum_{j}F_{j}\frac{\partial\hat{q}_{j}}{\partial q_{i}}\Big{]}}_{% \mbox{call it B}}$
 $\displaystyle A$ $\displaystyle=$ $\displaystyle\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\sum_% {j}\hat{q}_{j}\frac{\partial F_{j}}{\partial q_{i}}\Big{]}$ $\displaystyle=$ $\displaystyle\sum_{i}\Big{[}\frac{1}{h_{i}}\sum_{j}\underbrace{(\hat{q}_{i}% \circ\hat{q}_{j})}_{\delta_{ij}}\frac{\partial F_{j}}{\partial q_{i}}\Big{]}$ $\displaystyle=$ $\displaystyle\sum_{i}\frac{1}{h_{i}}\frac{\partial F_{i}}{\partial q_{i}}$
 $\displaystyle B$ $\displaystyle=$ $\displaystyle\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\sum_% {j}F_{j}\frac{\partial\hat{q}_{j}}{\partial q_{i}}\Big{]}$

Using the following equality22The proof of this identity is left as an exercise for the reader.

 $\displaystyle\frac{\partial\hat{q}_{j}}{\partial q_{i}}=\frac{\hat{q}_{i}}{h_{% j}}\frac{\partial h_{i}}{\partial q_{j}}\qquad\forall i\neq j$

we can write $B$ as

 $\displaystyle B$ $\displaystyle=$ $\displaystyle\sum_{i}\Big{[}\Big{(}\frac{1}{h_{i}}\hat{q}_{i}\Big{)}\circ\sum_% {j}F_{j}\Big{(}\hat{q}_{i}\frac{1}{h_{j}}\frac{\partial h_{i}}{\partial q_{j}}% \Big{)}\Big{]}\qquad\forall i\neq j$ $\displaystyle=$ $\displaystyle\sum_{i}\Big{[}\frac{1}{h_{i}}\sum_{j}F_{j}\underbrace{(\hat{q}_{% i}\circ\hat{q}_{i})}_{1}\frac{1}{h_{j}}\frac{\partial h_{i}}{\partial q_{j}}% \Big{]}\qquad\forall i\neq j$ $\displaystyle=$ $\displaystyle\sum_{i}\Big{[}\frac{1}{h_{i}}\sum_{j}F_{j}\frac{1}{h_{j}}\frac{% \partial h_{i}}{\partial q_{j}}\Big{]}\qquad\forall i\neq j$ $\displaystyle=$ $\displaystyle\sum_{i\neq j}\frac{F_{j}}{h_{j}h_{i}}\frac{\partial h_{i}}{% \partial q_{j}}$ $\displaystyle=$ $\displaystyle\sum_{i\neq 1}\frac{F_{1}}{h_{1}h_{i}}\frac{\partial h_{i}}{% \partial q_{1}}+\sum_{i\neq 2}\frac{F_{2}}{h_{2}h_{i}}\frac{\partial h_{i}}{% \partial q_{2}}+\sum_{i\neq 3}\frac{F_{3}}{h_{3}h_{i}}\frac{\partial h_{i}}{% \partial q_{3}}$
 $\displaystyle\Rightarrow\nabla\circ\vec{F}$ $\displaystyle=$ $\displaystyle A+B$ $\displaystyle=$ $\displaystyle\sum_{i}\frac{1}{h_{i}}\frac{\partial F_{i}}{\partial q_{i}}+\sum% _{i\neq 1}\frac{F_{1}}{h_{1}h_{i}}\frac{\partial h_{i}}{\partial q_{1}}+\sum_{% i\neq 2}\frac{F_{2}}{h_{2}h_{i}}\frac{\partial h_{i}}{\partial q_{2}}+\sum_{i% \neq 3}\frac{F_{3}}{h_{3}h_{i}}\frac{\partial h_{i}}{\partial q_{3}}$ $\displaystyle=$ $\displaystyle\Big{[}\frac{1}{h_{1}}\frac{\partial F_{1}}{\partial q_{1}}+\frac% {1}{h_{2}}\frac{\partial F_{2}}{\partial q_{2}}+\frac{1}{h_{3}}\frac{\partial F% _{3}}{\partial q_{3}}\Big{]}$ $\displaystyle+\Big{[}\frac{F_{1}}{h_{1}h_{2}}\frac{\partial h_{2}}{\partial q_% {1}}+\frac{F_{1}}{h_{1}h_{3}}\frac{\partial h_{3}}{\partial q_{1}}\Big{]}$ $\displaystyle+\Big{[}\frac{F_{2}}{h_{2}h_{1}}\frac{\partial h_{1}}{\partial q_% {2}}+\frac{F_{2}}{h_{2}h_{3}}\frac{\partial h_{3}}{\partial q_{2}}\Big{]}$ $\displaystyle+\Big{[}\frac{F_{3}}{h_{3}h_{1}}\frac{\partial h_{1}}{\partial q_% {3}}+\frac{F_{3}}{h_{3}h_{2}}\frac{\partial h_{2}}{\partial q_{3}}\Big{]}$
 $\displaystyle\nabla\circ\vec{F}$ $\displaystyle=$ $\displaystyle\Big{[}\frac{1}{h_{1}}\frac{\partial F_{1}}{\partial q_{1}}+\frac% {F_{1}}{h_{1}h_{2}}\frac{\partial h_{2}}{\partial q_{1}}+\frac{F_{1}}{h_{1}h_{% 3}}\frac{\partial h_{3}}{\partial q_{1}}\Big{]}$ $\displaystyle+\Big{[}\frac{1}{h_{2}}\frac{\partial F_{2}}{\partial q_{2}}+% \frac{F_{2}}{h_{2}h_{1}}\frac{\partial h_{1}}{\partial q_{2}}+\frac{F_{2}}{h_{% 2}h_{3}}\frac{\partial h_{3}}{\partial q_{2}}\Big{]}$ $\displaystyle+\Big{[}\frac{1}{h_{3}}\frac{\partial F_{3}}{\partial q_{3}}+% \frac{F_{3}}{h_{3}h_{1}}\frac{\partial h_{1}}{\partial q_{3}}+\frac{F_{3}}{h_{% 3}h_{2}}\frac{\partial h_{2}}{\partial q_{3}}\Big{]}$

If we define $\Omega\equiv\Pi_{i}h_{i}$, we can further write the above expression as

 $\displaystyle\nabla\circ\vec{F}$ $\displaystyle=$ $\displaystyle\Big{[}\frac{h_{2}h_{3}}{\Omega}\frac{\partial F_{1}}{\partial q_% {1}}+\frac{F_{1}h_{3}}{\Omega}\frac{\partial h_{2}}{\partial q_{1}}+\frac{F_{1% }h_{2}}{\Omega}\frac{\partial h_{3}}{\partial q_{1}}\Big{]}$ $\displaystyle+\Big{[}\frac{h_{1}h_{3}}{\Omega}\frac{\partial F_{2}}{\partial q% _{2}}+\frac{F_{2}h_{3}}{\Omega}\frac{\partial h_{1}}{\partial q_{2}}+\frac{h_{% 1}F_{2}}{\Omega}\frac{\partial h_{3}}{\partial q_{2}}\Big{]}$ $\displaystyle+\Big{[}\frac{h_{1}h_{2}}{\Omega}\frac{\partial F_{3}}{\partial q% _{3}}+\frac{h_{2}F_{3}}{\Omega}\frac{\partial h_{1}}{\partial q_{3}}+\frac{h_{% 1}F_{3}}{\Omega}\frac{\partial h_{2}}{\partial q_{3}}\Big{]}$ $\displaystyle=$ $\displaystyle\frac{1}{\Omega}\Bigg{(}\Big{[}\frac{\partial F_{1}}{\partial q_{% 1}}h_{2}h_{3}+F_{1}h_{3}\frac{\partial h_{2}}{\partial q_{1}}+F_{1}h_{2}\frac{% \partial h_{3}}{\partial q_{1}}\Big{]}$ $\displaystyle+\Big{[}h_{1}\frac{\partial F_{2}}{\partial q_{2}}h_{3}+\frac{% \partial h_{1}}{\partial q_{2}}F_{2}h_{3}+h_{1}F_{2}\frac{\partial h_{3}}{% \partial q_{2}}\Big{]}$ $\displaystyle+\Big{[}h_{1}h_{2}\frac{\partial F_{3}}{\partial q_{3}}+\frac{% \partial h_{1}}{\partial q_{3}}h_{2}F_{3}+h_{1}\frac{\partial h_{2}}{\partial q% _{3}}F_{3}\Big{]}\Bigg{)}$ $\displaystyle=$ $\displaystyle\frac{1}{\Omega}\Big{(}\frac{\partial}{\partial q_{1}}(F_{1}h_{2}% h_{3})+\frac{\partial}{\partial q_{2}}(h_{1}F_{2}h_{3})+\frac{\partial}{% \partial q_{3}}(h_{1}h_{2}F_{3})\Big{)}$

Hence,

 $\displaystyle\nabla\circ\vec{F}$ $\displaystyle=$ $\displaystyle\frac{1}{\Omega}\sum_{i}\frac{\partial}{\partial q_{i}}\Big{(}% \frac{\Omega}{h_{i}}F_{i}\Big{)}\quad\mbox{ where }\Omega=\Pi_{i}h_{i}$
Title Gradient and Divergence in Orthonormal Curvilinear Coordinates GradientAndDivergenceInOrthonormalCurvilinearCoordinates1 2013-03-11 19:26:23 2013-03-11 19:26:23 swapnizzle (13346) (0) 1 swapnizzle (0) Definition