Curvilinear Coordinate system, which in fact is the most general coordinate system used to describe the motion of any particle. It includes all our usual systems like Cartesian, Spherical and Cylindrical coordinate systems.
With one to one correspondence, it is always possible to define a set of transformation rules like, $$ x_1 = x_1(u_1, u_2, u_3) \\~\\ x_2 = x_2(u_1, u_2, u_3) \\~\\ x_3 = x_3(u_1, u_2, u_3)$$
to write each of the cartesian coordinates in terms of the general coordinates. We can also define inverse transformation rules like,
$$ u_1 = u_1(x_1, x_2, x_3) \\~\\ u_2 = u_2(x_1, x_2, x_3) \\~\\ u_3 = u_3(x_1, x_2, x_3) $$
to go from one system to another. These transformations are unique, since they have one to one correspondence.
The surfaces $ u_1 = const., u_2 = const., u_3 = const.$$ are called coordinate surfaces and the curve formed from the intersection of pair of two surfaces is called coordinate curves. The point where the tangent lines drawn to these coordinate curves intersect is chosen to be the origin of the coordinate system.
For the sake of simplicity, we often used to deal with the coordinate systems where the coordinate surfaces intersect at right angles. They are called "Orthogonal coordinate system".
Now, we can formulate the general rules for describing a point and its motion and to describe various vector operations in this new curvilinear coordinate system.
But the formulation is going to be more general to apply in any system at any point. It should apply to Cartesian, Cylindrical, Spherical, Paraboloidal, Ellipsoidal and etc.
Note: There are nearly more than 10 types of orthogonal coordinate systems we are using in mathematics.
To start, first we will consider the example of describing small differential element in 3D Euclidean Space. $$\vec{dr} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} ..... \ldots eq.(1)$$
Using the chain rule, we can write the same differential element as,
$$ \vec{dr} = \frac{\partial\vec{r}}{\partial{x}} dx + \frac{\partial\vec{r}}{\partial{y}} dy + \frac{\partial\vec{r}}{\partial{z}} dz \ldots... eq.(2)$$
Comparing eq.(1) and (2) we get that $$\frac{\partial\vec{r}}{\partial{x}} = \hat{e_x} \, , \,\frac{\partial\vec{r}}{\partial{y}} = \hat{e_y} \, , \, \frac{\partial\vec{r}}{\partial{z}} = \hat{e_z} $$ In a similar way, if the differential element is written in terms of curvilinear coordinates as $$ \vec{r} = \vec{r} (u_1, u_2, u_3)$$ Tangent vector to $u_1$ curve at some point P is, $ \frac{\partial\vec{r}}{\partial{u_1}}$
Therefore, the unit tangent vector in this direction given by, $$ \frac{\partial\vec{r}/\partial{u_1}}{|\partial\vec{r}/\partial{u_1}|} = \hat{e_1} $$
We call $$ |\partial\vec{r}/\partial{u_1}| = h_1 = scaling factor $$
Since these coordinates needn't necessarily have the dimension of distance, these parameters are used to make them all to same dimension - after all we cannot add mangoes and apples together to single count.
To complete, we write, $$\frac{\partial\vec{r}}{\partial{u_1}} = h_1 \hat{e_1} \ldots... eq.(3) \\ \frac{\partial\vec{r}}{\partial{u_2}} = h_2 \hat{e_2} \ldots... eq.(4) \\ \frac{\partial\vec{r}}{\partial{u_3}} = h_3 \hat{e_3} \ldots... eq.(5) $$
These basis vectors are tangent vectors to the curves. Similar to that, we can always form another basis whose unit vectors are normal to the coordinate surfaces, where the normal vectors are represented in terms of gradient operator $ \nabla{u_1} , \nabla{u_2}, \nabla{u_3}$
After normalizing, we get a new basis unit vectors represented as, $$ \hat{E_1} = \frac{\nabla{u_1}}{|\nabla{u_1}|} , \hat{E_2} = \frac{\nabla{u_2}}{|\nabla{u_2}|}, \hat{E_3} = \frac{\nabla{u_3}}{|\nabla{u_3}|} $$ It can be shown separately that, these two set of basis vectors constitute reciprocal system of vectors under coordinate transformation. It leads to the concept of Co-variant and Contra-variant vectors.
Thus, any vector can be expressed as either in terms of first set of basis vectors or in terms of second set of basis vectors.
The square of the magnitude of the differential element in terms of first set of unit basis vectors, $$ ds^2 = \vec{dr}\cdot\vec{dr} = {h_1}^2 {du_1}^2 + {h_2}^2 {du_2}^2 + {h_3}^2 {du_3}^2 $$ Since we got the basic things we need to work, now we can start defining the general relation for operations like Gradient, Divergence, Curl and Laplacian.
Gradient:
Gradient from the definition, $$ df = \nabla{f} \cdot \vec{dr} $$
Using the chain rule, $$ df = \frac{\partial{f}}{\partial{u_1}} du_1 + \frac{\partial{f}}{\partial{u_2}} du_2 + \frac{\partial{f}}{\partial{u_3}} du_3 \dots... eq.(6) $$ and we can also write the differential element $ \vec{dr} $ using eq. (3), (4), (5) as, $$ \vec{dr} = h_1 du_1 \hat{e_1} + h_2 du_2 \hat{e_2} + h_3 du_3 \hat{e_3} $$
Still we don't know what is the form for gradient operator, but we do know, from the definition of gradient that it would give "df" when dotted with $ \vec{dr}$ . So,
$$ \nabla{f} \cdot \vec{dr} = \nabla_1{f} h_1 du_1 + \nabla_2{f} h_2 du_2 + \nabla_{f} h_3 du_3 \ldots... eq.(7) $$
where $ \nabla_1{f} , \nabla_2{f} , \nabla_3{f} $ are the components of Gradient operator when it is written in terms of the general basis vectors $ \hat{e_1}, \hat{e_2}, \hat{e_3} $.
Again comparing eqs.(6) and (7), the components of the gradient operator found out to be, $$ \nabla_1{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}},\nabla_2{f} = \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}}, \nabla_3{f} = \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} $$
Hence the general form of Gradient operator in any curvilinear orthogonal coordinate system is given by,
$$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} \,\, \ldots...eq.(8)$$
Divergence:
Let us analyze the first term we will get, when we apply the divergence operator on any vector function $\vec{A} $,
$$ (\nabla\cdot\vec{A})_1 = \nabla \cdot (A_1\hat{e_1}) \ldots.....(9)$$
We don't know, what we will obtain when we apply the divergence operator on $ \hat{e_1} $. But, if we could write $\hat{e_1}$ in terms of some gradient operations, then there is a real possibility of obtaining the expression for Divergence with our prior knowledge of Gradient.
According to write the unit vectors in terms of gradient relations,
We apply the gradient operator for the functions $ u_1, u_2, u_3 $ in eq.(8) from which, we will get $$ \nabla{u_1} = \frac{\hat{e_1}}{h_1}\,, \,\nabla{u_2} = \frac{\hat{e_2}}{h_2}\, ,\, \nabla{u_3} = \frac{\hat{e_3}}{h_3} $$
The resultant unit vectors using gradient relations are,
$$ \hat{e_1} = h_1 \nabla{u_1} \, ,\, \hat{e_2} = h_2 \nabla{u_2} \, ,\, \hat{e_3} = h_3 \nabla{u_3} $$
But we need to relate it with $ \hat{e_1}$, So we apply the volume relation $$ \hat{e_1} = \hat{e_2} \times \hat{e_3} = h_2 h_3 \nabla{u_2} \times \nabla{u_3} $$
Applying this in eq.(9),
$$ \nabla \cdot (A_1\hat{e_1}) = \nabla\cdot[A_1h_2h_3 \nabla{u_2} \times \nabla{u_3}] \, \, \, \ldots...eq.(10)$$
Using the vector relation, $$ \nabla\cdot {f\vec{A}} = \nabla{f}\cdot\vec{A} + f \nabla\cdot\vec{A} $$
where f- scalar function, $\vec{A} = vector function $.
Eq.(10) becomes, $$ \nabla\cdot(A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) + A_1h_2h_3 \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) \, \, \, \ldots...eq.(11)$$
But using the vector identity, $$ \nabla\cdot(\vec{A}\times\vec{B}) = \vec{B}\cdot(\nabla\times \vec{A}) - \vec{A}\cdot (\nabla\times\vec{B})$$
$$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = \nabla{u_3}\cdot(\nabla\times\nabla{u_2}) - \nabla{u_2}\cdot(\nabla\times\nabla{u_3})$$
But, Curl of gradient is always zero for any scalar function, which implies $$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = 0 $$
Eq.(11) gives, $$\nabla \cdot (A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) \dots..eq.(12) $$
Again writing $ \nabla{u_2}\times\nabla{u_3} $ in terms of basis vectors that is $$ \nabla{u_2}\times\nabla{u_3} = \frac{\hat{e_2}\times\hat{e_3}}{h_2h_3} = \frac{\hat{e_1}}{h_2h_3} $$
Eq.(12) results into $$\nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3} \cdot \nabla(A_1h_2h_3) $$
Using our prior knowledge of Gradient, it can be expanded as,
$$ \nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3}\cdot\left[ \frac{\hat{e_1}}{h_1} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} + \frac{\hat{e_2}}{h_2} \frac{\partial(A_1h_2h_3)}{\partial{u_2}} + \frac{\hat{e_3}}{h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_3}}\right] $$ While, we are dealing with orthogonal basis, dot product between any two different basis gives zero and dot product of same vector gives unity.
Making using of the orthonormality, we finally arrive at the result,
$$ \nabla\cdot(A_1\hat{e_1}) = \frac{1}{h_1h_2h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} $$
Similar procedure gives the expression for other coordinates.
The final expression for the divergence operator in general curvilinear coordinates is,
$$\nabla\cdot\vec{A} = \frac{1}{h_1h_2h_3}\left[ \frac{\partial{A_1h_2h_3}}{\partial{u_1}} + \frac{\partial{A_1h_2h_3}}{\partial{u_2}} + \frac{\partial{A_1h_2h_3}}{\partial{u_3}}\right] \ldots...eq.(13)$$
Curl:
As the same, first we will take single component, write it in terms of gradient and apply the curl,
$$ \nabla \times (A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) \, \, \ldots...eq.(14)$$
Since there is already a curl operator, we don't need to use volume relation but we could just simply write $ \hat{e_1} $ in terms of its own gradient relation i.e. $ \hat{e_1} = h_1 \nabla{u_1} $
Using the vector identity, $$ \nabla \times (f\vec{A}) = f \nabla\times\vec{A} + \nabla{f}\cdot\vec{A} $$
Eq.(14) gives, $$ \nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} + A_1h_1\nabla \times \nabla{u_1} $$
but curl of gradient is zero.
So, Eq.(14) becomes, $$\nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} $$
With the help of eq.(8) we can rewrite the above into,
$$ \nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \nabla{u_1} $$
Again using the relation, $ \nabla{u_1} = \frac{\hat{e_1}}{h_1} $
$$\nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \frac{\hat{e_1}}{h_1} $$
Using the cross product rule for the positive volume element, we finally get,
$$\nabla \times (A_1\hat{e_1}) = \frac{1}{h_1h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{-e_3} + \frac{1}{h_1h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_2} \, \ldots... eq.(15)$$
Similar procedure gives for other components the following result,
$$\nabla \times (A_2\hat{e_2}) = \frac{1}{h_1h_2} \frac{\partial{A_2h_2}}{\partial{u_1}} \hat{e_3} + \frac{1}{h_2h_3} \frac{\partial{A_2h_2}}{\partial{u_3}} \hat{-e_1} \, \ldots... eq.(16)$$
$$\nabla \times (A_3\hat{e_3}) = \frac{1}{h_1h_3} \frac{\partial{A_3h_3}}{\partial{u_1}} \hat{-e_2} + \frac{1}{h_2h_3} \frac{\partial{A_3h_3}}{\partial{u_2}} \hat{e_1} \, \ldots... eq.(17)$$
Combining the components from eq.(15),(16),(17) we get the General expression for the Curl operator in Curvilinear coordinates as,
$$ \nabla\times\vec{A} = \frac{1}{h_2h_3} \left[\frac{\partial(A_3h_3)}{\partial{u_2}} - \frac{\partial(A_2h_2)}{\partial{u_3}}\right] \hat{e_1} + \\~\\ \frac{1}{h_1h_3} \left[\frac{\partial(A_1h_1)}{\partial{u_3}} - \frac{\partial(A_3h_3)}{\partial{u_1}}\right] \hat{e_2} + \\~\\ \frac{1}{h_1h_2} \left[\frac{\partial(A_2h_2)}{\partial{u_1}} - \frac{\partial(A_1h_1)}{\partial{u_2}}\right] \hat{e_3} \, \, \, \ldots...eq.(18)$$
Or simply we can write this in Matrix form as,
$$ \nabla \times \vec{A} = \frac{1}{h_1h_2h_3}\begin{vmatrix} h_1\,\hat{e_1} & h_2 \,\hat{e_2} & h_3 \, \hat{e_3} \\ \frac{\partial}{\partial{u_1}} & \frac{\partial}{\partial{u_2}} & \frac{\partial}{\partial{u_3}} \\ h_1\vec{A_1} & h_2 \vec{A_2} & h_3 \vec{A_3} \end{vmatrix} \,\,\, \dots...eq.(19) $$
Laplacian:
Unlike others, we don't need to find anything extra for Laplacian since it is just the combination of gradient and divergence.
Let us take a scalar function "f" and write its gradient from eq.(8), $$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} $$
Now, applying the Divergence operation for the resultant outcome, we get the General expression for Laplacian in curvilinear coordinates,
$$ \nabla^2f = \frac{1}{h_1h_2h_3} \left[ \frac{\partial\left(\frac{h_2h_3}{h_1}\frac{\partial{f}}{\partial{u_1}}\right)}{\partial{u_1}} + \frac{\partial\left(\frac{h_3h_1}{h_2}\frac{\partial{f}}{\partial{u_2}}\right)}{\partial{u_2}} + \frac{\partial\left(\frac{h_1h_2}{h_3}\frac{\partial{f}}{\partial{u_3}}\right)}{\partial{u_3}} \right] \ldots...eq.(20)$$
That is all we need to derive. We need to remember that, these derivations are done for general orthogonal curvilinear coordinate system.
For the most general curvilinear coordinate system (i.e. which are not orthogonal), we will need Tensors and its analysis.
to go from one system to another. These transformations are unique, since they have one to one correspondence.
The surfaces $ u_1 = const., u_2 = const., u_3 = const.$$ are called coordinate surfaces and the curve formed from the intersection of pair of two surfaces is called coordinate curves. The point where the tangent lines drawn to these coordinate curves intersect is chosen to be the origin of the coordinate system.
For the sake of simplicity, we often used to deal with the coordinate systems where the coordinate surfaces intersect at right angles. They are called "Orthogonal coordinate system".
Now, we can formulate the general rules for describing a point and its motion and to describe various vector operations in this new curvilinear coordinate system.
But the formulation is going to be more general to apply in any system at any point. It should apply to Cartesian, Cylindrical, Spherical, Paraboloidal, Ellipsoidal and etc.
Note: There are nearly more than 10 types of orthogonal coordinate systems we are using in mathematics.
To start, first we will consider the example of describing small differential element in 3D Euclidean Space. $$\vec{dr} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} ..... \ldots eq.(1)$$
Using the chain rule, we can write the same differential element as,
$$ \vec{dr} = \frac{\partial\vec{r}}{\partial{x}} dx + \frac{\partial\vec{r}}{\partial{y}} dy + \frac{\partial\vec{r}}{\partial{z}} dz \ldots... eq.(2)$$
Comparing eq.(1) and (2) we get that $$\frac{\partial\vec{r}}{\partial{x}} = \hat{e_x} \, , \,\frac{\partial\vec{r}}{\partial{y}} = \hat{e_y} \, , \, \frac{\partial\vec{r}}{\partial{z}} = \hat{e_z} $$ In a similar way, if the differential element is written in terms of curvilinear coordinates as $$ \vec{r} = \vec{r} (u_1, u_2, u_3)$$ Tangent vector to $u_1$ curve at some point P is, $ \frac{\partial\vec{r}}{\partial{u_1}}$
Therefore, the unit tangent vector in this direction given by, $$ \frac{\partial\vec{r}/\partial{u_1}}{|\partial\vec{r}/\partial{u_1}|} = \hat{e_1} $$
We call $$ |\partial\vec{r}/\partial{u_1}| = h_1 = scaling factor $$
Since these coordinates needn't necessarily have the dimension of distance, these parameters are used to make them all to same dimension - after all we cannot add mangoes and apples together to single count.
To complete, we write, $$\frac{\partial\vec{r}}{\partial{u_1}} = h_1 \hat{e_1} \ldots... eq.(3) \\ \frac{\partial\vec{r}}{\partial{u_2}} = h_2 \hat{e_2} \ldots... eq.(4) \\ \frac{\partial\vec{r}}{\partial{u_3}} = h_3 \hat{e_3} \ldots... eq.(5) $$
These basis vectors are tangent vectors to the curves. Similar to that, we can always form another basis whose unit vectors are normal to the coordinate surfaces, where the normal vectors are represented in terms of gradient operator $ \nabla{u_1} , \nabla{u_2}, \nabla{u_3}$
After normalizing, we get a new basis unit vectors represented as, $$ \hat{E_1} = \frac{\nabla{u_1}}{|\nabla{u_1}|} , \hat{E_2} = \frac{\nabla{u_2}}{|\nabla{u_2}|}, \hat{E_3} = \frac{\nabla{u_3}}{|\nabla{u_3}|} $$ It can be shown separately that, these two set of basis vectors constitute reciprocal system of vectors under coordinate transformation. It leads to the concept of Co-variant and Contra-variant vectors.
Thus, any vector can be expressed as either in terms of first set of basis vectors or in terms of second set of basis vectors.
The square of the magnitude of the differential element in terms of first set of unit basis vectors, $$ ds^2 = \vec{dr}\cdot\vec{dr} = {h_1}^2 {du_1}^2 + {h_2}^2 {du_2}^2 + {h_3}^2 {du_3}^2 $$ Since we got the basic things we need to work, now we can start defining the general relation for operations like Gradient, Divergence, Curl and Laplacian.
Gradient:
Gradient from the definition, $$ df = \nabla{f} \cdot \vec{dr} $$
Using the chain rule, $$ df = \frac{\partial{f}}{\partial{u_1}} du_1 + \frac{\partial{f}}{\partial{u_2}} du_2 + \frac{\partial{f}}{\partial{u_3}} du_3 \dots... eq.(6) $$ and we can also write the differential element $ \vec{dr} $ using eq. (3), (4), (5) as, $$ \vec{dr} = h_1 du_1 \hat{e_1} + h_2 du_2 \hat{e_2} + h_3 du_3 \hat{e_3} $$
Still we don't know what is the form for gradient operator, but we do know, from the definition of gradient that it would give "df" when dotted with $ \vec{dr}$ . So,
$$ \nabla{f} \cdot \vec{dr} = \nabla_1{f} h_1 du_1 + \nabla_2{f} h_2 du_2 + \nabla_{f} h_3 du_3 \ldots... eq.(7) $$
where $ \nabla_1{f} , \nabla_2{f} , \nabla_3{f} $ are the components of Gradient operator when it is written in terms of the general basis vectors $ \hat{e_1}, \hat{e_2}, \hat{e_3} $.
Again comparing eqs.(6) and (7), the components of the gradient operator found out to be, $$ \nabla_1{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}},\nabla_2{f} = \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}}, \nabla_3{f} = \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} $$
Hence the general form of Gradient operator in any curvilinear orthogonal coordinate system is given by,
$$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} \,\, \ldots...eq.(8)$$
Divergence:
Let us analyze the first term we will get, when we apply the divergence operator on any vector function $\vec{A} $,
$$ (\nabla\cdot\vec{A})_1 = \nabla \cdot (A_1\hat{e_1}) \ldots.....(9)$$
We don't know, what we will obtain when we apply the divergence operator on $ \hat{e_1} $. But, if we could write $\hat{e_1}$ in terms of some gradient operations, then there is a real possibility of obtaining the expression for Divergence with our prior knowledge of Gradient.
According to write the unit vectors in terms of gradient relations,
We apply the gradient operator for the functions $ u_1, u_2, u_3 $ in eq.(8) from which, we will get $$ \nabla{u_1} = \frac{\hat{e_1}}{h_1}\,, \,\nabla{u_2} = \frac{\hat{e_2}}{h_2}\, ,\, \nabla{u_3} = \frac{\hat{e_3}}{h_3} $$
The resultant unit vectors using gradient relations are,
$$ \hat{e_1} = h_1 \nabla{u_1} \, ,\, \hat{e_2} = h_2 \nabla{u_2} \, ,\, \hat{e_3} = h_3 \nabla{u_3} $$
But we need to relate it with $ \hat{e_1}$, So we apply the volume relation $$ \hat{e_1} = \hat{e_2} \times \hat{e_3} = h_2 h_3 \nabla{u_2} \times \nabla{u_3} $$
Applying this in eq.(9),
$$ \nabla \cdot (A_1\hat{e_1}) = \nabla\cdot[A_1h_2h_3 \nabla{u_2} \times \nabla{u_3}] \, \, \, \ldots...eq.(10)$$
Using the vector relation, $$ \nabla\cdot {f\vec{A}} = \nabla{f}\cdot\vec{A} + f \nabla\cdot\vec{A} $$
where f- scalar function, $\vec{A} = vector function $.
Eq.(10) becomes, $$ \nabla\cdot(A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) + A_1h_2h_3 \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) \, \, \, \ldots...eq.(11)$$
But using the vector identity, $$ \nabla\cdot(\vec{A}\times\vec{B}) = \vec{B}\cdot(\nabla\times \vec{A}) - \vec{A}\cdot (\nabla\times\vec{B})$$
$$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = \nabla{u_3}\cdot(\nabla\times\nabla{u_2}) - \nabla{u_2}\cdot(\nabla\times\nabla{u_3})$$
But, Curl of gradient is always zero for any scalar function, which implies $$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = 0 $$
Eq.(11) gives, $$\nabla \cdot (A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) \dots..eq.(12) $$
Again writing $ \nabla{u_2}\times\nabla{u_3} $ in terms of basis vectors that is $$ \nabla{u_2}\times\nabla{u_3} = \frac{\hat{e_2}\times\hat{e_3}}{h_2h_3} = \frac{\hat{e_1}}{h_2h_3} $$
Eq.(12) results into $$\nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3} \cdot \nabla(A_1h_2h_3) $$
Using our prior knowledge of Gradient, it can be expanded as,
$$ \nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3}\cdot\left[ \frac{\hat{e_1}}{h_1} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} + \frac{\hat{e_2}}{h_2} \frac{\partial(A_1h_2h_3)}{\partial{u_2}} + \frac{\hat{e_3}}{h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_3}}\right] $$ While, we are dealing with orthogonal basis, dot product between any two different basis gives zero and dot product of same vector gives unity.
Making using of the orthonormality, we finally arrive at the result,
$$ \nabla\cdot(A_1\hat{e_1}) = \frac{1}{h_1h_2h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} $$
Similar procedure gives the expression for other coordinates.
The final expression for the divergence operator in general curvilinear coordinates is,
$$\nabla\cdot\vec{A} = \frac{1}{h_1h_2h_3}\left[ \frac{\partial{A_1h_2h_3}}{\partial{u_1}} + \frac{\partial{A_1h_2h_3}}{\partial{u_2}} + \frac{\partial{A_1h_2h_3}}{\partial{u_3}}\right] \ldots...eq.(13)$$
Curl:
As the same, first we will take single component, write it in terms of gradient and apply the curl,
$$ \nabla \times (A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) \, \, \ldots...eq.(14)$$
Since there is already a curl operator, we don't need to use volume relation but we could just simply write $ \hat{e_1} $ in terms of its own gradient relation i.e. $ \hat{e_1} = h_1 \nabla{u_1} $
Using the vector identity, $$ \nabla \times (f\vec{A}) = f \nabla\times\vec{A} + \nabla{f}\cdot\vec{A} $$
Eq.(14) gives, $$ \nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} + A_1h_1\nabla \times \nabla{u_1} $$
but curl of gradient is zero.
So, Eq.(14) becomes, $$\nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} $$
With the help of eq.(8) we can rewrite the above into,
$$ \nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \nabla{u_1} $$
Again using the relation, $ \nabla{u_1} = \frac{\hat{e_1}}{h_1} $
$$\nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \frac{\hat{e_1}}{h_1} $$
Using the cross product rule for the positive volume element, we finally get,
$$\nabla \times (A_1\hat{e_1}) = \frac{1}{h_1h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{-e_3} + \frac{1}{h_1h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_2} \, \ldots... eq.(15)$$
Similar procedure gives for other components the following result,
$$\nabla \times (A_2\hat{e_2}) = \frac{1}{h_1h_2} \frac{\partial{A_2h_2}}{\partial{u_1}} \hat{e_3} + \frac{1}{h_2h_3} \frac{\partial{A_2h_2}}{\partial{u_3}} \hat{-e_1} \, \ldots... eq.(16)$$
$$\nabla \times (A_3\hat{e_3}) = \frac{1}{h_1h_3} \frac{\partial{A_3h_3}}{\partial{u_1}} \hat{-e_2} + \frac{1}{h_2h_3} \frac{\partial{A_3h_3}}{\partial{u_2}} \hat{e_1} \, \ldots... eq.(17)$$
Combining the components from eq.(15),(16),(17) we get the General expression for the Curl operator in Curvilinear coordinates as,
$$ \nabla\times\vec{A} = \frac{1}{h_2h_3} \left[\frac{\partial(A_3h_3)}{\partial{u_2}} - \frac{\partial(A_2h_2)}{\partial{u_3}}\right] \hat{e_1} + \\~\\ \frac{1}{h_1h_3} \left[\frac{\partial(A_1h_1)}{\partial{u_3}} - \frac{\partial(A_3h_3)}{\partial{u_1}}\right] \hat{e_2} + \\~\\ \frac{1}{h_1h_2} \left[\frac{\partial(A_2h_2)}{\partial{u_1}} - \frac{\partial(A_1h_1)}{\partial{u_2}}\right] \hat{e_3} \, \, \, \ldots...eq.(18)$$
Or simply we can write this in Matrix form as,
$$ \nabla \times \vec{A} = \frac{1}{h_1h_2h_3}\begin{vmatrix} h_1\,\hat{e_1} & h_2 \,\hat{e_2} & h_3 \, \hat{e_3} \\ \frac{\partial}{\partial{u_1}} & \frac{\partial}{\partial{u_2}} & \frac{\partial}{\partial{u_3}} \\ h_1\vec{A_1} & h_2 \vec{A_2} & h_3 \vec{A_3} \end{vmatrix} \,\,\, \dots...eq.(19) $$
Laplacian:
Unlike others, we don't need to find anything extra for Laplacian since it is just the combination of gradient and divergence.
Let us take a scalar function "f" and write its gradient from eq.(8), $$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} $$
Now, applying the Divergence operation for the resultant outcome, we get the General expression for Laplacian in curvilinear coordinates,
$$ \nabla^2f = \frac{1}{h_1h_2h_3} \left[ \frac{\partial\left(\frac{h_2h_3}{h_1}\frac{\partial{f}}{\partial{u_1}}\right)}{\partial{u_1}} + \frac{\partial\left(\frac{h_3h_1}{h_2}\frac{\partial{f}}{\partial{u_2}}\right)}{\partial{u_2}} + \frac{\partial\left(\frac{h_1h_2}{h_3}\frac{\partial{f}}{\partial{u_3}}\right)}{\partial{u_3}} \right] \ldots...eq.(20)$$
That is all we need to derive. We need to remember that, these derivations are done for general orthogonal curvilinear coordinate system.
For the most general curvilinear coordinate system (i.e. which are not orthogonal), we will need Tensors and its analysis.
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