Cartesian to Spherical coordinate system - Coordinate Transformation

In general orthogonal curvilinear coordinate system, general position in 3 dimension is given by, $$\vec{s} = \vec{s}(u_1, u_2, u_3)$$ and small displacement "ds" is can be written as, $$ \vec{ds} = \sum_i^3 \frac{\partial{\vec{s}}}{\partial{u_i}} du_i $$ where $ \frac{\partial{\vec{s}}}{\partial{u_i}} = h_i \hat{e_i} $ and $\hat{e_i} $ is the unit vector along the 'i'th direction. Rewriting it in simple form, we have, $$ \vec{ds} = \sum_i^3 h_i du_i \hat{e_i} $$
For cartesian coordinate system, $h_i = 1$
$\rightarrow$ 
$$ \vec{ds} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} $$
But there is no unique choice of coordinate system, we can also choose spherical coordinate system as, 
$$ \vec{ds} = dr \hat{e_r} + r \,d\theta\hat{e_\theta} + r \,sin\theta \,d\phi \hat{e_\phi} $$ with corresponding scaling factors. 

To go from one coordinate system to another, we use the relations, $$ x = x(r,\theta,\phi) \\~\\ y = y(r,\theta,\phi) \\~\\ z=z(r,\theta, \phi) $$ 
The only relation we know from our conventional assumption is that, $$ x = \,r \,sin\theta \,cos\phi \\~\\ y = \,r \,sin\theta \,sin\phi \\~\\ z = \,r \,cos\theta\, $$ 
Using the chain rule, $$ dx = \frac{\partial{x}}{\partial{r}}dr + \frac{\partial{x}}{\partial{\theta}} d\theta + \frac{\partial{x}}{\partial{\phi}}d\phi $$
$\rightarrow$ $$ dx = \,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi $$
Similarly, $$ dy = \, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi $$ and $$ dz = cos\theta \, dr - r\, sin\theta \, d\theta $$
Applying it in our Cartesian equation for "ds", we get, $$ ds =(\,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi )\hat{e_x} \\~\\+ (\, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi)\hat{e_y} \\~\\+ (cos\theta \, dr - r\, sin\theta \, d\theta) \hat{e_z} $$  

Thus we expressed the infinitesimal displacement in cartesian system using spherical measurements such $ r, \theta, \phi $. Now, from our definition of unit vector, $$ \hat{e_r} = \frac{{\frac{\partial{\vec{s}}}{\partial{r}}}}{|\frac{\partial{\vec{s}}}{\partial{r}}|} $$ and $$ \hat{e_\theta} = \frac{{\frac{\partial{\vec{s}}}{\partial{\theta}}}}{|\frac{\partial{\vec{s}}}{\partial{\theta}}|}$$ and $$\hat{e_\phi} = \frac{{\frac{\partial{\vec{s}}}{\partial{\phi}}}}{|\frac{\partial{\vec{s}}}{\partial{\phi}}|}$$ 
Using the above, equation, we can write the unit vectors along $r,\,\theta,\,\phi $ directions as follows, $$ \hat{e_r} = sin\theta\, cos\phi\, \hat{e_x} \,+ \,\sin\theta\,sin\phi\,\hat{e_y} \,+\, cos\theta\,\hat{e_z} $$ and $$ \hat{e_\theta} = cos\theta\, cos\phi \,\hat{e_x} \,+\, \cos\theta\, sin\phi \,\hat{e_y} - sin\theta\,\hat{e_z} $$ and $$ \hat{e_\phi} = \,-sin\phi \,\hat{e_x} \,+ \,cos\phi\,\hat{e_y} $$ From these three unit vectors we can solve for the other three unit vectors as following,
Solving the first two,
 $$ sin\theta\,\hat{e_r} + \, cos\theta\,\hat{e_\theta} = \,cos\phi \hat{e_x} + \,sin\phi \,\hat{e_y} $$

Combining with the third we can solve for x and y, $$ \hat{e_x} = sin\theta\,cos\phi\,\hat{e_r} + cos\theta\,cos\phi\,\hat{e_\theta} - \,sin\phi\,\hat{e_\phi} $$ and $$ \hat{e_y} = \,sin\theta\,sin\phi \hat{e_r} +\, cos\theta\, sin\phi\, \hat{e_\theta} + \,cos\phi\, \hat{e_\phi} $$ and finally solving for z, $$ \hat{e_z} = cos\theta\,\hat{e_r} - sin\theta\,\hat{e_\theta} $$

That is all we do need to derive for the coordinate transformation from cartesian to spherical. 

D'Alembert's principle and deriving Euler lagrange equation

This principle is just an alternative way of deriving the Euler Lagrange equation with new concepts like virtual displacement and virtual work. 

Virtual displacement is defined as the infinitesimal displacement in the coordinates of the system,with out any change in all the forces and constraints at same instant of time "t". Accordingly, virtual work is defined as, $$ \delta{W_{virtual}} = \vec{F_i}\cdot\vec{\delta{r_i}}$$ 

From the concept of system of particles, any force equation can be written as, $$ \frac{d^2}{dt^2}\sum_i m_i \vec{r_i} = \sum_i \vec{F_{i(e)}} + \sum_{i,j \,(i\neq j)\,} \vec{F_{ij}} \,\,\,\ldots...eq.(1)$$  

Or simply, $$ \vec{F_{net}} =  F_{external} + F_{internal} $$

Internal forces includes the constraints imposed on the system. 

Net work done on the "i"th particle is,
$$ \delta{W_i} =  \vec{F^{ext}_i}\cdot\delta{\vec{r_i}} + \vec{F^{int}_i}\cdot\delta{\vec{r_i}} $$ 

Further we assume that, the net virtual work done by the internal forces or the constraints is zero. 

Now comes the principle of virtual work,  If the system is in equilibrium, then the net force acting on the system is zero which will imply the net virtual work done on all the particles will be zero.  
This gives us, $$ \sum_i \vec{F^{ext}_i} \cdot \delta{\vec{r_i}} = 0 $$

But this principle works only for statics where the net force is zero. And also we need to write the virtual displacement in terms of generalized coordinates. Only then we can independently work with each equations and equalize it to zero to get the condition. 

Note: Now you may wonder, why do we need to define this new concept of virtual work? 
The concept of virtual work is to clearly understand the properties of an equilibrium point. We define equilibrium at a point where the object has no tendency to do any work. It is just as same as the least action principle. Virtual work measures the tendency to do any work and we minimize it to zero in the principle of virtual work. Thus we are doing the same thing in two different ways. 

To make a similar static situation in the moving objects, we put an additional effective reverse force in the principle of virtual work using Newton's second law, 

$$ \vec{F^{net}_i} = \frac{d\vec{p_i}}{dt} = \dot{\vec{p_i}} $$ $$ \vec{F^{net}_i} - \dot{\vec{p_i}} = 0 $$
Again splitting the force into net external and internal, and assuming the work done by internal forces is zero, we finally arrive at D'Alembert's principle, $$ \sum_i \left(\vec{F^{ext}_i} - \dot{\vec{p_i}}\right)\cdot \delta{\vec{r_i}} = 0 \,\,\,...eq.(2)$$

Now, we will proceed in generalized coordinates to get Euler Lagrange equation. 

Position coordinates in terms of generalized coordinates, $$ \vec{r_i} = \vec{r_i} (q_j,t) \,\,\, i= 1,2,..n \\~\\ j= 1,2,3..k \,\,\,...eq.(3)$$
Virtual displacement is given by, 
$$\delta{\vec{r_i}} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} + \frac{\partial\vec{r_i}}{\partial{t}} dt $$
But "dt = 0" from the definition of virtual work because virtual work is measured on the same instance of time. 
$$ \delta{\vec{r_i}} =  \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} \,\,\,...eq.(3)$$ 

Now the velocity of the "i"th particle is given by, $$ \vec{v_i} = \frac{d\vec{r_i}}{dt} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \frac{dq_j}{dt} + \frac{\partial{\vec{r_i}}}{\partial{t}} \,\,\,..eq.(4)$$

Lets just find out the expression for the first term in eq.(2) i.e. D'Alembert's principle, 
$$ \sum_i \vec{F_i}\cdot\delta{\vec{r_i}} = \sum_i \vec{F_i} \cdot \left(\sum_j \frac{\partial{\vec{r_i}}}{\partial{q_j}} \delta{q_j}\right) = \sum_i \sum_j\, \vec{F_i^{ext}} \frac{\partial\vec{r_i}} {\partial{q_j}} \delta{q_j} $$
We give it a name for the expression, $$\sum_i \vec{F^{ext}_i} \frac{\partial\vec{r_i}}{\partial{q_j}} = Q_j $$ where $Q_j$ is called the generalized force. 
Thus we end up with the simplified form, $$ \sum_i \vec{F_i^{ext}} \cdot \delta{\vec{r_i}} = \sum_j Q_j \delta{q_j} \,\,\,...eq.(5) $$

Now, taking the second part of the D'Alembert's principle, 

$$ \sum_i \dot{\vec{p_i}} \cdot \delta{\vec{r_i}} = \sum_i \sum_j\, \dot{\vec{p_i}} \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} = \sum_i\sum_j \,m_i \ddot{\vec{r_i}} \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} $$ where masses are assumed to constant over time. 

Making some alterations with the product rule gives,
$$ m_i \ddot{\vec{r_i}} \frac{\partial{r_i}}{\partial{q_j}}\delta{q_j} = \left[\frac{d}{dt}(m_i\dot{\vec{r_i}} \frac{\partial\vec{r_i}}{\partial{q_j}}) - m_i \dot{\vec{r_i}} \frac{d}{dt}(\frac{\partial\vec{r_i}}{\partial{q_j}}) \right] \delta{q_j} \,\,\,...eq.(6)$$

Making use of eq.(4) and commutation between differentiation we get $$ \frac{\partial\vec{v_i}}{\partial{\dot{q_j}}} = \frac{\partial{\vec{r_i}}}{\partial{q_j}} $$ and $$ \frac{d}{dt} (\frac{\partial\vec{r_i}}{\partial{q_j}}) = \frac{\partial}{\partial{q_j}}(\frac{d\vec{r_i}}{dt}) $$

Eq.(6) becomes, $$ \sum_i \sum_j\,\left[ \frac{d}{dt}\left( m_i \vec{v_i} \frac{\partial{\vec{v_i}}}{\partial\dot{q_j}}\right) - m_i \vec{v_i} \frac{\partial{\vec{v_i}}}{\partial{q_j}}\right] \delta{q_j} $$
again using the product rule,

$$ = \,\,\, \sum_i\sum_j \left\{ \frac{d}{dt}\left[\frac{\partial}{\partial{q_j}}(\frac{1}{2}m_i v_i^2)\right] - \frac{\partial}{\partial{q_j}} \left[\frac{1}{2}m_i v_i^2\right]\right\} \,\,\,...eq.(7)$$

Making use of the relation, $$ \sum_i \frac{1}{2}m_iv_i^2 = \sum_i T_i = T $$ where T is total Kinetic Energy of the system of particles. 

eq.(7) can be rewritten as, $$ \sum_i \dot{\vec{p_i}} \cdot \delta{\vec{r_i}} = \sum_j \left[\frac{d}{dt}\left(\frac{\partial{T}}{\partial\dot{q_j}}\right) - \frac{\partial{T}}{\partial{q_j}}\right] \delta{q_j}\,\,\,...eq.(8) $$

Substituting eq.(5) and eq.(8) in D'Alembert's principle, we get, 
$$ \sum_i \left[ \vec{F^{ext}_i} - \dot{\vec{p_i}}\right]\delta{\vec{r_i}} = \sum_j \left\{ Q_j - \left[\frac{d}{dt}\left(\frac{\partial{T}}{\partial{\dot{q_j}}}\right) - \frac{\partial{T}}{\partial{q_j}}\right]\right\} \delta{q_j} = 0 \,\,\,...eq.(9) $$ 

If we consider only conservative forces, then forces are derivable from the potential and it leads to, 
$$ \vec{F^{ext}_i} = -\nabla{V_i} $$

Eq.(5) becomes, $$ Q_j = \sum_i \vec{F^{ext}_i} \cdot \frac{\partial{\vec{r_i}}}{\partial{q_j}} =  \sum_j -\nabla{V} \frac{\partial{\vec{r_i}}}{\partial{q_j}} = -\frac{\partial{V}}{\partial{q_j}} $$

Substituting in eq.(9) gives, $$ \sum_j \left[\frac{d}{dt} \left(\frac{\partial{T}}{\partial\dot{q_j}}\right) -  \frac{\partial{T}}{\partial{q_j}}\right] \delta{q_j} = 0  $$ 
Further, if we consider only simple potentials that depends only on the coordinates then, $$ \frac{\partial{V}}{\partial{q_j}} = 0 $$  

Now we arrived at our final form,

$$ \sum_j \left[ \frac{d}{dt} \left(\frac{\partial (T-V)}{\partial{\dot{q_j}}}\right) - \frac{\partial(T-V)}{\partial{q_j}} \right] \delta{q_j} = 0 \,\,\,...eq.(10)$$

Since the equation is expressed in term of the displacements in generalized coordinates, they are all independent of each other. To make it zero, each of the coefficients should be zero. 

That gives us our desired Euler Lagrange equation, $$ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{q_j}}} - \frac{\partial{L}}{\partial{q_j}} = 0 \,\,\,,\,\,\, j = 1,2,...k \\~\\ L = T - V $$
where k - is the number of generalized coordinates, L - Lagrangian, T- Kinetic Energy of the system of particles and V - Potential Energy of the system of particles.

Linear Vector Space - Introduction

The concept of Vector space is not straight forward as it sounds. Unlike the usual ones, it doesn't have a perfect physical basis in reality starting with the question "Why". 

But the ideas are not completely abstract as you think,  it was created not from a specific topic in physics, but from the generalization of all the usual mathematical concepts. 

I will try to go with my own formal introduction, where everything could be started and understood from the beginning of Quantum Mechanics. 

When the idea of Wave function and operations of quantum mechanics are introduced, people really don't understand the insights. They just used all the arithmetical manipulations and concepts from the known classical mechanics and applied it into quantum mechanics in terms of operators. 

But they never know, why those operators behave in a classical form and why it explains the Nature so beautifully and so on with many philosophical questions. 

You may ask, then why people work with a mathematics for which, they themselves don't know the reason "why" it works.  

As scientists, they have other things to create and work with in real life instead of simply getting into the philosophical questions. 
After all, applications are more important than the complete reasoning.

A single Hydrogen atom in Earth gives a spectrum that is exactly as same as in the Jupiter. We can use this property to communicate, even if we don't know the answer for the question "why exactly it works the same?".

So they just said, "The mathematics works fine. What do we need extra other than that, to apply it in real life!!" . 

Eventually, they stopped about thinking "Why" and proceeded to define "How" things can be developed in this new mathematics with the help of introducing some new abstract concepts. 

This is the reason why, Quantum Mechanics looks as it has more postulates than any other field in Physics. 

From these numerous abstract postulates, they developed a whole lot of other concepts and succeeded with it in real life. And so, Quantum Mechanics was formulated. 

As we did go along, we found that these abstract definitions and operations plays crucial role not only in Quantum Mechanics but also in many other places. 
From this, it was believed that, may be there is some basic mathematics intrinsically hidden in Nature. 

To understand more, accordingly, they combined it together and found out some of the most common basic rules followed by all those abstract quantities and initiated the concept of Vector space. 

Linear Vector Space:   

[Note that, the following concepts are not the first and newly defined but they are just the compilation of basic concepts you can find it anywhere in physics.]


A vector is a mathematical notation or an entity used to denote a concept. As we used to express the whole of Nature itself using numbers, these concepts are also intrinsically related with one another with the help of numbers. 

The numbers can be either real, imaginary or complex, etc. These numbers form a field, i.e. just a new name to denote the set of numbers that is used to related these vectors. If they are scalar numbers, then it is called scalar field. 

Let us denote the set of vector elements by V and the set of field elements by F. They should obey some fundamental axioms to be defined as the Vector space. 

To be considered as a field, the set F should follow these axioms, 

Closure: For all two elements 'a' and 'b' , $ a\,,b\, \in F $ then $ a*b \in F $ where  *  denotes any binary operation. The most usual one is addition and multiplication.  

Associativity: For all three elements $a,\,,b\,,c \in F $ there exists an equality $$ a*(b*c) = (a*b)*c $$

Commutativity: For all two elements $ a,\,b \in F $ there exists an equality, $$ a*b = b*a $$ 

Existence of Identity: For all elements $ a \in F $ there exists an identity element "e" such that, $$ a * e = a $$

Existence of Inverse: For all element $ a \in F $ there exists an inverse element $ a^{-1} $ such that $$ a * a^{-1} = e $$ where "e" is the identity element. 

Distributivity: If two operations are considered i.e. addition and multiplication then distributivity is defined by the condition that for all three elements $ a,\,b\,,c \in F $ there is an equality $$ a(b+c) = ab + ac $$ 

The operations needn't be addition and subtraction but can be any binary operation. 

Once the above axioms are satisfied, the set is called a "field". Now, proceeding to the next set of vector elements "V" it has to satisfy the following axioms similar to the old one, but now we take two vector elements. 
Associativity, commutativity, identity and inverse are defined as the same as previous. 
The new properties are , 
compatibility with scalar multiplication with a field element. For $ a,b\in F $ and $ \vec{u} \in V $
$$ a(b\vec{u}) = (ab) \vec{u} $$
Distributivity with scalar multiplication with vector addition, For all $ a \in F $ and $ \vec{u}, \vec{v} \in V $ there exists, $$ a(\vec{u} +\vec{v}) = a\vec{u} + a\vec{v} $$ 

When I denote vector elements with vector notation, it doesn't mean it is the usual three dimensional vector. I am just using it for the notation consistency and nothing more!   

Idea of Hyperbolic trigonometric functions in complex analysis

Complex numbers complete the description of numbers as we know, by without losing or spoiling any of the known data in the physical world. It is just a developed notation to handle the extra numbers came along with the solutions of equations, and named to be the "so called" physically meaningless solutions. 

The notation used to represent the most general form of complex number is, $$ z = a + ib \,(or)\, x+iy $$ where "i" is the imaginary root i.e. the root of "-1". With this new notation, any known number in our Nature, can be written in this form. 

Once we expand the number system with this new notation, it eventually expands all the fundamental definitions used by those numbers in any field. It leads to subsequent changes in all of the functions that is defined over the real numbers. For example, we can analyze the effect in our usual well behaved functions such as trigonometric function, exponential functions, etc. 

We need to remember one thing that, the new real functions we are going to define in a complex domain is simply just the extension of the foreknown concepts and they are all just axiomatic definitions. So, it is not possible to ask for the proof of these definitions! 

For example, in the polar form, a complex number is denoted by $$ z = r\,e^{i\theta} $$ where $\theta = \theta_p \,+\, 2\pi\,k$ k= 1,2,3,..
$\theta_p$ is called the principle angle measured from the positive x-axis. 

Now, we extend this concept of trigonometric functions into complex functions by replacing the real x-values with new complex numbers. It is achieved with the help of the handful tool i.e. series expansions of all powers of x. 

Since complex part is in the form of addition, all these powers just adds extra terms into our real expansion series. So that, the essence of old functional forms are not affected in anyway due to this new definition transformation, except that it was just incorporated in a larger domain.  

Using Euler's formula, the general form of a complex number can be denoted as, $$ z\, = \, re^{i\theta} = r (cos{\theta} + i sin{\theta}) $$

From Euler's formula, we tend to write sin($\theta$) and cos($\theta$) in terms of exponential functions, where each exponential form is used to represent a complex number. 

Euler's formula also gives, $$ e^{-i\theta}\, = \,cos(\theta)\, -\, i\,sin(\theta)\,$$ Thus we get,
$$ cos(\theta) \,=\, \frac{e^{i\theta} + e^{-i\theta}}{2} $$ and $$ sin(\theta) \, =\, \frac{e^{i\theta} - e^{-i\theta}}{2i} \,$$ 
As the right hand side of this equation only deals with exponentials, if we replace the $\theta$ with z - complex number, the definition of sine function expands to wider regions which includes all the complex numbers. 
And so, a new name and definition is given, where the independent variable $\theta$ is replace by the complex number "z". And they are hyperbolic trigonometric functions. 
The complex number "iy" is used in the euler equation to give,    
$$ e^{\{i(iy)\}}\, = \,e^{-y} \,= \,cos(iy)\, + \,i \,sin(iy)\, $$ Similarly, $$ e^{\{-i(iy)\}}\, = \,e^y \,= \, cos(iy) \, - \, i\, sin(iy) \,$$ 

We get, $$ cos(iy) \,=\, \frac{e^y\, + \,e^{-y}\,}{2} = cosh(y) $$

We choose this cos(iy) as cosh(y) since it has the similar form of cosine in Euler formula.  

But sine is defined from, $$ sin(iy) = \,\frac{e^{-y}\,-\,e^{y}\,}{2i} = - (\frac{e^{y} - e^{-y}}{2i} = -i sinh(y)$$ so that the both structures of the equation will look similar. 

As we ourselves changed the basic definition, we cannot expect the same results of usual trigonometry in here. For example, the maximum value of sine and cos is equal to one when dealing real numbers, but with complex number it can have any value

That is all we need to know about definitions.. now we can proceed further to define all other identities from this basic concept and all other things can be sought out from those definitions. 

Rigid Body Motion - Euler Angles

A rigid body is defined from the idealized concept that, distance between any two mass points remains constant throughout the motion. 
     Since, it is satisfied by the most of the objects we use in real life (Not absolutely, but perfectly applicable), the kinematics of these rigid bodies plays significant role in many areas. 

The special property of this rigid body is that, we need only 6 independent coordinates to completely define the state of a rigid body in 3-Dimensional Space. No matter how many particle it contains, it can be always applied, entirely due to the constraints. 
Other than the distance constraints, it is also possible to add additional constraints in any rigid body motion, and so the number of independent coordinates will be more reduced. 

It is customary to use the first set of 3 independent coordinates as the "Space fixed coordinates" and the second set of 3 coordinates as the "Body fixed coordinates" but with the same origin. 
     So that, you can explain the state of the body (general position) with the Space fixed coordinates and its orientation relative to this Space fixed coordinates using the Body fixed coordinates. 

Specifying one set of coordinates relative to another needs the basic rules of "Coordinate Transformation".   

Let us say the $(x_1,x_2,x_3)$ and $(x'_1,x'_2,x'_3)$ are the components of same vector in two sets of orthogonal coordinate system with same origin. If $ \{ \hat{e_1}, \hat{e_2}, \hat{e_3} \}$ and $\{ \hat{e'_1} , \hat{e'_2} , \hat{e'_3}\}$ are the corresponding unit vectors,

Direction cosines are defined by,

$$ cos(\theta)_{ij} = cos (\hat{e'_i}\cdot\hat{e_j}) = \hat{e'_i}\cdot\hat{e_j} \,\,\,\ldots...eq.(1)$$

Using the direction cosines, new primed unit vectors in terms of old non primed unit vectors written by, 
$$ \hat{e'_i} = \sum_j (\hat{e'_i}\cdot\hat{e_j}) \hat{e_j} = \sum_j cos(\theta)_{ij} \hat{e_j} \,\,\ldots...eq.(2)$$

Let me write the general vector in cartesian coordinates as,
$$ \vec{r} = x_1\hat{e_1} + x_2 \hat{e_2} + x_3\hat{e_3} = x'_1\hat{e'_1} + x'_2\hat{e'_2} + x'_3\hat{e'_3} \,\,\ldots...eq.(3) $$ 

Then each of the new coordinates in general can be written using eq.(2),

$$ x'_i = \vec{r}\cdot\hat{e'_i} = \sum_j x_j\hat{e_j} \cdot \hat{e'_i} = \sum_j x_j cos(\theta)_{ij} \,\,\ldots...eq.(4)$$

The domain of this definition can be expanded for any general vector. 

Note: In three dimension, all indices run from 1 to 3

Accordingly in 3 dimension, we need 9 direction cosines to make the transformation from one to another namely,
$$ cos(\theta)_{ij} = \hat{e_j}\cdot\hat{e'_i} $$ where "i and j" each runs from 1 to 3 and so it gives 9 components. 

But only three of them are needed to specify the orientation. What about the others?
It so happens, there are quite few extra relations exists due to the orthogonality property of the coordinates. 

The orthogonality relations are given by, 
$$ \hat{e_i}\cdot\hat{e_j} = \delta_{ij} \\~\\ \hat{e'_i}\cdot\hat{e'_j} = \delta_{ij} \,\,\, eq.(5) $$
Expanding the unit vector in one system in terms of the unit vectors of other system using eq.(2) and making use of direction cosines, it can be rewritten as, 

$$ \sum_{j=1}^3 cos(\theta)_{i'j} cos(\theta)_{ij} = \delta_{ii'} \,\,\, eq.(6)$$  
where $\delta $ is the Kronecker-delta symbol.    

To make life simpler, we use the Einstein summation convention, and we denote the direction cosines $$ cos(\theta)_{ij} = a_{ij} $$

Now, eq.(4) becomes,
$$ x'_i = a_{ij} x_j \,\,\,\,\,\,\ldots...eq.(7) $$
where summation is assumed.
The magnitude of the vector $\vec{r} $ in both cases are the same. Using that property with the Pythagoras relation for orthogonal coordinates, we can write,

$$ x'_{i} x'_i = a_{ij} x_j a_{ik} x_j x_k $$
therefore,  $$  a_{ij} a_{ik} = \delta_{jk} \,\,\,\,\,\ldots...eq.(8)$$
where both j,k runs from 1 to 3.

This is the exact same condition obtained from orthogonality relation but in a new representation. 

Since both i and j runs from 1 to 3 in discrete sense, we can write the set of direction cosines in a general matrix form where i used for row and j for column. Let me call the matrix as A,
$$ A = \left[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right] $$

The general transformation from one system to other can be thought of as an Matrix operation on one system which results the coordinates of other system.

Using matrices, 
$$ \left[\begin{matrix} x'_1 \\ x'_2 \\ x'_3 \end{matrix}\right] = \left[\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right] \,\,\,\ldots...eq.(9)$$ 

Euler Angles: 

Lagrangian formalism is created based on the concept of degrees of freedom and constraints. Lagrangian itself is defined in terms of Independent coordinates. 

Finding the independent coordinates for a system in motion is essential for solving the Lagrangian. 

From orthogonality relations, we can be sure that, we don't need all the 9 direction cosines as independent coordinates. All we need is some three independent functions of these direction cosines. 

It will therefore necessary to define a set of three new independent functions to describe the orientation of rigid body in space using the Lagrangian. 

[But there is also another condition for our Matrices, which says the determinant of the matrix should be +1. Otherwise it would  be an inversion of the coordinates.]  

There are really a number of different independent functions obtained in a number of different ways to describe the Lagrangian. But Euler angles is the customary one, where we will make out transform from one to another by making three successive rotations in a specific way. 

The sequence in each step can be thought of a matrix operating on the coordinate system at that particular instant. 

Let me start the initial transformation from the $\{ \hat{e_1},\hat{e_2},\hat{e_3}\}$ coordinate system, by making angle $\phi$ counter clock wise about $\hat{e_3}$ axis. And let me denote the resultant coordinate system by, $\{\hat{e_1}^1, \hat{e_2}^1,\hat{e_3}^1\}$  

In the second stage, we make the transformation of $\{\hat{e_1}^1, \hat{e_2}^1,\hat{e_3}^1\}$ by rotating it about $\hat{e_1}^1 $ axis in the counter clock wise direction by an angle $\theta$ 
Consequently we arrive at a newer system denoted by $ \{\hat{e_1}^2, \hat{e_2}^2, \hat{e_3}^2\}$. 

Now, we make the final transformation by rotating $\{\hat{e_1}^2,\hat{e_2}^2,\hat{e_3}^2\}$ by an angle $\psi$ with respect to $\hat{e_3}^2$ axis.    

Thus, we finally arrived our desired transformation that is $$\{\hat{e_1}^3,\hat{e_2}^3,\hat{e_3}^3\} = \{\hat{e'_1},\hat{e'_2},\hat{e'_3}\} $$

In Matrix Notation, each transformation can be written as follows,

First transformation, $$ E_1 = A_1 E \,\,\,\,\ldots...eq.(10)$$

where "E" is the set of coordinate elements and "A" is the transformation Matrix.

Second transformation, $$ E_2 = A_2 E_1 \,\,\,\,\ldots...eq.(11)$$

Final transformation, $$ E_3 = A_3 E_2 = E' \,\,\,ldots...eq.(12)$$

Hence therefore, the combined transformation is denoted by the product of respective matrices,
$$ E' = A_3 A_2 A_1 E = R E \,\,\,\,\ldots...eq.(13) $$
where $ A_3 A_2 A_1 = R $

Writing each transformation in terms of its matrix element values i.e. direction cosine values, 
Since $A_1$ represents the counter clock wise rotation of "E" by an angle $\phi$ about $\hat{e_3}$ axis, it can be written in the matrix form as,

$$ A_1 = \left[ \begin{matrix} cos\phi &sin\phi &0\\ -sin\phi & cos\phi &0 \\ 0 & 0 & 1 \end{matrix}\right] \,\,\,\ldots...eq.(14) $$

Similarly, $A_2$ is the rotation of $E_1$ by an angle $\theta$ with respect to $\hat{e_1}^1$ axis, 

$$ A_2 = \left[ \begin{matrix} 1&0&0\\ 0& cos\theta &sin\theta\\ 0 & -sin\theta & cos\theta\end{matrix} \right] \,\,\,\ldots...eq.(15) $$

Finally, $A_3 $ is the rotation of $ E_2$ by angle $\psi$ with respect to $\hat{e_3}^2 $ axis,

$$ A_3 = \left[ \begin{matrix} cos\psi &\sin\psi & 0\\ -sin\psi & cos\psi & 0 \\ 0 &0&1 \end{matrix}\right] \,\,\,\ldots...eq.(16)$$

Combining together, three transformations can be written using a single matrix R as follows, 
$$ R = A_1A_2A_3 $$ 
subsituting for $ A_1, A_2, A_3 $ gives,
$$ R= \left[ \begin{matrix} cos{\psi} cos{\phi} - cos{\theta} sin{\phi} sin{\psi} & cos{\psi} sin{\phi}+ cos{\theta} cos{\phi} sin{\psi} & sin{\psi}sin{\theta}\\ -sin{\psi}cos{\phi} - cos{\theta}sin{\phi}cos{\psi} & -sin{\psi}sin{\phi}+cos{\theta}cos{\phi}cos{\psi} & cos{\psi}sin{\theta}\\ sin{\theta}sin{\phi} & -sin{\theta}cos{\phi} & cos{\theta} \end{matrix}\right] \,\, \ldots...eq.(17)$$

The inverse transformation from body fixed to space fixed coordinates is just given by the inverse of R i.e. $R^{-1}$ which is the transpose of R.

You can ask, why I choose this specific order of rotations. It needn't be. You can choose many other ways. But the only condition is, two consecutive rotations shouldn't be about the same axis. 

Quantum Mechanics - Postulates (Part -1) - Wave function, Hermitian Operators

I am not going to give the Postulates as it is in the books or anything, but I just want to postulate and speak its' mathematical importance, in a way I understood. 

From the Classical Physics of Lagrangian and Hamiltonian, we know that any system [it can be single particle or multi particle or anything] can be associated with a function so called Lagrangian or Hamiltonian, such that all the information about the system can be extracted from this function using the corresponding Equations of motion. 
Mathematically, we assume that the Lagrangian or Hamiltonian function contains all the necessary information we need to describe the system completely. 

In the same way, here we assume that "Every Quantum Mechanical System is completely described by an arbitrary State Vector or a Wave function $ \vert{\psi(t)}\rangle $ , read as "ket - psi" is an element of complex linear vector space called Hilbert Space. The State vector contains all the information about the system and it changes only with time.  

The state vector is an abstract concept and you can never measure this state vector or imagine it in a physical manner. 

From the concept of Vector space, we assume that it is always possible to define a set of vectors which are linearly independent and forms the basis for the Vector Space.  

Note: You needn't to panic on hearing the term Vector Space. Your Euclidean space follows the rules of Vector space. Whenever you get in trouble understanding vector space, you can always make a comparison with your 3 dimensional Euclidean space. 

The set of basis vectors needn't to be unique, but it is always possible to represent any vector in the Vector space as a linear combination of these basis vectors. 


As a consequence, you can imagine this arbitrary state vector as the linear combination of all the basis vectors.
We don't know what are these basis vectors, since there is many possible ways of choosing a set of basis vectors from different possible sets. Let us consider this as a general linear combination.  

It is represented as, $$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle = A_1(t) \vert{\phi_1}\rangle + A_2(t) \vert{\phi_2}\rangle + \ldots.....eq.(1)$$

All the basis vectors are ket vectors, after all left side should be equal to right side. And we can always make the time dependence of ket vector to come into the coefficients.  

We already said that, these are in abstract Hilbert space, so we cannot measure anything about them. 
To measure anything, we need to make the projection of this abstract quantities in the known space where we could describe the wave function completely. 

To measure the projection, we make the dot product of desired known parameter with this abstract Wave function. 
So that, the wave function and all its basis vectors are now described using our desired known parameter. 

For example, if the desired known parameter is position, then all the Wave function and its basis vectors will be projected into position space (where position is the parameter). And so, the new projected wave function is called "Position Space Wave function".

$$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle $$

Dotted with x to give the projection in Position space, 

$$ \langle{x}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{x}\vert{\phi_b}\rangle   \,\, \ldots...eq.(2)$$

Now, the new projection of Wave function in Position space, i.e. Position Space wave function is,
$$ \psi(x,t) = \sum_b A_b(t) \phi_b(x) $$

Where $\langle{x}\vert{\psi(t)}\rangle = \psi(x,t)$ and $ \langle{x}\vert{\phi}\rangle = \phi(x)$ 

If we choose momentum as the desired known parameter, then we can dot momentum with the general wave function. It will result into, $$ \langle{p}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{p}\vert{\phi_b}\rangle \,\,\ldots...eq.(3)$$ 
And the new wave function is called Momentum Space Wave function, $$ \psi(p,t) = \sum_b A_b(t) \phi_b(p) $$

That is all we can do with the first Postulate. 

The second Postulate is stated as, "Each dynamical variable that relates to the motion of the particle can be associated with a linear operator". 

An operator is called to be linear if it satisfies the condition, $$ \hat{Q}(c_1\psi_1 + c_2\psi_2) = c_1 \hat{Q}\psi_1 + c_2 \hat{Q}\psi_2 \,\,\,....\ldots.eq.(4)$$

With Each operator, it can be associated a linear eigen value equation such that $$ \hat{Q} \psi_i = \lambda_i \psi_i \,\,\,\,\ldots..eq.(5)$$
where $\psi_i $ is called the eigen state and 
$\lambda_i$ is called the eigen value. 

A linear operator is also an abstract concept, which is represented using a matrix. A linear operator is determined by how it acts on the basis vectors because any vector can be expanded as the linear combination of these basis vectors. 

If we know how an operator acts on the basis, then it gives us everything we need to know about the operator on that Vector Space. 

Let me represent the basis vectors as $$\vert{e_1}\rangle, \vert{e_2}\rangle, \ldots...$$

Therefore, $$ \hat{Q}\vert{\psi}\rangle = \hat{Q} \vert{e_1}\rangle + \hat{Q} \vert{e_2}\rangle + ... \,\,\ldots...eq.(6)$$


If we represent the linear operators with the matrix, knowing the matrix elements is knowing the operator itself. 

Let me take a basis vector $\vert{e_i}\rangle$ in the Hilbert Space. To understand how a linear operator works on this basis vector, we operate it on this basis and it will result some new vector. 

For example, you can consider the rotation of the coordinates as an operation that acts on the basis vectors. 

Due to linear property, $\hat{Q}\vert{e_i}\rangle$ - the new vector itself can be written again as a linear combination of the basis vectors, represented as $$ Q\vert{e_i}\rangle = \sum_k Q_{kj}\vert{e_k}\rangle $$

You can compare it with the coordinate transformation rules.   

Now, the third postulate says that, "Any observable in Quantum Mechanics is a linear Hermitian operator on the Hilbert space, where the eigenvalues are the only possible results of a precise measurement of that observable. 
Definition of a Hermitian operator:
$$ \int \psi_i^* (\hat{Q} \psi_i)\, dx = \int (\hat{Q}\psi_i)^* \psi_i \,dx \,\,\,\ldots...eq.(7)$$

Eq.(7) which gives a special property on expanding with eigenvalue eq.(5) as follows,
$$ \int \psi^* (\lambda_i \psi_i)\,dx =  \int (\lambda_i \psi_i)^* \psi_i\,dx $$
which gives, $$ \lambda_i \int \psi_i^*\psi_i \,dx = \lambda_i^* \int \psi_i^* \psi_i \,dx $$

$$(\lambda_i - \lambda_i^*) \int \psi_i^*\psi_i \,dx = 0 $$

But $\int\psi_i^*\psi_i \,dx = \int {|\psi_i|}^2 \,dx > 0 $ and it is equal to zero only when $|\psi_i| = 0 $ where wave function itself vanishes and that is not a desirable solution. 

So, the only solution is $$ \lambda_i - \lambda_i^* = 0 $$ or $$ \lambda_i = \lambda_i^* $$ It is only possible when $"\lambda_i"$ is a real number. This is a characteristic result of any Hermitian operator, which states that "The eigenvalues of an Hermitian Operator is always a real number". 
This is the reason why, eigenvalues of an operator is the only possible results on a precise measurement, because measurement should give a real number. 

There are much more things to talk about an operator and important relations like Completeness, Orthogonality, Hermiticity of an operator and etc. It should be dealt separately. 

Curvilinear Coordinate System and General expression for Gradient, Curl, Divergence and Laplacian

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. 
 

Functions and its conditions

Mathematics is the language of physics and Functions is the heart of mathematics. 
       Function is a mathematical representation of an artificial machine which gets some input and gives some output.

       But how functions were first formulated and how it was first defined by the people of physics and mathematics?  

The need of functions:

       In Physics, we always have the situation of relating ‘dependent and independent variables’.

Most of the physical terms are denoted using the relation between one or two or three or more variables.

Eg.

“Speed (s) = Distance (d) / Time (t)” or simply using variables as “s = d/t”

“Acceleration (a) = Velocity (v) / Time (t)” or simply using variables as “a= v/t”

“Gravitational force between two masses (F) = G*m1*m2 / r^2”, where G= gravitational constant, m1=mass of object1, m2=mass of object2, r^2=square of the distance between them. 

      

 We can give so many examples and all of the terms in Physics have the same formulation like above.

       Here Speed changes with the change in Distance and Time. Similarly Acceleration changes with the change in Velocity and Time. That is why Speed and Acceleration is called the dependent variables and distance, time are called the independent variables.  

       Physics itself is measured using these independent and dependent variables.

       The knowledge of the relations between these dependent and independent variables will give us a generalized study of Physics using Mathematics. Using the generalized study we can calculate any physical quantity using Mathematics.

       That is how people started to measure Physics using Mathematics in the name of functions.

       From all of the above physical terms, we can get some common thing.

       Speed changes with distance.

       Speed changes with time.

       Acceleration changes with velocity.

       Acceleration changes with time.

       The change in dependent variable makes change in independent variable.

       Let us look the first and the easiest physics term 'speed'. If we change the distance or time then speed changes.

If you take Speed as the output and time, distance as the input then the above statement changes to,

For the values of input [time, distance] there is an output value of speed.

       Change in input will give a change in output.

       In a mathematical way, it was stated as

Speed is a function of distance and time.

Sentences are not always worth to use and symbols were introduced for the sake of simple and compact representation,

s= f(d , t)

Here f( ) means “the function of” and “s= speed , d=distance, t=time”.  

       But before considering the functions of two independent variables, it should be taken care of the functions that varies with one single variable. Studying the single variable functions, gives practice to look into multivariable functions.

Single Variable functions:

For the generalized version, the variables were taken as “x” and “y”.

Let’s say “x” is the independent variable and y is the dependent variable.

Then, the functional statement is given as “y is the function of x” or

“y=f(x)”.

But all those things that vary with some variables cannot be considered as a function. What about its properties?

There should be some conditions imposed on them to proceed further. And so, there are two major conditions for any function and it is given by,

Condition 1:

To study anything in mathematics, the data should be clear.

In functions, it is necessary to have the complete data of the dependent and the independent variables. Only if the complete information of a function is known, we can use it for any physical or mathematical operations related with measurements.

       Without the complete knowledge, we cannot create any working machine or anything for any of our purpose.

       Let us imagine a machine that gives some output for every input. We should get output for every input.

       If there is no output for some input, it means the machine has some problems and it will give error to become useless.

       Similarly for the functions, there should be an output for every input. Otherwise it will not be said as a “function”.

In the E.g. of y=f(x)

“x” is the input and “y” is the output.

Thus the condition1 for functions was stated as,

For each value of input “x” there should be a value of output “y”.

 Condition2  

What if an input value of “x” has more than one output “y” values?

Does it have any problem? Is it allowed?

The same example of speed was taken again. As we are caring about the functions of one independent variable function, here we can take account of only speed and time variables.

Speed changes with time where time is the input value and speed is the output.

If we defined the function that “one input can take two outputs” then,

       It also allows the possibility of “one time input can take two speed outputs”.

How an object can have two speeds at one time? It is as same as two objects at one time.

How an object can be in two places at one time?

As far as we concerned it is not possible in our real world. So the possibility of “two or more outputs for a single input” is removed from the definition of function. 

That is how we arrived to the second condition for a function which can be stated as,

For each value of input “x” there should be one and only value of output “y”.

 Thus two conditions were formulated for a function. 

     

       Therefore only the mathematical relationships which follow the above conditions are called functions otherwise it is called Relations.  

       Still functions need to be defined in a complete mathematical way using Number system so that it could be quantified. Those kind of extra conditions and its structure is defined depending on its role in various situations. As far as concerned about the definition, these two conditions are the necessary and sufficient one.