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. 

Hamilton's equations from variational principle

As we used to derived the Lagrange's equations of motion from variational principle i.e.Hamilton's principle, we can derive Hamilton's equations. 

Hamilton's principle i.e. variational principle gives the condition that, $$\delta \int_{t_1}^{t_2} L \, dt = 0$$  
where L is the Lagrangian of the given system. In the previous derivation, it was considered various paths in the configuration space. Where else, now we will working in the paths which is the part of Phase space where q and p are the independent coordinates.
q - position and 
p - momentum

To go from Lagrangian to Hamilton, we just express the Lagrangian in terms of Hamiltonian where everything is the function of q and p.

$$\delta \int_{t_1}^{t_2} \left(p_i\dot{q_i} - H (q,p,t)\right) dt = 0$$  
Let us call this new function as L',
therefore, $$L' = p_i\dot{q_i} - H$$ and this new L' obeys the Lagrangian equations of motion that is derived earlier from the definition. 
Therefore, $$\delta \int_{t_1}^{t_2} L'(q, \dot{q}, p, \dot{p}) \,dt = 0$$
where the integral is defined over 2n dimensional phase space. And it gives symmetrically two sets of relations,
$$\frac{d}{dt}\left(\frac{\partial{L'}}{\partial\dot{q_j}}\right) - \frac{\partial{L'}}{\partial{q_j}} = 0  \,\,\,\,j = 1,2,...n$$ and
$$\frac{d}{dt}\left(\frac{\partial{L'}}{\partial\dot{p_j}}\right) - \frac{\partial{L'}}{\partial{p_j}} = 0 \,\,\,\, j = 1,2,....n$$  
Substituting for L' we get, $$\frac{\partial{H}}{\partial{q_j}} = -\dot{p_j}$$  and $$\frac{\partial{H}}{\partial{p_j}} = \dot{q_j}$$  
which is our desired Hamilton's equations of motion. 
The derivation is nothing but just the result of our Legendre transformation in variational principle. 

Energy stored in the Capacitor and the effect of Dielectric

A capacitor is usually charged by connecting it to a battery. To charge up the capacitor, we take a small elemental positive charge "+dq" from a neutral system and move it a distance "d" so, that the system gets "-dq". The work is done against the Electric field which directs opposite to the motion. 
Let say, the amount of charge piled in the positive plate is +q, so that the potential difference between the plates is "q/C". 
Now, the work required to do move the next "dq" charge is simply just the potential multiplied by the amount of charge [The definition of potential is work done per unit charge].
So, $$dW = \frac{q}{C} dq$$  
Then the total work done to charge up the plate to +Q charge is, $$W = \int_0^Q \frac{q}{C} dq = \frac{Q^2}{2C}$$ Using the relation, $Q =  CV$ we get, $$W = \frac{1}{2} CV^2$$ which is the electrostatic potential energy stored in the system. 

For example, in the case of parallel plate capacitor, the energy stored can be calculated as, where $C = \frac{\epsilon_0 A}{d}$ and V = Ed , $$W = \frac{CV^2}{2} = \frac{\epsilon_0 A E^2 d }{2}$$  
Here, "Ad" represents the volume of the in between region where the energy is stored. So, we can define a new term as, 
Electrostatic Energy density = $\frac{1}{2} \epsilon_0 E^2$ 

As an example, if we try to calculate the electrostatic energy density of air at the break down voltage which is about $E = 3\times 10^6 Vm^{-1}$ and the energy density can be calculated as, 39.825 $Jm^{-3}$, the value is so high. 
One Joule is defined as the work required to move one kilogram object to one meter distance. It is nearly 40 Joules of energy in a volume one meter cube. 

Now, we will give a slight attention to dielectric system. We know there are two types of charges are produced in polarization. First one is the bound charge, $\rho_b = - \nabla\cdot\vec{P}$ and the surface charge, $\sigma_b = \vec{P}\cdot \hat{n}$ 
To apply Gauss law, the total charge inside the material is identified as the free charge and the bound charge. $$\rho _{total}= \rho_{bound} + \rho_{free}$$  
Using gauss law, $$\rho_{total} = \epsilon_0\nabla\cdot\vec{E} = -\nabla\cdot\vec{P} +\rho_{free}$$
Calling, $$\vec{D}= \epsilon_0\vec{E} + \vec{P}$$ we get, $$\nabla\cdot\vec{D} = \rho_{free}$$  
In most of the cases, when Electric field is not so much high, Polarization is given by, $$\vec{P} = \epsilon_0 \chi_e \vec{E}$$
where $\chi_e$ is called the electric susceptibility. 
 Then, $$\vec{D} = \epsilon_0 (1+\chi_e) \vec{E}$$ $$\vec{D}= \epsilon \vec{E}$$ and $$\epsilon_r = \frac{\epsilon}{\epsilon_0}$$ is called relative permitivity or dielectric constant. 

For vacuum, $$\vec{D} = \epsilon_0 \vec{E_{ext}}$$ since there is no polarization to take place. In a dielectric medium $$\vec{D} = \epsilon \vec{E_{in}} = \epsilon_r\epsilon_0 \vec{E_{in}} = \epsilon_0 \vec{E_{ext}}$$
Thus it gives, $$\vec{E_{ext}} = \epsilon_r \vec{E_{in}} \rightarrow \vec{E_{in}} = \frac{\vec{E_{ext}}}{\epsilon_r}$$ which says that the Electric field is reduced to "$\frac{1}{\epsilon_r}$" factors than the Electric field applied (in the vacuum). 

Then, we can see that, if a dielectric medium is introduced in the between region of capacitor plates, capacitance becomes $$C'= \frac{Q'}{V'} = \frac{Q}{V'}$$ since the charge is the same in both the cases. But, $$V' = \frac{E'}{d} = \frac{E}{\epsilon_r{d}}$$ and $$C' = \epsilon_r C_{vacuum}$$
which means the capacitance of the capacitor is increased by the factor of dielectric constant value of the inserted dielectric medium.

Using this, we can find the expression for the new capacitance of a parallel plate capacitor where dielectric material of thickness "t" is introduced. 
To find the new potential difference, $$V' = - \int_-^+ \vec{E}\cdot \vec{dl} = \int_0^{d-t}E_{vacuum} dl + \int_0^t E_{dielectric} dl$$
which gives, $$V' = E_{vacuum}(d-t) + \frac{E_{vacuum}}{\epsilon_r} t$$
Electric field between the plates is given by, $$E_{vacuum} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$$
So, $$V' = \frac{Q}{A\epsilon_0} [ (d-t) + \frac{t}{\epsilon_r}] = \frac{Q}{A\epsilon_0} [d-t(1-\frac{1}{\epsilon_r})]$$
And the capacitance is given by, $$C' = \frac{Q}{V'} = \frac{A\epsilon_0}{d-t(1-\frac{1}{\epsilon_r})}$$    

Capacitance for different types of capacitors [Parallel plate, cylindrical and spherical capacitors]

A capacitor is the combination of two metal plates (conductors) having opposite charges separated by distance "d" apart. Don't assume that the metal plates should be rectangular plates with some infinite distance. 
Our usual capacitor is just so small enough within the size of a finger, where the conducting material rolled down in a cylindrical shape with some non conducting or dielectric material in between them. 

Let us discuss with mathematics, 
The two plates have opposite charges, let say, "+Q" on the first plate and "-Q" on the second plate. So, the potential difference between these plates can be calculated as the work done required to bring a unit positive charge from -Q to +Q [from the definition of potential] 
$$\int_{V_-}^{V_+}\,dV = V = V_+ - V_- = - \int_{-}^{+} \vec{E}\cdot \vec{dl} \,\,\,...eq.(1)$$  
We don't assume anything about the Shapes of the plates, so Electric field is just given by the definition as, $$\vec{E} = \frac{1}{4\pi\epsilon_0} \int \rho \frac{\hat{r}}{r^2} d\tau$$ where $\rho$ is the volume charge density and integral is over the volume "V".

Electric field is proportional to both Potential "V" and the charge on the each capacitor "Q" and the ratio is some constant which is defined as the capacitance. $$C = \frac{Q}{V}$$

Let us try to calculate the capacitance of some simple known shapes, where assumptions are easy to make. 
Capacitance of Parallel Plate Capacitor

First one is the parallel plate capacitor, where Electric field is directed from positive charge to negative charge plate and assumed to be uniform in the direction. So, it can be taken out from the integral in equation (1) and integral sign is canceled by the negative sign and the equation becomes, $$V = E \int_0^d \,dl = Ed$$ From our definition of Capacitance, $$C = \frac{Q}{V} = {Q}{E d}$$ And, the Electric field in the region between parallel plate capacitors is given by, $$E = \frac{\sigma}{\epsilon_0}$$ where $\sigma = \frac{Q}{A}$ is the surface charge density. 
Then, $$C = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{A\epsilon_0}{d}$$ which is determined only by the sizes, shapes, and separation distance of the two conductors. 

Capacitance of Cylindrical capacitor

Second is the cylindrical capacitors, where inside is solid cylinder with positive charge with radius "a" and outside is hollow cylinder with bigger radius "b" (b>a). Let us take a point "r" in the "in between region a<r<b "
The Electric field in this region is given by making use of Gauss law in cylindrical symmetry as, $$\vec{E} = \frac{\lambda}{2\pi r \epsilon_0} \hat{r}$$
Then potential difference is calculated by, $$- V =  - \int_a^b \vec{E} \cdot \vec{dr}$$ where $\vec{dr}$ is directed from "a" to "b" i.e. radially outwards. So, it becomes $$V = \frac{\lambda}{2\pi \epsilon_0} \int_a^b \,\frac{1}{r} dr = \frac{\lambda }{2\pi\epsilon_0}\ln{(\frac{b}{a})}$$  
where again the integral sign canceled by the negative sign(slightly different from the previous - Here we started with finding $-V = V_ - -  V_+$ because the limits will be easy and all are measured from the centre - otherwise $\vec{dl}$ will be directed from -Q to +Q which will have limits 0 to "b-a" where logarithmic function not finite). 
Thus we get the capacitance of the cylindrical capacitor as, $$C = Q/V = \frac{\lambda{l} 2\pi\epsilon_0}{\lambda \ln{(\frac{b}{a})}} = \frac{2\pi\epsilon_0{l}}{\ln{(\frac{b}{a})}}$$  

Capacitance of spherical capacitor

Finally, let us consider a spherical capacitor with two concentric spherical metal shells with radii a and b. Inner shell with +Q and outer shell with -Q charge. 
In the same way, the Electric field in the region is given by, $$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r}$$ vector "r" points radially outwards. 
Potential difference is calculated by, (integration technique is similar to cylindrical capacitor), $$- V = - \frac{Q}{4\pi\epsilon_0} \int_a^b \frac{1}{r^2} dr$$ Thus, $$V = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{a} -\frac{1}{b}\right)$$  
 And the capacitance is given by, $$C = 4\pi\epsilon_0 \frac{ab}{b-a}$$

Thus we can find the capacitance for various shapes. 

Angular Momentum Operator in spherical cooridnate system - derivation

Classical definition of angular momentum is given by, $$\vec{L} =\vec{r} \times \vec{p} \,\,\,...eq.(1)$$
Writing the components in terms of Cartesian coordinates we get,
$$L_x = yp_z - zp_y \\~\\ L_y =  zp_x -  xp_z \\~\\ L_z = xp_y - yp_x$$

In quantum mechanics, the momentum operator in position basis for x component is given by $- \hbar \frac{\partial}{\partial{x}}$ using this, we rewrite the above equations as,   $$L_x = -i\hbar ( y\frac{\partial}{\partial{z}} - z\frac{\partial}{\partial{y}}) \\~\\ L_y = -i\hbar ( z\frac{\partial}{\partial{x}} - x\frac{\partial}{\partial{z}}) \\~\\ L_z = -i\hbar ( x\frac{\partial}{\partial{y}} - y\frac{\partial}{\partial{x}})$$

But, spherical symmetric potentials are often useful in quantum mechanics where we might need to express the equations in spherical coordinate system for simplicity. So, we do need to make the coordinate transformation rules to these equations to write this in terms of spherical polar coordinate system.

From our previous post of Cartesian to Spherical Coordinate transformation, we know that $$dx = \,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi$$
$$dy = \, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi$$  
$$dz = cos\theta \, dr - r\, sin\theta \, d\theta$$
where $$x = \,r \,sin\theta \,cos\phi \\~\\ y = \,r \,sin\theta \,sin\phi \\~\\ z = \,r \,cos\theta\,$$
Solving for $dr, \,d\theta\,, d\phi$ as follows,
first we solve for $$d\phi = -\frac{1}{r}\frac{sin\phi}{sin\theta} dx + \frac{1}{r}\frac{cos\phi}{sin\theta} dy$$ Now we solve for $$dr = sin\theta\,cos\phi\, dx + sin\theta\, sin\phi\,dy + cos\theta\, dz$$ and finally we get, $$d\theta = \frac{1}{r} cos\theta\, cos\phi\,dx + \frac{1}{r} cos\theta\,sin\phi \, dy - \frac{1}{r}sin\theta\, dz$$
Using the above three equations with chain rule,
$$\frac{\partial}{\partial{x}} = \frac{\partial}{\partial{r}}\frac{\partial{r}}{\partial{x}} +\frac{\partial}{\partial{\theta}}\frac{\partial{\theta}}{\partial{x}} +\frac{\partial}{\partial{\phi}}\frac{\partial{\phi}}{\partial{x}}$$  
or $$\frac{\partial}{\partial{x}} = sin\theta\,cos\phi\, \frac{\partial}{\partial{r}} + \frac{1}{r} cos\theta \,cos\phi\, \frac{\partial}{\partial{\theta}} - \frac{1}{r}\frac{sin\phi}{\sin\theta} \frac{\partial}{\partial{\phi}}$$  Similary for others, $$\frac{\partial}{\partial{y}} = sin\theta\,sin\phi\, \frac{\partial}{\partial{r}} + \frac{1}{r} cos\theta \,sin\phi\, \frac{\partial}{\partial{\theta}} + \frac{1}{r}\frac{cos\phi}{\sin\theta} \frac{\partial}{\partial{\phi}}$$ $$\frac{\partial}{\partial{z}} = cos\theta\,\frac{\partial}{\partial{r}} - \frac{1}{r} sin\theta \, \frac{\partial}{\partial{\theta}}$$
Applying our equations in our definition of angular momentum, the angular momentum operator in terms of spherical coordinate system can be written as,
$$L_x = i\hbar \left( sin\phi \,\frac{\partial}{\partial{\theta}} + cot\theta\,cos\phi\, \frac{\partial}{\partial{\phi}}\right) \\~\\ L_y = i\hbar \left( -cos\phi\, \frac{\partial}{\partial{\theta}} + cot\theta\, sin\phi\,  \frac{\partial}{\partial{\phi}} \right) \\~\\ L_z = -i\hbar \frac{\partial}{\partial{\phi}}$$ and $$\vec{L}^2 = L_x^2 + L_y^2 +L_z^2 = -\hbar^2 \left[ \frac{1}{sin\theta} \frac{\partial}{\partial{\theta}}\left(sin\theta\frac{\partial}{\partial{\theta}}\right) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial{\phi^2}}\right]$$  

We will be often using this expression in quantum mechanics when we deal hydrogen atom, spherically symmetrical potential problem, etc. 

Canonical Transformations

In this section, we are primarily concerned about the problems where the Hamiltonian is independent of all the $q_i$ coordinates i.e. cyclic coordinates. In those cases, the canonical conjugate momentum of the system is constant over time. Let us call those constants as, $$p_i =  c_i$$ Then the Hamiltonian is rewritten as the function of these constants, $$H = H ( c_1, c_2,... c_n )$$   In this simple form, the corresponding equations of motion are linear and easy to solve. In general all of the coordinates in the problem is not cyclic in Nature. 

But it was known that, the number of cyclic coordinates in a given system depends on the choice of generalized coordinates. That is why there are various coordinates systems are defined like spherical, cylindrical, ellipsoidal and etc. So we try to find a coordinate system such that all of the coordinates are cyclic. And the procedure is popularly known as canonical transformation which is the transformation procedure from our initial coordinate system to the desired cyclic coordinate system. 

Before going to the transformation, we prefer to maintain the structure of Hamilton's equations i.e. we would like to get a new Hamiltonian with new set of Q and P but with old structure which follows the variational principle and its results. 

From variational principle we know that the Lagrangian is not unique and can differ from one another about the differential of a function that vanishes at the end points, given by,
$$L' = L + \frac{dF}{dt}$$ Similarly, the variation principle gives us the freedom, $$p_i\dot{q_i} - H = P_i \dot{Q_i} - H' + \frac{dF}{dt} \,\,\,...eq.(1)$$ with necessary scale transformations. 

Depending on the functional form of F, there are various ways of obtaining the transformation rules. Let us look the at the four basic functional forms. 
First let us assume that F is a function of old and new coordinates i.e.q and Q. $$F _1= F(q,Q,t) \,\,\,...eq.(2)$$  
Applying it in eq.(1) we get, 
$$p_i\dot{q_i} - H = P_i \dot{Q_i} - H' + \frac{dF(q, Q, t)}{dt}$$
using the chain rule, 
$\rightarrow$ $$p_i\dot{q_i} - H = P_i\dot{Q_i} - H' + \frac{\partial{F}}{\partial{q_i}} \dot{q_i} + \frac{\partial{F}}{\partial{Q_i}} \dot{Q_i} + \frac{\partial{F_1}}{\partial{t}} \,\,\,...eq.(3)$$
$$p_i \dot{q_i} - H = \left( P_i + \frac{\partial{F}}{\partial{Q_i}} \right) \dot{Q_i} + \frac{\partial{F}}{\partial{q_i}} \dot{q_i} + \frac{\partial{F}}{\partial{t}} -H'$$
Equation the coefficients, (since all q,p,Q,P are independent variables) we get, $$\frac{\partial{F}}{\partial{q_i}} = p_i \,\,\,...eq.(4)$$ and $$P_i + \frac{\partial{F}}{\partial{Q_i}} = 0 \,\,\,...eq.(5)$$ and $$H' = H + \frac{\partial{F}}{\partial{t}} \,\,\,...eq.(6)$$

From these relations we can solve for 2n new coordinates and momentum provided we know the generating function. Similarly, from the inverse method that is if we know the transformed equation we can seek the corresponding generating function. 

Sometimes, it needn't to be the generating function of the above type that completely solves the problem but rather be in a different form like $$F_2 = F (q,P,t)\,\,\,...eq.(7)$$ But the function should depend only on q, Q,t and not on p or P because as a whole it behaves like Lagrangian and so the derivatives of p and P should not exist. 

Thus we make the corresponding Legendre transformation such that the dP element should be canceled out,
  $$dF = f_1 dq_i + f_2 dP_i + f_3 dt \,\,\,...eq.(8)$$  
we know $$d((f_2)_iP) = P (df_2)_i + f_2 dP_i \,\,\,...eq.(9)$$
Subtracting eq.(9) from eq.(8) we get, $$dF - d((f_2)_iP) = f_1dq_i - P (df_2)_i + f_3 dt \,\,\,...eq.(10)$$  If we call $F - (f_2)_iP$ as F' then, F' is the function of only f_2, q and t. There is a special condition that $$\frac{\partial{F}}{\partial{P_i}} = f_2$$  
Let us test the condition for this new function F' by again substituting this in our initial definition eq.(1),

$$p_i\dot{q_i} - H  = P_i\dot{Q_i} - H' + \frac{dF'}{dt}$$
$\rightarrow$ 
$$p_i\dot{q_i} - H  = P_i \dot{Q_i} - H' + \frac{dF}{dt} - (f_2)_i \dot{P} - P_i\dot{(f_2)_i} \,\,\,...eq.(11)$$
If we make the further assumption that, $$(f_2)_i = \frac{\partial{F}}{\partial{P_i}} = Q_i$$
$\rightarrow$
$$p_i\dot{q_i} - H = -Q_i\dot{P_i} - H' + \frac{\partial{F}}{\partial{q_i}} \dot{q_i} + \frac{\partial{F}}{\partial{P_i}} \dot{P_i} + \frac{\partial{F}}{\partial{t}}\,\,\,...eq.(12)$$

Comparing the coefficients we get the necessary conditions, $$\frac{\partial{F}}{\partial{P_i}} = Q_i \\~\\ \frac{\partial{F}}{\partial{q_i}} = p_i  \\~\\ H' = H + \frac{\partial{F}}{\partial{t}}$$

Similarly we can derive other generating functions and corresponding relations by suitable legendre transform.  

Let us assume the third generating function as the function of p, Q, t. 
dF = f dp + g dQ + h dt
and
d(wp) = w dp + p dw 
Subtracting with the assumption $\frac{\partial{F}} {\partial{p}} = f = -w = -q$, we get,  d(F + qp) = g dQ + p dq + h dt 

Substitution in eq.(1)  where $$F_3 = F(p,Q,t) + p_iq_i$$ because subtraction or negative sign would result contradiction. 
Finally we will get, $$\frac{\partial{F}}{\partial{p_i}} = q_i \\~\\ \frac{\partial{F}}{\partial{Q_i}} = -P_i \\~\\ H' = H + \frac{\partial{F}}{\partial{t}}$$

Finally, the fourth generating function is given by, 
$$F_4 = F(p,P,t) + q_ip_i - Q_iP_i$$  
and the resultant solutions are, $$\frac{\partial{F}}{\partial{p_i}} = -q_i \\~\\ \frac{\partial{F}}{\partial{P_i}} = Q_i  \\~\\ H' = H + \frac{\partial{F}}{\partial{t}}$$  

Central force Problem - Two body problem - Part:1

Since Planetary motions plays a crucial role in our life, we would like to study about two body problems in a elaborate way. 

Let us consider two masses $m_1$ and $m_2$ at a distance $\vec{r_1}$ and $\vec{r_2}$ from the origin of the coordinate system. Let say $\vec{r}$ is the relative position vector of $m_2$ from $m_1$. The centre of mass of these two masses are described by vector $\vec{R}$ We would like to reduce this problem to one body problem, because it is easy to work with one body and its motion other than working two masses at same time. And we would like to solve this using Lagrangian principle. 

Lagrangian for a conservative system is equal to the difference between Kinetic energy and Potential Energy i.e. L = T - V where T- total kinetic Energy and V - total potential Energy. 

Kinetic energy of the system of two particles is, $$T = \frac{1}{2} m_1 \dot{\vec{r_1}}^2 + \frac{1}{2} m_2 \dot{\vec{r_2}}^2 \,\,\,...eq.(1)$$

Using the centre of mass concept, $$\dot{\vec{R}} = \frac{m_1\dot{\vec{r_1}} + m_2\dot{\vec{r_2}}}{ m_1+m_2} \,\,\,...eq.(2)$$
and $$\dot{\vec{r_1}} + \dot{\vec{r}} = \dot{\vec{r_2}} \,\,\,...eq.(3)$$  

With the help of eq.(3) and eq.(2), 
$$\dot{\vec{r_1}} = \dot{\vec{R}} - \frac{m_2}{m_1+m_2} \dot{\vec{r}} \,\,\,...eq.(4)$$
and $$\dot{\vec{r_2}} = \dot{\vec{R}} + \frac{m_1}{m_1+m_2} \dot{\vec{r}} \,\,\,..eq.(5)$$

Substituting eq.(4) and (5) in eq.(1) we get, 
$$T = \frac{1}{2} m_1 \left(\dot{\vec{R}} - \frac{m_2}{m_1+m_2} \dot{\vec{r}}\right)^2 + \frac{1}{2} m_2 \left(\dot{\vec{R}} + \frac{m_1}{m_1+m_2}\dot{\vec{r}}\right)^2$$

Doing all simplifications, we will get, $$T = \frac{1}{2} (m_1+m_2) \dot{\vec{R}}^2 + \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{r}}^2 \,\,\,...eq.(6)$$  

The first term involving the position of centre of mass doesn't play any role in further discussion, if we are considering the potential that depend on the relative distance between the masses. For, example you can always choose your origin of the coordinate system at the centre of mass position, since it just moves at constant velocity or rest. 

The reason is so simple that if you consider the two masses as a whole system, then there is no any external force acting on this system, and so the linear momentum of the system always remains constant i.e. the linear momentum of centre of mass.

Thus we finally get our Lagrangian for two particle system as,
$$L = \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{r}}^2 - V(r) \,\,\,...eq.(7)$$
taking $\frac{m_1m_2}{m_1+m_2} = \mu$ as reduced mass, the problem simplifies into simple lagrangian that looks like the lagrangian for a single particle, $$L = \frac{1}{2} \mu\dot{\vec{r}}^2 - V(r)$$  
Since central force acts only in the radial direction, the problem has spherical symmetry and so the total angular momentum is conserved. One of the coordinate can be chosen to by cyclic and the motion is confined to the plane perpendicular to angular momentum vector. All central force motions are confined to planar motion. 
  Due to the simplification of the problem, we just need two polar coordinates to completely describe the motion of the system. 

Lagrangian again simplifies in the polar coordinates as,
$$L =  \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - V(r)\,\,\,...eq.(8)$$ [we can use m for $\mu$]
$$\frac{\partial{L}}{\partial{\theta}} = 0$$ where $\theta$ is a cyclic coordinate. 
Therefore, $$\frac{d}{dt}\left(\frac{\partial{L}}{\partial\dot{\theta}}\right) = 0$$ or $$\frac{\partial{L}}{\partial{\dot{\theta}}} = const. = l \,\,\,...eq.(9)$$
$$\rightarrow\,\,\,\,\, l =  m r^2 \dot{\theta}$$  
From, this we can prove that the Areal velocity is constant in central force motion, which is exactly the Kepler's second law. 

For radial component, Euler lagrange equation gives,
$$\frac{\partial{L}}{\partial{r}} =  \frac{d}{dt} \left(\frac{\partial{L}}{\partial{\dot{r}}}\right)$$ or $$\frac{d}{dt} (m\dot{r}) = mr\dot{\theta}^2 - \frac{\partial{V}}{\partial{r}}$$  
Applying for $\dot{\theta}$ 
$$m\ddot{r} = \frac{l^2}{mr^3} - \frac{\partial{V}}{\partial{r}} \,\,\,...eq.(10)$$  multiplying on both sides by $\dot{r}$ we get.
$$m\ddot{r}\dot{r} = \left(\frac{l^2}{mr^3} - \frac{\partial{V}}{\partial{r}}\right) \frac{dr}{dt}$$
or $$\frac{d}{dt} \left(\frac{1}{2}m\dot{r}^2\right) = \frac{\partial}{\partial{r}} \left( \frac{-l^2}{2mr^2} - V\right) \frac{dr}{dt} = 0$$ Taking out the differential with repect to time, 
$$\frac{d}{dt} \left(\frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} + V(r)\right) = 0$$  
$$\rightarrow\,\,\,\,\, \frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} + V(r) = const. = E \,\,\,\,...eq.(11)$$  
Where this const. E is the total Energy of the system. 

Eq.(11) is our desired differential equation. From this equation we can derive all the result that we need to understand the system completely. 

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. 

Moment of Inertia Tensor

We found that from the mechanics of system of particles, Total kinetic energy can be split into two parts as the sum of kinetic energy of the all system of particles with its total mass at the centre of mass position plus the kinetic energy of the all the system of particles about the the centre of mass point. 

It is like splitting the motion into pure translational and pure rotational motion. Not only in mechanics, but in many places of physics, this kind of separation into purely translational and rotational is possible. 

So, it is always possible in mechanics, choosing our coordinate system at the centre of mass or at the point that is stationary in a rotational motion. 

About this stationary point, all the measurements will include only rotational quantities. 

Let us determine the angular momentum of this system of particles about this stationary point, 
$$\vec{L} = m_i \vec{r_i}\times \vec{v_i}$$ The summation convention is implied. 
Applying the known relation of $$\vec{v_i} = \vec{\omega}\times\vec{r_i}$$
Therefore, $$\vec{L} = m_i [\vec{r_i} \times (\vec{\omega}\times \vec{r_i}) ]$$
Using the cross product rule, $$\vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\cdot\vec{C})\vec{B} - (\vec{A}\cdot\vec{B})\vec{C}$$  
$$\vec{L} = m_i [r_i^2\vec{\omega} - (\vec{r_i}\cdot\vec{\omega})\vec{r_i}] \,\,\,...eq.(1)$$

Or, writing a general component of angular momentum vector in Einstein notation, $$L_a = m_i [ \omega_a r_b^2 - (r_{ib}\omega_{ib})r_a] \,\,\,...eq.(2)$$ with Einstein summation convention. 
If I consider only 3 dimensional space, then b - runs from 1 to 3 (or) x,y,z .      

From the equation of angular momentum, we can know that the angular momentum does depend not only on the angular momentum about its axis lets say \omega_a but depends on all other axis where in \omega_b , b - runs from all a,b,c (if it is 3 dimension).

If I need to take out $\omega$ then eq.(2) becomes, $$L_a = I_{ab} \omega_b$$
where $$I_{ab} = m_i (r_i^2 \delta_{ab} - r_a r_b)$$ If we want to write the mass in terms of integration, then, $$I_{ab} = \int_V \rho(\vec{r}) (r_i^2\delta_{ab} - r_ar_b) \,dV$$

Writing in terms of vectors, $$\vec{L} = I\vec{\omega}$$  
where "I" is called moment of Inertia Tensor. 

Velocity Dependent Potentials and Dissipation function

When we derive Euler Lagrange equation, we considered only simple potentials that depends only on the positional coordinates. But there are other kinds of potentials, for e.g. Potential which depends on the velocity of the particle. Let us deal with those potential to derive a general Euler Lagrange Equation.  

Let us denote velocity dependent potentials $V(q_j, \dot{q_j})$ where generalized force is defined by, $$Q_j = -\frac{\partial{V}}{\partial{q_j}} + \frac{d}{dt}\left(\frac{\partial{V}}{\partial{q_j}}\right) \,\,\,...eq.(1)$$  

But we know that, if  L = T - V the generalized force is given by, $$Q_j =  \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_j}}} - \frac{\partial{L}}{\partial{q_j}} \,\,\,...eq.(2)$$  

where L contains only the Potential that arise only from the gradient of generalized coordinates i.e. only conservative potential, and Q_j is the generalized force that doesn't include any part from the conservative forces. It includes only forces that are not derivable from a conservative potentials, Eg. Frictional forces. 

But if we could denote the frictional forces itself in a perfect differential form, there is a hope we could make it simpler looking with Euler Lagrange equations. 

If frictional forces could be written in the form, $$F^{frictional}_{j} = -k_{j}v_{j}$$

where "j" the generalized coordinate.

These kind of frictional forces can be derived from a general function U, also known as Rayleigh's dissipation function. 

It is given by, $$U = \frac{1}{2} \sum_i \sum_j k_j v_{ij}^2$$

where "i" is the summation over each particle.

From this definition, the forces are derived from the potential as, $$F^{frictional}_j = -\frac{\partial{U}}{\partial{v_j}}$$  

Then the generalized force arising from this frictional force is given by, 

$$Q_j =  \sum_i \vec{F^{frictional}_i}\cdot \frac{\partial{\vec{r_i}}}{\partial{q_j}} = - \sum_i \frac{\partial{U}}{\partial{v_i}} \frac{\partial{\vec{r_i}}}{\partial{q_j}}$$

From eq.(4) in D'Alemberts principle and derivation of Euler Lagrange equation, we know that, 

$$\frac{\partial{\vec{v_i}}}{\partial{\dot{q_j}}} = \frac{\partial{\vec{r_i}}}{\partial{q_j}}$$  

Applying it in our equation, we get, 

$$Q_j = -\sum_i \frac{\partial{U}} {\partial{v_i}} \frac{\partial{\vec{v_i}}}{\partial{\dot{q_j}}} \\~\\ = - \frac{\partial{U}}{\partial{\dot{q_j}}}$$   

Substituting in eq.(2) we get our desired result, 

$$\frac{d}{dt} \left(\frac{\partial{L}}{\partial{\dot{q_j}}}\right) - \frac{\partial{L}}{\partial{q_j}} + \frac{\partial{U}}{\partial{\dot{q_j}}} = 0 \,\,\,...eq.(3)$$

If we are initially given with L and U i.e. Lagrangian and Rayleigh dissipation function, then everything about the problem can be easily solved using the above Euler - Lagrange equation (with dissipation). 

Evoultion of Wave Function - Splitting the Box

I was struggling with the structure of Schrödinger's equation over time when we consider a real physical process. The problem can be stated in two parts as follows,

Is it possible to construct a wave function for a system with a new introduced potential from the previously known wave function solution of the same system before the introduction of the potential.  Eg: I want to get the solution of an infinite square well where the dirac delta potential is placed at the center from the usual square well problem without potential. We are not talking about entirely two things. Basically, it is a process of introducing an infinite square well and slowly putting the delta potential at the center over time.

    The second part is where we don't introduce any kind of bizarre potentials, instead we change the boundary conditions over time. It is like expanding the square well or making some regions in space as inaccessible to the wave function.

    Note:

        There is a difference between wave function going to zero at some place and the space itself becomes inaccessible to the wave function.

        One should not to be confused these things with the concept of perturbation even though it looks similar in some places. In perturbation, there is no change in boundary conditions or any introduction of completely arbitrary potentials. It is a very small change in the initial potential with same conditions maintained at

all the time. 

    To understand this, I took a series of problems with different levels in ascending order. First, we try to solve the general schrodinger equation for one dimensional infinite square well with a very little change by moving it to length "a" and "2a". Except for the limits, the solution is written in normal convention of "k" and "E" , we have $$\psi = A e^{ikx} + B e^{-ikx}$$   and applying the left, right limits, $$A e^{ika} + Be^{-ka} = 0 \,\,\,...(1)\\ A e^{i2ka} + B e^{-i2ka}\,\,\,...(2)$$ (1) gives $$A e^{i2ka}= -B$$ provided $e^{ika}$ don't become zero. $$(2) \rightarrow \,\,\, B (e^{-i2ka}-1) = 0 \\ B = 0 \,\,or\,\, e^{-2ika} = 1$$   B = 0 gives A=0 i.e. trivial solution where there is no wave function. It leaves us with $$e^{-i2ka} = 1 \\ A = -B\\ cos2ka = 1 \,\,and\,\, sin 2ka = 0$$ both conditions need to be satisfied if, $$sin 2ka = 0  \,\,and \,\,cos2ka = 1 \\ 2ka = n\pi \,\,and\,\, 2ka = 2n\pi$$ larger condition is the common one i.e. $$ka = n\pi$$ which gives us the exact solution $$\psi = A e^{ika} - A e^{-ika} = C sin 2ka$$ where "C" is again an overall constant to be determined by normalization and $ka = n\pi$ Normalization gives again the same constant [ since, $$sin^2 2kx = \frac{1 - cos2kx}{2}$$ where integration is from "a" to "2a" i.e. sin 2ka = sin ka = 0 ] i.e. $\sqrt{\frac{2}{a}}$ giving us the complete solution, $$\psi (x) = \sqrt{\frac{2}{a}} sinkx \,\,\,\,and\,\, k = \frac{n\pi}{a}$$

It was quite a bit surprise for me, that there is no any kind of reflection about the shift of the coordinate system or the shift about the coordinate syystem. Even though the wave function depends on the position, it doesn't reflect anything i.e. independent of coordinate system.

But don't think they are exactly the same in every way because now the limits have changed to give negative values for all x values in between "a" and "2a". But, it doesn't matter since the measurable quantities depend on the square of the wave function - saves the day.

    With this information, if I take the a box of length "2a" and split it into two parts "0" to "a" and "a" to "2a" with an imaginary line drawn at x = a . Actually, there is no anything physical about this problem, but I just wanted to see how the problem will evolve from one to two.

    As we have two parts, the solution is, $$\psi(x) = Ae^{ikx} + Be^{-ikx} \,\,\,\,0\leq{x}\leq{a} \\ Ce^{ikx} + De^{-ikx} \,\,\,\,{a}\leq{x}\leq{2a}$$ with the boundary condition that the first part should vanish at the left x= 0 and the right should vanish at x=2a, we have, $$A+B = 0 \\ C e^{2ika} + De^{-i2ka} = 0 \\ \rightarrow \,\,\,\, Ce^{i4ka} = -D$$

    If we seek the continuity at x = a, by comparing independent sine and cosine terms $$A - B = C - D \\ (C+D) coska = 0$$ Putting all the conditions together and working out gives, $$A = C \\ B = D \\  A= -B\\ k = \frac{n\pi}{2a}$$ which gives the wave function as, $$\psi(x) = F sin(kx)$$ After Normalizing, $$\psi(x) = \frac{1}{\sqrt{a}} sin(kx) \,\,\,\,or\\ \frac{1}{\sqrt{a}} i sin(kx) \\ k = \frac{n\pi}{2a} \,\,\,\, n= 0, \pm{1},\pm{2},..$$

    I put 'i' because it is the exact solution I got and I just wanted to show "i" doesn't affect our wave function in anyway. The physical results are the same.

    Actually we can talk a little about this 'i' . Any wave function, even though it is normalized, there is no unique form for its' structure. It is always possible to multiply it by 'i' or '-i' like we can do it in classical sense, multiplying by '1'. Hence, there is no way of telling whether your wave function solution is real or complex by just looking at it.

    This itself is a justification for why the modulus square of the wave function that all matters for any physical result (which is called the probability density) and not the value of the wave function (probability amplitude).