Monopoles - 5 - Dirac Monopoles in Quantum Mechanics - Part - 1

We know, Magnetic vector potential plays the crucial part in the Hamiltonian of an Electromagnetic system where the Hamiltonian formulation is the basis definition of transformation of equations from classical to Quantum.   
And Experimental results like Aharonov-Bohm effect make it look like the magnetic vector potential is inevitable in Quantum mechanics. Together, they state the importance of vector potential

But, if we consider the possibility of monopoles, then we may have to reject the concept of vector potential and need to redefine our Hamiltonian with new potentials, which in turn may result into contradictions with unexpected results. 

To prevent this, we reject at the beginning itself the possibility of an isolated magnetic monopole in our Universe, so that everyone can live in their happy little world with conventional equations.
But, Dirac first took a completely different approach with his new mathematical treatment along with vector potential and proposed a new magnetic monopole that can even co-exist with the vector potential. i.e.without any change in our old potential formalism. 
Moreover, the vector potential itself allows for a such particle to exist in nature without any violations. 

Let us look at that approach  from his 1931 paper..
First we introduce the wave function in the usual form as, $$ \vert\psi\rangle = Ae^{i\gamma} $$ where $\gamma$ is the function of x,y,z,t and also we take A - amplitude as the function of position and time since we take the general case. 
Now, the indeterminacy in this wave function can be regarded as the possible addition of any constant to the phase. That is equivalent to $$\psi = A e^{i\gamma + \chi} = Ae^{i\gamma} e^{\chi}$$ where $\chi$ is some constant. It doesn't change anything physical about the wave function, because we know that from superposition, the wave function will not change from the possible multiplication of any arbitrary real or complex number. 
So, you can never determine any wave function up to an arbitary constant which can be chosen arbitrarily to normalize the wave function. 

From this fact, the wave function cannot have a definite phase at all the points but can have only definite difference. All it does matter is the difference in $\gamma$ value. But we are not sure whether this difference is unique for any two arbitrary points, as there are many paths from going from one point to another. We are not even sure whether the closed integral of this phase change will vanish or not.  

Away from this, the definitions of our phase change in any sense should not give rise to ambiguity in the applications of the theory.       
First it is seen that the concept of phase doesn't change anything in the density function as, $$ \langle\psi\vert\psi\rangle = A^2 e^{i\gamma} e^{-i\gamma} = A^2 $$ which is a pure real number that doesn't depend on the value of phase (as the phase vanishes by its complex conjugate). 

But, it is no longer the case if we take two different wavefunctions.
$$ \langle\psi_m\vert\psi_n\rangle = \langle\psi_m\vert[c_1\psi_1 + c_2\psi_2 + ...+ c_m\psi_m + ...]\rangle = c_m $$
where I expanded $\psi_n$ in terms of the eigen functions $\psi_m$. 

We know from the Quantum Mechanics postulates that, $\vert{c_m}\vert^2$ gives the probability of $\psi_m$ state in $\psi_n $ state which termed by dirac as, probability of agreement of the two states. 

The limits of the integral in the bra-ket notation needn't to be from $(-\infty,\infty)$.
I think, this may be the only disadvantage in Dirac notation. The limits are not represented explicitly.

Since the integral does depend on the end points, even though the wave functions don't have any definite phase, they should have definite phase difference between two points (because the integral is a number).

The same physical argument gives us that, the change in phase round the closed path should be zero. 

Let us look at it like this by saying, 

the phase of the wave function $\psi_m$ is $\gamma_m$ and for the second $\psi_n$ is $\gamma_n$ So, the phase of $ \langle\psi_m\vert\psi_n\rangle $ is $e^{i(\gamma_n - \gamma_m)}$

When we go along the path from one point to another, there is a corresponding change in phase for each wave function respectively $\chi_m\,,\,\chi_n$. 

And so, the new phase difference at this point is $$ e^{i(\gamma'_n -\gamma'_m)} = e^{i[(\gamma_n +\chi_n)-(\gamma_m+\chi_m)]}$$ For different set of two points, the phase difference can have definite values. But, we know that from the physical fact that, if we come again at the same initial point, we should have the same probability and so the same integral value on closed path integral, i.e. $$ e^{i(\gamma'_n-\gamma'_m)} = e^{i(\gamma_n -\gamma_m)} $$ or $$ \gamma_n + \chi_n -\gamma_m-\chi_m = \gamma_n - \gamma_m \\ \rightarrow \chi_n = \chi_m $$ 
Which says that the change in phase of $\psi_m$ and $\psi_n$ should be the same and opposite around a closed path. 

Since it is a general result, it can be stated as in Dirac's paper, 
The change in phase of a wave function round any closed curve must be the same for all the wave functions. 

The change in phase doesn't talk anything about the nature of wave function or concerned with any specific system. So, the change in phase should be something a property of the dynamical system or the force field in which it moves. 

For the mathematical treatment, it was expressed the wave function as, $$ \psi = \psi_1 e^{i\beta}$$ $\psi_1$ being the usual wave function with definite phase and the uncertainty in phase is put in the factor $ e^{i\beta}$ where $beta$ is the same as $\chi$ we used in the previous. This $\beta$ having definite values at each point is not a function of x,y,z,t. (because different paths at the same point will possibly give different values of phase change). But it has definite derivatives at each point (x,y,z,t). 

We represent its derivatives as, $$ k_x =\frac{\partial\beta}{\partial{x}} \\ k_y =\frac{\partial\beta}{\partial{y}}\\k_z =\frac{\partial\beta}{\partial{z}}\\k_0 =\frac{\partial\beta}{\partial{t}}$$ In general these derivatives needn't be integrable following the condition $$ \frac{\partial\beta}{\partial{y}\partial{x}} = \frac{\partial\beta}{\partial{x}\partial{y}}$$ 

Now, using Stokes' theorem, we try to calculated the change in phase around a closed path as, $$ \oint \vec{K}\cdot\vec{dl} = \int (\nabla\times\vec{K})\cdot\vec{dS} $$ 
where the length and area element is considered in four dimensions, since K has four components. 

Here is one essential point we shouldn't assume, that is all the wave functions should have the same phase factor $e^{i\beta}$ because of the fact that all wave functions have same phase difference along a closed path. 

The reason is because only the change in phase depend on the curl of K vector. We can still change the components by the gradient of any scalar function, so that same phase difference can be obtained to be the same.
   
We can start from here in the next post.    
      

Monopoles - 4 - Thomson dipole

Let us start with a new problem that involves an imaginary magnetic monopole, producing magnetic field that follows inverse square law similar to the electrostatic case. 
As a consequence, we can exploit some more information about this system.
A combination of electric monopole and a magnetic monpole is known to be Thomson dipole. 
Here, we assume the magnetic field of the magnetic charge is equal to , $$ \vec{B} = \frac{q_m}{r^2}\hat{r}$$ Just from the Lorentz force, it is known that Magnetic force can never do work on a electric charge since it is always perpendicular to the velocity. Similarly, the electric force can never do work on magnetic charge. 
Mathematically, work done by the Magnetic field (of the magnetic charge) on electric charge is, $$ dW = \int \vec{F_{em}}\cdot\vec{dl} = q_e \int(\frac{\vec{v}}{c}\times\vec{B})\cdot\vec{dl} \\~\\ = q_e \int (\vec{v}\times\vec{B})\cdot \vec{v}\, dt = 0 $$
Similarly, work done by Electric field on Magnetic charge is always zero, $$dW = \int\vec{F_{me}} \cdot\vec{dl} = \frac{-\vec{v}}{c}\times\vec{E}\cdot\vec{v}\, dt = 0 $$ 
From this fact, we can be sure that the particles will not gain any kinetic energy from Work-Energy theorem. So, the velocity of the particles should be a constant. 

Other than this, you can also determine the basic constants of motion you can find in any mechanics problem i.e.Total Energy, total linear and angular momentum. 

Except now, the definition of momentum is upgraded to Electromagnetic momentum which has wider applicability than the former definition.    

There are  some astonishing results about this system because of the above fact. 
First, if you solve the system completely you will find out that the motion of the particle will lie only along the surface of cone centered about an axis passing through the direction of a constant vector quantity i.e. $$\vec{L'} = (\vec{r}\times{m\vec{v}}) - q_eq_m\hat{r} $$. 

where the spherical coordinates are measured from where the z-axis lie along the direction of vector Q.

And the Second which is very peculiar that, 
this system has some intrinsic angular momentum stored in its fields which is independent of the distance between the charges. This in fact happens to be the basis for the idea of quantization of the Electric and Magnetic charge in Quantum mechanics - one of the famous ideas of Dirac.

It is derived as follow, 
The magnetic field, $$ \vec{B} = \frac{q_m}{r^3}\vec{r}$$ and the electric field is placed at a distance "d" from this origin. So,
the Electric field, $$ \vec{E} = \frac{q_e}{r'^3} \vec{r'} $$ 
Using vector addition rule, $$ \vec{r'} = \vec{r} + \vec{d} $$ where $\vec{d}$ is directed from electric charge to magnetic charge. 
Henceforth, $$ \vec{E} = \frac{q_e(\vec{r}+\vec{d})}{(r^2 + d^2+ 2\,r\,d\, cos\theta)^{3/2}} $$ Then linear momentum density stored in the fields is calculated as,  $$ \vec{P} = \frac{1}{4\pi{c}}\vec{E}\times\vec{B}= \frac{q_eq_m}{4\pi{c}}\frac{((\vec{r}+\vec{d})\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}\\~\\ = \frac{q_eq_m}{4\pi{c}}\frac{(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$ 
Now, finding the angular momentum density stored in the fields,
$$ \vec{l} = \vec{r}\times\vec{P} = \frac{q_eq_m}{4\pi{c}}\frac{\vec{r}\times(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$

Using the vector relation, $$ \vec{r}\times(\vec{d}\times\vec{r}) = \vec{d} (\vec{r}\cdot\vec{r}) - \vec{r} (\vec{r}\cdot\vec{d}) $$
$$ \vec{l} = \frac{q_eq_m}{4\pi{c}}\frac{(r^2 \vec{d} - r\,d\,cos\theta\, \vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$

To get the total angular momentum, we integrate this all over the space using spherical coordinate system where we assume $r\,cos\theta$ lie along the direction $\vec{d}$ and we can split the vector into its components as $\vec{r} = r cos\theta \hat{d} + \,\,components\,\, perpendicular\,\, to\,\, the \,\,direction\,\,\hat{d}$$ 
so that  $$ \vec{L} = \frac{q_eq_m}{4\pi{c}}\int_{space}\frac{(r^2 \,d - r^2\,d\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} d\tau $$
The other perpendicular components will integrate to the value zero. $$ \vec{L} = \frac{q_eq_md}{4\pi{c}}\int_{space}\frac{(r^2  - r^2\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} r^2 sin\theta \,d\theta \,d\phi\, dr $$
On integration we will get, $$ \vec{L} = \frac{q_eq_m}{c} $$ 

When we go to quantum mechanics, we have studied the angular momentum is quantized in terms $$ L = n\frac{\hbar}{2} $$
Comparing the results we get, $$ L = \frac{q_eq_m}{c} = n\frac{\hbar}{2} $$
Therefore, $$ \frac{2q_eq_m}{\hbar{c}} = Integer $$ 

which is the exact result obtained by Dirac. From this condition, even if one magnetic charge exists in nature, it would imply the quantization of all the electric charges in the Universe. 

Monopoles - 3 - Consequence of Duality transformation

The major consequence of this duality transformation is that, you can never say it for sure that whether any charged particle in Nature has only electric charge or magnetic charge or both. 
For example, let us say there are 3 planets separated very far from each other  with 3 different kind of aliens, where the laws of physics are the same and so the Maxwell's equations are equally applicable anywhere in these 3 planets. 

They will predict exactly the same results for any Electromagnetic phenomena in their universe using the common Maxwell's equations. But they needn't to have the same form. If they vary their definition of Electric and magnetic fields according to duality transformation, there is no way finding which one is true. (It is not correct use the word "true" - after all their definition are different but they will conclude the same results).


It is a possibility for the first one (let us say humans are the first type of aliens) to describe any EM phenomena with our usual definition of Electric and Magnetic field where $q^m = 0$ magnetic monopole charge is zero, and the second planet is our inverse where they define only the magnetic charge with no electric charge by choosing $q^e = 0$
Unlike it so happens that, the third Planet define their electron with both electric and magnetic charge!

So, let us leave the first two planets and go to the third planet, where we will try to get some intrinsic physical understanding of their definitions.  

In this planet, we will first take two positive electric charges with respect to our conventional Maxwell's equations $Q_1$ and $Q_2$. 
The electrostatic repulsion force is given by coulomb's law as,
$$\vec{F_{21}}=\frac{Q_1Q_2}{r^2}\hat{r_{12}}$$
But, if the aliens define it in a such a way that it has both the electric and magnetic charge as, $$ Q_1 = q_1^e + q_1^m $$ and $$ Q_2 = q_2^e + q_2^m $$
Then the two charges as we study in Electrostatics and Magneto statics (no changing Electric or Magnetic fields), the force on charge 2 due to charge 1 is given by, $$ F_{Q_2Q_1} = q_2^e \left[\vec{E} + \frac{(\vec{v}\times\vec{B})}{c}\right] + q_2^m \left[ \vec{B} - \frac{(\vec{v}\times\vec{E})}{c}\right] $$

Since we are working with static conditions, $\vec{v} = 0$.
So, $$F_{Q_2Q_1} = q_2^e\vec{E} + q_2^m\vec{B}$$ where E and B due to $Q_1$ is given by, 
$$ \vec{E} = \frac{q_1^e}{r^2}\hat{r_{12}}$$ and $$ \vec{B} = \frac{q_1^m}{r^2} \hat{r_{12}}$$ because, now we just consider the problem as the combination of Electric and Magnetic charge placed closed together at the same point.

The Net force, $$ F_{Q_2Q_1} = q_2^e\frac{q_1^e}{r^2}\hat{r_{12}} + q_2^m \frac{q_1^m}{r^2}\hat{r_{12}}$$
or simply, $$ F_{21} = \frac{(q_2^eq_1^e+q_2^mq_1^m)}{r^2}\hat{r_{12}} $$ Since, they both direct along the same direction, we will just see some Net force acting as a repulsion force. 

So, you will always see the same result independent of the assigned electric or magnetic charge to the charged particle. Physically observable results are invariant under different definitions. 

Then, how do we decide the truth? what we really mean by a magnetic monopole?

The definition Magnetic monopole is,
Given the usual conventional Maxwell's equations, where there is only Electric charge, we haven't found any particle in Nature with pure magnetic charge i.e.the particle that transformed with $\frac{\pi}{2}$ angle in duality transformation equations.  

But, still it doesn't answer whether I have an Electromagnetic charge or pure electric charge in my hand!! 

If I say, I have pure electric charge and expect to find in Nature a new particle with pure magnetic charge, then I can also expect for another particle with both Electric and Magnetic charge (i.e. Electromagnetic charge).

After all there is no any kind of specification about the quantization or anything about the charge in Maxwell's equations. It can just assume any arbitrary value of unlike the reality where it can have only discrete values (also in energy, angular momentum, etc.).

Thus, the role of Quantum Mechanics is inevitable when you talk about any subatomic particle in reality. It is the reason why, subsequent development about Magnetic monopoles were first made by Paul Dirac with his new concept of Dirac string.

Monopoles - 2 - Duality Transformation

We can derive, how an arbitrary vector transforms under the rotation of coordinate system. For example, if a point in x-y plane is given by P(x,y). The same point in a rotated coordinate frame (conventionally we take - anticlockwise as a positive angle with respect to x axis i.e. angle $\alpha$). The new coordinates can be denoted as P(x',y').

If we want to go from one system to another, the relation between old and new coordinates axes is imminent. It can be easily verified these relations, x' = x cos$\alpha$ (projection of old x-axis on new x'-axis) + y sin$\alpha$ (projection of old y-axis with new x'-axis)  and y' = x cos(90+$\alpha$) ( projection of old x-axis on new y'-axis) + y cos$\alpha$ ( projection of old y-axis on new y'-axis).

Simply they are written as, $$ x' = x cos\alpha + y sin\alpha \\ y' = -xsin\alpha + y cos\alpha $$
where x,y are measured in same units. In a similar way, Electric field and Magnetic field is the only thing you need to know, when you are dealing with Electrodynamics, which is analogues to our usual coordinate system. 

Instead of any point, Any EM phenomena can be pointed in a plane as a point where we can put Electric field on the x-axis and Magnetic field on the y-axis. 

Once we made this analogy, all the equations and condition we derive for coordinate axes can be transferred here with careful analysis. Now, we just need  the above coordinate rotation property where we change variables (x,y) $\rightarrow$ (E,B)
E,B should be in the same units. So, we prefer Gaussian system. In SI we just need to use "cB" instead of B where "c" is the speed of light used for pure dimensional reasons.

In our new system, the rotation of coordinate system is given by, $$ E' = E cos\alpha +  B sin\alpha \\ B' = -E sin\alpha + B cos\alpha $$ with  the corresponding transformation of charge densities $$ \rho_e' = \rho_e cos\alpha + rho_m sin\alpha \\ \rho_m' = -\rho_e sin\alpha + \rho_m cos\alpha $$ 
This transformation specifically known as Duality transformation. 

As in the previous case, Rotation of coordinate system doesn't change any physical fact about the location of the point, Maxwell's equations are invariant under this duality transformation. We can check it as follows, 
Maxwell's equations before transformation,


$$\nabla\cdot \vec{E} = 4\pi{\rho_e} \\ \nabla\cdot\vec{B} = 4\pi\rho_m \\ \nabla\times \vec{E} = -\frac{4\pi}{c}\vec{J_m}- \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \frac{4\pi}{c}\vec{J_e}+\frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}}$$

After the transformation, 

(1)
$$ \nabla\cdot\vec{E'} = (\nabla\cdot\vec{E}) cos\alpha + (\nabla\cdot\vec{B}) sin\alpha = 4\pi\rho_e cos\alpha + 4\pi\rho_m sin\alpha = 4\pi\rho'_e $$  

(2)$$ \nabla\cdot\vec{B'} = (\nabla\cdot\vec{B}) cos\alpha - (\nabla\cdot\vec{E}) sin\alpha = 4\pi\rho_m cos\alpha - 4\pi\rho_e sin\alpha = 4\pi\rho'_m $$ 

(3) $$ \nabla \times \vec{E'} = (\nabla\times\vec{E}) cos\alpha + (\nabla\times\vec{B}) sin\alpha \\~\\= - \frac{4\pi}{c} \vec {J_m} cos\alpha + \frac{4\pi}{c} \vec{J_e} sin\alpha - \frac{1}{c} \frac{\partial{\vec{B}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}} sin\alpha \\~\\= - \frac{4\pi}{c} \vec{J_m'} - \frac{1}{c}\frac{\partial\vec{B'}}{\partial{t}} $$

(4)    
$$\nabla \times \vec{B'} = (\nabla\times\vec{B}) cos\alpha - (\nabla\times\vec{E}) sin\alpha \\~\\= \frac{4\pi}{c} \vec {J_e} cos\alpha + \frac{4\pi}{c} \vec{J_m} sin\alpha + \frac{1}{c} \frac{\partial{\vec{E}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} sin\alpha \\~\\=  \frac{4\pi}{c} \vec{J_e'} + \frac{1}{c}\frac{\partial\vec{E'}}{\partial{t}}$$ 

Also, the Lorentz force, 
(5)
$$ F' = q_e' \left[ \vec{E'} + \frac{(\vec{v}\times\vec{B'})}{c}\right] + q_m' \left[\vec{B'} - \frac{(\vec{v}\times\vec{E'})}{c} \right] \\~\\ = (q_e cos\alpha + q_m sin\alpha) \left[\vec{E} cos\alpha + \vec{B} sin\alpha + \frac{(\vec{v} \times \vec{B}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{E}) sin\alpha}{c}\right] +\\~\\ (q_m cos\alpha - q_e sin\alpha ) \left[ \vec{B} cos\alpha - \vec{E} sin\alpha - \frac{(\vec{v}\times\vec{E}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{B}) sin\alpha}{c}\right] $$

After doing the arithmetic manipulations (8 terms will cancel out), the remaining 8 terms are, $$ F' = q_e\vec{E} cos^2\alpha + q_m\vec{B} sin^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c} cos^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} sin^2 \alpha +\\~\\ q_e \vec{E} sin^2\alpha + q_m \vec{B} cos^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c}sin^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} cos^2 \alpha \\~\\ = q_e \left[\vec{E} + \frac{(\vec{v}\times\vec{B})}{c}\right] + q_m \left[ \vec{B} - \frac{(\vec{v}\times\vec{E})}{c}\right] = F $$

Thus, we proved all the four Maxwell's equations with the Lorentz force is invariant under duality transformation. 

We should note that, the angle $\alpha$ can vary arbitrarily. There is no any restriction to the values of $\alpha$ in the classical sense.

We will see the consequence in the next post. 

Monopoles -1 - Introduction

I just want to start from the basics where the idea of mono poles come into play in Classical Electrodynamics. We can straightly start from Maxwell's equations given by, $$ \nabla\cdot \vec{E} = \frac{\rho_e}{\epsilon_0} \\ \nabla\cdot\vec{B} = 0 \\ \nabla\times \vec{E} = \frac{-\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \mu_0\vec{J_e}+\mu_0\epsilon_0\frac{\partial{\vec{E}}}{\partial{t}} $$ with conventional notation of charge and current density. 
These equations will transform into a symmetrical set of equations in vacuum where there is no charge or current as, $$ \nabla\cdot\vec{E} = 0 \\ \nabla\cdot\vec{B} = 0 \\ \nabla\times \vec{E} = \frac{-\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \mu_e\epsilon_0 \frac{\partial{\vec{E}}}{\partial{t}} $$
We don't need to put much attention towards the constant factors that shows on the front. But, these are just a matter of unit system. If you take Gaussian system, all these complexities will disappear where E and B will be measured in same units. 

From this symmetry, it will arise a question whether we can prevail this symmetry even when charges and currents are present. 

In a pure mathematical perspective, Maxwell's equations are symmetrical when it is introduced magnetic charges and currents. The new Maxwell's equations are given by, 
$$\nabla\cdot \vec{E} = \frac{\rho_e}{\epsilon_0} \\ \nabla\cdot\vec{B} = \mu_0\rho_m \\ \nabla\times \vec{E} = -\mu_0\vec{J_m}- \frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \mu_0\vec{J_e}+\mu_0\epsilon_0\frac{\partial{\vec{E}}}{\partial{t}}$$

From this symmetry, mathematically we can never differentiate Electric fields from Magnetic fields. 
The equations are invariant when you make a transformation such that $$ \vec{E} \rightarrow \vec{B} \\ \vec{B} \rightarrow -\mu_0\epsilon_0\vec{E} $$

It shows that, if there is an alternate universe where the Electric and Magnetic fields are related to the Electric and Magnetic fields in our universe in such a way as above transformation relation then both observers will explain the same result of Physics from their Maxwell's equations. 

From this fact, it can be deduced that the laws of Nature (from Maxwell's equations) does allow any stable particle with pure magnetic charge or the particle with both Electric and Magnetic charge. It will not affect our Mathematical formalism in anyway.

For the first time, I believed in its existence, but when I did learn this next line - I really got confused. 

It was mentioned that, from Duality transformation between Electric and Magnetic fields, we can never say whether an electron has electric charge or magnetic charge or both. It is just a convention, not a condition that electron should have electric charge. 

So, which transformation I should use? 
What should I choose - whether electric or magnetic or both for electrons? (to make the Maxwell's equations look more symmetric). 
Why should I expect and search specifically for magnetic mono poles with pure magnetic charge? How do I confirm?

To answer these, probably I should do elaborately on the scale transformation with the analogue of coordinate transformation rules. 

For any future reference, in a way to make the equations in much simpler form, we will use Gaussian system of Units instead of SI units. 


$$\nabla\cdot \vec{E} = 4\pi{\rho_e} \\ \nabla\cdot\vec{B} = 4\pi\rho_m \\ \nabla\times \vec{E} = -\frac{4\pi}{c}\vec{J_m}- \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \frac{4\pi}{c}\vec{J_e}+\frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}}$$

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. 

Effect of magnetic field on Atomic orbits

Effect of magnetic field in atomic level can be quantified in a classical level with some assumptions such as, the atomic orbit is circular and electron revolves around the nucleus at radius R and the current produced is assumed to steady. 
The current is given by, $$ I = \frac{q}{t} = \frac{-ev}{2\pi{R}} $$ where $$ T = \frac{2\pi}{\omega} = {2\pi{R}}{v} \\~\\ v = R\omega $$ 
The orbital dipole moment of this configuration is given by, $$ \vec{m} = I \vec{a} = \frac{-ev}{2\pi{R}}\pi{R^2} \hat{z} = \frac{-evR}{2}\hat{z} $$ where $\hat{z} $ points in the direction perpendicular to the area of the loop when current flows by the usual right hand thumb rule direction. When it is placed in the magnetic field, this dipole moment experiences a torque which tries to align it along the magnetic field direction. 
Without magnetic field, there is only electrostatic interaction, therefore, the force equation gives, $$ \frac{e^2}{4\pi\epsilon_0R^2} = \frac{m_ev^2}{R}$$ 
If suppose we assume the magnetic field is in the direction of $\hat{z} $ , then the centripetal force can be written as, $$ \frac{e^2}{4\pi\epsilon_0R^2} + ev'B = \frac{m_e v'^2}{R} $$ 
where v' is the new velocity. If we assume $v'\simeq v $ then, $$ ev'B = \frac{m_e(v'^2 -v^2)}{R} = \frac{m_e}{R} (v'+v)(v'-v) $$
then, $$ v'-v = \frac{eRB}{2m_e} = \delta{v} $$ 
e,R,B, m are all positive quantities. So, the electron will speed up when the magnetic field is turned on. Similarly, the change in orbital speed develops a change in magnetic moment by, $$\delta\vec{m} = \frac{-e\delta{v} R} \hat{z} = \frac{-e^2R^2}{4m_e} \vec{B} $$ which directed in the opposite direction of applied magnetic field. 

That is it! But all this proved to be wrong with quantum mechanics where better explanation is given!   

Dipole Moment

     It is very important to understand the mathematical beauty of the terms occurring in the Multipole expansion. Since higher order terms vanish faster than the monopole, dipole terms, they are important only when the need of the accuracy is high. 
     As a consequence, in most of the common problems, dipole terms plays the most significant role after the monopole term . 
    
    The dipole term in the expansion is , 
$$ V(\vec{r})_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int_{V'} r'cos\theta' \rho(\vec{r'}) \,dV'  ...\ldots eq.(1)$$
To make the integrand a vector quantity, we know that
 $\hat{r} \cdot \vec{r'} = r' cos\theta' $
and so the dipole term becomes, $$ V(\vec{r})_{dipole} = \frac{\hat{r}}{4\pi\epsilon_0 r^2}\cdot \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV'$$ Terms they depend only on r' is separated and called as dipole moment, $$ \vec{p} = \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV'         ...\ldots eq.(2)$$
and the dipole potential becomes, $$ V_{dipole}(\vec{r}) = \frac {\vec{p}\cdot \hat{r}}{4\pi\epsilon_0 r^2}           ...\ldots eq.(3)$$ From eq.(2) it is known that the dipole moment defined only in terms of r' and it implies that, 'dipole moment of a volume charge depends only on it distribution'.
For point charges, $$ \vec{p} = \sum_{i=1}^N q_i \vec{r'_i}$$ and for a physical dipole it becomes, $ \vec{p} = q(\vec{r'_+} - \vec{r'_-}) = q\vec{d}$ where d is the vector from -q to +q.
    But we should remember that dipole moment doesn't mean there should be only two charges. Our definition is general for any charge distribution. It so happens the physical dipole has similar kind of representation. It is always possible to ask for the dipole moment of any number of charges e.g. three charges in a triangle.
    Some properties of the dipole moment are, Change in coordinate system usually changes the dipole moment except when the total charge is zero. And if we place a physical dipole in a uniform Electric field E, it will experiences a torque and if it is non-uniform it will experience an additional force other than the torque given by, $  \vec{F} = (\vec{p}\cdot\nabla)\vec{E} $
   Each problem will give more insight. 
Similar kind of multipole expansion for a vector potential reveals that the dipole term for a vector potential as,
$$ \vec{A_{dipole}(\vec{r})} = \frac{\mu_0}{4\pi} \frac {\vec{m}\times\hat{r}}{r^2} $$ where $\vec{m}$ is the magnetic dipole moment $$ \vec{m} = I \int \,d{\vec{a}} = I\vec{a} $$
'a' is the area enclosed by the loop and 'I' is the current. Magnetic dipole moment is always independent of the coordinate system since they don't play any role. 
   As we did in Electric dipole, certain properties for a magnetic dipole are obtained and are,
In a uniform field, the net force on any loop is zero. In a non-uniform magnetic field, an infinitesimal loop of dipole moment 'm' will experience a force, $$ \vec{F} = \nabla (\vec{m}\cdot\vec{B})$$ Thus, the concept of polarization explains the newer ideas namely, bound charges and bound currents. 

Multipole Expansion and its symmetry with common dipole

       Concept of dipole - You may ask why do we need to study about these dipoles, quadruples or etc. ?

      All of the objects around us are neutral in Nature but we know that everything is made of positive and negative charges. That is it... 

      A system with total charge zero but having separate positive and negative charge is what we need to study. You can think of quadrupole, octopole, etc. Simplest of those system is a dipole.

But the real mathematical definition of dipole is not made from this physical fact. It is a completely mathematical abstraction derived from the so called Multipole expansion. 

Note: To get a feel, analyze the following sentence,
A combination of three or more or any volume distribution of charges possess dipole moment from its definition.

All our definitions are more general. It has nothing to do with dipole or quadrupole that we used to imagine in our lower classes. This derivation is applicable for all 1/r potential.  

     For general discussion, let us consider the potential for an arbitrary volume charge distribution $\rho(r')$ at a point r from the origin and r' is the distance to the source from origin. 
Coulomb potential $$ V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int_{V'} \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|} \,dV' \ldots.. eq.(1)$$  but we know that from the vector addition,$$|\vec{r}-\vec{r'}|=R=(r^2-2\vec{r}\cdotp\vec{r'}+r'^2)^{\frac{1}{2}} = r \left(1-2\frac{\hat{r} \cdotp\vec{r'}}{r}+\frac{r'^2}{r^2}\right)^{\frac{1}{2}}$$
Hence $$ \frac{1}{R} = \frac{1}{r} \left[\frac{1}{(1-2\frac{\hat{r}\cdotp\vec{r'}}{r}+\frac{r'^2}{r^2})^{\frac{1}{2}}}\right].... eq.(2)$$
From special functions, the generating function of Legendre polynomials are given by,
$$\frac{1}{(1-2xz+z^2)^\frac{1}{2}} = \sum_{n=0}^\infty P_n(x) z^n \ldots.. eq.(3) \\~\\ \\~\\  for -1\leq x \leq1  and   |z|<1$$ 
In eq.(2) we know that $\hat{r}\cdotp\vec{r'} = r' cos\theta$ where $-1 \leq cos\theta \leq 1$ and we assumed here that source charges are near to origin and the potential is comparatively calculated very far from the origin and so, $r>>r' or \frac{r'}{r}<< 1$.

      Thus comparing eq(2) and (3) we get the general expansion of 1/R in terms of legendre polynomial as, $$\frac{1}{R} = \frac{1}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \ldots.. eq.(4)$$
Substituting this at our primary eq(1) we get, $$ V(r) = \frac{1}{4\pi {\epsilon}_0} \int_{V'} \frac{\rho(\vec{r'})}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \,dV' \ldots... eq.(5)$$which is the general expansion for multipole expansion.  

      But what else this expansion tells us..? 

      I think it could be examined by looking at the first few terms of the expansion of eq.(5).. $$ V(\vec{r})= \frac{1}{4\pi\epsilon_0} \left[ \frac{1}{r} \int_{V'} \rho(\vec{r'}) \,dV' + \frac{1}{r^2} \int_{V'} r' cos\theta' \rho(\vec{r'}) \,dV' + \\~\\ \frac{1}{r^3} \int_{V'} r'^2 \left(\frac{3}{2}cos^2\theta'-\frac{1}{2}\right) \rho(\vec{r'}) \,dV' + \ldots.. \right] \ldots.. eq.(6)$$

   The first term is just the familiar Electric monopole term from the definition. Where else the second term is the dipole and the third is the quadrupole term and so on. The higher order terms are useful for better approximations. 

  [We know it is a dipole and the quadrupole term because it is possible to arrive at the same resultant potential by separately taking the case of a single dipole or quadruple system]. 

  But wait.. It is quite mysterious because in the whole derivation I haven't used any distinct physical input about dipoles or quadrupoles or octopoles, etc. Then how it comes into play automatically? What is the physical significance and from where it arise into this picture?

 I think the reason maybe as simple as follows, 
 1 = 1 + [1 - 1] + [2 - 2] + [4 - 4] + ... 
The reason for the non existance of tripole term in the expansion is due to its lack of physical symmetry in its distribution [because dipole moments follows vector addition and it plays the role as the integrand].

It looks there is a much more mathematical connection than anticipated with the physical structure of these multi poles in Nature.   

Self working Light Bulb

This idea did somehow strike to me a few days ago. In my house we often used to charge the emergency light. I have been studying about Electricity and Magnetism chapter.

         

          I thought for a while to get a good way to solve this problem of charging. I don’t want to spend my money on this small light. I thought for a good physics way to get light with the lowest cost. But I finally found a light not only with lowest cost but also can give light for a long time.

         

The idea is very simple. It just deals with the faraday’s law of inductive effect that is changing magnetic field produces Electromotive force in a closed loop. But all we need is a strong magnet and a coil of many closed loop wires and a stand.

         

          It is just like a simple pendulum that is oscillating inside a closed loop. The model will be similar to the below diagram,

         

          The continuously changing magnetic field due to the magnet induces current in the closed loop. To get a maximum current, the magnet should have maximum strength and the coil should have maximum closed loops.

          To get rid of the resistance effect from the environment, the whole setup can be brought into a box of vacuum inside. So the external effects can be neglected. But this will not behave as a 100% efficient light source since there are resistive effects inside the closed loop itself. And there is resistance present in it and continuous current will heat up the coil and the resistance itself will increase with respect to temperature.

         

         The current is Alternating Current by means of the direction change in each period of the oscillation. But it will serve as a good light source.    

Electric Potential

         I spend little time on thinking about Electric Potential while reading about Electrodynamics. It is really a great amazement the concept of Field and Potential is just created for the sake of Mathematical Simplicity. But it was realized later only that the field concept is a real Phenomenon.  

Electric Potential is defined as the work required per unit charge to move it in some linear path.

The definition and the sign of the potential functions are based only on our conventional sign systems. Eg. The potential of a positive charge is chosen to be a positive quantity with respect to the reference frame of point at infinity.

It is just our convention to use that the potential of positive charge is hill and the potential of a negative charge is a valley. But it can be chosen alternatively if we desire. For the sake of conventional mathematics we chose some sign conventions. These sign conventions also enter in Thermodynamics.

In mathematical form Potential difference between two points is defined as

Va – Vb = - ∫ E . dl

where E is Electric field vector

dl is linear integral path vector

V is called the Potential. The integral is from point “a” to point “b”.

The interesting relation between the potential function and the Electric field is that they are related in the following way

E = - ∂V/∂r

or                                            E  = - ΔV

From this equation we can simply deduce the following result that

if Electric field in any space is zero then it implies the potential at those points is const. And that is why equipotential surfaces are always perpendicular to the direction of the Electric field. If it is not then it will be contradictory to the above Equations.

But the great thing about the potential is, it is a scalar quantity. A scalar quantity in Nature simply turns out to be vector quantity in Nature by a simple mathematical formulation. Isn’t it just great?

It says that though the potential is scalar one, it always follows some rule from the Nature itself. And the rule is just expressed as the “gradient” operator “Δ” in mathematics.

This is the reason why Mathematics serves as the language for Physics.  

Electric flux

Yesterday I watched some of the videos of Walter Lewin lectures on Electricity and Magnetism. He is really amazing especially when he is doing experiments. Thought the concepts are a familiar one it gave me a good time pass. I spend a little time on thinking about Electrostatics.

We need to know that Electric field lines are not real. It is just a simple representation invented by Michael Faraday to visualize Electric field. We should not simply accept it as the real fact.

Electric flux is the measure of the number of field lines passing through the surface that is chosen. But it is not the number of the absolute number of field lines.

The electric flux “ϕ” over a surface S is therefore given by the surface integral:

ϕ = ∬ E. ds

       s

where “E” is the electric field and “ds” is a differential area on the surface S with the normally outward facing direction.

And the electric flux around the closed surface “S” is given by:

ϕ = ∯ E. ds = q / ϵ

But be careful that it is not absolute that, there is exactly 

“q / ϵ” electric flux around a point charge “q” [In SI unit system].

Electric flux is not really the measurement of the number of field lines around a charge. It is just a calculating tool of Electric strength on the basis of proportionality arguments.

We know that we can draw infinitely many electric field lines as much as we want. But for the sake of simplicity in our calculations, we can’t use “infinite” as a number.

[Note: We need to be very careful that “∞” is not a number. It is just a symbol given for concepts.

It is like instead of saying “Newton’s Laws of Motion”, if I made a code “£” in a book, it says that where ever I use the symbol “£” it really means “Newton’s laws of Motion”.

In the same manner, the symbol “∞” is also used just to represent a Mathematical and Physical concept. The “∞” symbol just says that “it gets bigger and bigger where there is no end”. Simply it means “no limit or end”.]  

So we made a proportionality factor that is related with the number of field lines. Though the number of lines is infinite it is a proportional and dependent quantity.  

It was defined the Electric flux as it is proportional to the number of field lines.

It is already known from the definition that charge “q” is also proportional to the number of field lines.

Therefore we made calculations from the conclusion; Electric flux is proportional to “q”

It is first formulated by Carl Freidrich Gauss that is

The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface.

If the number of field lines was found out to be increasing that implies the charge “q” inside the closed surface is increased.

In SI units the proportionality constant is taken to be “1/ϵ” and so the electric flux  

Thus we are just finding the proportional increase in Electric flux from “q/ϵ” but it is not the measure of the absolute electric flux around the space of the charge “q”. 

Contemplation of the positive and the negative

Charge comes in two varieties, which we call “positive” and “negative”.

The reason we called it as “positive” and “negative” because their effect tends to cancel each other naturally. Initially there were no charges on the glass or ebonite. When we rubbed the glass with silk and ebonite with fur they get charged. And if you again touch the positive glass rod with the negative ebonite rod [Note that “positive” and “negative” is chosen by the humans and not by the nature], they cancel their effects and comes to neutral which is the initial condition. So, the fact is that positive and negative charges occur in exactly equal amounts in all matter, so that their effects are almost completely neutralized.

What if the two kinds of charge did not tend to cancel?

Were it not for this, we would be subjected to enormous electric shocks from all of the objects in Nature.    

Thus we assumed that the electric charges are there already in these objects (glass, ebonite, silk, fur, etc).

It shows their properties [effects] while rubbing with other objects.

Question and think:

Why did we choose the charge on the glass as “positive” and the charge on the ebonite as “negative”?

What will happen if I chose the charge on the glass as “negative” and the charge on the ebonite as “positive”?

Maybe in some other alien planet, people have been calling the charge on the glass is “negative” when rubbed with silk and ebonite is “positive” when rubbed with fur. 

Maybe they will be studying that “the protons in the nucleus of an atom have “negative” charge and the electrons surrounding the nucleus have “positive” charge.

But it will not make any problems in their physics. Physics holds true in every place in this universe.

No matter whether protons have positive charge or negative charge, it doesn’t cause any problem with physics. The most important thing is that protons should have the opposite charge of the electrons because these things are just the matter of sign. Signs are the matter of our choice. Nature doesn’t have any positive or negative. It has only the effect of “opposite”.

If you chose one as “positive” then the other is “negative” or if you chose one as “negative” then the other should be “positive” [It is what the effect of “opposite”].

It is important to be aware of this sign confusion because the same “positive” and the “negative” confusion will come, when you are studying about the direction of the

conventional current.  

Statement:

We observed,

 “Negative” ebonite rod repels another “Negative” ebonite rod.

            “Negative” ebonite rod repels the charge on the “amber” that is rubbed with fur.

So we said the charge on the “amber” is “negative” when rubbed with fur. Thus we choose “positive” and “negative”.

But why it should be like that?

Think:

            This may seem too obvious to the opposite comment of the statement of choosing “positive” and “negative”, but I want to contemplate other possibilities also.

What if there were 8 or 10 or infinite different species of charge?

                         

Instead of the above statement, we can also think it in another way,

Say the charge on the ebonite as “charge A” and the charge on the amber as “charge B”

and the statement of “positive” and “negative” changes to,

“Charge A” repels “charge A” and

“Charge A” repels “charge B”.

But now we don’t know whether the “charge A” and “charge B” is same. It can also be different type of charges.

So,

Why it should be only the positive and the negative electric charge? 

   

There are also the possibilities of 8 or 10 or 100 or 1000 or infinite types of different charges like charge A, charge B, charge C, charge D, charge α, charge β, charge 1, charge 2, charge 100, charge 1000, etc. 

         

Think about it.