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.   

History of Spin

After too many attempts, I just understood the correct way to go through Quantum Mechanics. 

    

      "Don't waste time on the postulates of Quantum Mechanics". 

because for me it is just a great mathematical technique to understand the results of Nature. Understanding the reason for the technique should be handled separately only after you have some practice on it. 

       Currently, it is the spin that I concern the most. We know that the development of quantum mechanics begins from the field of spectroscopy.

       When any element is heated it produces energy in the form of light. This so called emission spectrum has distinct properties varies for different elements. By passing the light through a spectroscope [Instrument prepared to separate components of light since wavelengths of different light are different e.g.prism] its spectrum is analyzed. 

      The above is for just simple idea, but spectroscopy is really a big deal in Physics. 

      The developments in spectroscopy lead to more knowledge about the fundamental elements in Nature. The astounding result of these spectrum analysis is that they are not continuous but discrete in Nature. 

     Lets get through this..

I am heating an element and it produces light but in discrete units. But I know all these phenomena are due to proton and neutron because change in nucleus will result transmutation in element. So, it is entirely due to the electrons in the atom. From heating, we are just giving energy to this system. 

    Thus some kind of transition between electrons produce the energy and the essential point is that the transitions are discrete. Not everything is allowed in Nature. Nature is quantized.  

    This is the question that led to quantum world and  models were created to explain this. 

     But the concept of spin has to do more with the analysis of the spectrum in the presence of Electric and Magnetic fields. It has its own name respectively, Stark effect and Zeeman effect. Especially Zeeman effect arises due to magnetic properties of atom. 

     It is observed that before the field is applied there is only one line but when it is applied it splits into more than one line. This shows that magnetic field has somewhat to do with the energy levels of the electron. 

      From Electrodynamics we know that moving charges produces magnetic field and so it implies that there are some intrinsic rotational motions in the atom to produce the magnetic effects. Note that it is not the motion we are accustomed with day to day life. Only the mathematical formulation is similar to the one in classical.   

      

    Hence the concept of angular momentum aroused automatically in the quantum view. Careful analysis of the spectrum showed there is always an addition of angular momentum in terms of discrete units. And this extra term is known to be the result of spin of electrons. It is just a name given to the property of these fundamental particles. It is not the real spin. 

    

    Why it is not the real spin? 

How angular momentum plays the major role?

All these could be understood from the concept of magnetic dipole moment. 

Equations of motion

I don't know how many times I think about this idea, but each and every time I get some new perspective on its definition. And also it comes with some more questions.

I was just deriving the equations of motion from the basic definitions of mechanics that is from the definition of acceleration, $$ a = \frac{dv}{dt} \\ $$ On Integrating from initial velocity u to final velocity v in time t we get the first equation 
$$ \\ \int_0^t a \,dt = \int_u^v \,dv \\ v= u + at$$  
And velocity is rate of change of displacement,   $ \\ \frac{dx}{dt} = v = u + at \\ \int_{x_0}^x \,dx = \int_0^t u+at \,dt$  gives $$ x - x_0 = ut+ \frac{1}{2} at^2$$ and by squaring the first equation and applying the second result we get,
$$ v^2 = [u+at]^2 \\ v^2 = u^2 + a^2t^2+ 2uat \\ v^2 = u^2+ 2a[ut + \frac{1}{2}at^2] \\ v^2 = u^2 + 2a [x-x_0]$$
Thus we obtain the three fundamental equations of motion and this is the usual method of derivation. 
          But unfortunately my mind was struggling and thinking about the Taylor series and its derivation. 
  With surprise!! It just made me to realize that these equations are nothing but the simple Taylor expansion of respective variables. 
Taylor expansion of a function is,
$$ f(x) = f(x_0) + \frac{df}{dx}|_{x_0} x + \frac {d^2f}{dx^2}|_{x_0} \frac {x^2}{2!} + ...$$
Now, if we expand the velocity as function of time,
$$ v(t) = v(t_0)+ \frac{dv}{dt}|_{t_0} t + \frac {d^2v}{dt^2}|_{x_0} \frac {t^2}{2!} +...$$

Thus we get the firs equation in the general form,
$$v = u + at + \frac{a't^2}{2!} + \frac{a''t^3}{3!} + ...$$
where $a', a'',..$ are the higher derivatives of acceleration. When higher derivatives become zero this general form results into the above equation where acceleration is constant over time.

        Similarly the Taylor expansion for displacement follows as,
$$x = x_0 + \frac{dx}{dt}|_{x_0}t + \frac{d^2x}{dt^2}|_{x_0}\frac{t^2}{2} +\frac{d^3x}{dt^3}|_{x_0}\frac{t^3}{6} +...$$
As we derived in the first case, by squaring the first expansion and combining it with the second, the general derivation for the third equation of motion can be obtained.  

        Really it is astounding for me that why I didn't see this for these many days. But eventually there are questions and that goes with the understanding of the higher derivatives and the simplicity in Newton's laws. 

Question is simple, 
Why Newton's law is defined that 

$$F = ma$$ and not $$ F = ma^2$$ or $$ F = ma'+ma''+...$$}

The most probable answer is, "It is the way the law is defined."

Yes.. of course. It is the axiomatic definition. But from Taylor expansion, you will find that function at a point is determined by its neighbourhood points since derivatives depends on continuity between the points. 
             Only when the higher derivatives are negligibly small, you will arrive at simple solution. All these things itself appropriately tells us that there are intrinsic conditions in our definitions. 
             To understand those conditions, it should be tried out by changing the fundamental definitions and rules.