Path Integral formulation - Part - 3 - Free Particle and Schrodinger's time evolution equation

Of  course, for any problem all the paths are not going to be counted one by one, but it will be used a simple plausible way.
To understand how this works, it is conventional to start with the most general free particle problem whose Lagrangian is \[L = \frac{m\dot{x}^2}{2}\,\,\tag{2.1}\]
The technique is to split the action into two parts of which one of them is the classical action and the other is treated as the variational part. For this, it is defined the arbitrary path as \[x(t) = x_c(t)+y(t)\,\,\,\tag{2.2}\] where $x_c(t)$ represents the actual classical path. Substituting this in our Lagrangian and expanding it in terms of Taylor expansion, \[L(\dot{x}) = L(\dot{x}_c(t)+\dot{y}(t)) = L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\,\tag{2.3}\] The expansion is exact since L is quadratic in $\dot{x}$. Thus, it is possible to write the action integral as, \[S = \int_{t_1}^{t_2} \,dt \left(L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\right)\,\,\tag{2.4}\]
The first term can be denoted as,
\[ S_c = \int_{t_1}^{t_2}dt\,L(\dot{x}_c)\,\,\tag{2.5}\]
Using (2.1)
\[\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}}\vert_{\dot{x}_c} = mass = const.\,\tag{2.6}\]
and using integral by parts
\[\int_{t_1}^{t_2}dt\,\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} = \left[\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c}y(t)\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}dt\, \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right)y \]
for the lagrangian
\[ \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right) = m\dot{x}_c = 0 \]
Which results finally,
\[S = S_c + \frac{m}{2} \int_{t_1}^{t_2} dt \,\dot{y}^2\,\tag{2.7}\]
and
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}S_c} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.8}\]
where it is changed the integration variable to $dy(t)$ using (2.2)
The classical action is calculated to be
\[S_c = \frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}\,\tag{2.9}\]
which finally yields,
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.10}\]
The integral limit points out the fact that the deviation from the classical path at the end points is zero.
The integral over $y(t)$ is independent of $x_1$ and $x_2$. Its value depends only on $t_1$ and $t_2$ , since the entire problem is time translation invariant,
\[A(t_2-t_1) = \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.11}\]
and
\[K(x_2,t_2;x_1,t_1)= A(t_2-t_1)e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}}\,\tag{2.12}\]
To determine A(t) for $t_1=0$ making use of the group property (1.23) and (1.19),
\[ \delta(x_2-x_1) = K(x_2,t;x_1,t) = \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K(x,0;x_1,t)\] \[\delta(x_2-x_1) =  \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K^*(x_1,t;x,0) \,\tag{2.13}\]
Substituting for K and using (2.12),
\[K(x_2,t;x,0)= A(t)e^{\frac{i}{\hbar}S_c(x_2,t;x,0)}\]
\[K^*(x_1,t;x,0)= A^*(t)e^{\frac{-i}{\hbar}S_c(x_1,t;x,0)}\]
and
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} dx \,\left\vert{A(t)}\right\vert^2 e^{\frac{i}{\hbar}\left(S_c(x_2,t;x,0)-S_c(x_1,t;x,0)\right)}\,\tag{2.14}\]
When $x_2 = x_1+\Delta{x}$ the argument of exponential can be modified with Taylor expansion as,
\[ S_c(x_2,t;x,0)-S_c(x_1,t;x,0) = \frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} \Delta{x_1}\]
Considering $\Delta{x_1}\rightarrow\,0$ all higher order terms are neglected.
Substituting for $S_c$,
\[\frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} = \frac{\partial(\frac{m\,(x_1-x)^2}{2\,t})}{\partial{x_1}} = \frac{m\,(x_1-x)}{t} = \gamma(x) \,\tag{2.15}\]
Since $\gamma(x)$ is linear function of $x$, its derivative $\frac{d\gamma}{dx}$ is independent of $x$. Using this information to write,
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} d\gamma\,\left\vert\frac{dx}{d\gamma}\right\vert \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)\left(x_2-x_1\right)}\,\tag{2.16}\]
From Fourier transform,
\[\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\beta(x_2-x_1)} \,d\beta = \delta(x_2-x_1)\,\tag{2.17}\]
for $\beta = \frac{\gamma}{\hbar}$, 
\[\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} e^{\frac{i}{\hbar}\gamma(x_2-x_1)} \,{d\gamma} = \delta(x_2-x_1)\,\tag{2.18}\]
Thus, multiplying and dividing by the factor $2\pi\hbar$ in (2.16),
\[ \delta(x_2-x_1) = \int_{-\infty}^{\infty} \frac{d\gamma}{2\pi\hbar} \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)(x_2-x_1)}\frac{2\pi\hbar}{\left\vert\frac{d\gamma}{dx}\right\vert}\,\tag{2.19}\]
Comparing both the left and right side, 
\[\delta(x_2-x_1) = \delta(x_2-x_1)\frac{2\pi\hbar|A(t)|^2}{\left\vert\frac{d\gamma}{dx}\right\vert}\]
so that,
\[|A(t)|^2 = \frac{1}{2\pi\hbar}\left\vert\frac{d\gamma}{dx}\right\vert = \frac{1}{2\pi\hbar}\left\vert\frac{-m}{t}\right\vert =  \frac{1}{2\pi\hbar}\left\vert\frac{\partial^2S_c(x_1,t;x,0)}{\partial{x}\partial{x_1}}\right\vert\,\tag{2.20}\]
The phase can be chosen such that,
\[A(t) = \sqrt{\frac{m}{2i\pi\hbar{t}}}\]
Thus, it is determined the propagator for a free particle as,
\[K(x_2,t_2;x_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[S_c = \frac{m\,(x_2-x_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.21}\]
In general for three dimensions,
\[K(r_2,t_2;r_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(r_2-r_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.22}\]
For fixed $x_1=0\,\,,t_1=0$ using (3.5) the propagator for a free particle should reduce to the Schr\"{o}dinger wave function for a free particle. Thus, the probability amplitude of the free particle from the propagator is,
\[\psi(x,t) = K(x,t,0,0) = \sqrt{\frac{m}{2i\pi\hbar{t}}}e^{\frac{i}{\hbar}\frac{m\,x^2}{2\,t}}\,\tag{2.23}\]
If it is considered a specific point $(x_0,t_0)$, then the classical momentum at this point is $$p_0 = mv_0 = m\frac{x_0}{t_0}$$ with energy $$E = \frac{mv_0^2}{2}= \frac{mx_0^2}{2t_0^2}$$ The change in phase in the vicinity of $(x_0,t_0)$ is then using again Taylor expansion,
\[
\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} exp\frac{i\,m}{\hbar\,2}\left[\frac{x_0^2}{t_0}+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{x}}\right\vert_{(x_0,t_0)}(x-x_0)+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{t}}\right\vert_{(x_0,t_0)}(t-t_0)+...\right]
\]
The bracketed term reduces to (neglecting higher order terms),
\[\left[\frac{x_0^2}{t_0}+ \frac{2x_0}{t_0}(x-x_0) -\frac{x_0^2}{t_0^2}(t-t_0)\right]\] Simplifying with arithmetic, 
\[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,exp{\frac{i}{\hbar}\left[m\frac{x_0}{t_0}x - \frac{m}{2}{\frac{x_0^2}{t_0^2}t}\right]}\]
Thus, the wave function varies in the immediate vicinity of $(x_0,t_0)$ according to \[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,e^{\frac{i}{\hbar}\left[p_0x - E_0t\right]}\,\tag{2.24}\]
which is the well-known Einstein-de Broglie relation, according to which a particle with momentum $p$ and energy $E$ is assigned a wave function with wave length and wave number respectively $\lambda = \frac{h}{p}\,\,\rightarrow\,\, k = \frac{2\pi}{\lambda}$ Similarly with frequency and angular frequency $\nu = \frac{E}{h}\,\,\rightarrow\,\,\omega = {2\pi\nu}$
So,
\[ e^{i\left(kx- \omega{t}\right)}= e^{i\left(\frac{2\pi}{\lambda}x- {2\pi\nu}t\right)} = e^{\frac{i}{\hbar}(px-Et)} \tag{2.25}\]
which is equivalent to (2.24).
 
Schr\"{o}dinger’s time evolution equation

So far, everything is discussed in position basis. Whereas it needs to be in momentum or energy basis for explicit derivation of Schr\"{o}dinger’s time evolution equation. This is achieved using the group property of our propagators,
\[K(x,t;p,0) = \int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,K(x',0;p,0)\,\tag{2.26}\]
For fixed initial momentum $p$ at time $t$,
\[K(x,t;p,0) = \chi_{p,0}(x,t)=\int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,\chi_{p,0}(x',0)\,\tag{2.27}\]
Let us take this as an ansatz for the transformation amplitude,
\[\chi_{p,0}(x,0) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}xp}\,\tag{2.28}\]
Substituting this in (2.27) with the corresponding substitution for $K$, 
\[\chi_{p,0}(x,t) =\int_{-\infty}^{\infty}\sqrt{\frac{m}{2i\pi\hbar{(t)}}}e^{\frac{i}{\hbar}\frac{m\,(x-x')^2}{2\,t}} \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{i}{\hbar}x'p} \,dx'\tag{2.29}\]
As $x'$ is the integrating variable, the x' terms are combined together and simplified using arithmetic, 
\[\chi_{p,0}(x,t) = e^{\frac{i}{\hbar}\left(xp-\frac{p^2}{2m}t\right)}\sqrt{\frac{m}{2i\pi\hbar{t}}}\sqrt{\frac{1}{2\pi\hbar}} \int_{-\infty}^{\infty}e^{\frac{i}{\hbar}\frac{m}{2\,t}\left[x'-\left(x-\frac{pt}{m}\right)\right]^2}\,\tag{2.30}\]
changing the variable to $u = \left[x'-\left(x-\frac{pt}{m}\right)\right]$ and making use of Gaussian Integral, 
\[\chi_{p,0}(x,t) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[xp-\frac{p^2}{2m}t\right]}\,\tag{2.31}\]
In three dimensions,
\[\chi_{p,0}(x,t) = \left(\frac{1}{2\pi\hbar}\right)^{\frac{3}{2}} e^{\frac{i}{\hbar}\left[r.p-\frac{p^2}{2m}t\right]}\tag{2.32}\]
Again using the group property with momentum arguments,
\begin{align*}
K(p_2,t;p_1,0) &=\int_{-\infty}^{\infty} dx'\,K(p_2,t;x,t) \,K(x,t;p_1,0) \\ &= \int_{-\infty}^{\infty} dx\,K(p_2,0;x,0) \,\chi_{p_1,0}(x,t) \\  & = \int_{-\infty}^{\infty} dx\,\chi^*_{p_2,0}(x,0) \,\chi_{p_1,0}(x,t) \\ &= \int_{-\infty}^{\infty}dx\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{-i}{\hbar}\left[p_2x\right]}\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[p_1x-\frac{p_1^2}{2m}t\right]}
\end{align*}
Using Fourier transform,$$ \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} dx \,e^{\frac{-i}{\hbar}x(p_2-p_1)} = \delta(p_2-p_1)$$ which results,
\[K(p_2,t;p_1,0) = \delta(p_2-p_1) \,e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}}\,\tag{2.33}\]
From this, it can be shown that $K(p_2,t;p_1,0)$ satisfies Schr\"{o}dinger equation,
\[i\hbar\frac{\partial{K(p_2,t;p_1,0)}}{\partial{t}} = \delta(p_2-p_1)\frac{p_1^2}{2m}e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}} = \frac{p_2^2}{2m} K(p_2,t;p_1,0) \,\tag{2.34}\]
Thus, it is shown the equivalence between Schr\"{o}dinger formulation and path integral formulation.
It can also be checked that,
\[i\hbar\frac{\partial}{\partial{t}}\chi_{p,0}(x,t) = \frac{p^2}{2m}\chi_{p,0}(x,t)\,\tag{2.35}\]
and
\[i\hbar\frac{\partial}{\partial{t}}K(x,t;x',0) = \frac{-\hbar^2}{2m}K(x,t;x',0)\,\tag{2.36}\]
Reference: Classical and Quantum Dynamics - W.Dittrich, M.Reuter

Path Integral formulation - Part - 2 - Quantum Paths

Things become complicated when it is tried to introduce the concept of paths in the domain of Quantum Mechanics. Though the position in Quantum Mechanics is completely a measurable quantity, the position in consecutive time intervals is not a determinate one.

In the sense, even if it observed a perfect value by making a position measurement on a quantum mechanical system at time $t_1$, there is no way of predicting, what would be the result of a position measurement at time $t_2$. All one can talk about is the average value of the position of the particle [known as the expectation value of position operator].

Nevertheless the wave function in quantum mechanics is absolutely defined in terms of probability. Because of this, even for a single particle there is a non-zero probability of the particle to be found at any point in the entire three dimensional space.

The particle can be found anywhere in the universe in any two consecutive position measurements. This indeterminacy makes it extremely difficult to apply the concept of paths for a particle in the Quantum World. 

To put forth the idea, it is to be started with a simple definition and expanded in terms of probability arguments.

For instance, a classical path in position space is defined as the consecutive value of the position of the particle over a time interval. If a particle is found at position $x_1$ at time $t_1$ and found at a later time in position $x_2$ at time $t_2$ then it is described as, the particle travels from the position $x_1$ at time $t_1$ to position $x_2$ at time $t_2$ in the specific path determined by the extremum principle of action. This same classical path by incorporating the concept of Probability can be restated as,
the path of the classical particle is the one where the quantum wave function reduces to Dirac delta function at every point.

Similar to the above, first it is started with the restated definition of well known classical concepts in terms of probability argument and then the concepts are extrapolated to the Quantum domain.
This way of extrapolation of the 'concept of classical paths' to the quantum particles was first done by Richard Feynman in 1948. The idea is basically described in the simplest form as,

the quantum particles can follow any path as well as every path in the three dimensional Euclidean space.
This gives rise to an infinite number of possible paths for a quantum particle even if the particle wants to go the most nearest point. And the mathematical model consists of two types of strategies for the summation procedure of these paths in the form of integrals. To demonstrate, let us consider the case of a free particle going from region one to region two, where it is restricted with an infinite wall with only two slits for the particle to cross between the regions [Figure (1)].
 

 

It is known for sure that, to reach the point $(r_2,t_2)$ from $(r_1,t_1)$, the particle can only take either of the two slits. But just from making a single observation at $(r_2,t_2)$, one cannot say anything about the initial point from where the particle has started its path. The particle could have started from anywhere within the left side region of the wall.

From the figure(1), it could be from $(r_1,t_1)$ or $(r_1^*,t_1^*)$ or from some other point. But the essential point is that, a particle found at position $r_2$ at time $t_2$ will never tell anything about the initial position $r_1$ from where it started its motion.

This lack of information leads to the first of the two strategies that needs to be addressed in the general theory.
The first one is,

In the general motion of a Quantum particle in the three dimensional space, if the particle is observed at a specific position [It is possible to take out the wall by considering infinite number of slits instead of two], then the particle is said to have come from every possible initial point in the entire three dimensional space.
And each of those paths contribute to the net probability amplitude.
So, it is introduced an integral such that it is carried over every possible initial position of the particle.

The second strategy is,

Once the initial position and the final position of the particle is fixed, the next difficulty comes from the fact that the particle can now follow any path as well as every path between those two points.

In the figure(1), once the initial position is fixed, the particle can take any one of the infinite possible paths via slit 1 or slit 2 or through any one of the infinite number of slits.
To account for this new fact, it is introduced a second integral within the first integral to take into account the contribution from every possible paths.

Technically, using the definition of wave function, $$ \psi_1(r_1, t_1) $$ is the probability amplitude of finding the particle at the position ${r_1}$ at time $t_1$ and $$ \psi_2({r_2},t_2)$$ is the probability amplitude of finding the particle at ${r_2}$ at time $t_2$. Using the first integral,  one would like to obtain $\psi_2({r_2},t_2)$ from $\psi_1({r_1},t_1)$ by defining a correlating function called transition amplitude $$ K (r_2,t_2,r_1, t_1)$$ which is the probability amplitude for finding a particle at $({r_2},t_2)$ when it was initially found at $({r_1},t_1)$.

Hence forth, the fundamental dynamical equation is stated as, \[\psi_2({r_2},t_2) = \int_{space} K({r_2},t_2;{r_1},t_1) \psi({r_1},t_1) d^3{r_1} \,\,\,\tag{1.14}\]  
The only unknown term in this equation is $K({r_2},t_2;{r_1},t_1)$. It is also called the Feynman propagator. Once the explicit form of this propagator is known, it is then possible to determine how it controls the dynamical development of the Schr\"{o}dinger wave function.

To obtain the Propagator, one can make use of the second integral where it is summed over all possible paths between the two points A$({r_1},t_1)$ and B$({r_2},t_2)$. So, it is taken a general point C between A and B and considered the motion of the particle along this point. i.e. Path from A to B via C. The probability amplitude for this path A-C-B is denoted as $\phi_{BA}(C)$ and the propagator is obtained by integrating through all possible A-C-B paths, \[K(B,A) = \int_{all\,possible\, paths} dC \,\phi_{BA}(C) \,\,\,\tag{1.15}\]  

Determining this integral for $({r_1}, t_1) \rightarrow ({r_2},t_2)$ in general consists of an infinite number of possible paths with their corresponding probability amplitudes. And obviously there is no fundamental physical principle that determines this amplitude $\phi_{BA}(C)$ directly.

This difficulty was first overcame by Dirac, who postulated that each path contributes same amount of probability amplitude to the final result with different phase factors given by,  \[\phi_{BA}(C) = e^{\frac{i}{\hbar}S(C)}\,\,\,\tag{1.16}\] where S is the classical action integral.
Thus one finally obtains the formula for the Feynman propagator as, \[ K({r_2},t_2; {r_1},t_1) = \int_{{r}(t_1)={r_1}}^{{r}(t_2)={r_2}} d{r(t)}\, e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,L({r}(t), \dot{{r}}(t),t)}\,\tag{1.17}\]
It is easily seen that, in the classical limit when $\hbar\rightarrow\,0$ or $S/\hbar>>1$ the exponential factor oscillates rapidly for all regions except where "S" remains stationary. Thus, the corresponding amplitudes of rapidly oscillating factor will be washed out by destructive interference and the major contribution will come from stationary "S" value which occurs for the classical action. Thus consistent with the correspondence principle, classical results are obtained as a limiting case of the quantum theory. 

In addition, if it is considered $K(x_2,t_2,x_1,t_1)$  (for simplicity it is considered in one dimension) , and $(x_1,t_1)$ kept fixed, one obtains \[K(x,t;x_1,t_1) = K_{(x_1,t_1)}(x,t)\,\,\,\tag{1.18}\] which is a function of $x$ and $t$ alone.

From (1.14) for $t=t_1$ as, $$ \psi(x,t)=\int_{-\infty}^{\infty} dx_1 K(x,t;x_1,t_1) \psi(x_1,t_1)$$ and
$$\psi(x,t)=\int_{-\infty}^{\infty} dx_1 K(x,t;x_1,t) \psi(x_1,t)$$
And from the definition of Dirac delta function, our new function reduces as, \[K(x,t;x_1,t) = K_{(x_1,t_1)}(x,t)\vert_{t=t_1} = \delta(x-x_1)\,\tag{1.19}\]

It is seen from the above results that the propagator function $K_{(x_1,t_1)}(x,t)$,
at the initial point $(x_1, t_1)$ reduces to Dirac delta function where the particle was found certainly without the amplitude being smeared out. And at later times, the propagator is just the probability amplitude for finding the particle at a variable point $(x,t)$.

Comparing the above properties with the postulates of Quantum Mechanics, it is understood that the propagator function is exactly what it is meant by the Schr\"{o}dinger wave function. 
Thus, we have found an intrinsic way to determine the general schrodinger wave function from the propagator function.  

Another interesting property of the propagator is that, from the definition (1.14), \[\psi(x_3,t_3)=\int_{-\infty}^{\infty} dx_2 K(x_3,t_3;x_2,t_2) \psi(x_2,t_2)\,\tag{1.20}\] and \[\psi(x_2,t_2)=\int_{-\infty}^{\infty} dx_1 K(x_2,t_2;x_1,t_1) \psi(x_1,t_1)\,\tag{1.21}\] and \[\psi(x_3,t_3)=\int_{-\infty}^{\infty} dx_1 K(x_3,t_3;x_1,t_1) \psi(x_1,t_1)\,\tag{1.22}\] substituting (1.21) in (1.20) and equating with (1.22) one obtains an important group property, \begin{align*}
& \int_{-\infty}^{\infty} dx_1 K(x_3,t_3;x_1,t_1) \psi(x_1,t_1) \\ & = \int_{-\infty}^{\infty} dx_2 K(x_3,t_3;x_2,t_2) \int_{-\infty}^{\infty} dx_1 K(x_2,t_2;x_1,t_1) \psi(x_1,t_1)
\end{align*}
which gives,  \[K(x_3,t_3;x_1,t_1) = \int_{-\infty}^{\infty} dx_2 K(x_3,t_3;x_2,t_2)K(x_2,t_2;x_1,t_1)\,\tag{1.23}\] In general, for $f(x_f,t_f)$ and $i(x_i,t_i)$ one can write \[K(f;i) = \int_{-\infty}^{\infty} dx_{N-1}.....\int_{-\infty}^{\infty}dx_1 K(f;N-1)\,K(N-1;N-2)\,...K(2;1)K(1;i)\,\tag{1.24}\] It is to be noted that, the intermediate times are not integrated over.
 
Reference: Classical and Quantum Dynamics - W.Dittrich, M.Reuter

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

Curvilinear Coordinate system, which in fact is the most general coordinate system used to describe the motion of any particle. It includes all our usual systems like Cartesian, Spherical and Cylindrical coordinate systems. 

     With one to one correspondence, it is always possible to define a set of transformation rules like, $$ x_1 = x_1(u_1, u_2, u_3) \\~\\ x_2 = x_2(u_1, u_2, u_3) \\~\\ x_3 = x_3(u_1, u_2, u_3)$$

to write each of the cartesian coordinates in terms of the general coordinates. We can also define inverse transformation rules like,

$$ u_1 = u_1(x_1, x_2, x_3) \\~\\ u_2 = u_2(x_1, x_2, x_3) \\~\\ u_3 = u_3(x_1, x_2, x_3) $$

to go from one system to another. These transformations are unique, since they have one to one correspondence. 

The surfaces $ u_1 = const., u_2 = const., u_3 = const.$$ are called coordinate surfaces and the curve formed from the intersection of pair of two surfaces is called coordinate curves. The point where the tangent lines drawn to these coordinate curves intersect is chosen to be the origin of the coordinate system. 

For the sake of simplicity, we often used to deal with the coordinate systems where the coordinate surfaces intersect at right angles. They are called "Orthogonal coordinate system". 

Now, we can formulate the general rules for describing a point and its motion and to describe various vector operations in this new curvilinear coordinate system.  

But the formulation is going to be more general to apply in any system at any point. It should apply to Cartesian, Cylindrical, Spherical, Paraboloidal, Ellipsoidal and etc. 

Note: There are nearly more than 10 types of orthogonal coordinate systems we are using in mathematics. 

To start, first we will consider the example of describing small differential element in 3D Euclidean Space. $$\vec{dr} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z} ..... \ldots eq.(1)$$ 

Using the chain rule, we can write the same differential element as,
$$ \vec{dr} = \frac{\partial\vec{r}}{\partial{x}} dx + \frac{\partial\vec{r}}{\partial{y}} dy + \frac{\partial\vec{r}}{\partial{z}} dz \ldots... eq.(2)$$  

Comparing eq.(1) and (2) we get that $$\frac{\partial\vec{r}}{\partial{x}} = \hat{e_x} \, , \,\frac{\partial\vec{r}}{\partial{y}} = \hat{e_y} \, , \, \frac{\partial\vec{r}}{\partial{z}} = \hat{e_z} $$   In a similar way, if the differential element is written in terms of curvilinear coordinates as $$ \vec{r} = \vec{r} (u_1, u_2, u_3)$$ Tangent vector to $u_1$ curve at some point P is, $ \frac{\partial\vec{r}}{\partial{u_1}}$ 

Therefore, the unit tangent vector in this direction given by,  $$ \frac{\partial\vec{r}/\partial{u_1}}{|\partial\vec{r}/\partial{u_1}|} = \hat{e_1} $$

We call $$ |\partial\vec{r}/\partial{u_1}| = h_1 = scaling factor $$

Since these coordinates needn't necessarily have the dimension of distance, these parameters are used to make them all to same dimension - after all we cannot add mangoes and apples together to single count. 

To complete, we write, $$\frac{\partial\vec{r}}{\partial{u_1}} = h_1 \hat{e_1} \ldots... eq.(3) \\ \frac{\partial\vec{r}}{\partial{u_2}} = h_2 \hat{e_2} \ldots... eq.(4) \\ \frac{\partial\vec{r}}{\partial{u_3}} = h_3 \hat{e_3} \ldots... eq.(5) $$


These basis vectors are tangent vectors to the curves. Similar to that, we can always form another basis whose unit vectors are normal to the coordinate surfaces, where the normal vectors are represented in terms of gradient operator $ \nabla{u_1} , \nabla{u_2}, \nabla{u_3}$ 

After normalizing, we get a new basis unit vectors represented as, $$ \hat{E_1} = \frac{\nabla{u_1}}{|\nabla{u_1}|} , \hat{E_2} = \frac{\nabla{u_2}}{|\nabla{u_2}|}, \hat{E_3} = \frac{\nabla{u_3}}{|\nabla{u_3}|} $$  It can be shown separately that, these two set of basis vectors constitute reciprocal system of vectors under coordinate transformation. It leads to the concept of Co-variant and Contra-variant vectors.

Thus, any vector can be expressed as either in terms of first set of basis vectors or in terms of second set of basis vectors. 

The square of the magnitude of the differential element in terms of first set of unit basis vectors, $$ ds^2 = \vec{dr}\cdot\vec{dr} = {h_1}^2 {du_1}^2 + {h_2}^2 {du_2}^2 + {h_3}^2 {du_3}^2 $$ Since we got the basic things we need to work, now we can start defining the general relation for operations like Gradient, Divergence, Curl and Laplacian. 

Gradient:
Gradient from the definition, $$ df = \nabla{f} \cdot \vec{dr} $$

Using the chain rule, $$ df = \frac{\partial{f}}{\partial{u_1}} du_1 + \frac{\partial{f}}{\partial{u_2}} du_2 + \frac{\partial{f}}{\partial{u_3}} du_3 \dots... eq.(6) $$  and we can also write the differential element $ \vec{dr} $ using eq. (3), (4), (5) as, $$ \vec{dr} = h_1 du_1 \hat{e_1} + h_2 du_2 \hat{e_2} + h_3 du_3 \hat{e_3} $$

Still we don't know what is the form for gradient operator, but we do know, from the definition of gradient that it would give "df" when dotted with $ \vec{dr}$ . So, 

$$ \nabla{f} \cdot \vec{dr} = \nabla_1{f} h_1 du_1 + \nabla_2{f} h_2 du_2 + \nabla_{f} h_3 du_3 \ldots... eq.(7) $$

where $ \nabla_1{f} , \nabla_2{f} , \nabla_3{f} $ are the components of Gradient operator when it is written in terms of the general basis vectors $ \hat{e_1}, \hat{e_2}, \hat{e_3} $. 
Again comparing eqs.(6) and (7), the components of the gradient operator found out to be, $$ \nabla_1{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}},\nabla_2{f} = \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}}, \nabla_3{f} = \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} $$  

Hence the general form of Gradient operator in any curvilinear orthogonal coordinate system is given by,
$$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} \,\, \ldots...eq.(8)$$  

Divergence:

Let us analyze the first term we will get, when we apply the divergence operator on any vector function $\vec{A} $,
$$ (\nabla\cdot\vec{A})_1 = \nabla \cdot (A_1\hat{e_1})  \ldots.....(9)$$

We don't know, what we will obtain when we apply the divergence operator on $ \hat{e_1} $. But, if we could write $\hat{e_1}$ in terms of some gradient operations, then there is a real possibility of obtaining the expression for Divergence with our prior knowledge of Gradient.     

According to write the unit vectors in terms of gradient relations, 

We apply the gradient operator for the functions $ u_1, u_2, u_3 $ in eq.(8) from which, we will get $$ \nabla{u_1} = \frac{\hat{e_1}}{h_1}\,, \,\nabla{u_2} = \frac{\hat{e_2}}{h_2}\, ,\, \nabla{u_3} = \frac{\hat{e_3}}{h_3} $$ 

The resultant unit vectors using gradient relations are,
$$ \hat{e_1} = h_1 \nabla{u_1} \, ,\, \hat{e_2} = h_2 \nabla{u_2} \, ,\, \hat{e_3} = h_3 \nabla{u_3} $$  

But we need to relate it with $ \hat{e_1}$, So we apply the volume relation $$ \hat{e_1} = \hat{e_2} \times \hat{e_3} = h_2 h_3 \nabla{u_2} \times \nabla{u_3} $$
Applying this in eq.(9),
$$ \nabla \cdot (A_1\hat{e_1}) = \nabla\cdot[A_1h_2h_3 \nabla{u_2} \times \nabla{u_3}] \, \, \, \ldots...eq.(10)$$
Using the vector relation, $$ \nabla\cdot {f\vec{A}} = \nabla{f}\cdot\vec{A} + f \nabla\cdot\vec{A} $$
where f- scalar function, $\vec{A} = vector function $.

Eq.(10) becomes, $$ \nabla\cdot(A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) + A_1h_2h_3 \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) \, \, \, \ldots...eq.(11)$$

But using the vector identity, $$ \nabla\cdot(\vec{A}\times\vec{B}) = \vec{B}\cdot(\nabla\times \vec{A}) - \vec{A}\cdot (\nabla\times\vec{B})$$
$$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = \nabla{u_3}\cdot(\nabla\times\nabla{u_2}) - \nabla{u_2}\cdot(\nabla\times\nabla{u_3})$$

But, Curl of gradient is always zero for any scalar function, which implies $$ \nabla\cdot(\nabla{u_2}\times\nabla{u_3}) = 0 $$  

Eq.(11) gives, $$\nabla \cdot (A_1\hat{e_1}) = (\nabla{A_1h_2h_3})\cdot(\nabla{u_2}\times\nabla{u_3}) \dots..eq.(12) $$  

Again writing $ \nabla{u_2}\times\nabla{u_3} $ in terms of basis vectors that is $$ \nabla{u_2}\times\nabla{u_3} = \frac{\hat{e_2}\times\hat{e_3}}{h_2h_3} = \frac{\hat{e_1}}{h_2h_3} $$  

Eq.(12) results into $$\nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3} \cdot \nabla(A_1h_2h_3) $$
Using our prior knowledge of Gradient, it can be expanded as,
 $$ \nabla\cdot(A_1\hat{e_1}) = \frac{\hat{e_1}}{h_2h_3}\cdot\left[ \frac{\hat{e_1}}{h_1} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} + \frac{\hat{e_2}}{h_2} \frac{\partial(A_1h_2h_3)}{\partial{u_2}} + \frac{\hat{e_3}}{h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_3}}\right] $$  While, we are dealing with orthogonal basis, dot product between any two different basis gives zero and dot product of same vector gives unity. 

Making using of the orthonormality, we finally arrive at the result,
$$ \nabla\cdot(A_1\hat{e_1}) = \frac{1}{h_1h_2h_3} \frac{\partial(A_1h_2h_3)}{\partial{u_1}} $$

Similar procedure gives the expression for other coordinates. 

The final expression for the divergence operator in general curvilinear coordinates is,

$$\nabla\cdot\vec{A} = \frac{1}{h_1h_2h_3}\left[ \frac{\partial{A_1h_2h_3}}{\partial{u_1}} + \frac{\partial{A_1h_2h_3}}{\partial{u_2}} + \frac{\partial{A_1h_2h_3}}{\partial{u_3}}\right] \ldots...eq.(13)$$  

Curl:

As the same, first we will take single component, write it in terms of gradient and apply the curl,

$$ \nabla \times (A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) \, \, \ldots...eq.(14)$$
Since there is already a curl operator, we don't need to use volume relation but we could just simply write $ \hat{e_1} $ in terms of its own gradient relation i.e. $ \hat{e_1} = h_1 \nabla{u_1} $

Using the vector identity, $$ \nabla \times (f\vec{A}) = f \nabla\times\vec{A} + \nabla{f}\cdot\vec{A} $$

Eq.(14) gives, $$ \nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} + A_1h_1\nabla \times \nabla{u_1} $$
but curl of gradient is zero.

So, Eq.(14) becomes, $$\nabla \times(A_1\hat{e_1}) = \nabla \times (A_1h_1\nabla{u_1}) = \nabla (A_1h_1)\times \nabla{u_1} $$  

With the help of eq.(8) we can rewrite the above into,
$$ \nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \nabla{u_1} $$

Again using the relation, $ \nabla{u_1} = \frac{\hat{e_1}}{h_1} $

$$\nabla \times (A_1\hat{e_1}) = \left[\frac{1}{h_1} \frac{\partial{A_1h_1}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_3}\right] \times \frac{\hat{e_1}}{h_1} $$


Using the cross product rule for the positive volume element,  we finally get, 

$$\nabla \times (A_1\hat{e_1}) = \frac{1}{h_1h_2} \frac{\partial{A_1h_1}}{\partial{u_2}} \hat{-e_3} + \frac{1}{h_1h_3} \frac{\partial{A_1h_1}}{\partial{u_3}} \hat{e_2} \, \ldots... eq.(15)$$ 

Similar procedure gives for other components the following result,
$$\nabla \times (A_2\hat{e_2}) = \frac{1}{h_1h_2} \frac{\partial{A_2h_2}}{\partial{u_1}} \hat{e_3} + \frac{1}{h_2h_3} \frac{\partial{A_2h_2}}{\partial{u_3}} \hat{-e_1} \, \ldots... eq.(16)$$

$$\nabla \times (A_3\hat{e_3}) = \frac{1}{h_1h_3} \frac{\partial{A_3h_3}}{\partial{u_1}} \hat{-e_2} + \frac{1}{h_2h_3} \frac{\partial{A_3h_3}}{\partial{u_2}} \hat{e_1} \, \ldots... eq.(17)$$

Combining the components from eq.(15),(16),(17) we get the General expression for the Curl operator in Curvilinear coordinates as,

$$ \nabla\times\vec{A} = \frac{1}{h_2h_3} \left[\frac{\partial(A_3h_3)}{\partial{u_2}} - \frac{\partial(A_2h_2)}{\partial{u_3}}\right] \hat{e_1} + \\~\\ \frac{1}{h_1h_3} \left[\frac{\partial(A_1h_1)}{\partial{u_3}} - \frac{\partial(A_3h_3)}{\partial{u_1}}\right] \hat{e_2} + \\~\\ \frac{1}{h_1h_2} \left[\frac{\partial(A_2h_2)}{\partial{u_1}} - \frac{\partial(A_1h_1)}{\partial{u_2}}\right] \hat{e_3} \, \, \, \ldots...eq.(18)$$ 

Or simply we can write this in Matrix form as,

$$ \nabla \times \vec{A} = \frac{1}{h_1h_2h_3}\begin{vmatrix} h_1\,\hat{e_1} & h_2 \,\hat{e_2} & h_3 \, \hat{e_3} \\ \frac{\partial}{\partial{u_1}} & \frac{\partial}{\partial{u_2}} & \frac{\partial}{\partial{u_3}} \\ h_1\vec{A_1} & h_2 \vec{A_2} & h_3 \vec{A_3} \end{vmatrix} \,\,\, \dots...eq.(19) $$

Laplacian:

Unlike others, we don't need to find anything extra for Laplacian since it is just the combination of gradient and divergence. 

Let us take a scalar function "f" and write its gradient from eq.(8), $$ \nabla{f} = \frac{1}{h_1} \frac{\partial{f}}{\partial{u_1}} \hat{e_1} + \frac{1}{h_2} \frac{\partial{f}}{\partial{u_2}} \hat{e_2} + \frac{1}{h_3} \frac{\partial{f}}{\partial{u_3}} \hat{e_3} $$
Now, applying the Divergence operation for the resultant outcome, we get the General expression for Laplacian in curvilinear coordinates, 

$$ \nabla^2f = \frac{1}{h_1h_2h_3} \left[ \frac{\partial\left(\frac{h_2h_3}{h_1}\frac{\partial{f}}{\partial{u_1}}\right)}{\partial{u_1}} + \frac{\partial\left(\frac{h_3h_1}{h_2}\frac{\partial{f}}{\partial{u_2}}\right)}{\partial{u_2}} + \frac{\partial\left(\frac{h_1h_2}{h_3}\frac{\partial{f}}{\partial{u_3}}\right)}{\partial{u_3}} \right] \ldots...eq.(20)$$  




That is all we need to derive. We need to remember that, these derivations are done for general orthogonal curvilinear coordinate system. 
For the most general curvilinear coordinate system (i.e. which are not orthogonal), we will need Tensors and its analysis. 
 

Lorentz Transformation Equations from the fundamental Postulates of Special theory of Relativity

The Two fundamental Postulates are,

               

           1. The laws of Physics are same in all inertial reference frames.

           2. The speed of light is always a constant independent of the motion of the observer. 

      As a student When I heard these postulates for the first time, I really never thought that there would be so much in it to understand. But the implications are quite significant to explain the Nature. Let us see, how it changes the universe...
      The most important term is that "the speed of light". We know the velocity follow Galilean addition rules withing one frame to another frame of reference. 

And we know how to make this Galilean transformation between two inertial reference frames.     
      Now, with surprise the postulate is given as the speed of light is constant in all reference frames, which means we cannot apply our previous Galilean transformations. It should be changed!!
     
      But we are already using these transformation and getting best results. That means we can be sure that these laws can never proved to be completely wrong. 

     All it can be done is some alteration without affecting the known phenomena. For the mathematics, let us take a case..  

     As usual we take two inertial reference frames S and S' with relative speed "v".  We want to keep the structure of Galilean equations at speed lower than c. 
     So, we assume that there is an extra factor comes into play with Galilean equations, which has the special property that it gets the value "1" when v<<c. Let us assume the factor is $ \gamma $

Rewriting the known Galilean transformation with the new factor "gamma".. $$ x' = \gamma (x - vt)..............eq.(1) $$ For convenience we always used to take the motion to be only in x-direction. And it will never going to change anything in y or z- direction. It is always y'=y and z'=z. 

    Inverse transformation similarly given as, $$ x = \gamma (x'+vt)..............eq.(2) $$ So far, we never used any information from the postulates. 
    Now from the second postulate, for a light wave speed is always the constant - c. Let us imagine that a light wave is allowed to propagate in the positive x direction, when the both reference frames starts from the common origin. 

[ Why I am taking the case of Light wave?

We assume the Universe is same for everything else whether it is a light or a particle. As the postulate defined at light speed, if we could find the value of the gamma factor at light speed, it will eventually obey for all velocities. Intrinsically, we believe that the Universe won't have difference structure at different velocities. 

Simply, it is finding the function by applying the boundary conditions.]   

    For the light wave, the equations become, x' = ct' and x = ct 
[We now thrown away the concept of space and time being flat and universal to everyone]    
Applying in eq.(1) and (2).. $$ ct' = \gamma (ct-vt)...............eq.(3)$$ 
and $$ ct = \gamma (ct'+vt')................eq.(4)$$  eq.(3) becomes $$ t' = \gamma \frac{(c-v)}{c}t $$ and eq.(4) becomes $$ t = \gamma \frac{(c+v)}{c} t' $$
   Combining the (4) with (3) we can cancel the time variables..
$$ t = \gamma \frac {(c+v)}{c} ( \gamma \frac{(c-v)}{c}t ) $$
$$ t = \gamma^2 \frac{(c^2 - v^2)}{c^2} t $$
Cancelling the 't' on both sides..
$$ \gamma^2 = \frac{1}{1 - (v^2/c^2)}$$ Thus we finally get the value of "gamma" as $$ \gamma = \frac {1}{\sqrt{1-(v^2/c^2)}}$$

    Remember, we took a special case at light speed and got the value for gamma - $\gamma$ factor. Now, we need to substitute this in the normal case i.e. eq.(1) and (2) which is given a new name - Lorentz Transformation equations. $$ x' = \frac{x - vt}{\sqrt{1 - (v^2/c^2)}} $$ and the inverse transformation is $$ x = \frac{x' + vt'}{\sqrt{1 - (v^2/c^2)}} $$

    To derive the general transformation equation for time, substitute eq.(2) in eq.(1) $$ x' = \gamma ((\gamma (x' + vt')) - vt) \\~\\ x' = \gamma^2 (x' + vt') - \gamma vt $$ Taking t in one side and the remaining in the other gives..$$ \gamma vt = (\gamma^2 - 1) x' + \gamma^2 vt' $$ Finally we get t in terms of x' and t'.. $$ t = \frac{\gamma^2 x' (v^2/c^2)} {\gamma v} + \frac{\gamma^2 vt'}{\gamma v} \\~\\ t = \gamma (t' + (x'v/c^2) = \frac{t' + (x'v/c^2)}{\sqrt{1 - (v^2/c^2)}} $$

Similarly, the inverse tranformation can be obtained as, $$ t' = \frac {t - (xv/c^2)}{\sqrt{1 - (v^2/c^2)}} $$

I think.. we finished getting the most basic Lorentz transformation equations from one inertial frame of reference to other which moves at a relative speed "v". 

From these transformation equations we may able to understand all those amazing, significant physical nature of Space and Time.  

Inertial Frame of Reference

     This abstract started with the fundamental concept called "the frame of reference". 
     We know that the laws of physics are not same for all observers from the simple fact that a moving observer would explain things in a different way from a relative stationary observer (unless and until they both move relatively at constant speed). 
     Try it for yourself by explaining the motion of objects around you i.e. for example one from the train and one from the ground.   

Since physics is the most compact mathematical tool we can afford.. We don't want to create new and separate laws for moving observers and stationary observers. 
     And so, Galilean transformation rules are invented for the transformation of physical quantities from one reference frame to another. 

      Galilean transformation rules are just basic arithmetic thing. Let us consider two reference frame relatively moving at constant speed i.e. S and S' moving with relative speed "v" where else the movement considered only to be in the x-direction. 
     *Don't bother about the direction of motion. We can always chose the co-ordinate system such a way that x-axis lies in the direction of motion. 

     If a position is to be specified in the universe in "S- frame" the components are (x,y,z,) at time - t. Similarly the components measured in " S' -frame " are (x',y',z') at time - t. In Galilean time, time was considered to be the same for everyone. 

    If S' is thought to be moving in the positive x-direction relative to S then the components observed by two frames are related by,
                           x' = x -vt
                           y'= y
                           z'= z
the reverse is of course x = x' + v t ; y = y'; z = z'

    Of course there should be primed coordinates on the right [because we are relating the measurements made from one frame to other.] 
i.e.  x = x' + v t'
but we assumed t = t' and so the problem solved. 

   You may ask why I am concerned about just positions?
After all, positions and their change with respect to time are just needed to explain any motion from Newton's laws of motion.

   Now, Let us check how Newton's laws are behaving in these reference frames. In S - frame of reference
$$ \vec{F} = m \frac{d^2x}{dt^2} $$ Applying the same law in S' - frame, it gives $$ \vec{F'} = m \frac{d^2x'}{dt^2}$$ 
To compare the S' frame with S , substituting the value of x' in terms of co-ordinates observed in S-frame..$$ \vec{F'} = m \frac{d^2(x-vt)}{dt^2}$$ which gives $$ \vec{F'} = \frac {d^2x}{dt^2} - \frac{d^2(vt)}{dt^2}$$ "vt" term cancels on double integration $$ \rightarrow                    \vec{F'} = m \frac{d^2x'}{dt^2} = \vec{F} = \frac{d^2x'}{dt^2}$$

   As we can see that, both the observers obtain the same Newtonian force and will explain any phenomena exactly in the same way.  They can't even know, which one is moving from the observed motion. 

   Newton's laws alone cannot determine exactly either which one is moving when they move at relatively constant speed. 

Thus it was realized that reference frames moving with relatively constant speed has special meaning than the usual ones and they are called "inertial reference frames". 

How GPS works?

Global Positioning System

       The cost for launching some 12 to 24 satellites into the space from our atmosphere with those big satellite launching vehicles needs a huge amount of money, hundreds of scientists for tremendous amount of calculations, good engineering in the sense mechanical, electrical, computational, etc. and finally some hard work.   

      Thinking about all these things, we may think for a while that the GPS system in our mobile does really needs a huge amount of technology. But it is really amazing is that the physics behind this small GPS thing in our smartphones is not more than the level of 10th standard physics. 

       The principle is just,  Velocity [1] =  distance / time . It is all needed to determine the position of an object (ofcourse for the most accurate results we will need some of general and special relativity).

The concept is simple as we determine the distance between two stations in a highway. For example, if you go from station-1 to station-2 with some constant Velocity , the distance between them can be calculated as, $$ Distance(d)_{1-2}= velocity(v) \times time(t)$$
where t - time taken for you to reach from station 1 - 2.

If you don't want to go, you can send a messenger(msg) to station-2 at constant velocity and make him to travel back at the same velocity. Now, you can calculate the time taken for him to travel twice the distance ($d_{1-2}$) with velocity "v". Hence, $$ d_{1-2-1} = 2* d_{1-2} = v_{msg}* t_{msg} $$ or simply $$ d_{1-2} = \frac {v_{msg}*t_{msg}} {2}$$ That is all we need to determine the position of station-2. 

In an exact similar way, satellite determines your position by sending a messenger to your smartphone and receiving it.
And the messenger, in god's grace always travel with same speed in vacuum, Electromagnetic waves. All these things happened just because of EM waves and its constant fundamental speed.

We get all the information so fast because of the tremendous speed of light which is about $3\times 10^8 m/s$
[Think: If it is about some 1 km/s, then before knowing your GPS coordinates, you would have gone to hell.] 

Ok.. Now I can determine the position with light signals. 
But why do you need some 12 to 24 satellites in the space to determine your position?
It is because unfortunately our satellites don't have any eyes. They can just only receive the signals but can never determine accurately "from which direction the signals are coming".

Your position on earth is with respect to the satellite is not a fixed straight line as we did it in the highway problem. 

From the measurements, satellites will determine the position of your cellphone but on the surface of giant hollow sphere centered about the satellite. Since, it knows the signal is coming from earth, it just focuses on the circle that is on earth which is the common intersection hollow sphere about satellite on earth. 




















Thus it reduces the problem of determining the position from infinite space to a small circle on earth. 

We know from geometry that, minimum three circles are needed to intersect to create a point. And so, to determine a position on earth, at least we need three satellites which is visible to that place. You cannot receive and determine the position from the satellite that is on the other side of the earth. 

One or three satellites are not enough to cover the entire globe of sphere. That is why, we are sending so many satellites such that at any point on earth at any time should be covered by at least three satellites in visible range so that all the single points and positions on earth can be located most accurately. 

That is all for the basic lever. 
         But it may not so simple as it sounds, if you are going to need too much accurate and perfect solutions such as determining the distance between your couch and your T.V. from the far distance orbiting satellites. 

        For those things, you will need to finish your degree in Physics, since it handles with synchronization of clocks with the help of special and general relativity.  

[To understand how Relativity plays the crucial role in GPS tracking, we will need some calculations].

The Universe with or without Sense

I did an experiment in my house where I just tried to walk in a same manner by closing my eyes. 
I couldn’t last for few minutes. But I got something of a bizarre idea.

          If I don’t have only eyes, I can’t see anything but I can hear, touch, smell, and speak. For a visually challenged person, the world will look black. But he can understand something about the Nature and about humans, etc. We can make him feel by other four senses. But he will never know the world of colours. He don’t know what is light or how does object look. He will just know the shape, and whether it is heat or cold or whether it smells good or bad and whether it can make sound or not.

          But he will never know what is light.

          Then I thought for a while and started to eliminate the senses in human body one by one.

          For a person of both blind and deaf,

You can’t communicate with him. He can only feel by his skin and smell and may be can taste some food.

          But for a person who is blind, deaf, dumb and neither can he smell nor taste, how will be the Universe?

           He can just feel by touching anything.

But if he was isolated in the mid space, then?!!

          I couldn’t feel how that will be. It is just a black. He will never know what he is or where he is or answer for any question?

         

Think for a little time.

Think how will you explain the Universe to a person without five senses?

But don’t stop with that, now imagine what if you have more senses?

The world of a person without eyes is different from the usual person.

Applying the same thing in a normal person, I wonder whether we are also blind, deaf and dumb to understand the Universe.

Do we need some more senses to understand the real Universe?

Is it enough the senses to feel the real world?

          How many extra senses we need?

Just think and feel it. It is really amazing on thinking about these.

It is not philosophical but something we need to understand.

Think, what is the real Universe? if we are also just like a dumb, deaf, blind when comparing to the sufficient or really essential senses to sense the Universe. 

Faster Than Light? or Newton's Basic Laws are Wrong!

I was thinking about this problem in my first year of college. It is that we know.. No signal can travel faster than Light Speed and Newton’s Laws are valid in all reference frames i.e. When a force applied to a body, no matter how big is, from Newton’s first law the body will get an acceleration that is equal to a = F / m

Consider a body that is perfectly rigid i.e. neither incompressible nor stretchable. Assume that the body has a length of 5*10^8 meters which is more than one light year. It was also assumed that the experiment is done in vacuum so that there will not be any external effects. Now a force “F” is applied in “x” direction so that the body is displaced to a displacement of “dx” in x direction.

From Newton’s Laws of Motion,

Acceleration in x direction = Force / mass of the object (M)

                   a = F/M

The body is considered to be made of system of particles with infinitesimal mass “dm”. From the Centre of Mass Principle it is known that the path and the acceleration of the centre of mass is as same as the path of the point with mass “M” and acceleration “a”. Centre of Mass of an object that has uniformly distributed mass depends on the geometry of the object. We choose the object such that its centre of mass lies in that object.

From calculations we get,

Acceleration of Centre of Mass = F/M

So, it is sure that when it is applied an external force F to the object of mass M then the centre of mass of the object will get an acceleration of magnitude F/M.

From Electrostatics theory

When an external force is given to an object like rope, the motion of the rope is can be described as follows,

Consider that the rope has the molecules only arranged in linear order. Let’s say 5 molecules.

When the external force is applied, the first molecule will move in the direction of the force.

Let us take the case of pulling a rope and the Force Body Diagram of the first molecule is

Only if   “ Fext ” is greater than the net electrostatic attraction or repulsive force on the first molecule due to all other molecules namely -

“ Felectro-1 ”  , the first molecule will get an acceleration in the direction of “ Fext ” and will move distance "dx".

When the first molecule is moved through the distance “dx” in x direction, the second molecule will also move in the “x” direction to maintain the stable configuration. 

So the force on second molecule will increase to“ Fext ”  and it will be greater than the net attraction or repulsive force one second molecule due to all other molecules namely“ Felectro-2 ” . So, the second molecule will also move by distance “dx”.

Note: The body is considered to be perfectly rigid so that each molecule will be displaced by same distance “dx”. And so the external force will be transmitted to each particle.

In this way, throughout there will be electrostatic interaction between the molecules. And the interaction will make the last molecule that is at the other tip of the rope to move by the same displacement “dx”.

Thus we can explain the movement with Electrostatics Interaction.   

Here comes the question:

When the same situation is dealt with the object of such large length like 5*10^8 meters, it comes the problem.

The problem is,

From Special Theory of Relativity, “No kind of Electrical interaction can travel faster than Light Speed”.

But from Newton’s Laws of Motion, “The object that is the centre of mass of the object should instantaneously get acceleration

a = F/M and it is independent of time. The processes are instantaneous. ”. 

Which will happen in real life?

The question is whether the body will immediately move or not.

If it behaves by following the Special theory of relativity,

Then the other end of the object will move after some time interval of applied force. But it will be completely against the Newton’s laws of Motion. And it looks like a paradox.