General Solution of Klein Gordon Equation

Let us discuss something about the general solution of the Klein Gordon equation which will be later useful when we adopt for the field formulation of our theory.  
Assuming our solution as a general waveform solution in the form,$$ \phi(x) = \frac{1}{(2\pi)^{2}}\int \,d^4k \tilde{\phi}(k) \,e^{-(ikx = ik_{\mu}x^{\mu})}$$ where 4 vector notation is generally implied unless it is specified. 
We substitute this general form in our equation to obtain the conditions to be satisfied from the KG equation. It gives, $$ \frac{1}{(2\pi)^{2}}\int \,d^4k \,e^{-ikx}\left(-k^2+m^2\right)\tilde{\phi}(k) $$ [Normalization factor $\sqrt{2\pi}$ for four integrations]. 

From the above equation, the solution should be of the form such that, if $k^2=m^2$, $\tilde{\phi}(k)$ can take any arbitrary value and if $k^2\neq {m^2}$ then $\tilde{\phi}(k)$ should be zero, 
therefore it should be in the form, $$\tilde{\phi}(k) = \delta(k^2-m^2)\,\tilde{f}(k)$$ 
Thus, our general wave solution cannot take any arbitrary form but should always satisfy the energy momentum relation - which is imposed by the dirac delta function. It is equivalent to saying that the energy can take only positive and negative values of $\omega_k$ given by $\delta(k^2-m^2) = \delta(k_0^2-|\vec{k}|^2-m^2) = \delta(k_0^2-\omega_k^2)$ where $\omega_k^2=|\vec{k}|^2+m^2$ 
Using the identity, $$\delta(f(x)) = \sum_{k}\frac{\delta(x-x_k)}{f'(x_k)}$$, we can rewrite our solution as,
$$\phi(x) = \frac{1}{(2\pi)^{2}}\int \,d^4k \left[e^{-ikx}\delta(k_0-\omega_k)\tilde{f}(k) + e^{-ikx}\delta(k_0+\omega_k)\tilde{f}(k)\right]$$ Now, integrating the zeroth component of momentum vector and using the fact that the integration is symmetric under $+\vec{k}$ and $-\vec{k}$ we can arrive at the form, $$\phi(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{-ikx}\frac{\tilde{f}(\omega_k,\vec{k})}{\sqrt{4\pi\omega_k}}+ e^{ikx}\frac{\tilde{f}(-\omega_k,-\vec{k})}{\sqrt{4\pi\omega_k}}\right]$$
Now we identify the terms other than exponential as $ a(k), b^*(k)$ just for a general convenience. $$\phi(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{-ikx} a(k) + e^{ikx} b^*(k)\right]\vert_{k_0=\omega_k} = \phi_+ +\phi_-$$ 
The complex conjugate is, $$\phi^*(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{ikx} a^*(k) + e^{-ikx} b(k)\right]\vert_{k_0=\omega_k} $$ 

For real fields, one can show that, $a(k) = b(k)$ and $a^*(k) = b^*(k)$

Similarly, the momentum space terms can be written as, $$a(p) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3x}{\sqrt{2\omega_p}} \left[\omega_p \phi(x) +i\partial_0\phi(x)\right]e^{ipx}\vert_{p_0=\omega_p} $$ and $$b^*(p) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3x}{\sqrt{2\omega_p}} \left[\omega_p \phi(x) - i\partial_0\phi(x)\right]e^{ipx}\vert_{p_0=\omega_p}$$ 
Hint: One can prove from RHS to LHS by integrating with respect to $d^3x$, making use of the definition of Dirac delta function, integrating again with respect to $d^3k$ and finally using the fact that $\omega_k=\omega_{-k}=\omega_p=\omega_{-p}$ - the energy is independent of the direction of the momentum vector. 
It is always a good practice to use different symbols to discuss it in a general form of the derivation. 
Finally, one can obtain $a^*(p), b(p)$ by taking the complex conjugate. 

Klein Gordon Equation from Correspondence Principle in Relativistic domain

The Klein Gordon equation is developed from the general relativistic energy momentum relation by the substitution of corresponding operators as, $$ E = i\hbar\frac{\partial}{\partial{t}}$$ $$ \vec{p} = -i\hbar\nabla$$
acting on $\phi(x,t)$ to give, 
$$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$
where the d'alembertian operator is defined here as $$ \Box =  \frac{1}{c^2}\partial_t^2 - \nabla^2$$

We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ \eta_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have,
$$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.
To make life simple, we adopt to the convention,
where we equate the planck's constant and the speed of light equal to 1 (dimensionless number). The consequences are, the dimension of Length and Time are the same and the dimension of Mass is just the inverse of Length or Time. In addition, Energy and Momentum is measured in the same unit as of the Mass. 

And, contravariant vectors are $ A^{\mu}= (A_0, \vec{A})$ and corresponding covariant transformation is $ A_{\mu}= \eta_{\mu\nu}A^{\nu} = (A_0, -\vec{A})$ where the differential is defined in reverse way 
$ \partial_\mu = \left(\partial_0,\vec{\nabla}\right) $
[$\partial_0 = \partial_t$] and 
$ \partial^{\mu}= \eta^{\mu\nu}\partial_{\nu} = ({\partial}_0,-\vec{\nabla})$
With these substitution we get our KG equation as, $$\left(\Box + m^2\right)\phi = 0 $$ where $p^\mu{p}_\mu = m^2 \\ \,\, E^2 = \omega^2 = m^2 + {|\vec{p}|^2} $
Since $m^2$ and d'Alembertian operator is invariant under Lorentz transformation, if the function $\phi$ satisfies the condition $ \phi'(x') = \phi(x)$ then the whole KG equation is invariant under Lorentz transformation. 
But, there are some problems with this equation in the basic definition of $\phi(x)$. First of all it cannot be the wave function of the particle as it is in the case of Non-relativistic Schr\"{o}dinger equation.

The problem arises because of the probability statements. To understand its significance, let us just assume that $\phi(x)$ is the usual wave function. 

Then, it says that the probability of finding the particle at a position x is the same as of the finding the particle in $x'$ position in some other reference frame [Lorentz invariance condition]. It implies the wave function should behave like a scalar quantity and independent of direction, which is not true in general for spin half particles. The properties of spin half particles depends on from which direction it is measured and change with respect to different orientations of the reference frame. 

And the second and the most important problem is that the probability density definition changes completely and it allows for the weird possibility of the negative probability. 
The continuity equation with the time component becomes, $$\partial_0\rho+\vec{\nabla}\cdot\vec{J} = 0 = \partial_\mu{J}^\mu$$ where $ \rho = \frac{1}{2} \left[\phi^*(\partial_0\phi) - (\partial_0\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $ 
and \newline
$\vec{J} = \frac{-1}{2}\left[\phi^*(\nabla\phi) - (\nabla\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
where $\rho$ is the equivalent probability density as defined in non-relativistic case can now have positive as well as negative values, that is not compatible with the definition of probability - you can check it for monochromatic wave of the for $\phi = Ae^{\pm{ikx}}$. Thus, either we will have to abandon KG equation or should find an alternative way of description for its definition.

Relativistic Field theory - Euler Lagrange equation

To understand some of the advanced theories in monopoles and etc., we will take a look on the relativistic field theory.

Classically, the motion of a particle is understood by solving for the position of the particle as a function of time. Where else the fields are defined over a region in space as $\phi(x_i)\,\,\,i=0,1,2,3$ where $x_0$ is the time component taken together with the space coordinates.These fields give all the information we would like to know about the system. 

In classical method, the Lagrangian is defined as a function of coordinates and its time derivatives. Since, we take here time as one of the components, the new Lagrangian for our fields is a function of both $\phi$ and its derivatives about each of the components. 
We can obtain our new Euler Lagrange equation as,
$$ \delta \int\,L(\partial_\mu\phi,\phi) d\tau = 0 $$ where the L here is called the Lagrangian density and $d\tau = dx_0 \,dx_1 \,dx_2 \,dx_3$
On expansion, $$ \int\,\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\delta(\partial_\mu\phi)+\frac{\partial{L}}{\partial\phi}\delta(\phi)\right) d\tau = 0$$ using $\delta(\partial_\mu\phi) = \partial_\mu(\delta\phi)$ and product rule we get, $$ \int \left(\frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi) }\delta\phi\right) - \frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right)\delta\phi + \frac{\partial{L}}{\partial\phi}\delta{\phi}\right) d\tau = 0 $$
The first term on integrating and applying the condition $\delta\phi$ becomes zero at the end points becomes zero. Taking out the negative sign, $$\int\left(\frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right)-\frac{\partial{L}}{\partial\phi}\right)\delta\phi\,d\tau = 0$$
gives us the final Euler-Lagrange equation for relativistic field theory, $$ \frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial{L}}{\partial\phi} = 0 $$
It needn't to be only one function $\phi$ but more functions of $\phi_j$.
Let us consider in specific, the Klein Gordon equation which describes the motion of particles with spin zero. The equation is developed from the general relativistic energy momentum relation by the substitution of corresponding operators as, $$ E = i\hbar\frac{\partial}{\partial{t}}\\ \vec{p} = -i\hbar\nabla$$ acting on a scalar field $\phi(x,t)$ to give, $$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$ where the d'alembertian operator is defined here as $$ \Box =  \frac{1}{c^2}\partial_t^2 - \nabla^2$$

We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ g_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have, $$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.

This can be derived from the Lagrangian, $$ L = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \frac{1}{2}\left(\frac{mc}{\hbar}\right)^2\phi^2 $$ 

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].

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. 

Signals faster than Light Speed

Light Speed cannot be achieved by any means of Physics. No signal can be transmitted faster than light Speed. I used to hear these statements often when studying relativity theory.

Keeping this in mind I started to think about signals. Why signals can’t be transmitted faster than light speed.  

I suddenly thought about an experiment with a rod.

Let us say I have a rod that is 5, 00,000 km - five lakhs kilometers of length. Assume that the stick is incompressible and rigid.

Now I am going to pull one of its ends to small displacement “dx” or small rotation “dθ”

Everyone knows from our experience that the other end of the rod spontaneously without any time gap would have been moved by the same displacement “dx” or rotation “dθ”.

I don’t think signals cannot be transmitted with this rod.

I will use the following way to transmit the signal to other end of the rod that is five lakh kilometers away from this end. It was already announced to my friend that if the rod moves below it indicates the number zero and if the rod moves above then it indicates the number one. In this way my friend on the other side can easily note down the numbers using a machine and can easily understand the messages as in the same way how a computer does.

You can also imagine two computers instead of two humans. The computers can interact in the following way though they are separated with the distance of five lakhs kilometers.

I am not arguing with Special theory of Relativity. I know the velocity defined in Special theory of relativity defines different signal velocity. Here I am not talking about signal velocity but I am just trying to get a way how we can transmit a signal for such a long distances. And I just want to simply show that it is not impossible for us to transmit signals to such large distances.

Using this method we can transmit signals between the stars and the galaxies by slowly spreading throughout the entire universe.

Here it was already assumed that my friend is there on the other side. It is easy to make him go on the other side by simply make him to sit on the other point of the rod. After making him to sit on the other side we can start to build the rod and move it in the space slowly.

This is just an idea but not the absolute way. I am just trying to prove the possibility of transmitting these signals. Using perfect Engineering we can find a more suitable method. But I don’t want go into that Engineering.    

Do you think there is anything wrong with this method?

Time travel to the Past

I used to watch science programs and sci-fi films related with time travel. In those programs they used to describe the time travel as people going into the past or the future using a portal.

Going to the future is possible with Einstein’s theory of Relativity if you are in a rocket that travels near to the speed of light.

But what about travelling back in time? It looks impossible using the basic concepts of Physics.

It is used to say that we can travel through time by Wormholes.

A Wormhole, also known as an Einstein–Rosen bridge, is a hypothetical topological feature of space-time that thought to be a "shortcut" through space-time. It is much like a tunnel with two ends each in separate points in space-time. We don’t know how we are going to travel in the past. But let us assume that we found a way to travel back in time. With a portal like worm hole, if one person can travel through time, then following questions arises in everybody’s mind.

Will there be the past of the man who is travelling?

I watched the Stephen Hawking’s program where he used to say that,

What if I killed myself by travelling back in time?  

But I am not going to ask the same question, but I want to ask questions rather something more related with the basic laws of physics i.e. The Conservation of Energy.

The Conservation of Energy is the fundamental tool we are using all over the Physics. Neither Classical Mechanics nor Quantum Mechanics can work without this Conservation of Energy. Energy can neither be created nor can be destroyed.

If it is assumed that with a time travel portal, if a person could travel through time, then it implies that he is suddenly vanishing from the Universe that we are living.

The human body is simply nothing but the Energy from Einstein’s E = mc^2 equation. That is we are just a form of energy.

If the person vanished through time then it means that Energy is destroyed in that frame of reference.  And if he come suddenly out of the portal by travelling through the past then,

In the reference frame of the Past time that is if somebody is doing experiments in the past and if he just happened to see the person who had been sent from the future, then it implies that in his reference frame Energy is created from nothing. In both of these reference frames Conservation Laws are violated.

May be you can argue that, if I see the Space and time as a unique one that is a reference frame outside the Space and time, then

The Conservation laws are not violated since the Energy disappeared in the Future time is immediately appears in the Past time. So energy is linked and so you will conclude that, “Neither the energy is destroyed nor created”.

But I strongly disagree this argument since from Einstein’s theory of Relativity; Newton’s laws are valid in all reference frames independent of Space and time.

So, Newton’s laws of Motion or Conservation laws should not be violated in both of the reference frames.

Any Energy or any force can’t be created out of nothing. Time travel in backwards that is to the past would violate all of the Conservation Laws. 

Basic Outline of Relativity

We can understand the basic outline of theory of relativity in a simple following way from Fermat’s principle of least time.

First you need to understand Fermat’s Principle of least time.

It states that the path taken between two points by “a ray of light” is the path that can be traversed in the least time. So it implies light will always travel in the shortest path between two points.

After believing in this principle, we can take an example of moving rocket with 9.8 m/s^2 of acceleration and earth.

From Newton’s Gravitational law, it was known that all objects near to the earth will fall at a rate of acceleration 9.8 m/s^2. Many of them had learnt this in their school days. Consider that the rocket is at first on earth at rest. If you are inside the rocket, you will see the things inside the rocket will fall at a rate of 9.8 m/s^2 when it is subjected to fall from a height not comparable to the radius of the earth.

[Since gravitational field of earth will be different at different places]

Now, I made you to sleep in the rocket by giving you anesthesia. After you had slept, the rocket was started and made to fly into the space with an acceleration of 9.8 m/s^2 .

From Newton’s law it is known that a man inside the rocket will not feel any difference between these two situations. So, he will think he is still at earth, since the experiments done in the rocket will be as same as it is at earth.

Now if you take torch light and pointed it on the horizontal direction, from very careful measurements you can see that light beam bends exactly same amount as it is in earth.

Since all the experiments done with other objects gives same results in earth as well as in rocket, it implies Light should also give same results independent of the place. 

Thus we came to know that Light indeed bend near the gravitational field. 

From this we can get two information.

One is,

• Newton’s Gravitational law says that only masses are attracted by gravitational force. But here light is attracted towards the earth, where light is thought to be pure energy. It shows that energy and mass has some relation.

The second information is derived from Fermat’s principle.

• If light travels in the shortest path between two points,

In horizontal direction the shortest path between the two points is the straight line drawn between those two points. But light here is actually travels in a bend path not following Fermat’s principle.

To make Fermat’s principle the true one, it was just made a small adjustment in the statement that follows as,

The shortest path between the two points in horizontal direction near the gravitational field is really the bend path.

That says that the space itself is a curved one. And light travels in the shortest path in that curved space where the path of the light is curved. Thus we can get two aspects of Relativity theory.

I am explaining here just the simple outline of Relativity theory and not the complete one.

This is to get just a little basic knowledge of Relativity theory to common people.