Special theory of relativity : Explained in simple words

First of all you will imagine why this is named as special theory of relativity?

Well that is because this theory deals with a special case, Special case that is this theory deals with particles moving at constant velocity. That is why this is called special theory of relativity. Einstein’s other theory General theory of relativity deals with accelerated and decelerated motion which is called General theory of relativity.

Space and time

We all know that we require co-ordinates in order to pin point the exact location of a point is space if it is stationary with respect to co-ordinate system. If the point is moving then we will need an additional parameter to determine the location of the particle that is time.

4 DIMENSIONS OF SPACE

As shown in above image after big bang our known universe came into existence. If we desire to define one point in this reference frame assuming our reference frame is stationary, we will need 4 parameters that will determine the exact location of that point that is (X,Y,Z,t).

From this we can imagine time as 4th dimension. As we can move in 3 dimensional space with our will, we can’t move in 4th dimension with our will, till now. That is the only difference between time and other dimensions.

Measurement of length

Now consider yourself on a railway platform. that is our 1st frame of reference we will call it ‘K1’ and moving train which is moving at constant velocity ‘v’ is our 2nd frame of reference and we will call it ‘K2’

Now let’s perform an experiment according to Einstein’s instructions.

Take a rod which has length equal to unity. that rod is traveling with the train at constant velocity ‘v’. Now if we try to measure the length of the rod from platform that is from reference frame K1 the result we will get will be smaller that unity. whose proof is explained below.

Lorentz Transformation

Here we will try to establish relation between our co-ordinate systems K1 and K2

Two frame of references K1 and K2(K2 moving at constant velocity ‘v’)

Now assume a light signal travelling along positive X axis is transmitted according to equation

x=ct       or         x-ct=0 . . . . . . . . .(1)

Since the signal has to be transmitted relative to K2 with velocity c, the propagation relative to the system K2 will be represented by the analogous formula

x’-ct’=0 . . . . . . . .(2)

(Because speed of light is constant in vacuum doesn’t depends on the speed of the reference frame in which it is travelling.

Now those points who satisfy (1) must also satisfy (2), which will give us a relation (3) which is fulfilled in general, where M and N is a constant.

(x’-ct’)=M(x-ct). . . . . . . .(3)

If we apply quite similar considerations to the light travelling in negative X direction we obtain the condition

(x’+ct’)=N(x+ct). . . . . . . (4)

By adding and subtracting equations (3)and(4)

x’-ct’=Mx-Mct                                                     x’-ct’=Mx-Mct

+                                                                                 –

x’+ct’=Nx+Nct                                                      x’+ct’=Nx+Nct

——————-                                       ——————-

2x’= (M+N)x – (M-N)ct                                          2ct’= (M+N)ct-(M-N)x

Assuming,

a=(M+N)/2                                                                    b=(M-N)/2

we will have following equations

x’=ax-bct . . . . . . . . . . (5a)

ct’=act-bx. . . . . . . . . . (5b)

Here as we already know x’ is position co-ordinate of K2 frame’s as observed from K1

that is co-ordinates of a point as observed from platform travelling at constant velocity inside the train.

Now to solve our problem we will need the values of a and b. for that origin of K2 we have permanently x’=0 and hence according to equation (5)

x=(bc)t/a

If we call v the velocity with which the origin of K2 is travelling relative to K1 then re writing above equation will give us velocity v

x/t=(bc)/a                             therefore     v=(bc)/a . . . . . . . . (6)

Now here comes our rod of unit length

Relativity teaches us that as judged from K the length of a unit measuring rod which is at rest with respect to K2 must be exactly the same as length judged from K1. In order to see how the points of the x’ axis appear as viewed from K1 we require to tale snapshot of K2 from K1. That means we have to insert a particular value of t (time of K1) e.g. t=0

from equation (5a)and (5b)

x’=ax

Two points of x’ axis (Moving train’s frame) which are separated by the distance Δx’=1 (Length of rod) when measured in K2(measured inside moving train)are thus separated in our instantaneous photograph (Taken from platform) by the distance

Δx=1/a

to understand this equation here are some notes

x’1=ax1                                            x’2=ax2

Δx’=x’1-x’2

Δx’=a(x1-x2)

Now,  Δx’=1 (Length of the rod as measured from train)

and (x1-x2)= Δx             ∴1=a Δx      ∴ Δx=1/a

But now if snapshot taken from K2(t’=0), and if we eliminate t from the equations (5) Taking into account the expression (6)

0=act-bx                                                 ∴ t=(b/ac)x

Now, in equation (5a)

x’=ax-bct

x’= ax- bc (bx/ac)                               ∴ x’=ax- (bc/a)(bx/c)                      ∴x’=ax-vx(bc/c²)

Because equation (6)

∴ x’=a(x-(v/c²)(bc/a)x)

 

Now,  Δx= Δx’  which will imply that

Solving for “a” we will get

 

Now, Put the values of a and b in equations (5a) and (5b) we will obtain values of x’ and t’

Calculation of length

Assuming our rod is placed in such a way that one end of the rod is at origin(x’=0) and other end of the rod is at x’=1. Now, what is the length of the rod as observed from platform ?

For that we will calculate from above results that where our rod’s both ends lie with respect to stationary frame of reference K1 (Railway platform) using equations for x’ and finding x from that relation by putting x’=0 or 1

so the distance between the points is    But the meter rod is moving with velocity “v”  relative to K1 so apparent length will be shorter for the rod as seen from platform.

So, according to this if we assume that peed of the train v=c/2 then we can say that apparent length of the rod as observed from the platform will be 0.8660 meters.

  Behavior of time

Consider a seconds clock situated inside that train (K2) at x’=0

Two ticks of the clock are at t’=0 and t’=1

Now, again using x’ and t’ equations we can calculate that

at x’=0 t’=0  ⇒ t=0            and               at x’=0 t’=1

As we know that train is moving at velocity v time elapsed at platform is more than one second while time elapsed inside the train is one second.

This implies that Bodies who travel fast enough experience slower time with respect to bodies who are stationary.

For our example v=c/2 observed time dilation is 1.15 min

i.e. while train will observe 1 min platform will observe 1.15 min

which means if you got off the train after 1 year you will be 0.15 years younger then you should have been.

The historic equation E=mc²

In accordance with the theory of relativity the kinetic energy of a material of point mass m is no longer given by the well known equation

E= mv²/2 But by the expression                                

Now, according to maxwell : A body moving with the velocity v, which absorbs an amount of energy E(as judged from reference frame moving with the body)in form of radiation without suffering an alteration in velocity in the process, has as a consequence, it’s energy increased by an amount.

 Now total energy of the body can be obtained by combining both the equations

Thus we can say that if body absorbs the energy E without changing in velocity it will increase it’s inertial mass by E/c².

This equation formulates to the historic equation E=mc²

With the advent nuclear transformation processes, which result from the bombardment of elements by α- particles, protons, deuterons, neutrons or ϒ-rays, the equivalence of mass and energy expressed by the relation E=mc² has been amply confirmed. the sum of the reacting masses, together with the mass equivalent of the kinetic energy of the bombarding particle(Photon), is always greater than the sum of the resulting masses. The difference is the equivalent mass of the kinetic energy of the particles generated, or of the released electromagnetic energy ( ϒ-photons).In the same way , the mass of a spontaneously disintegrating radioactive atom is always greater than the sum of the masses of the resulting atoms by the mass of equivalent of the kinetic energy of the particles generated (or the photonic energy).

So, this was the gist of special theory of relativity. I tried to explain it as briefly as i could, Hope it helped.

 

 

Bibliography

1. 1st Image source : https://universe-review.ca/I15-71-4Dsupernova.jpg
2. 2nd Image source : http://www.cv.nrao.edu/course/astr534/LorentzTransform.html
3. Equation png generated at : http://www.codecogs.com/latex/eqneditor.php
4. Reference book used:  Relativity : the special and general theory (100th anniversary edition)-Albert Einstein