Can there be something faster than speed of light? This is the question that occasionally come into mind of physics students. The speed of light is 3 x 100000000(meter/sec. Basically the answer of this question lies in the Einstein postulates of theory of relativity. Continue reading “Can someone travel faster than speed of light”

## Common queries related to theory of relativity-1

Q1. Who proposed the theory of relativity?

Ans: Albert Einstein.

Q2. What is the difference between general theory of relativity and special theory of relativity?

Ans: General Theory of relativity: This theory deals with the the accelerated objects.

Special theory of relativity: This theory of relativity deals with the non-accelerated objects.

Out of the above two, the special theory of relativity is easier to understand because it deals with the non-accelerated objects. Continue reading “Common queries related to theory of relativity-1”

## No signal can travel faster than the speed of the light

Proof that no signal can travel faster than the speed of the light.

**Solution**: **Method 1**

As relativistic addition of velocity relation is already derived and from Addition of velocity relation

u = (u’ + v)/ (1 + u’(v/c^{2}))

Suppose there is a signal which travels equal to the speed of light that is put u’ = c and then try to solve, the answer will be

u = c

If we put u’ = c and v = c then solve, we get

u = c

It proves that no signal can travel faster than the speed of the light.

**Method 2:**

Suppose there is a signal which travels faster than the speed of light, that is

v > c

By relation of length contraction in relativity

l = l’(√1 – v^{2}/c^{2})

if we put v > c, then l become imaginary but length can not be imaginary. Therefore it prove that no signal can travel faster than the speed of the light.

## Lorentz transformation equations for space and time

Results of Galilean Transformation equations can not be applied for the objects moving with a speed comparative to the speed of the light.

Therefore new transformations equations are derived by Lorentz for these objects and these are known as Lorentz transformation equations for space and time.

Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.

Let an event happen at position P in the frame S’. The coordinate of the P will be x’ according to the observer in S’ and it will be x according to O in S.

The frame S’ has moved a distance “vt” in time t (refer figure).

What should be the relation between x and x’? As we can see from the figure that from frame S’

x’ α x – vt

or x’ = k (x – vt) (1)

where k is constant of proportionality that we will determine.

Similarly from frame S

x = k(x’ + vt’) (2)

Put equation (1) in (2)

x = k[k(x – vt) + vt’]

or x/k = kx – kvt + vt’

or vt’ = x/k – kx + kvt

or t’ = x/kv – kx + kvt

or t’ = kt – kx (1 – 1/k^{2})/v (3)

Similarly from frame S, time t will be

t = kt’ + kx’ (1 – 1/k^{2})/v (4)

(This equation can be derived by putting equation 2 in 1 and then solving.)

**Calculation of k**: Continue reading “Lorentz transformation equations for space and time”

## Michelson-Morley experiment

**Aim of the Michelson-Morley experiment**: The Michelson-Morley experiment was done to confirm the presence of hypothetical medium called ether.

Therefore, one question should be there what was ether? Yes, I have used “was ether” not “is ether”. Let us discuss why?

As we have already discussed in earlier articles that there is nothing like absolute rest. Thus the scientists in the 19^{th} century assumed that our universe is filled with hypothetical medium called ether. Ether was supposed to be transparent and highly elastic.

The main objective of this Michelson-Morley experiment was to check the presence of this medium called ether. The aim was supposed to be fulfilled by measuring the velocity of the earth with respect to the ether. If earth is supposed to be propagating through the stationary ether with a uniform velocity and if a beam of light is sent from source to observer towards the direction of the motion of the earth, then it should take more time if sent through the opposite direction. If this time difference can be measured then velocity of earth with respect to ether can be measured.

**Experimental arrangement of Michelson-Morley experiment**: Continue reading “Michelson-Morley experiment”

## Real life example of time dilation

**Real example of time dilation:**

As we have already discussed the concept of time dilation. Let us discuss its example:

**Decay of µ- mesons: **

µ- mesons are the particles formed in the earth atmosphere. The half life time of µ- mesons is 3.1 microseconds.

They travel with the speed of 0.9c.

where c is the speed of the light.

So they must covered the distance d = vt

d = 3.1 x 10^{-6} x 0.9c = 840m

It means there population should become half after this distance. But this does not happen. Population remains much higher than the half value.

Why this happened? This is because the time here should be dilated time and 3.1 microseconds should be the proper time. This is because the µ- mesons are travelling with speed comparable to the speed of the light.

So t = t’/(1 – v^{2}/c^{2})

Here t’ = 3.1 microseconds

After solving, we get

t = 7.2 microseconds

thus distance traveled by µ- mesons will be

d = vt

or d = 7.2 x 10^{-6} x 0.9c

or d = 1920 m

Now when the population is measured after this distance it was approximately half.

It proves that the time dilation is a real effect.

## Variation of mass with velocity and its derivation

Do you know that there is variation of mass with velocity in relativity that is mass varies with the velocity when the velocity is comparable with the velocity of the light. Let us derive and discuss the variation of mass with the velocity relation:

Let there are two inertial frames of references S and S1. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.

Suppose there are two particles moving in opposite direction in frame S’. velocity of particle A will be u’ and of B will be –u’ according to the observer O’. Continue reading “Variation of mass with velocity and its derivation”

## Relativistic addition of velocity

Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.

Suppose a particle P is place in frame S’ and it is moving.

The velocity component of particle P from observer O’ in frame S’ will be:

u’_{x} = dx’/dt’ (1a)

u’_{y} = dy’/dt’ (1b)

u’_{z} = dz’/dt’ (1c)

The velocity component of particle P from observer O in frame S will be:

u_{x} = dx/dt (2a)

u_{y} = dy/dt (2b)

u_{z} = dz/dt (2c)

From Lorentz transformation equations: Continue reading “Relativistic addition of velocity”

## Relativistic energy-momentum relation derivation

Relativistic energy momentum relation:

From Einstein mass energy relation

E = mc^{2 } (1)

Also from variation of mass with velocity relation

m = m_{0}/(1 – v^{2}/c^{2})^{1/2} (2)

Where m_{0} is the rest mass of the object

Put value of m in equation (1) and then square both sides, we get

E^{2}= m_{0}^{2}c^{4}/(1 – v^{2}/c^{2}) (3)

As momentum is given by

p = mv

Put equation (2) and square

p^{2} = m_{0}^{2}v^{2}/(1 – v^{2}/c^{2})^{ }

Multiply both sides by c^{2}

p^{2}c^{2} = m_{0}^{2}v^{2} c^{2}/(1 – v^{2}/c^{2}) (4)

Subtract equation (4) from (3) and solve, we get

E^{2} – p^{2}c^{2} = m_{0}^{2}c^{4}

Or E = (p^{2}c^{2 }+^{ }m_{0}^{2}c^{4})

This is Relativistic energy momentum relation

^{ }

## Simultaneity in relativity

Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.Let two events in frame S occur simultaneously at positions P1 and P2. The coordinates of the P1 will be (x1,y1,z1,t1) and of P2 will be (x2,y2,z2,t2). The events will be simultaneous (occur at the same time) according to the observer in frame S. Therefore

t1 = t2 (1) Continue reading “Simultaneity in relativity”