Mr. Einstein gave his theory of relativity. According to one of his postulates that nothing can travel faster than the speed of light and speed of light is constant for free space or vacuum.

This was his theoretical assumption and the he was so great and of course he is still great that no one till date can prove him wrong. Scientists tried so many times to disapprove this concept but always failed. Recent example is some scientists proved through their research that neutrinos can travel faster than the speed of light. But at the end they also failed to prove this to the world. I will discuss about neutrinos later on. For today, I can just tell that neutrinos are elementary subatomic particles and they are everywhere and travel near to the speed of light.

But no one was able to prove Einstein wrong means till that no one can travel faster than the speed of light.

If you know something interesting science concept, you can share. If you like the article, then  post in the comment section.

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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.

Q3: With what kinds of objects or particles are studied in the theory of relativity?
Ans: The objects or particles moving with a relativistic speed that is the speed comparative to the speed of the light are covered under the theory of the relativity.

Q4. Will there be any change in the length of the object, if it moves with relativistic speed?
Ans: Yes, the appeared length will be contracted.

Q5. Will there be any change in the time of the object, if it moves with relativistic speed?
Ans: Yes, the time will be dilated that is it will be increased or in other words the clock will then move slow.
For example, as there are 60 seconds in 1 minute, and if the clock moves with relativity speed then suppose its time will be now more than 60 sec in 1 minute, thus the clock will be slower then.

Note: If you have any query related to relativity, kindly post in the comment section.

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Solution: Method 1

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

u = (u’ + v)/ (1 + u’(v/c²))

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 – v²/c²)

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.

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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/k²)/v                                (3)

Similarly from frame S, time t will be

t = kt’ + kx’ (1 – 1/k²)/v                                               (4)

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

Calculation of k:

Let us suppose a flash of light is emitted from the common origin of S and S’ at time t=t’=0. From Einstein’s 2^nd second postulate, the flash of light travels with the velocity of light c and which remains same in both the frames.

After sometime, the position of the flash of the light as seen from observer O will be

x = ct

And as seen from O’ will be

x = ct’      (Here the form of Physics law is same that is position = (velocity)(time) from Einstein 1^st postulate)

Put these two values in equation (1) and (2) respectively, we get

ct’ = k (ct – vt) = kt (c –v)

and ct = kt’ (c + v)

Multiply above two equations

c²tt’ = k²tt’(c² – v²)

or k² = c²/(c² – v²)

or k² = 1/  (1- v²/c²)                             (5)

or k = 1/√(1- v²/c²)                              (6)

The k is known as relativistic factor.

Substitute equation (6) in (1), we get

x’ = (x – vt)/(√1 – v²/c²)                                   (7)

As it is assumed that frame S’ is moving only along x direction, therefore along y and z direction

y’ = y                                                   (8)

And z’ = z                                            (9)

Equations 7-9 are known as Lorentz transformation equations for space.

Let us derive Lorentz transformation equation for time:

Cross-multiply equation (5)

1/k² = 1 – v²/c²

Or 1 – 1/k² = v²/c²

Put the above equation in equation (3)

t’ = kt – kx(v²/c²)/v

or t’ = k (t – kxv/c²)

Put value of k from equation 5 in above equation, we get

t’ = (t – kxv/c²)/ (√1 – v²/c²)                 (10)

Equation (10) is Lorentz transformation equation for time.

Equations 7 -10 are known as Lorentz transformation equations for space and time. These are again rewritten below:

x’ = (x – vt)/(√1 – v²/c²)

y’ = y

z’ = z

t’ = (t – xv/c²)/(√1 – v²/c²)

If the frame is changed (that is from S), then the equations are known as Lorentz inverse transformation equations for space and time. These are given as:

x = (x’+ vt’)/(√1 – v²/c²)

y = y’

z = z’

t = (t’ + x’v/c²)/(√1 – v²/c²)

Special case:

If v <<< c

Then Lorentz equations will become Galilean by neglecting v²/c² or v/c² wherever necessary as shown below:

x’ = x – vt

y’ = y

z’ = z

t’ = t

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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: The light is emitted from source S and is incident on a collimating lens L (Refer figure 1). The L will make the light parallel and the beams will now incident on a plate P which is inclined at an angle of 45⁰. The P will divide the light into two parts, one reflected part and other transmitted beam. The reflected beam will incident on a mirror M1 and transmitted on M2.

The separation between the P and M1 and P and M2 is same and that is equal to l (suppose) and this separation is called length of the arm. The light will reflected back from mirrors M1 and M2 respectively and will interfere at P. This interference pattern is noticed by Telescope T.

Derivation and Discussion of Michelson-Morley experiment:

(Special Note to students: The Michelson and Morley started the experiment in his laboratory and they themselves tried to derive the results theoretically using certain physics laws. If the theory and experiments results matched, then it is alright. If it does not, then either theory is wrong or experiment.

So students, now assume you are Michelson and Morley and you are doing the theoretical calculations in your respective copies. Then we will match the result with the experiment. Therefore, Let us start and wait what will happen?

As the observer is assumed to be on ether and he is studying the motion of the earth wit respect to ether. Thus the light beam when incident on P and reflected towards M1. It will catch the M1 at new position M’1 at B (according to Michelson-Morley). Then the light reflected back to P at C (Refer Figure).

Here the all the apparatus are assumed to be moving with the speed of the earth that is v and speed of light is c. As A to B is more distance than A to P’, even then the light will catch the M’1 at the same time as from A to P’. This is because c > v. Thus the time taken for light to reach at M’1 and for P to reach at P’ will be taken same as t.

Calculation for time taken for reflected path that is from A to B and then B to C:

Take triangle ABP’ and apply Pythagoras theorem:

(AB)² = (BP’)² + (AP’)²

Where AB is the path covered by light in time t and AP’ is path covered by the Plate P in same time t. BP’ is the length of the arm that is equal to l.

Thus the equation becomes

c²t² = l² + v²t²

or t²(c² – v²) = l²

or t² = l²/(c² – v²)

or t² = l²/ c² (1 – v²/ c²)

t = l/ c [(1 – v²/ c²)]^1/2

or t = l[(1 – v²/ c²)]^-1/2/c

Apply Binomial theorem and neglect higher terms

t = l[(1 + v²/2 c²)]/c

This is the time taken by light from A to B. Same time will be taken from B to C. Therefore, the total time for the reflected path will be

t1 = 2t

or t1 = 2l[(1 + v²/2 c²)]/c                                                          (1)

Calculation for time taken for transmitted path that is from A to M’2 and then M’2 to A:

As the apparatus and the light both are moving in same direction that is when light is going towards M’2. Thus the relative velocity will be c – v. After reflection, the apparatus and the light both are moving in the opposite direction that is when light is going towards P. Thus the relative velocity will be c + v.

Thus the time taken from A to M’2 and from M’2 to C will be

t2 = l/(c – v)   +  l/(c + v)

Further solving, we get

2l(1 – v²/ c²)^-1/c

Apply binomial theorem to the RHS and neglect higher terms, we get

t2 = 2l(1 + v²/ c²)/c                                                       (2)

Therefore the time difference between the transmitted and reflected rays will be

∆t = t2 – t1

Put equations (1) and (2)

∆t = 2l(1 + v²/ c²)/c  –  2l[(1 + v²/2 c²)]/c

Or ∆t = lv²/c³

Path difference for the transmitted and reflected rays will be

∆x1 = c∆t

∆x1 = c lv²/c³

Or ∆x1 =  lv²/c² (3)

After this the apparatus is rotated through 90⁰ so that mirrors will exchange their positions and path difference is again calculated. It will come out

∆x2 =  -lv²/c² (4)

Therefore the total path difference will be

∆x = ∆x2 – ∆x1

∆x = 2 lv²/c² (5)

Then the number of fringe shift is calculated by following relation (fringes are the pattern obtained by interference of two more rays that is by constructive and destructive interference):

N = path difference/wavelength of light

N = ∆x/λ

Put equation (5)

N = 2 lv²/c² λ                           (6)

Then N is calculated by putting l = 11m, v = 3 x 10⁴m/s, c = 3 x 10⁸m/s and λ = 5800 angstroms

Thus N = 0.37 fringes

But experimentally N = 0

Thus the theory and experiment results are not matched.

But the experimental were right. So there was some problem in theory calculation.

Different scientists tried to explain these negative results of Michelson-Morley experiment.

The explanation of negative results of Michelson-Morley experiment is given in separate article here.

(The main reason lies in equation (2), when c – v and c + v is done. Can you add any velocity in c or subtract any velocity from c? The answer is given in another article in the explanation of negative results of Michelson-Morley experiment.)

The post Michelson-Morley experiment first appeared on Winner Science.

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 – v²/c²)

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.

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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’.

Let us study the velocities and mass of these particles from frame S.

Velocity of A is u1 and B is u2 from frame S and these are given by  relativistic addition of velocity relation respectively:

u1 = (u’ + v)/(1 +u’v/c²)                                   (1)

u2 = (-u’ + v)/(1 -u’v/c²)                                  (2)

Let m1 and m2 are the mass of A and B from frame S respectively.

As the particles are moving to each other, at certain instant they will collide and momentarily came to rest. But even when they came to rest, they travel with the velocity of the frame S’ that is with v.

According to the law of conservation of momentum:

Momentum before collision = momentum after collision

Thus  m1u1 + m2u2 = (m1 + m2)v = m1v + m2v

Or  m1(u1 – v) = m2(u2 – v)

Put equations (1) and (2) in above equations, we get

m1[(u’ + v)/(1 +u’v/c²) – v] = m2[v – (-u’ + v)/(1 -u’v/c²)]

Then take LCM of terms in the bracket and solve, we get

m1[1/(1 +u’v/c²)] = m2[1/(1 -u’v/c²)]

or m1/m2 = (1 +u’v/c²)/ (1 -u’v/c²)                               (3)

Now square equation (1), then divide both sides by c² and subtract both sides by 1, we get

1 – u1²/c² = 1 – [(u’ + v)/c/(1 +u’v/c²)]²

By taking LCM on RHS and solving, we get

1 – u1²/c² = (1 + u’²v²/c⁴ – u’²/c² –v²/c²)/ (1 +u’v/c²)² (4)

Similarly by squaring equation (2), then dividing both sides by c² and subtracting both sides by 1, we get

1 – u2²/c² = (1 + u’²v²/c⁴ – u’²/c² –v²/c²)/ (1 -u’v/c²)² (5)

On dividing equation (5) by (4), we get

(1 – u2²/c²)/(1 – u1²/c²) = (1 +u’v/c²)²/(1 -u’v/c²)²

Take square root on both sides

(1 – u2²/c²)^1/2/(1 – u1²/c²)^1/2 = (1 +u’v/c²)/(1 -u’v/c²)    (6)

Now compare equations (3) and (6), we get

m1/m2 = (1 – u2²/c²)^1/2/(1 – u1²/c²)^1/2 (7)

This is more of a complicated result. To make this result simple, let us assume that the particle B is in the state of rest from frame S that is it has zero velocity before collision

Thus  u2 = 0

And m2 = m0

Where m0 is the rest mass of the particle,

Therefore equation (7) becomes

m1/m0 =1 /(1 – u1²/c²)^1/2

Also assume u1 = v and m1 = m

Therefore above equation becomes

m/m0 =1 /(1 – v²/c²)^1/2

or m =mo /(1 – v²/c²)^1/2 (8)

This equation represents the equation of the variation of mass with the velocity.

It shows that if a particle or anything moves with a speed comparable to the speed of the light then its mass will appeared increased because the factor /(1 – v²/c²)^1/2 will always be in decimal and something divided by decimal will be more. For example on dividing 1 by .1, we get 10.

Numerical: An object is moving with relativistic speed and it has mass equals to 3 times its rest mass.   Calculate its velocity.

Solution:

Given m = 3m0

Put this is in equation (8)

3m0 =mo /(1 – v²/c²)^1/2

Or 3 = 1/(1 – v²/c²)^1/2

Squaring and solving, we will get c. Try yourself.

I have also derived and discussed earlier the following very important relativistic relations:

1.      Length contraction

2.      Time dilation

3.      Addition of velocity

4.      Einstein mass energy relation

Note from winnerscience: If you want the e-notes/e-book of relativity or laser or both then contact us at winnerscience@gmail.com

The post Variation of mass with velocity and its derivation first appeared on Winner Science.

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:

x’ = (x – vt)/(√1 – v²/c²)                       (3a)

y’ = y                                       (3b)

z’ = z                                        (3c)

t’ = (t – xv/c²)/(√1 – v²/c²)        (3d)

Differentiate equations (3)

dx’ = (dx – vdt)/(√1 – v²/c²)     (4a)

dy’ =d y                                               (4b)

dz’ = dz                                                (4c)

dt’ = (dt – dxv/c²)/(√1 – v²/c²)  (4d)

Now substitute equations (4a) and (4d) in equation (1a), we get

u’_x = (dx – vdt)/ (dt – dxv/c²)

Divide numerator and denominator of R.H.S with dt, we get

u’_x = (dx/dt – v)/ (1 – dx/dt(v/c²))

Put equation (2a) in above equation,

u’_x = (u_x – v)/ (1 – u_x(v/c²))       (5)

Similarly by putting equations (4b) and 4d) in equation (1b) and then dividing the numerator and denominator of R.H.S with dt, and then putting equation (2b), we get

u’_y = u_y(√1 – v²/c²)/ (1 – u_x(v/c²))                     (6)

Similarly by putting equations (4c) and 4d) in equation (1c) and then dividing the numerator and denominator of R.H.S with dt, and then putting equation (2c), we get

u’_z = u_z(√1 – v²/c²)/ (1 – u_x(v/c²))                      (7)

Equations (5, 6 and 7) represent the addition of velocity relations as observed by observers O’ from frame S’.

From the observer O in frame S, the relations (5, 6 and 7) will become:

u_x = (u’_x + v)/ (1 + u’_x(v/c²))                             (8)

u_y = u’_y(√1 – v²/c²)/ (1 + u’_x(v/c²))                    (9)

u_z = u’_z(√1 – v²/c²)/ (1 + u’_x(v/c²))                    (10)

Generally equation (8) is written as

u = (u’ + v)/ (1 + u’(v/c²))                                 (11)

Special case: If v <<< c, then v/c² will get neglected and the equation 11 will become

u = u’ + v

The post Relativistic addition of velocity first appeared on Winner Science.

From Einstein mass energy relation

E = mc² (1)

Also from variation of mass with velocity relation

m = m₀/(1 – v²/c²)^1/2 (2)

Where m₀ is the rest mass of the object

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

E²=  m₀²c⁴/(1 – v²/c²)                (3)

As momentum is given by

p = mv

Put equation (2) and square

p² = m₀²v²/(1 – v²/c²)

Multiply both sides by c²

p²c² = m₀²v² c²/(1 – v²/c²)         (4)

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

E² – p²c² = m₀²c⁴

Or E = (p²c² + m₀²c⁴)

This is Relativistic energy momentum relation

The post Relativistic energy-momentum relation derivation first appeared on Winner Science.

t1 = t2                                                              (1)

The question arises here will the event be simultaneous from frame S’? Let us discuss it?

From Lorentz transformation equations of time:

t’1 = (t1 – x1v/c²)/(√1 – v²/c²)                          (2)

and t’2 = (t2 – x2v/c²)/(√1 – v²/c²)                   (3)

Subtracting equation (2) from equation (3), we get

t’2 – t’1 = (t2 – x2v/c²)/(√1 – v²/c²) – (t1 – x1v/c²)/(√1 – v²/c²)

or t’2 – t’1 = (t2 – t1)/ )/(√1 – v²/c²) – v/c²(x2 – x1)/(√1 – v²/c²)

Put equation (1) in above equation, we get

t’2 – t’1 = – v/c²(x2 – x1)/(√1 – v²/c²)        (4)

As the two events occur at different positions, that is

x2 ≠x1

Therefore L.H.S of equation (4) will not be zero, thus

t’2 – t’1 ≠ 0

or t’1 ≠ t’2

It proves that the same events will not be simultaneous from frame S’, that is it will not appear to occur at the same time from S’.

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