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

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