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:

ux = dx/dt                     (2a)

uy = dy/dt                     (2b)

uz = dz/dt                      (2c)

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

Share and Like article, please:

Relativistic energy-momentum relation derivation

Relativistic energy momentum relation:

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

Share and Like article, please: