Question

How to prove `m=m_0/(sqrt(1-(v^2/c^2)))?`

Answer 1

Let us consider two inertial frames of reference `S and S^'`. Let `S` is a stationary frame of reference with a stationary observer in it and `S^'` is the moving frame of reference. In the beginning, suppose these are at the same position so that respective observers `O` and `O^'` coincide. After time `t=t^'=0` frame `S^'` begin to move with a uniform velocity `v` along `x`-axis.

Suppose there are two particles moving towards each other in the frame `S^'`. According to the observer `O^'`, velocity of particle `A` will be `u^'` and of particle `B` will be `–u^'` .

As observed from frame `S`, Velocity of `A` is `u_1` and `B` is `u_2`. These velocities are given by relativistic addition of velocity expressions as below

`u_1 = (u^' + v)/(1 +(u^'v)/c^2)` .............(1)
`u_2 = (-u^' + v)/(1 -(u^'v)/c^2)` ..............(2)

Let `m_1 and m_2` be respective masses of `A and B` as observed from frame `S`.

Let the particles collide with each at a certain instant and momentarily come to rest. Even while at rest, these travel with the velocity `v` of the frame `S^'`.

According to the Law of conservation of momentum we have

`m_1u_1 + m_2u_2 = (m_1 + m_2)v = m_1v + m_2v`
`=>m_1(u_1 – v) = m_2(-u_2 + v)`

Using equations (1) and (2) in above we get

`m_1[(u^' + v)/(1 +(u^'v)/c^2) – v] = m_2[v -(-u^' + v)/(1 -(u^'v)/c^2)]`

`m_1[(u^' + v-v(1 +(u^'v)/c^2))/(1 +(u^'v)/c^2) ] = m_2[(v(1 -(u^'v)/c^2) +u^' - v)/(1 -(u^'v)/c^2)]`
`=>m_1[(u^' - (u^'v^2)/c^2)/(1 +(u^'v)/c^2) ] = m_2[( (-u^'v^2)/c^2 +u^' )/(1 -(u^'v)/c^2)]`
`=>m_1[1/(1 +(u^'v)/c^2) ] = m_2[ 1/(1 -(u^'v)/c^2)]`
`=>m_1/m_2 = (1 +(u^'v)/c^2)/ (1 -(u^'v)/c^2)` .............(3)

Now writing LHS of (1) as

`1 - u_1^2/c^2`

Similarly RHS can be written as

`1 - ((u^' + v)/(1 +(u^'v)/c^2))^2/c^2`
`=> 1 - (((u^' + v)/c)/(1 +(u^'v)/c^2))^2`
`=> ((1 +(u^'v)/c^2)^2 - ((u^' + v)/c)^2)/(1 +(u^'v)/c^2)^2`

Simplifying and equating with rewritten (1) we get

`1 - u_1^2/c^2 = (1 +(u^'v)^2/c^4 - (u^')^2/c^2 - (v^2)/c^2)/(1 +(u^'v)/c^2)^2` ................(4)

Similarly we get from (2)

`1 - u_2^2/c^2 = (1 +(u^'v)^2/c^4 - (u^')^2/c^2 - (v^2)/c^2)/(1 -(u^'v)/c^2)^2` ............(5)

On dividing equation (5) by (4) and taking square root of both sides we get

`sqrt(1 - u_2^2/c^2)/sqrt(1 - u_1^2/c^2) = (1 +(u^'v)/c^2)/(1 -(u^'v)/c^2)` .............(6)

Comparing (3) and (6) we get

`m_1/m_2=sqrt(1 - u_2^2/c^2)/sqrt(1 - u_1^2/c^2)` .......(7)

Let us assume that the particle `B` is at rest as seen from frame `S` that is it has zero velocity before collision

`=>u_2 = 0`

also writing `m_2 = m_0` where `m_0` is the rest mass of this particle,
With this equation (7) reduces to

`m_1/m_0=1/sqrt(1 - u_1^2/c^2)`

Using common notation as `m_1=mand u_1 = v`, above equation becomes

`m/m_0 =1 /sqrt(1 – v^2/c^2)`
`=>m = m_0/sqrt(1 – v^2/c^2)` .......(8)
This is the required derivation.