Question

Question #b3589

Answer 1

Start with the relativistic momentum equation:
`p = (m_0 v)/sqrt(1-v^2/c^2` square and multiple top and bottom by `c^2`
`p^2c^2 = (m_0^2v^2c^2)/(1-v^2/c^2) = (m_0^2v^2c^4/c^2)/(1-v^2/c^2` re-arranger add and subtracting a term and write:
`= m_0^2c^4[v^2/c^2-1]/(1-v^2/c^2) + (m_0^2c^4)/(1-v^2/c^2)`
`= -m_0^2c^4[cancel(1-v^2/c^2]/cancel(1-v^2/c^2)] + cancel(m_0^2/(1-v^2/c^2))^(m^2) c^4`
`= -m_0^2c^4 + color(red)((mc^2)^2) = -m_0^2c^4 + color(red)(E^2)`
bring the negative term to the left rearrange and you have:
`color(red)(E^2) = (pc)^2 + (m_0c^2)^2`
`m_0 ne m` OK?!
You should note that `=>m^2=m_0^2/(1-v^2/c^2)`
Also I want to point out that this effectively a pythagorean identity with hypotenuse of `color(red)(E)` and the cateti `pc and m_0c^2`

Cheers!

Answer 2

Follow the explanation.

Explanation:

`E = (mc^2)/sqrt(1-(v/c)^2)`

`so, E^2 = (m^2c^4)/(1-(v/c)^2)=(m^2c^6)/(c^2-v^2)`

In the same way

`p = (mv)/sqrt(1-(v/c)^2)`

`so, p^2c^2 = (m^2v^2c^2)/(1-(v/c)^2)=(m^2v^2c^4)/(c^2-v^2)`

So,

`E^2-p^2c^2 = (m^2c^6)/(c^2-v^2)- (m^2v^2c^4)/(c^2-v^2) = m^2c^4 *((c^2-v^2)/(c^2-v^2)) =m^2c^4 = (mc^2)^2`

`=>E^2-p^2c^2= (mc^2)^2`

`=>E^2=p^2c^2 + (mc^2)^2`