# Question b3589

Apr 20, 2016

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:
$= {m}_{0}^{2} {c}^{4} \frac{{v}^{2} / {c}^{2} - 1}{1 - {v}^{2} / {c}^{2}} + \frac{{m}_{0}^{2} {c}^{4}}{1 - {v}^{2} / {c}^{2}}$
$= - {m}_{0}^{2} {c}^{4} \left[\frac{\cancel{1 - {v}^{2} / {c}^{2}}}{\cancel{1 - {v}^{2} / {c}^{2}}}\right] + {\cancel{{m}_{0}^{2} / \left(1 - {v}^{2} / {c}^{2}\right)}}^{{m}^{2}} {c}^{4}$
$= - {m}_{0}^{2} {c}^{4} + \textcolor{red}{{\left(m {c}^{2}\right)}^{2}} = - {m}_{0}^{2} {c}^{4} + \textcolor{red}{{E}^{2}}$
bring the negative term to the left rearrange and you have:
$\textcolor{red}{{E}^{2}} = {\left(p c\right)}^{2} + {\left({m}_{0} {c}^{2}\right)}^{2}$
${m}_{0} \ne m$ OK?!
You should note that $\implies {m}^{2} = {m}_{0}^{2} / \left(1 - {v}^{2} / {c}^{2}\right)$
Also I want to point out that this effectively a pythagorean identity with hypotenuse of $\textcolor{red}{E}$ and the cateti $p c \mathmr{and} {m}_{0} {c}^{2}$

Cheers!

Apr 20, 2016

#### Explanation:

$E = \frac{m {c}^{2}}{\sqrt{1 - {\left(\frac{v}{c}\right)}^{2}}}$

$s o , {E}^{2} = \frac{{m}^{2} {c}^{4}}{1 - {\left(\frac{v}{c}\right)}^{2}} = \frac{{m}^{2} {c}^{6}}{{c}^{2} - {v}^{2}}$

In the same way

$p = \frac{m v}{\sqrt{1 - {\left(\frac{v}{c}\right)}^{2}}}$

$s o , {p}^{2} {c}^{2} = \frac{{m}^{2} {v}^{2} {c}^{2}}{1 - {\left(\frac{v}{c}\right)}^{2}} = \frac{{m}^{2} {v}^{2} {c}^{4}}{{c}^{2} - {v}^{2}}$

So,

${E}^{2} - {p}^{2} {c}^{2} = \frac{{m}^{2} {c}^{6}}{{c}^{2} - {v}^{2}} - \frac{{m}^{2} {v}^{2} {c}^{4}}{{c}^{2} - {v}^{2}} = {m}^{2} {c}^{4} \cdot \left(\frac{{c}^{2} - {v}^{2}}{{c}^{2} - {v}^{2}}\right) = {m}^{2} {c}^{4} = {\left(m {c}^{2}\right)}^{2}$

$\implies {E}^{2} - {p}^{2} {c}^{2} = {\left(m {c}^{2}\right)}^{2}$

$\implies {E}^{2} = {p}^{2} {c}^{2} + {\left(m {c}^{2}\right)}^{2}$