# Can you replace c with velocity in E = mc^2?

## If the question is about the energy of a neutron moving at a certain fraction of the speed of light. mass of neutron is given.

Jun 23, 2018

see below

#### Explanation:

It really depends upon what you mean by "certain fraction".

Generalising that, the relativistic expressions for energy and momentum of a free particle moving in 1-D are:

$\left\{\begin{matrix}E = m {c}^{2} = \gamma {m}_{o} {c}^{2} \\ p = m v = \gamma {m}_{o} v q \quad \square\end{matrix}\right. q \quad q \quad \left\{\begin{matrix}{m}_{o} = \text{rest mass" \\ gamma = 1/sqrt(1-v^2/c^2) = "Lorentz factor}\end{matrix}\right.$

• With rest energy: $q \quad {E}_{o} = {m}_{o} {c}^{2}$

So the additional energy in the rest frame of a relativistic free particle due to its motion can be explored as:

${E}^{2} - {E}_{o}^{2} = \left({\gamma}^{2} - 1\right) {\left({m}_{o} {c}^{2}\right)}^{2}$

$= \left(\frac{1}{1 - {v}^{2} / {c}^{2}} - 1\right) {\left({m}_{o} {c}^{2}\right)}^{2}$

$= \left(\frac{{v}^{2} / {c}^{2}}{1 - {v}^{2} / {c}^{2}}\right) {\left({m}_{o} {c}^{2}\right)}^{2}$

$= {\gamma}^{2} {v}^{2} / {c}^{2} {m}_{o}^{2} {c}^{4}$

• From $\square$: $q \quad {v}^{2} = {p}^{2} / \left({\gamma}^{2} {m}_{o}^{2}\right)$

$= {\gamma}^{2} \frac{{p}^{2} / \left({\gamma}^{2} {m}_{o}^{2}\right)}{c} ^ 2 \setminus {m}_{o}^{2} {c}^{4}$

$= {p}^{2} {c}^{2} q \quad \left[= {E}^{2} - {E}_{o}^{2}\right]$

Therefore:

${E}^{2} = \textcolor{b l u e}{{p}^{2} {c}^{2}} + {\left({m}_{o} {c}^{2}\right)}^{2}$

In the real world, we work with the blue term