# Mass-Energy Equivalence

While going through the derivation of $$E=mc^2$$, I've stumbled across this integral and don't know how the integration takes place. A few quickies:

How does integral 1 transform to integral to 2 and that to integral 3?

Thank you

Note by John Muradeli
3 years, 9 months ago

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It is pretty straight forward. So, the proof begins by stating that kinetic energy is equal to the work done by all the forces on the body. (They have perhaps assumed the to be initially at rest). So,

$$\displaystyle K = \int \limits_0^s F \text{d}s$$
$$\displaystyle K = \int \limits_0^s \frac{\text{d}p}{\text{d}t} \text{d}s \text{ Since, } F = \frac{\text{d}p}{\text{d}t}$$
$$\displaystyle K = \int \limits_0^s \frac{\text{d}(mv)}{\text{d}t} \text{d}s$$ , which is the first integral. But $$\displaystyle \frac{\text{d}s}{\text{d}t} = v$$, therefore,
$$\displaystyle K = \int \limits_0^{mv} v \text{ d}(mv)$$ Since, at a displacement $$s$$, the particle has a momentum $$p = mv$$, the limits have been changed accordingly, giving the second integral.
$$\displaystyle K = \int \limits_0^{v} v \text{ d}\bigg[ \frac{m_0 v }{\sqrt{1 - v^2/c^2}} \bigg]$$ . Simply substituting for relativistic mass, we obtain the third integral. Since variable is just $$v$$, the limits have been modified accordingly. The integral is a cakewalk using integration by parts.

$$\begin{array} \displaystyle \int \limits_0^{v} v \text{ d}\bigg[ \frac{m_0 v }{\sqrt{1 - v^2/c^2}} \bigg] & = \frac{m_0 v^2 }{\sqrt{1 - v^2/c^2}} \bigg|_0^v - \int \limits_0^{v} \frac{m_0 v }{\sqrt{1 - v^2/c^2}} \text{d}v \\ & = \frac{m_0 v^2 }{\sqrt{1 - v^2/c^2}} -(- m_0 c^2 \sqrt{1 - v^2/c^2} ) \bigg|_0^v \\ & = \frac{m_0 v^2 }{\sqrt{1 - v^2/c^2}} - m_0 c^2 \\ & = m c^2 - m_0 c^2 \\ \end{array}$$

Therefore, $$\displaystyle K = m c^2 - m_0 c^2$$ and $$\displaystyle E = K + m_0 c^2 \Rightarrow E = m c^2$$.

Hope this helps.

- 3 years, 9 months ago

I have a proof without integration Will post later.

- 3 years, 9 months ago