Proving $e = (\Delta m) c^2$ by linear approximation

Question

In relativity the mass is $\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ at velocity $u$. By Problem 20 this is near $m_0 + $ _____ for small $v$. Show that the kinetic energy $\frac 1 2 mv^2$ and the change in mass satisfy Einstein's equation $e = (\Delta m) c^2$.

We start with $m \approx m_0 + \frac 1 2 \frac {v^2}{c^2}$. Then, we have $\Delta m \approx \frac 1 2 \frac {v^2}{c^2}$. This equation looks strange as there is mass on the left side and no unit on the right side. However, we ignore it and proceed. We now get $(\Delta m) c^2\approx \frac 1 2 v^2$. Why are we are missing an $m$ on the right side?

Answer 1

Recall from the binomial theorem that we have

$$\left(1-x\right)^{-1/2}=1+\frac12 x+O(x^2)$$

Now, using $x=v^2/c^2$ we have

$$m\approx m_0\left(1+\frac{v^2}{2c^2}\right) $$

so that

$$m-m_0=\frac12 m_0v^2/c^2$$

Finally, let $\Delta m=m-m_0$. Then, we obtain

$$\Delta mc^2=\frac12 m_0v^2$$

Answer 2

The correct equation is $m\approx m_0(1+\frac{v^2}{2c^2})\implies \Delta m_0=\frac{m_0v^2}{2c^2}$ Then, the dimensions are matched.

Answer 3

According to relativity, if $m_0$ is the mass of a body at rest then the mass $m$ of a body at a velocity $v$ $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=m_0\left(1-\frac{v^2}{c^2}\right)^{-1/2}$$
apply binomial theorem for $v^2/c^2<<1$, $$m=m_0\left(1-\left(-\frac12\right)\frac{v^2}{c^2}\right)$$
$$m=m_0\left(1+\frac{v^2}{2c^2}\right)$$
$$m=m_0+\frac{m_0v^2}{2c^2}$$
$$m-m_0=\frac{m_0v^2}{2c^2}$$
$$\Delta m=\frac{m_0v^2}{2c^2}$$
$$\Delta m c^2=\frac{1}{2}m_0v^2$$