Special relativity energy

Question

How can I get $E=\gamma mc^2$ from $E^2-(pc)^2=(mc^2)^2$?

Answer 1

Start with the rest frame so $p_0=0$ which gives $E_0=mc^2$ where $E_0$ is the energy in the rest frame and $m$ is the invariant mass.

Now, form the four-momentum in the rest frame $(E/c,\vec p)=(mc,\vec 0)$ and then boost to a frame moving at $v$ to get $(E/c,\vec p)=(\gamma m c, \gamma m v, 0,0)$ so then the timelike component gives $E=\gamma mc^2$

Answer 2

we know that: $\;\;E^{2}=(mc^{2})^{2}+p^{2}c^{2}\;\;\;,$ $\;\;\;p=\gamma mv$

$E^{2}=m^{2}c^{4}+\gamma^{2}m^{2}v^{2}c^{2}=m^{2}c^{4}\left(1+\gamma^{2}\frac{v^{2}}{c^{2}}\right)$

$E^{2}=\gamma^{2}m^{2}c^{4}\left(\frac{1}{\gamma^{2}}+\frac{v^{2}}{c^{2}}\right)=(\gamma\,m c^{2})^{2}$

so $\;\;\;\;E=\gamma\,m c^{2}$

Answer 3

From the rapidity approach (where $(v/c)=\tanh\theta$ and $\gamma=\cosh\theta$),
the question reads:
How can I get $E=mc^2\cosh\theta$ from $E^2−(pc)^2=(mc^2)^2$?

Akin to what @TheTiler wrote, we know $p=mc\sinh\theta$.
So, \begin{align} E^2 &=(mc^2)^2+(pc)^2\\ & =(mc^2)^2+(mc^2\sinh\theta)^2\\ & =(mc^2)^2\left(1+\sinh^2\theta\right)\\ & =(mc^2)^2\left(\cosh^2\theta\right)\\ E &= mc^2\cosh\theta\qquad\qquad \mbox{(taking the positive root)} \end{align} In other words, \begin{align} E^2−(pc)^2&=(mc^2)^2\\ (mc^2)^2\cosh^2\theta−(mc^2)^2\sinh^2\theta&=(mc^2)^2\\ \cosh^2\theta−\sinh^2\theta &=1 \end{align} (so the above is the Minkowski version of the Pythagorean theorem for the components of the "energy-momentum 4-vector")
and $$\frac{pc}{E}=\frac{c\cdot mc\sinh\theta}{mc^2 \cosh\theta}=\tanh\theta=\frac{v}{c}.$$ (and a ratio of the two components of an "energy-momentum 4-vector" is Minkowski's version of the slope).