## Wednesday, 14 July 2021

### Why does E=mc²?

Trying to follow the second worked example in Box 2.2 of Misner, Thorne, Wheeler I realised that I did not know how Einstein proved $E=mc^2$ and, when I checked in my books, both MWT and Carroll gloss it over! Using Einstein's 1905 paper, an internet video and the relativistic Doppler shift formulas given by Wikipedia I now do. Here's my version based on those.
We take two scenarios
1) observe the object, in its rest frame, emitting two pulses of light with energy $E/2$ in opposite directions so that it does not change velocity and then we take off in a rocket with velocity $v$ so the object has some kinetic energy $K_1$

2) measure its kinetic energy $K_2$ from a frame moving with velocity $v$ then it emits the light which will have energy doppler shifted according to $$E_r=E\left(1+\frac{v^2}{2c^2}\right)$$After the emission we are now in the final state of scenario 1 and the object must have kinetic energy $K_1$ and total energy $K_1-E$.

In both scenarios the object lost energy by radiation and gained kinetic energy (because the observer started moving) and we end up at the same place. The total energy must be the same in either scenario therefore$$K_1-E=K_2-E\left(1+\frac{v^2}{2c^2}\right)\Rightarrow\frac{Ev^2}{2c^2}=K_2-K_1$$We know how to calculate the kinetic energy ($mv^2/2$) and the velocities of the kinetic energies are the same so the masses must have changed. So$$\frac{Ev^2}{2c^2}=m_2\frac{v^2}{2}-m_1\frac{v^2}{2}\Rightarrow E=\Delta Mc^2$$where $\Delta M$ is the change in the mass of the object. Furthermore we conclude that if one could get all the energy out of an object of mass $m$ that energy would be
$$E=mc^2$$For more details including the derivation of $E_r$ see the very short 2.2 Box E=mc2.pdf.

Coming soon. More on that Box 2.2.