I was following a derivation of $E = mc^2$ in this video: <https://www.youtube.com/watch?v=KZ8G4VKoSpQ> The relevant parts are from 18:00 - 23:18.

I was perplexed at how the narrator introduced $E$. He seemed to have pulled it out of thin air. The process that he detailed is as follows:

The kinetic energy is the work done. Expressed using the particle's momentum and Newton's second law, we have

$KE = \int\_{x\_1}^{x\_2} \frac{dp}{dt} dx$ for a particle moving along a straight line, where $t$ is the time measured by a moving observer. I am already a little confused as to why $F \neq \frac{dp}{dt\_0},$ where $t\_0$ is the time measured in the rest frame. Particularly because he goes on to say that

$p = mv = m \frac{dx}{dt\_0}$. Okay, so this one gets derived with respect to $t\_0$, while the change in momentum is derived with respect to $t$? Seems inconsistent, but moving on...

Using a chain rule on $p$ by chaining through $t$, we find:

$p = \gamma m v$, where $v = \frac{dx}{dt}$. I'm confused again because I thought we started out by saying $v = \frac{dx}{dt\_0}$.

He goes on to compute $\frac{dp}{dt}$ to insert it into the integral. After some massaging, he gets $\frac{dp}{dt} = \gamma^3 m \frac{dv}{dt}$. Inserting this into the integral, and doing a change of variables to integrating over $dv$, he finds:

$KE = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2$. *Now for my main question.*

He goes on to say, "The expression $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} = KE + mc^2$ is simply equal to $E$." **Where is he getting this from, what is the definition of E**?

What is $E$? I know from classical mechanics that $E$ is the total mechanical energy, the sum of kinetic plus potential. It is a conserved quantity in problems where only conservative forces do work. Is this the angle he is coming at it from? He simply sees $KE + \text{something}$ and is trying to say that that $\text{something} = mc^2$ is a potential energy? I know that in classical mechanics, potential energy means that you can derive a conservative force from it. Is that the case here with $mc^2$?