Derivation of $E = mc^2$. Difference between $KE$ and $E$?

Question

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 $mc^2$?

Answer 1

In the video $E$ is the total energy of the object (but not including any potential energies).

In relativity theory every mass has an energy associated with it, $mc^2$, so $E$ is the sum of K.E. and the internal rest mass energy $mc^2$.

Answer 2

If a particle is moving with velocity $v$ in a frame, then its momentum and force are defined as :

$$ \overrightarrow p = m_0\frac{d \overrightarrow x}{d\tau}$$ $$F=\frac{dp} {dt}$$

where $\tau$ is the proper time (the time measured by the clock in the particle). You can show that if momentum is conserved in all frame for a closed system of particles, then the quantity $\sum_i \gamma m_0c^2$ is conserved as well which have the dimension of energy as well.(if a component of a four-vector vanishes in all frame, then other components vanish as well.) Hence, this quantity is called the energy of the system. Now, if you calculate the total energy when the velocity is zero, you get $m_0c^2$ which is the rest energy. Therefore, we can define $K.E = (\gamma-1)mc^2$. I would also recommend you to read the article by Ehlers , Penrose and Rindler (https://aapt.scitation.org/doi/10.1119/1.1971205)