I wish to derive the relativistic energy-momentum relation $E^2 = p^2c^2 + m^2 c^4$ following rigorous mathematical steps and without resorting to relativistic mass.
In one spatial dimension, given $p := m \gamma(u) u$ with $\gamma(u) := (1 – \frac{|u|^2}{c^2})^{-1/2}$, the energy would be given by
$$E = \int{ \frac{d}{dt}p \space dx}$$
I'm having a hard time with this this integration.
How is the relation $E^2 = p^2c^2 + m^2 c^4$ rigorously derived starting from relativistic momentum, without resorting to relativistic mass?
To give an idea of the rigour I expect in an answer, in example, an answer I'd accept for the derivation of $ E = \frac{1}{2} m v^2$ in classical mechanics would have been as follows:
We seek to integrate the differential form $F \space dx$. Parametrising $x$ by $t$, we obtain $dx = \frac{d}{dt} x \space dt$.
The integral of interest is $\int F \space dx = m \int \frac{d^2}{dt^2}x \space dx = m \int (\frac{d^2}{dt^2}x) (\frac{d}{dt} x) dt$ after changing variables.
We recognize the integrand as $\frac{d}{dt} \left( \frac{1}{2} \left(\frac{d}{dt}x \right)^2 \right) $, and so the result $E = \frac{1}{2} m v^2$ follows from the fundamental theorem of calculus.
Again, as an example, a derivation of $E = \frac{1}{2} m v^2$ I would definitely not accept would be as follows:
$ \int F \space dx = m \int a \space dx = m \int \frac{dv}{dt} \space dx = m \int dv \frac{dx}{dt} $ = $ m \int v \space dv = \frac{1}{2}m v^2$.
Please carry out rigorous mathematical manipulations only.
Best Answer
Since $P = Fv$ we have $$\frac{dE}{dt} = \frac{dp}{dt} v$$ by Newton's second law. Integrating both sides with respect to $t$ gives $$\int \frac{dE}{dt} \, dt = \int v \frac{dp}{dt} \, dt = \int v \, dp$$ by the chain rule, aka ordinary $u$-substitution. We have $$p = \gamma m v = \frac{m v}{\sqrt{1-v^2}} \quad \Rightarrow \quad dp = \frac{m \, dv}{(1-v^2)^{3/2}}$$ where I set $c = 1$ for convenience and used the quotient rule. Integrating with initial and final velocities zero and $v_0$ gives $$E(v_0) - E(0) = \int_0^{v_0} \frac{mv}{(1-v^2)^{3/2}} \, dv = \frac{m}{\sqrt{1 - v_0^2}} - m.$$ At this point we cannot proceed further since we don't know the constant of integration. One can show by physical arguments that $E(0) = m$. Thus $$E(v) = \frac{m}{\sqrt{1-v^2}}$$ as desired. This isn't a hard derivation, but you're right: a lot of textbooks botch it.