Proof of conservation of energy with the relativistic definition of force Professor Susskind proved conservation of energy in one of his lectures by taking the classical definition of force ($F = ma$) and by showing that its time derivative is zero. How can we do that with the modern, relativistic definition of force ($F = dP/dt$)? Can we do it the same way, or do we need to use a different method?
 A: SECTION A : Non-relativistic conservation of energy 
The work done by the non-relativistic force $\:\mathbf{f}\:$ per time unit, that is the power  produced or consumed, on a particle moving with velocity $\:\mathbf{v}=d\mathbf{r}/dt\:$ is
\begin{align}
\dfrac{dW}{dt}=\mathbf{f}\circ \mathbf{v}=&\dfrac{d\mathbf{p}}{dt}\circ \mathbf{v}=\\ \dfrac{d\left(m\mathbf{v}\right)}{dt}\circ \mathbf{v}=
& m \left(\dfrac{d\mathbf{v}}{dt}\circ \mathbf{v}\right)=\dfrac{1}{2}m\dfrac{d\left(\mathbf{v}\circ \mathbf{v}\right)}{dt}=\dfrac{1}{2}m\dfrac{d\Vert\mathbf{v}\Vert^{2}}{dt}
\tag{A-01} 
\end{align}   
In above equation the mass $\:m\:$ is considered constant (in the observer's inertial system) and invariant (an other observer in an other inertial system "feels" the same quantity $\:m\:$ as the mass of the particle).
The law that "force is the time rate of change of the linear momentum" is valid
\begin{equation}
\mathbf{f}=\dfrac{d\mathbf{p}}{dt}=\dfrac{d \left(m \mathbf{v}\right)}{dt}=m\dfrac{d\mathbf{v}}{dt}
\tag{A-02} 
\end{equation}   
So, the non-relativistic conservation of energy is expressed as
\begin{equation*}
\dfrac{dW}{dt}=\dfrac{d}{dt}\left(\dfrac{1}{2}m\upsilon^{2}\right)=\dfrac{dE_{kin}}{dt}
\end{equation*}
or
\begin{equation}
\dfrac{d}{dt}\left(-W\;+\;E_{kin}\right)=0
\tag{A-03} 
\end{equation}
where $\:E_{kin}\:$  the kinetic energy of the particle
\begin{equation}
E_{kin}=\dfrac{1}{2}m\upsilon^{2}
\tag{A-04} 
\end{equation}
If the force $\:\mathbf{f}\:$ is conservative, that is comes from a potential $\:U_{pot}\:$
\begin{equation}
\mathbf{f}=-\boldsymbol{\nabla} U_{pot} 
\tag{A-05} 
\end{equation}
then
\begin{equation}
\dfrac{dW}{dt}=\mathbf{f}\circ \mathbf{v}=\left(-\boldsymbol{\nabla} U_{pot}\right)\circ \dfrac{d\mathbf{r}}{dt}=-\dfrac{\boldsymbol{\nabla} U_{pot}\circ d\mathbf{r}}{dt}==-\dfrac{d U_{pot}}{dt}
\tag{A-06} 
\end{equation} 
and (A-03) yields
\begin{equation}
\dfrac{d}{dt}\left(U_{pot}\;+\;E_{kin}\right)=0
\tag{A-07} 
\end{equation}
SECTION B : Relativistic conservation of energy 
In Special Relativity the relation (A-02) is valid but for the differentiation in the last equality we must take account that the mass $\:m\:$ is no longer constant but depends explicitly on velocity, so implicitly on time
\begin{equation}
   m=\gamma m_{o}= m_{o}\left(1-\frac{\upsilon^2}{c^{2}}\right)^{-\frac{1}{2}}=\dfrac{m_{o}}{\sqrt{1-\dfrac{\upsilon^2}{c^{2}}}}\\   
\tag{B-01}   
\end{equation}
The mass $\:m_{o}\:$ is the rest mass of the particle, that is the mass seen by an observer moving with the particle. We consider that $\:m_{o}\:$ is constant in the system moving with the particle (if you add up material or energy of any kind, heat for example, to the particle in its rest frame, then $\:m_{o}\:$ is not constant).
So,
\begin{equation}
\mathbf{f}=\dfrac{d\mathbf{p}}{dt}=\dfrac{d \left(m \mathbf{v}\right)}{dt}=m\left( \dfrac{d\mathbf{v}}{dt}\right)+\left( \dfrac{dm}{dt}\right)\mathbf{v}
\tag{B-02} 
\end{equation}
and for the work done by the relativistic force $\:\mathbf{f}\:$ per time unit
\begin{equation}
\dfrac{dW}{dt}=\mathbf{f}\circ \mathbf{v}= m \left(\dfrac{d\mathbf{v}}{dt}\circ \mathbf{v}\right)+\dfrac{dm}{dt}\left(\mathbf{v}\circ \mathbf{v}\right)
\tag{B-03} 
\end{equation} 
Now
\begin{equation}
m \left(\dfrac{d\mathbf{v}}{dt}\circ \mathbf{v}\right)=\dfrac{1}{2}m\dfrac{d\left(\mathbf{v}\circ \mathbf{v}\right)}{dt}=\dfrac{1}{2}\gamma  m_{o}c^{2}\dfrac{d}{dt}\left(\dfrac{\upsilon^2}{c^{2}}\right)
\tag{B-04} 
\end{equation}
From (B-01)
\begin{equation}
 \dfrac{dm}{dt}=\dfrac{d\left(\gamma m_{o}\right)}{dt}= m_{o}\dfrac{d\gamma}{dt}=
m_{o}\dfrac{d}{dt}\left[\left(1-\frac{\upsilon^2}{c^{2}}\right)^{-\frac{1}{2}}\right]=\dfrac{1}{2} \gamma^{3} m_{o}\dfrac{d}{dt}\left(\frac{\upsilon^2}{c^{2}}\right)
\tag{B-05} 
\end{equation}
and so
\begin{equation}
 \dfrac{dm}{dt}\left(\mathbf{v}\circ \mathbf{v}\right)=\dfrac{dm}{dt}\upsilon^{2} =\dfrac{1}{2} \gamma^{3} m_{o}c^{2}\left(\frac{\upsilon^2}{c^{2}}\right)\dfrac{d}{dt}\left(\frac{\upsilon^2}{c^{2}}\right)
\tag{B-06} 
\end{equation}
Adding (B-04 )and (B-06) equation (B-03) yields
\begin{equation}
\dfrac{dW}{dt}=\dfrac{d m}{d t}\left(\mathbf{v}\circ \mathbf{v}\right) + m \left(\dfrac{d\mathbf{v}}{dt}\circ \mathbf{v}\right) =\dfrac{1}{2}\gamma  m_{o}c^{2}\dfrac{d}{dt}\left(\dfrac{\upsilon^2}{c^{2}}\right)\underbrace{\left[\gamma^{2}\left(\frac{\upsilon^2}{c^{2}}\right)+1\right]}_{\gamma^{2}}
\tag{B-07} 
\end{equation} 
That is
\begin{equation}
\dfrac{dW}{dt}=\underbrace{\dfrac{1}{2}\gamma^{3}  m_{o}\dfrac{d}{dt}\left(\dfrac{\upsilon^2}{c^{2}}\right)}_{(B-05):\dfrac{dm}{dt}}\cdot c^{2}=\dfrac{d\left( mc^{2}\right)}{dt}
\tag{B-08} 
\end{equation}
The quantity 
\begin{equation}
E_{total}= mc^{2}
\tag{B-09} 
\end{equation}
is the total energy of the particle and may be expressed as the sum of the internal energy and the kinetic energy
\begin{equation}
E_{total}= mc^{2}=\underbrace{m_{o}c^{2}}_{internal} + \underbrace{\left( m-m_{o}\right)c^{2}}_{kinetic}
\tag{B-10} 
\end{equation}
A: I will try to give you here the proof of work-energy theorem.
We have that $\bar p = {m \bar u \over \sqrt{1- {u^2 \over c^2} } } $ the relativistic momentum. We define the force as $\bar F ={dp \over dt} $. So the work is: $W= \int \bar F \cdot d \bar l =\int {d \bar p \over dt} \cdot d \bar l = \int  {d \bar p \over dt} {\cdot d \bar l \over dt } dt = \int {d \bar p \over dt} \cdot \bar u dt$. We calculate the inner product:
${d \bar p \over dt} \cdot \bar u ={d \over dt} ({m \bar u \over \sqrt{1- {u^2 \over c^2} } }) \cdot u  = {m \bar u \over (1 - {u^2 \over c^2})^{3/2} } \cdot {d \bar u \over dt} = {d \over dt} ({mc^2 \over \sqrt{1 - {u^2 \over c^2}} })= {dE \over dt}$
Thus, our integral is: $W= \int {dE \over dt} dt = ΔE $
Hope this helps.
