Mass-Energy Equivalence

Let's consider a particle of mass m₀ measured in the two inertial frames at rest oxyz and o'x'y'z'.
If the mass m₀ is fixed in the frame o'x'y'z' now moving at the speed u relative to the frame oxyz, the mass of the particle is still m₀ in o'x'y'z', but becomes m = γ m₀ for the observer at rest oxyz.

In the frame oxyz at rest:
If a force F (doing a work, then changing the kinetic energy), is applied on the particle, along the x-axis, we can write:
KE = ∫ F dx
F = dP/dt
P = mu = γm₀u (P is the linear momentum of the particle, and u is the speed of the particle measured in oxyz. (m₀ ix fixed in o'x'y'z')
Hence:
F = d(mu)/dt = mdu/dt + udm/dt, so
KE = ∫ Fdx = ∫ (mdudx/dt + udmdx/dt)
as u = dx/dt, thus:
KE = ∫ (mudu + u²dm) = ∫ ( mdu/dm + u)udm
But,
du/dm = 1/(dm/du) , then:
dm/du = m₀d[γ]/du
d[γ]/du = - d/du([ 1 - u²/c²]^1/2)/[ 1 - u²/c²]
d/du([ 1 - u²/c²]^1/2) = (1/2)[ 1 - u²/c²] ^{- 1/2} ( - 2u/c²) = [1 - u²/c²] ^{- 1/2} ( - u/c²)
Therefore:
d[γ]/du = (u/c²) /[ 1 - u²/c²]^3/2
dm/du = m₀d[γ]/du = (m/γ)d[γ]/du = m (u/c²) /[ 1 - u²/c²] = mu /[ c² - u²]
dm/du = mu /[ c² - u²] Then:
mdu/dm = [ c² - u²]/u
It follows that:
KE = ∫ ( mdu/dm + u)udm = ∫ ([ c² - u²]dm + u²dm) = ∫ c²dm
The mass varies from m₀ to m,
KE = ∫ c²dm = c²∫ dm = c²[m](m₀ → m)
KE = c²[m - m₀]
We do not have potentia;l energy here, we can just write:
KE + m₀c² = mc²
and call the term m₀c² energy at rest and E the total energy.
and then : E = E₀ + KE
With:

E = mc²

The total energy of a particle of mass at rest m₀, moving at velocity u is:
E = mc²
with m = m₀/[1 - u²/c²]^1/2.