First we show the work and energy in the linear motion of a particle with space-time (see J.M. Knudsen and P.G. Hjorth’s Elements of Newtonian Mechanics, 1995, p.51). Consider a particle of mass m moving along the r axis of r so all quantities are scalars. Newton’s second law is then
m d²r/dt² = F, or
m dv/dt = F,
with mass m, force F, and velocity v. Multiply both sides by v = dr/dt:
m (dv/dt) v = F (dr/dt), or
mv dv = F dr := dW,
where W is called the work done by the force F over the segment dr. Define T as
T = mv²/2,
which is called the kinetic energy of the particle. Then
dT = dW.
That is, the change in the kinetic energy of the particle over the segment dr equals the work done by the force F.
If F = F(r) does not depend on time, then define the potential energy U = U(r) through
dU(r) := –dW = –F(r)dr.
That is, the change in the potential energy U(r) over the segment dr is equal to minus the work done by the external force F. Since
dT = –dU(r),
and upon integrating,
T + U(r) = E,
where the constant E is called the total mechanical energy of the system.
Next we show the repose and lethargy in the linear motion of a tempicle (vass-instant) with time-space. Consider a tempicle of vass n moving along the t axis of w so all quantities are scalars. Newton’s second law for time-space is then
n d²t/dr² = Γ, or
n du/dr = Γ,
with vass n, release Y, and lenticity u. Multiply both sides by u = dt/dr:
n (du/dr) u = Y (dt/dr), or
nu du = Y dt := dX,
where X is called the repose done by the release Y over the time segment dt. Define V as
V = nu²/2,
which is called the kinetic lethargy of the tempicle. Then
dV = dX.
That is, the change in the kinetic lethargy of the tempicle over the segment dt equals the repose done by the release Y.
If Y = Y(t) does not depend on position, then define the potential lethargy Z = Z(t) through
dZ(t) := –dX = –Y(t)dt.
That is, the change in the potential lethargy V(t) over the time segment dt is equal to minus the repose done by the external release Y. Since
dV = –dZ(t),
and upon integrating,
V + Y(t) = M,
where the constant M is called the total mechanical lethargy of the system.