Here we show the *work* and *energy* in the linear motion of a particle in 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* d²*r*/d*t*² = *F*, or

*m* d*v*/d*t* = *F*,

with mass *m*, force *F*, and velocity *v*. Multiply both sides by *v* = d*r*/d*t*:

*m* (d*v*/d*t*) *v* = *F* (d*r*/d*t*), or

*mv* d*v* = *F* d*r* := d*W*,

where *W* is called the *work* done by the force *F* over the segment d*r*. Define *T* as

*T* = *mv*²/2,

which is called the *kinetic energy* of the particle. Then

d*T* = d*W*.

That is, the change in the kinetic energy of the particle over the segment d*r* 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

d*U*(*r*) := –d*W* = –*F*(*r*)d*r*.

That is, the change in the potential energy *U*(*r*) over the segment d*r* is equal to minus the work done by the external force *F*. Since

d*T* = –d*U*(*r*),

and upon integrating,

*T* + *U*(*r*) = *E*,

where the constant *E* is called the total mechanical energy of the system.

Here we show the *invork* and *invertegy* in the linear motion of a transicle (point vass) in time-space. Consider a transicle 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* d²*t*/d*r*² = *Γ*, or

*n* d*u*/d*r* = *Γ*,

with vass *n*, rush *Γ*, and legerity *u*. Multiply both sides by *u* = d*t*/d*r*:

*n* (d*u*/d*r*) *u* = *Γ* (d*t*/d*r*), or

*nu* d*u* = *Γ* d*t* := d*X*,

where *X* is called the *invork* done by the *rush **Γ* over the time segment d*t*. Define *V* as

*V* = *nu*²/2,

which is called the *kinetic invertegy* of the transicle. Then

d*V* = d*X*.

That is, the change in the *kinetic invertegy* of the transicle over the segment d*t* equals the *invork* done by the *rush **Γ*.

If *Γ* = *Γ*(*t*) does not depend on position, then define the *potential invertegy* *Y* = *Y*(*t*) through

d*Y*(*t*) := –d*X* = –*Γ*(*t*)d*t*.

That is, the change in the potential invertegy *V*(*t*) over the time segment d*t* is equal to minus the *invork* done by the external *rush **Γ*. Since

d*V* = –d*Y*(*t*),

and upon integrating,

*V* + *Y*(*t*) = *Z*,

where the constant *Z* is called the total mechanical invertegy of the system.