This post is about the conservation of (space) energy and time energy. I wrote about the conservation of fulmentum *here*. See also the post on *Work, effort and energy*.

First, here is a derivation of the *conservation of (space) energy* from classical physics:

The law of the conservation of (space) energy states that the total (space) energy in an isolated system remains constant over time (distime). The total (space) energy over an arbitrary length of distime, Δ*t*, is constant. Let the total (space) energy at two times be *E*_{1} and *E*_{2}. Then:

(*E*_{2} – *E*_{1})/Δ*t* = 0.

Since the total energy equals the *kinetic space energy* (*KSE*) plus the *potential space energy* (*PSE*), we have

(*KSE*_{2} + *PSE*_{2} – *KSE*_{1} – *PSE*_{1})/Δ*t* = 0

= (*KSE*_{2} – *KSE*_{1})/Δ*t* + (*PSE*_{2} – *PSE*_{1})/Δ*t* = 0

= (Δ*KSE* – Δ*PSE*)/Δ*t*.

Now *KSE* = ½*mv*² with mass, *m*, and velocity, *v*, so

(½*mv*_{2}² – ½*mv*_{1}²)/Δ*t* = ½*m*(*v*_{2}² – *v*_{1}²)/Δ*t*

= ½*m*(*v*_{2} – *v*_{1})(*v*_{2} + *v*_{1})

= *m*(*v*_{2} – *v*_{1})/Δ*t* + (*v*_{2} + *v*_{1})/2.

This is mass times the average acceleration times the average velocity:

= *mav* = F*v*

because of Newton’s second law, F = *ma.*

As for *potential space energy*, a change in potential space energy is related to work as follows:

Δ*PE* = –*W* = –FΔ*x*

where Δ*x* is a change in position. So we have

Δ*PE*/Δ*t* = –F(Δ*x*/Δ*t*) = –F*v*

since Δ*x*/Δ*t* is the average velocity. Putting this together we have

F*v* – F*v* = 0

which completes the derivation of the conservation of space energy over an arbitrary distime.

Next is a derivation of the *conservation of time energy* similar to that above:

The law of the conservation of time energy states that the total time energy in an isolated system remains constant over space (distance). The total time energy over an arbitrary distance, Δ*s*, is constant. Let the total time energy at two distances be *E*_{1} and *E*_{2}. Then:

(*E*_{2} – *E*_{1})/Δ*s* = 0.

Since the total time energy equals the *kinetic time energy* (*KTE*) plus the *potential time energy *(*PTE*), we have

(*KTE*_{2} + *PTE*_{2} – *KTE*_{1} – *PTE*_{1})/Δ*s* = 0

= (*KTE*_{2} – *KTE*_{1})/Δs + (*PTE*_{2} – *PTE*_{1})/Δ*s* = 0

= (Δ*KTE* – Δ*PTE*)/Δ*s*.

Now *KTE* = ½*nu*² with vass, *n*, and legerity, *u*, so

(½*nu*_{2}² – ½*nu*_{1}²)/Δ*s* = ½*n*(*u*_{2}² – *u*_{1}²)/Δ*s*

= ½*n*(*u*_{2} – *u*_{1})(*u*_{2} + *u*_{1})

= *n*(*u*_{2} – *u*_{1})/Δ*s* + (*u*_{2} + *u*_{1})/2.

This is vass times the average expedience times the average legerity:

= *nbu* = Γ*u*

because of the dual of Newton’s second law, Γ = *nb*.

As for *potential time energy*, a change in potential time energy is related to effort as follows:

Δ*PTE* = –*X* = –ΓΔ*t*

where Δ*t* is a change in time position. So we have

Δ*PTE*/Δ*s* = –Γ(Δ*t*/Δ*s*) = –Γ*u*

since Δ*t*/Δ*s* is the average legerity. Putting this together we have

Γ*u* – Γ*u* = 0

which complete the derivation of the conservation of time energy over an arbitrary distance.

Note that there is a relation between kinetic space energy and kinetic time energy:

*KTE* = ½*nu*² = ¼(1/½*mv*²) = ¼/*KSE*.

There is also a relation between F and Γ:

Γ*u* = *nbu* = *b*/*mv* = *ab*/*mav* = *ab*/F*v*.

(*u*/*b*)Γ = (*a*/*v*)/F.