iSoul Time has three dimensions

Conservations of energy

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, Δt, is constant. Let the total (space) energy at two times be E1 and E2. Then:

(E2E1)/Δt = 0.

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

(KSE2 + PSE2KSE1PSE1)/Δt = 0

= (KSE2KSE1)/Δt + (PSE2PSE1)/Δt = 0

= (ΔKSE – ΔPSE)/Δt.

Now KSE = ½mv² with mass, m, and velocity, v, so

mv2² – ½mv1²)/Δt = ½m(v2² – v1²)/Δt

= ½m(v2v1)(v2 + v1)

= m(v2v1)/Δt + (v2 + v1)/2.

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

= mav = Fv

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:

ΔPE = –W = –FΔx

where Δx is a change in position. So we have

ΔPEt = –F(Δxt) = –Fv

since Δxt is the average velocity. Putting this together we have

Fv – Fv = 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, Δs, is constant. Let the total time energy at two distances be E1 and E2. Then:

(E2E1)/Δs = 0.

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

(KTE2 + PTE2KTE1PTE1)/Δs = 0

= (KTE2KTE1)/Δs + (PTE2PTE1)/Δs = 0

= (ΔKTE – ΔPTE)/Δs.

Now KTE = ½nu² with vass, n, and legerity, u, so

nu2² – ½nu1²)/Δs = ½n(u2² – u1²)/Δs

= ½n(u2u1)(u2 + u1)

= n(u2u1)/Δs + (u2 + u1)/2.

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

= nbu = Γu

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:

ΔPTE = –X = –ΓΔt

where Δt is a change in time position. So we have

ΔPTEs = –Γ(Δts) = –Γu

since Δts is the average legerity. Putting this together we have

Γu – Γu = 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 = ½nu² = ¼(1/½mv²) = ¼/KSE.

There is also a relation between F and Γ:

Γu = nbu = b/mv = ab/mav = ab/Fv.

(u/b)Γ = (a/v)/F.

Post Navigation