This post is about the conservation of energy and lethargy. I wrote about the conservation of levamentum here. See also the post on Work, effort and energy.
First, here is a derivation of the conservation of energy from classical physics:
It is required that forces must be conservative: F(t2) − F(t1) = 0 for all t.
The law of the conservation of energy states that the total energy in an isolated system remains constant over time (distime). The total energy over an arbitrary length of distime, Δt, is constant. Let the total energy at two times be E1 and E2. Then it is required that:
(E2 – E1)/Δt = 0.
Since the total energy equals the kinetic energy (KE) plus the potential energy (PE), we have
(KE2 + PE2 – KE1 – PE1)/Δt = 0
= (KE2 – KE1)/Δt + (PE2 – PE1)/Δt = 0
= (ΔKE – ΔPE)/Δt.
Now KE = ½mv² with mass, m, and velocity, v, so
(½mv2² – ½mv1²)/Δt = ½m(v2² – v1²)/Δt
= ½m(v2 – v1)(v2 + v1)
= m(v2 – v1)/Δt ⋅ (v2 + v1)/2.
This is the mass times the average acceleration times the average velocity:
= mav = Fv
because of Newton’s second law, F = ma.
As for potential energy, a change in potential energy is related to work as follows:
ΔPE = –W = –FΔz
where Δz is a change in vertical position. So we have
ΔPE/Δt = –F(Δz/Δt) = –Fv
since Δz/Δt is the average velocity. Putting this together we have
Fv – Fv = 0
which completes the derivation of the conservation of energy over an arbitrary distime.
Next is a derivation of the conservation of lethargy similar to that above:
It is required that releases must be conservative: R(s2) − R(s1) = 0 for all s.
The law of the conservation of lethargy states that the total lethargy in an isolated system remains constant over distance. The total lethargy over an arbitrary distance, Δs, is constant. Let the total lethargy at two distances be L1 and L2. Then it is required that:
(L2 – L1)/Δs = 0.
Since the total lethargy equals the kinetic lethargy (KL) plus the potential lethargy (PL), we have
(KL2 + PL2 – KL1 – PL1)/Δs = 0
= (KL2 – KL1)/Δs + (PL2 – PL1)/Δs = 0
= (ΔKL – ΔPL)/Δs.
Now KL = ½nu² with vass, n, and lenticity, u, so
(½nu2² – ½nu1²)/Δs = ½n(u2² – u1²)/Δs
= ½n(u2 – u1)(u2 + u1)
= n(u2 – u1)/Δs ⋅ (u2 + u1)/2.
This is the vass times the average relentation times the average lenticity :
= nbu = Ru
because of the dual of Newton’s second law, R = nb.
As for potential lethargy, a change in potential lethargy is related to effort as follows:
ΔPL = –X = –RΔζ
where Δζ is a change in vertical time position. So we have
ΔPL/Δs = –R(Δζ/Δs) = –Ru
since Δζ/Δs is the average lenticity. Putting this together we have
Ru – Ru = 0
which complete the derivation of the conservation of lethargy over an arbitrary distance.
Note that there is a relation between kinetic energy and kinetic lethargy:
KL = ½nu² = ¼(1/½mv²) = ¼/KE.
There is also a relation between F and R:
Ru = nbu = b/mv = ab/mav = ab/Fv.
(u/b)R = (a/v)/F.