iSoul Time has three dimensions

Work, effort and energy

First we show the work and (space) 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

mr/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 space energy of the particle. Then

dT = dW.

That is, the change in the kinetic space 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 space energy U = U(r) through

dU(r) := –dW = –F(r)dr.

That is, the change in the potential space 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 space energy of the system.


Next we show the effort and time energy 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

nt/dr² = Γ, or

n du/dr = Γ,

with vass n, rush Y, and legerity 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 effort done by the rush Y over the time segment dt. Define V as

V = nu²/2,

which is called the kinetic time energy of the transicle. Then

dV = dX.

That is, the change in the kinetic time energy of the transicle over the segment dt equals the effort done by the rush Y.

If Y = Y(t) does not depend on position, then define the potential time energy Z = Z(t) through

dZ(t) := –dX = –Y(t)dt.

That is, the change in the potential time energy V(t) over the time segment dt is equal to minus the effort done by the external rush Y. Since

dV = –dZ(t),

and upon integrating,

V + Y(t) = M,

where the constant M is called the total mechanical time energy of the system.

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