This post is related to the one on circular orbits. I’ll continue to follow the exposition in Elements of Newtonian Mechanics by J.M. Knudsen and P.G. Hjorth (Spriner, 1995), this time starting with page 33. As before, the point is to derive a temporo-spatial theory that is symmetric with the spatio-temporal one. Although the parallel theory is shown for 1D space + 1D time, it may be expanded to 1D space + 3D time.
A harmonic oscillator is like the spring below with a bob (weight):
The force is an example of Hooke’s law. Let’s first look at the horizontal case:
Fx = –kx,
where k is a constant determined from the properties of the spring. From Newton’s second law:
m d²x/dt² = –kx,
where m is the mass of the bob, x is the displacement, and t is the time. The solution to this differential equation can be written in the form:
x(t) = As cos(ωt + θs),
where ω = √(k/m), A is the spatial amplitude, and θ is the spatial phase angle. The period T must satisfy
cos(ωt + θ) = cos(ω(t + T) + θ).
Since the cosine is periodic with the period 2π, we have
ωT = 2π,
or
T = 2π/ω = 2π √(m/k).
Thus the period is independent of the amplitude. Note that
d²x/dt² = –ω²x.
Now consider a bob (weight) on a vertical spring under gravity. The total force acting on the bob is:
F = –ky + mg,
where k is the spring constant. The equilibrium position is y0 such that:
y0 = mg/k.
The equation of motion is
m d²y/dt² = –ky + mg,
which can be written in the form
d²y/dt² + (k/m)(y – y0) = 0.
The solution to this differential equation is
y – y0 = As cos(ωt + θ).
This shows that the bob will oscillate around the equilibrium position with the period, T, which is the distime:
T = 2π √(m/k).
Now let’s look at simple temporo-spatial harmonic motion. Hooke’s law in the horizontal case is this:
Γt = –k′t,
where k′ is a constant determined from the properties of the spring. From Newton’s second law:
ℓ d²t/dx² = –k′t,
where ℓ is the vass of the bob, t is the time, and x is the displacement. The solution to this differential equation can be written in the form:
t(x) = At cos(ψx + φt),
where ψ = √(k/m), At is the temporal amplitude, and φt is the temporal phase angle. The displacement extent S must satisfy
cos(ψx + φ) = cos(ψ(x + S) + φ).
Since the cosine is periodic with the period 2π, we have
ψS = 2π,
or
S = 2π/ψ = 2π √(ℓ/k′).
Thus the extent S is independent of the temporal amplitude. Note that
d²t/dx² = –ψ²t.
Now consider a bob (weight) on a vertical spring under levity. The total release acting on the vass is:
Y = –k′t + ℓh,
where k′ is a spring constant and h is the levitational relentation. The equilibrium time position is t0 such that:
t0 = ℓh/k′.
The equation of motion is
ℓ d²t/dy² = –k′t + ℓh,
which can be written in the form
d²t/dy² + (k′/ℓ)(t – t0) = 0.
The solution to this differential equation is
t – t0 = At cos(ψy + φt).
This shows that the bob will oscillate around the equilibrium time position with the displacement extent, S:
S = 2π √(ℓ/k′).