Classical Newtonian harmonic oscillator...
but in the final Friedmann equation, the units have switched...
Affirmative, the units have changed in the relativistic harmonic oscillator potential energy well.
Non-relativistic classical Newtonian harmonic oscillator potential energy:
U = U_k + U_{\Lambda} = - \frac{kx^2}{2} + \frac{\Lambda m r^2}{6}} = - \frac{kx^2}{2} + \frac{\Lambda m a^2 x^2}{6} = - \left( \frac{kx^2}{2} - \frac{\Lambda m a^2 x^2}{6} \right)
\boxed{U = - \left( \frac{kx^2}{2} - \frac{\Lambda m a^2 x^2}{6} \right)} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}
Relativistic harmonic oscillator potential energy:
\boxed{U = - mc^2 \left( \frac{kx^2}{2} - \frac{\Lambda a^2 x^2}{6} \right)} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dL^2}}
When the harmonic oscillator switches from a classical Newtonian harmonic oscillator to a relativistic harmonic oscillator, the Systeme International units change for the cosmological constant.
Newton-Einstein relative unit conversion:
\boxed{\frac{\Lambda_N}{\Lambda_E} = \frac{dL^2}{dt^2} = c^2} \; \; \; \; \; \; \boxed{\frac{k_N}{k_E} = dF \cdot dL = \frac{mc^2}{a}} \; \; \; \; \; \; \boxed{\frac{\rho_N}{\rho_E} = 2}
If you were to model a classical circular orbit, which accounted for the Lambda force, would the following form be correct?
Negative.
Newton's universal law of gravitation:
F_g = m \frac{d^2 r}{dt^2} = - \frac{G m^2}{r^2}
Newtonian centripetal force:
F_c = m \frac{d^2 r}{dt^2} = \frac{mv^2}{r}
Harmonic oscillator force:
F_k = m \frac{d^2 r}{dt^2} = - kx
Newtonian cosmological constant force:
F_{\Lambda} = m \frac{d^2 r}{dt^2} = \frac{\Lambda m r}{3} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}
Newtonian force theorem:
\boxed{F_g + F_c = F_k + F_{\Lambda}}
\boxed{F_g = -F_c + F_k + F_{\Lambda}}
F = m \frac{d^2 r}{dt^2} = - \frac{Gm^2}{r^2} = - \frac{mv^2}{r} - kx + \frac{\Lambda m r}{3}
\boxed{\frac{Gm^2}{r^2} = \frac{mv^2}{r} + kx - \frac{\Lambda m r}{3}}
Multiply completely through by \frac{r}{2}:
\frac{Gm^2}{2r} = \frac{mv^2}{2} + \frac{k r x}{2} - \frac{\Lambda m r^2}{6}
Kinetic energy:
\boxed{E_k = \frac{mv^2}{2} = \frac{Gm^2}{2r} - \frac{k r x}{2} + \frac{\Lambda m r^2}{6}}
Multiply completely through by \frac{2}{m}:
v^2 = \frac{G}{r} \left( \frac{4 \pi \rho r^3}{3} \right) - \frac{k r x}{m} + \frac{\Lambda r^2}{3} = \frac{4 \pi G \rho r^2}{3} - \frac{k r x}{m} + \frac{\Lambda r^2}{3}
\boxed{v^2 = \frac{4 \pi G \rho r^2}{3} - \frac{k r x}{m} + \frac{\Lambda r^2}{3}} \; \; \; \; \; \; \boxed{a = \frac{r}{x}} - co-moving coordinates
Non-relativistic Newtonian Hubble parameter equation:
\boxed{H^2 = \frac{v^2}{r^2} = \frac{4 \pi G \rho}{3} - \frac{k}{ma} + \frac{\Lambda}{3}} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}
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there is another way to get them, you can arrive by using Newtonian Mechanics, and when finally you have the constant then you have to compare with the real one and then you substitute your constant by the one that appears in the original one. That's the way I was asking for.