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Orion1
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#9
Jan5-09, 11:40 AM
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P: 991

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:
[tex]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)[/tex]

[tex]\boxed{U = - \left( \frac{kx^2}{2} - \frac{\Lambda m a^2 x^2}{6} \right)} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}[/tex]

Relativistic harmonic oscillator potential energy:
[tex]\boxed{U = - mc^2 \left( \frac{kx^2}{2} - \frac{\Lambda a^2 x^2}{6} \right)} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dL^2}}[/tex]

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:
[tex]\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}[/tex]

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:
[tex]F_g = m \frac{d^2 r}{dt^2} = - \frac{G m^2}{r^2}[/tex]

Newtonian centripetal force:
[tex]F_c = m \frac{d^2 r}{dt^2} = \frac{mv^2}{r}[/tex]

Harmonic oscillator force:
[tex]F_k = m \frac{d^2 r}{dt^2} = - kx[/tex]

Newtonian cosmological constant force:
[tex]F_{\Lambda} = m \frac{d^2 r}{dt^2} = \frac{\Lambda m r}{3} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}[/tex]

Newtonian force theorem:
[tex]\boxed{F_g + F_c = F_k + F_{\Lambda}}[/tex]

[tex]\boxed{F_g = -F_c + F_k + F_{\Lambda}}[/tex]

[tex]F = m \frac{d^2 r}{dt^2} = - \frac{Gm^2}{r^2} = - \frac{mv^2}{r} - kx + \frac{\Lambda m r}{3}[/tex]

[tex]\boxed{\frac{Gm^2}{r^2} = \frac{mv^2}{r} + kx - \frac{\Lambda m r}{3}}[/tex]

Multiply completely through by [tex]\frac{r}{2}[/tex]:
[tex]\frac{Gm^2}{2r} = \frac{mv^2}{2} + \frac{k r x}{2} - \frac{\Lambda m r^2}{6}[/tex]

Kinetic energy:
[tex]\boxed{E_k = \frac{mv^2}{2} = \frac{Gm^2}{2r} - \frac{k r x}{2} + \frac{\Lambda m r^2}{6}}[/tex]

Multiply completely through by [tex]\frac{2}{m}[/tex]:

[tex]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}[/tex]

[tex]\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}}[/tex] - co-moving coordinates

Non-relativistic Newtonian Hubble parameter equation:
[tex]\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}}[/tex]