Thread: Friedmann Equation View Single Post

 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:
$$\boxed{H^2 = \frac{v^2}{r^2} = \frac{4 \pi G \rho}{3} - \frac{k}{ma} + \frac{\Lambda}{3}} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}$$