1. 1. Since
      $$\frac{dt}{d\tau} = \gamma\ ,$$
      we may convert $\tau$-derivatives into $t$- derivatives:
      $$\frac{d\ }{d\tau} = \frac{dt}{d\tau} \ \frac{d\ }{dt} = \gamma \ \frac{d\ }{dt} \ .$$
      This is useful for thinking of a Newtonian trajectory described by $(ct, \vec{x}\, (t))$.

      Applying this to the force four-vector gives

      $$F^\mu = m\ \frac{d^2 x^\mu}{d\tau^2} \ = m \gamma \ \frac{d\ }{dt} \ \left(\ \gamma \ \frac{dx^\mu }{dt} \ \right)\ \ ,$$
      which gives
      $$F^\mu = m \gamma \ \left(\ \dot{\gamma} \ \frac{dx^\mu }{dt} + {\gamma} \ \frac{d^2 x^\mu }{dt^2} \ \right)\ \ .$$
      Note that
      $$\gamma = \left(\ 1 - \vert\vec{v}\vert^2/c^2 \right )^{-1/2}$$
      gives
      $$\dot{\gamma} = \left(\ 1 - \vert\vec{v}\vert^2/c^2 \right )^{-3/2... ...ot \dot{\vec{v}}\ /\ c^2 = \gamma^3\ \vec{v}\ \cdot \dot{\vec{v}}\ /\ c^2 \ \ ,$$
      which vanishes when $\vec{v}= 0$ (the Lorentz frame is instantaneously at rest). Thus, when $\vec{v}= 0$, we obtain
      $$F^\mu = m \gamma^2 \ \frac{d^2 x^\mu }{dt^2} \ \ .$$
      Plugging in $\gamma = 1$, $x^\mu = (ct, \vec{x})$ gives
      $$F^\mu = m\ (0, \ddot{\vec{x}}) = (0, \vec{F})\ ,$$
      using the Newtonian definition $\vec{F}= m \ddot{\vec{x}}$. $\qedsymbol$

   2. For vanishing force $F^\mu$, the covariantized Newton's second law gives
      $$0 \ \ = \ \ \frac{Dp^\mu}{D\tau} \ \ =\ \ \frac{dx^\nu}{d\tau}\ ... ...\ p^\mu\ \ =\ \ m\ \frac{dx^\nu}{d\tau}\ \nabla_\nu\ \frac{dx^\mu}{d\tau}\ \ .$$
      That is, for the covariant proper time derivative of $p^\mu$ to vanish, the tangential directional covariant derivative of $p^\mu$ must vanish, but this means that the tangent vector ${dx^\nu}/{d\tau}$ itself must have vanishing tangential directional derivative. That is, the tangent vector ${dx^\nu}/{d\tau}$ must be parallel transported along the trajectory. This is exactly the requirement defining a geodesic.

   3. Plugging in for the covariant derivative $\nabla_\nu$ above,
      

      $$= m\ \frac{dx^\nu}{d\tau}\ \nabla_\nu\ \frac{dx^\mu}{d\tau}$$ 0

      $$= m\ \frac{dx^\nu}{d\tau}\ \left(\ \partial_\nu\ \frac{dx^\mu}{d\tau} + \Gamma^\mu_{\ \ \nu\rho}\ \frac{dx^\rho}{d\tau}\ \right)$$

      $$= m\ \left(\ \left(\ \frac{dx^\nu}{d\tau}\ \partial_\nu\ \right)\ ... ...Gamma^\mu_{\ \ \nu\rho}\, \frac{dx^\nu}{d\tau}\, \frac{dx^\rho}{d\tau}\ \right)$$

      $$= m\ \left(\ \frac{d^2\,x^\mu}{d\tau^2} + \Gamma^\mu_{\ \ \nu\rho}\, \frac{dx^\nu}{d\tau}\, \frac{dx^\rho}{d\tau}\ \right) \ \ ,$$

      
      
      Where in the last step we note that $\partial_\nu\, x^\mu = 0$. Note this is simply $m$ times the geodesic equation.$\qedsymbol$

2. 1. For
      $$T^{\mu\nu} = (\rho + p)\ u^\mu\, u^\nu - p \,\eta^{\mu\nu}\ \ ,$$
      $\partial_\mu\, T^{\mu\nu} = 0$ gives
      

      $$= \partial_\mu\ \left(\ (\rho + p)\ u^\mu\ \right)\ u^\nu + (\rho + p)\ u^\mu\ \partial_\mu\, u^\nu - \partial^\nu\, p\ \ .$$ 0 (1)

      
      
      We simplify things by taking the component of both sides parallel to $u^\nu$:
      $$0 = u_\nu\ \partial_\mu\, T^{\mu\nu} = \partial_\mu\ \left(\ (\rho + p)\ u^\mu\ \right) - u_\nu\,\partial^\nu\, p\ \ ,$$
      where we have used the fact that $u_\nu\, u^\nu = 1$ and hence $u_\nu\ \partial_\mu \, u^\nu = 0$ (since it is proportional to $\partial_\mu\ (\ u_\nu\, u^\nu\ )$. Doing the algebra reduces things to
      

      $$= \left(\ \partial_\mu \rho + \partial_\mu p\ \right) \ u^\mu + (\rho + p)\ \partial_\mu\ u^\mu - u_\nu\,\partial^\nu\, p$$ 0

      $$= \partial_\mu \rho \ u^\mu + (\rho + p)\ \partial_\mu\ u^\mu \ \ .$$

      
      
      $\qedsymbol$ We can also take components perpendicular to $u^\nu$ by dotting equation (1) with $\partial_\rho\,u_{\nu}$. This gives
      $$0 = \partial_\rho\,u_{\nu}\ \left(\ (\rho + p)\ u^\mu\ \partial_\mu\, u^\nu - \partial^\nu\, p\ \right)$$
      or simply
      $$0 = (\rho + p)\ u^\mu\ \partial_\mu\, u^\nu - \partial^\nu\, p\ \ .$$
      $\qedsymbol$

   2. In the nonrelativistic limit, $p << \rho$ so the parallel equation becomes
      $$0 = \partial_\mu \rho \ u^\mu + \rho\ \partial_\mu\ u^\mu = \partial_\mu\ (\ \rho \ u^\mu\ )\ \ .$$
      Plugging in $u^\mu = (1, \vec{v}/c)$ gives $c^{-1}$ times the continuity equation for mass,
      $$\frac{\partial{\rho}}{\partial{t}} + {\boldsymbol {\nabla}} \cdot (\rho\, \vec{v}) = 0\ \ .$$
      $\qedsymbol$ The perpendicular equation has vanishing time component, since $u^0$ is constant and $(\partial_t\, p)/c \approx 0$ in the nonrelativistic limit. The spatial components give (again using $p << \rho$, and noting the sign from the raised spatial index $\partial^\nu$ )
      

      $$= \rho \ u^\mu\ \partial_\mu\, \vec{u}+ {\boldsymbol {\nabla}}\, p$$ 0

      $$= \frac{\rho}{c^2} \ \left(\ \frac{\partial{\vec{v}}}{\partial{t}} ... ...t\,{\boldsymbol {\nabla}})\ \vec{v}\ \right) + {\boldsymbol {\nabla}} \, p\ \ ,$$

      
      
      using $u^\mu = (1, \vec{v}/c)$. This reproduces Euler's equation for the motion of an inviscid (nonviscous) fluid,
      $$\frac{\rho}{c^2} \ \left(\ \frac{\partial{\vec{v}}}{\partial{t}} ... ...{\boldsymbol {\nabla}})\ \vec{v}\ \right) = - {\boldsymbol {\nabla}} \, p \ \ .$$
      $\qedsymbol$

3. We consider the first Friedmann equation,
   $$\frac{3\dot{R}^2}{\ R^2}= 8\pi G \rho - \frac{3k}{R^2}\ \ ,$$
   in the case where $w -1/3$, so $R(t)$ always decelerates and $\rho$ decays for large $R$ faster than $R^{-2}$.
   1. We know that $\dot{R} 0$ now. For $k=0$ or $k= -1$, the right hand side of the first Friedmann equation is strictly postive. Thus $\dot{R}$, which continuously decreases, can never pass through the value zero, and must remain positive forever. The universe thus never begins contracting. $\qedsymbol$

      Thus we have large values of $R$ at late times, making $\rho$ become negligible compared to the curvature term. Keeping the curvature term only, we find

      $$\begin{array}{ll} \mbox{for\ } k=0 &\begin{array}{l} \dot{R} ... ...1\\ H = \frac{\dot{R}}{R} \rightarrow 0\end{array}\end{array}$$
      $\qedsymbol$

   2. Again, $\dot{R} 0$ now, and continuously decreases. For $k = +1$, it reaches the value zero when the right hand side of the Friedmann equation vanishes, at
      $$8\pi G \rho_o \ \left(R_o/R\right)^{3(1+w)} = \frac{3}{R^2}\ \ ,$$
      or
      $$R^{(1+3w)} = \frac{ 8\pi G \rho_o}{3}\ R_o^{3(1+w)}\ \ .$$
      $\qedsymbol$ At that point the expansion turns over into a contraction, with $\dot{R}$ continuing to decrease through negative values. (Note that this contraction phase gives a symmetric solution $\dot{R} \rightarrow - \dot{R}$ about the turnover point in $R$, and so the contraction is symmetric to the expansion.)$\qedsymbol$

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