Energy and the Schwarzschild Metric In the Schwarzschild metric, the geodesic Lagrangian is 
$$L=\frac{1}{2} \left[ \left( 1-\frac{2m}{r} \right) \dot{t}^2-\frac{\dot{r}^2}{1-2m/r}-r^2(\dot{\theta}^2+\sin^2\theta\dot{\psi}^2) \right]$$
Therefore the $t$ equation for the geodesic motion of a free particle is
$$\frac{d}{d\tau}\left(\frac{\partial L}{\partial \dot{t}}\right)=0$$
because the Lagrangian is independent of t. Consequently
$$E=(1-2m/r)\dot{t}$$
is constant along the particle worldline.... Suppose that the particle has four-velocity $V$ and unit mass. Then relative to an observer at rest at some point in the particle's history, the particle has speed $v$ given by
$$\gamma(v) = \frac{1}{\sqrt{1-v^2}}=U^aV_a=g_{00}U^0V^0=\dot{t}\sqrt{1-2m/r}$$
1) How do we get $t$ equation so obviously?
2) How do we get the equation for $E$? There was no justification in my book for this step
3) What is the interpretation of the quantity $U^aV_a$? Is it like a dot product?
4) How do we know that $U^aV_a$ equals $g_{00}U^0V^0$? Where have all the other components of the metric gone?
 A: The world lines that extremize the proper time between A and B are those which satisfy the Lagrange equation:
\begin{equation}
-\frac{d}{d \sigma} \left(\frac{\partial L}{\partial({dx^{\alpha} / d \sigma})} \right)+\frac{\partial L}{\partial x^{\alpha}}=0
\end{equation}
Where the Lagrangian is:
\begin{equation}
L= \left(-g_{\alpha \beta} \frac{d x^{\alpha}}{d \sigma}  \frac{d x^{\beta}}{d \sigma} \right)^{\frac{1}{2}}
\end{equation}
The Schwarzschild metric in geometrical units is
\begin{equation}
\left(
\begin{array}{cccc}
 -(1-\frac{2 m}{r}) & 0 & 0 & 0 \\
 0 & (1-\frac{2 m}{r})^{-1} & 0 & 0 \\
 0 & 0 & r^2 & 0 \\
 0 & 0 & 0 & r^2 \sin ^2(\theta ) \\
\end{array}
\right)
\end{equation}
The parameter t is therefore called world time. If m = 0 the Schwarzschild metric reduces to the Minkowski metric; if $m \neq 0$ it is singular for $r = 0$. For r large with respect to m the Schwarzschild metric coincides with the Newtonian approximation, since $2m/r$ is the Newtonian potential outside a spherical body of mass m centered at $r = 0$. In addition, when 
$1−(2m/r)$ vanishes for $r = 2m$, the metric appears to be singular there. This singularity can be removed by choosing a different coordinate system.
The geodesic equations based on the Schwarzschild metric above are
\begin{equation}
\left(
\begin{array}{ccc}
\frac{d u_0}{d \tau} =  \frac{2 m u_0 u_1}{2 m r-r^2} \\
\frac{d u_1}{d \tau}  =  \frac{m (2 m-r) u_0^2}{r^3}-\frac{m u_1^2}{2 m r-r^2}-(2 m-r) u_2^2-(2 m-r) \sin ^2(\theta) u_3^2 \\
\frac{d u_2}{d \tau} =  \cos (\theta) \sin (\theta ) u_3^2-\frac{2 u_1 u_2}{r} \\
\frac{d u_3}{d \tau} =  -\frac{2 (u_1+r \cot (\theta ) u_2) u_3}{r} \\
\end{array}
\right)
\end{equation}
where $u_{\alpha}$ are the component of the four proper velocity vector while $\tau$ is the proper time. When looking at the metric you can see that is independent of $t$, thus there is a Killing vector associated with this symmetry under displacement of the time coordinate t with components $\psi=(1,0,0,0)$. The first equation can be written as:
\begin{equation}
\frac{d}{d \tau} \left((1-\frac{2 m}{r}) u_0(\tau) \right) = 0
\end{equation}
which is integrated to give
\begin{equation}
\left((1-\frac{2 m}{r}) u_0(\tau) \right) =- g_{00} \frac{dt}{d \tau}= E
\end{equation}
with E constant.
Now you have to remember that the dot product is obtained by computing the first fundamental form.The first fundamental form is the restriction of the usual dot product in $\mathbb{R}^3$ to the tangent plane $T_p S$. Given two vector a and b
the first fundamental form $Ip (a, b)$ at the point $p$ can be expressed as the bilinear form $a^T g b$, where g is the matrix. In the case of the xy-plane or a cylinder $g$ is a diagonal matrix with diag(g) = (1,1). In this case the dot product of $a$ and $b$ is 
\begin{equation}
(a_1,a_2) \left( 
\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}
\right) (b_1,b_2)^T= a_1 b_1+ a_2 b_2
\end{equation}
However is the matric is not flat we have to account for the terms $g_{\alpha, \beta}$. So yes, $U^aV_a$ is a dot product, with the index a running from 1 to the 4, since you are in a 4 dimensional space. This dot product can be written as
\begin{equation}
g_{00} u_0 v_0-g_{00}^{-1} u_1 v_1+r^2 u_2 v_2+ r^2 \sin^2(\theta)u_3 v_3
\end{equation}
However, The 4-velocity of a stationary observer is $(u_0,0,0,0)$, so for a stationary observer we have only the term $g_{00} u_0 v_0$.
Consider now the lapse of proper time τ between two events at a fixed spatial point (differentiation with respect the position is zero) in Schwarzschild space-time:
\begin{equation}
ds^2=-d \tau^2=-g_{00} dt^2
\end{equation}
hence
\begin{equation}
d \tau=\sqrt{g_{00}} dt
\end{equation}
The element of proper time dτ is measured by a clock at the particular point, while the element of world time dt is fixed for the whole manifold. Since $g_{00}<1$, so $d \tau < dt$ i.e. clocks go slower in a gravitational field. In fact time itself goes slower in a gravitational field. In flat spacetime $m$ turn down to zero so that $dt=d \tau$, and a clock records the coordinate time $t$. 
