Geodesics from variational principle with respect to coordinate? I know you can find geodesic equations with respect to proper time $\tau $ using the variational principle, i.e. using Euler-Lagrange equations 
$$ \frac{\partial}{\partial x^{\mu}}L-\frac{d}{d\tau}\frac{\partial}{\partial\dot{x}^{\mu}}L=0 \tag{1} $$ 
for Lagrangian $$
L=g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}.\tag{2}
$$
Now can you use this same method to find geodesic equations with respect to coordinate time $t$ using Euler-Lagrange equations for Lagrangian 
$$
L=g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}~?\tag{3}
$$
More generally do you get equations that gives geodesics for solutions when you use Euler-Lagrange equations for a Lagrangian of the form
$$
L=g_{\mu\nu}\frac{dx^{\mu}}{dx^{\sigma}}\frac{dx^{\nu}}{dx^{\sigma}},\tag{4}
$$ 
where $x^{\sigma}$ is one of the coordinates?
 A: A geodesic can be parameterized by any parameter that increases monotonically along the geodesic. For timelike geodesics, which is what we have in mind when we derive them by minimising $\int d\tau$, we can use a coordinate $t$ as the parameter, provided the corresponding vector field $\partial/\partial t$ is everywhere timelike. This ensures that $t$ increases monotonically along any timelike worldline. The catch is that the geodesic equation comes from extremizing of the integral of $\sqrt{L}$, which doesn't necessarily give the usual Euler-Lagrange equation for $L$.
To see how this works, start with
$$
 L(\lambda) = g_{ab}\dot x^a \dot x^b
$$
where $\dot x^a \equiv dx^a/d\lambda$, where $\lambda$ is any parameter that increases monotonically along timelike worldlines. The quantity
$$
 \int \sqrt{L(\lambda)}\,d\lambda
$$
is reparameterization-invariant (intuitively, the $d\lambda$'s cancel), so it's the same no matter what parameter $\lambda$ we use. In particular, it's the same whether or not $\lambda$ ends up being the worldline's proper time; it doesn't even need to be an affine parameter. In particular, it can be one of the coordinates if that coordinate increases monotonically along all timelike worldlines.
The geodesic condition is
$$
 0=\delta \int \sqrt{L(\lambda)}\,d\lambda.
$$
Use
\begin{align}
 \delta \int \sqrt{L(\lambda)}\,d\lambda
 &\propto
 \int \frac{1}{\sqrt{L}}\left(
 \frac{\partial L}{\partial x^a}\delta x^a
 +
 \frac{\partial L}{\partial \dot x^a}\delta \dot x^a
 \right)d\lambda
\\
 &=
 \int \left(
 \frac{1}{\sqrt{L}}\frac{\partial L}{\partial x^a}
 -
 \frac{d}{d\lambda}\left[\frac{1}{\sqrt{L}}
  \frac{\partial L}{\partial \dot x^a}\right]
 \right)\delta x^a\,d\lambda
\end{align}
to conclude that the geodesic equation is
$$
 \frac{\partial L}{\partial x^a}
 -
 \sqrt{L}\frac{d}{d\lambda}\left[\frac{1}{\sqrt{L}}
  \frac{\partial L}{\partial \dot x^a}\right]
 =0.
$$
In the special case where the parameter $\lambda$ is set equal (in hindsight) to the wordline's proper time, the equation simplifies because $L=1$ in that case. More generally, for an affine parameter, we have (by definition) $L=$ constant, so the equation again simplifies in the same way, leaving the usual Euler-Lagrange equation. But for a general parameter, such as a timelike coordinate, the equation does not simplify in that way; the derivative with respect to $\lambda$ will act nontrivially on the factor $\sqrt{L}$, leading to an extra term compared to the usual form of the geodesic equation. Despite the extra term, the result is correct. The reparameterization-invariance of $\int \sqrt{L(\lambda)}\,d\lambda$ implies that the resulting equation selects the same set of worldlines (the ones we call geodesics) no matter what monotonic parameterization we used.
A: Comments to the post (v2):


*

*Note that one cannot use proper time $\tau$ (or arclength) as the independent parameter $\lambda$ before applying the principle of stationary action to find geodesics. This is explained in my Phys.SE answer here, which also explains the connection to the corresponding square root Lagrangian. 

*For the non-square root Lagrangian, only after the variation is performed, a stationary solution is then affinely parametrized wrt. proper time $\tau$ (if the geodesic is timelike, i.e. if the point particle is massive).
