A Deriving Equations of Motion in GR

[Matter_Matters]

Question Background:
I'm considering the Eddington-Robertson-Schiff line element which is given by
(ds)^2 = \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) dt^2 - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (dr^2 + r^2 d\theta^2 + r^2 \sin^2{\theta} \;d\phi^2 ),
where \mu = GM = \text{const.} and r=|\mathbf{r}|.
I'm interested in determining the equations of motion for such a line element which can be obtained from the least action principle. The classical action S is the integral along the particle trajectory
S = \int ds,
which can be equivalently expressed as
S = \int \left( \frac{ds}{dt} \right) dt \equiv \int L \; dt.
We can see from the above that
L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right]^{1/2},
where L is the associated Lagrangian over time.
Problem and question
The associated equations of motion are given by (Eq. 20)
\frac{d^2\mathbf{r}}{dt^2} = \frac{\mu}{r^3} \left[ \left(4 \frac{\mu}{r} - v^2 \right) \mathbf{r} + 4 (\mathbf{r}\cdot \mathbf{\dot{r}} ) \mathbf{\dot{r}}\right].
I cannot for the life of me obtain this using the Euler-Lagrange equations.
Attempt at a solution:
The Euler-Lagrange equations are given by
\frac{d}{dt} \left( \frac{\partial L}{\partial \mathbf{\dot{r}}} \right) - \frac{\partial L}{ \partial \mathbf{r}} =0.
I note that the equations of motion should be equivalent for either
L = \sqrt{g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu},
or
L = g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu.
Bearing this in mind and working through the process using
L = \left[ \left( 1 - 2 \left(\frac{\mu}{r}\right) + 2 \left(\frac{\mu^2}{r^2}\right) \right) - \left( 1 + 2 \left( \frac{\mu}{r} \right) \right) (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}}) \right],
I find
\frac{d}{dt} \left( \frac{\partial L}{ \partial \mathbf{\dot{r}}} \right) = -2 \left[ \left( 1 + 2 \frac{\mu}{r} \mathbf{\ddot{r}} \right) - 2 \frac{\mu}{r^3} (\mathbf{r}\cdot \mathbf{\dot{r}}) \mathbf{\dot{r}} \right],
and
\left( \frac{\partial L }{\partial \mathbf{r}} \right) = 2\frac{\mu}{r^3} \mathbf{r} - 4 \frac{\mu^2}{r^4} \mathbf{r} + 2 \frac{\mu}{r^3} \mathbf{r} (\mathbf{\dot{r}} \cdot \mathbf{\dot{r}} ).
Clearly, adding these together does not give the desired result. Any suggestions?

 

[Orodruin]

I note that the equations of motion should be equivalent for either

L=√gμν˙xμ˙xμ,L=gμνx˙μx˙μ,​

L = \sqrt{g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu},
or

L=gμν˙xμ˙xμ.L=gμνx˙μx˙μ.​

L = g_{\mu\nu} \dot{x}^{\mu}\dot{x}^\mu.

This is only true if the geodesic is parametrised by an affine parameter. The coordinate $t$ is in general not affine.

 

[Matter_Matters]

This is only true if the geodesic is parametrised by an affine parameter. The coordinate $t$ is in general not affine.

O wow! If this was the issue the whole time I will be very pleased but also annoyed at my ignorance!

 

[Matter_Matters]

This is only true if the geodesic is parametrised by an affine parameter. The coordinate $t$ is in general not affine.

Hmmm I am confused now! So, normally in GR we can set the Lagrangian = $\pm c$ depending on the signature of the line element. Is this only true when the geodesic is parametrised by an affine parameter also? Even using the $L = \sqrt{ }$, I can't seem to manage to get the correct expression!

 

[Orodruin]

So, normally in GR we can set the Lagrangian = ±c depending on the signature of the line element. Is this only true when the geodesic is parametrised by an affine parameter also?

In general, it will just give you a requirement that gives an affine parametrisation (constant length tangent vector). It tells you nothing about whether or not you found a geodesic.