Deriving the Equation of Motion out of the Action

[JD_PM]

Homework Statement

Given the action (note $G_{ab}$ is a symmetric matrix, i.e. $G_{ba} = G_{ab}$):

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)\Big)$$

Show (using Euler Lagrange's equation) that the following equation holds:

$$\ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c = -\sum_{a} F^{da}\partial_a V$$

Relevant Equations

The action

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)\Big)$$

Exercise statement:

Given the action (note $G_{ab}$ is a symmetric matrix, i.e. $G_{ba} = G_{ab}$):

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)\Big)$$

Show (using Euler Lagrange's equation) that the following equation holds:

$$\ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c = -\sum_{a} F^{da}\partial_a V$$

Where $F^{ab}$ is the inverse of $G_{ab}$.

Also note that:

$$\sum_{b} F^{ab}G_{ab} = \delta_c^a$$

$$\partial_{a} = \frac{\partial}{\partial q_a}$$

What I have done:

We know that the action functional corresponds to the Lagrangian (for the time interval $[t_0, t_1]$):

$$S[q] = \int_{t_0}^{t_1} L(q, \dot q, t)dt$$

Thus:

$$L = \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)$$

Euler Lagrange's equation is:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big) = \frac{\partial L}{\partial q_k}$$

Let's go step by step:

1) We compute the term $\frac{\partial L}{\partial \dot q_k}$ (which turns out to be the definition of generalized momentum):

$$p_k = \frac{\partial L}{\partial \dot q_k} = \sum_{ab} \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = \sum_{b} G_{kb} \dot q^b + \sum_{a} G_{ak} \dot q^a = 2\sum_a G_{ak} \dot q^a$$

2) We now compute the term $\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big)$

NOTE: I know that the symmetric matrix $G_{ab}$ only depends on $q_k$. By the chain rule (for the sake of clarity: $G_{ab} (q_k)$ notation means that the matrix $G_{ab}$ is a function of $q_k$):

$$\frac{d}{dt} \sum_a G_{ab} (q_k) = \sum_{a} \partial_k G_{ab} \dot q_k$$

That being said, let's go through the calculation:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big) = \frac{d}{dt}\Big( 2\sum_a G_{ak} (q_k) \dot q^a \Big) = 2\sum_a G_{ak} \ddot q^a + 2\sum_a \partial_k G_{ak} \dot q^a \dot q^k$$

3) We now compute the term $\frac{\partial L}{\partial q_k}$

$$\frac{\partial L}{\partial q_k} = \sum_{ab} \partial_k G_{ab} \dot q^a\dot q^b - \partial_k V(q)$$

Here's where I get stuck: I do not see why the following holds:

$$2\sum_a F^{da}G_{ak} \ddot q^a + 2\sum_a F^{da}\partial_k G_{ak} \dot q^a \dot q^k -\sum_{ab} F^{da}\partial_k G_{ab} \dot q^a\dot q^b + F^{da}\partial_k V(q) = \ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c + \sum_{a} F^{da}\partial_a V$$

To sum up: In the steps I have shown I computed the Lagrangian and I am left to convert this Lagrangian into the asked form.

Any help is appreciated, as I have also asked the teacher assistant and we both got stuck here.

I also asked on PSE but I got little attention:

https://physics.stackexchange.com/questions/518630/deriving-an-equation-of-motion-out-of-an-action

I do think this exercise is helping me a lot to get the hang of Lagrange formalism. That is why I am being so persistent with it.

Thanks.

 

[TSny]

Show (using Euler Lagrange's equation) that the following equation holds:

$$\ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c = -\sum_{a} F^{da}\partial_a V$$

I think that there should be an overall factor of $\frac{1}{2}$ on the right-hand side.

We now compute the term $\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big)$

NOTE: I know that the symmetric matrix $G_{ab}$ only depends on $q_k$.

Be careful here. You have already chosen the index $k$ to be a particular index corresponding to the particular degree of freedom for which you are finding the equation of motion. Each $G_{ab}$ is generally a function of all of the $q_c$ , not just the one particular $q_k$ .

By the chain rule (for the sake of clarity: $G_{ab} (q_k)$ notation means that the matrix $G_{ab}$ is a function of $q_k$):

$$\frac{d}{dt} \sum_a G_{ab} (q_k) = \sum_{a} \partial_k G_{ab} \dot q_k$$

In taking the time derivative of $G_{ab}$ you have to consider the time derivatives of all of the $q_c$.

 

[samalkhaiat]

I will not use summation sign: repeated PAIR of (upper and lower) indices are summed over: \sum_{a} A_{a} B^{a} \equiv A_{a}B^{a} = A_{c}B^{c} (summed over indices are dummy indices so you can rename them as you like), and expressions like A_{k}B^{k}C_{k} are meaningless. In any term an index should not be repeated more than 2 time.
L = g_{ab}(q) \ \dot{q}^{a}\dot{q}^{b} - V(q) .
\frac{\partial L}{\partial \dot{q}^{c}} = 2 g_{ac}(q) \ \dot{q}^{a} ,
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{c}} \right) = 2 g_{ac} \ \ddot{q}^{a} + 2 \partial_{b}g_{ac} \ \dot{q}^{b}\dot{q}^{a} . \ \ \ \ (1) Now pay attention to the trick which solve the problem: In the second term of (1), the indices (a,b) are dummy indices, so you write that term as the sum of two equal terms with (a,b) \leftrightarrow (b,a) is done in the second term:
2 \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} .
So, rewrite (1) as \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^{c}} \right) = 2 g_{ac} \ \ddot{q}^{a} + \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{a}\dot{q}^{b} . \ \ \ (2)
The rest is easy stuff: \frac{\partial L}{\partial q^{c}} = \partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b} - \partial_{c}V. \ \ \ \ \ \ (3) Now, construct the E-L equation [i.e., (2) = (3) ], then contract the equation with inverse metric f^{cd}

 

[JD_PM]

Thank you very much for your answers!

I think that there should be an overall factor of $\frac{1}{2}$ on the right-hand side.

I also think so.

Be careful here. You have already chosen the index $k$ to be a particular index corresponding to the particular degree of freedom for which you are finding the equation of motion. Each $G_{ab}$ is generally a function of all of the $q_c$ , not just the one particular $q_k$ .

Oh I see it now!

In taking the time derivative of $G_{ab}$ you have to consider the time derivatives of all of the $q_c$.

Oh so it would be:

$$\frac{d}{dt} \sum_a G_{ab} (q) = \sum_{a} \partial_c G_{ab} \dot q_c$$

 

[TSny]

Oh so it would be:

$$\frac{d}{dt} \sum_a G_{ab} (q) = \sum_{a} \partial_c G_{ab} \dot q_c$$

Yes.

 

[JD_PM]

Now, construct the E-L equation [i.e., (2) = (3) ], then contract the equation with inverse metric f^{cd}

Alright let me go step by step here and include a lot of details (I am going to use Einstein's notation from now on):

1) Construct the E-L Equation:

$$2 g_{ac} \ \ddot{q}^{a} + \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{a}\dot{q}^{b} - \partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b} = - \partial_{c}V.$$

2) Transform it in such a way that it looks identical to the given form:

2.1) OK So the first term on the LHS is:

$$2 f^{cd}g_{ac} \ \ddot{q}^{a} = 2 f^{dc}g_{ca} \ \ddot{q}^{a} = 2\ddot{q}^{a} = 2\ddot{q}^{d}$$

Where:

$$F^{dc}G_{ca} = \delta_a^d$$

NOTE for Math lovers: $f^{dc} = f^{cd}$ (i.e. the inverse of a symmetric matrix is also symmetric). Proofs: https://math.stackexchange.com/questions/325082/is-the-inverse-of-a-symmetric-matrix-also-symmetric.

2.2) OK so the second, third and fourth terms on the LHS are:

$$f^{cd}\partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} - f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b} = f^{cd}\partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + f^{cd}\partial_{c}g_{ab} \ \dot{q}^{b}\dot{q}^{a} - f^{cd}\partial_{a}g_{bc} \ \dot{q}^{a}\dot{q}^{b}$$

One may say 'but this is not equal to what you're given!' It is. We just have to notice that a and b are dummy indices. For instance, for the third term on the LHS we get: (let me drop the $f^{cd}$ term for the sake of clarity):

$$\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = \partial_{a}g_{cb} \ \dot{q}^{c}\dot{q}^{a} = \partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{c} = \partial_{c}g_{ab} \ \dot{q}^{b}\dot{q}^{c} . \ \ \ \ (EQ.2.2) $$

Doing the same that above with the fourth term we get: $\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{c}$

Thus the second, third and fourth terms on the LHS can be compacted as follows:

$$f^{cd}\Big(\partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c}$$

2.3 Conclusion:

$$2\ddot{q}^{d} + f^{cd}\Big(\partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c} = -f^{cd}\partial_{c} V$$

Note that $c$ on the RHS equation is a dummy index, so we can call it whatever we want. To match the answer, let $c = a$

Rearranging we get the final result:

$$\ddot{q}^{d} + \frac{1}{2} \ f^{da}\Big( \partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c} = -\frac{1}{2} \ f^{da}\partial_{a} V$$

NOTE: I am pretty confident with the result but there's one more thing: do you completely agree on my $EQ.2.2$ and the reasoning behind it?

 

[TSny]

2.2) OK so the second, third and fourth terms on the LHS are:

$$f^{cd}\partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} - f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b} = f^{cd}\partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + f^{cd}\partial_{c}g_{ab} \ \dot{q}^{b}\dot{q}^{a} - f^{cd}\partial_{a}g_{bc} \ \dot{q}^{a}\dot{q}^{b}$$

I don't follow this. Are you saying that the second term on the left equals the second term on the right? I don't see it. Similarly for the third terms on the left and right.

I believe you can get the desired result by just an appropriate relabeling of the summation indices $a$ and $c$ in the expression on the left side.

$$\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = \partial_{a}g_{cb} \ \dot{q}^{c}\dot{q}^{a} = \partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{c} = \partial_{c}g_{ab} \ \dot{q}^{b}\dot{q}^{c} . \ \ \ \ (EQ.2.2) $$

Consider the first equality: $$\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = \partial_{a}g_{cb} \ \dot{q}^{c}\dot{q}^{a} $$ Note that the index $c$ is not being summed on the left side, but $c$ is being summed on the right side. So, this can't be correct.

 

[JD_PM]

I don't follow this. Are you saying that the second term on the left equals the second term on the right?

Yes, and I am also saying that the third term on the left equals the third term on the right. More explicitly:

$$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

$$- f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b} = - f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$$

Why? It is based on the property @samalkhaiat showed:

$$2 \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} . $$

Consider the first equality: $$\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = \partial_{a}g_{cb} \ \dot{q}^{c}\dot{q}^{a} $$ Note that the index $c$ is not being summed on the left side, but $c$ is being summed on the right side. So, this can't be correct.

Mmm you're right. I knew there was something sloppy on my $EQ.2.2$.

Let me think a bit about it.

 

[JD_PM]

I believe you can get the desired result by just an appropriate relabeling of the summation indices $a$ and $c$ in the expression on the left side.

Yes, I also think it is possible.

I think I got it.

The main issue I have is to convert the result I got:

$$f^{cd}\Big(\partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{a}\dot{q}^{b}$$

into the desired:

$$f^{cd}\Big(\partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c}$$

In this expression $a$, $b$ and $c$ are all summed over, so they are all dummy indices. Thus, I could simply swap $a$ by $c$.

Note this is not like the case you pointed out, where we just had $a$ and $b$ as dummy indices:

$$\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$$

 

[TSny]

Yes, and I am also saying that the third term on the left equals the third term on the right. More explicitly:

$$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

$$- f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b} = - f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$$

Why? It is based on the property @samalkhaiat showed:

$$2 \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} . $$

The last equation is true. But I still don't see how that can be used to show the first two equations. I don't believe the first two equations are valid.

For example, take the first equation $$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

Suppose that we have only two degrees of freedom so that the indices take on only the values 1 and 2. Suppose at some instant of time $\dot q^1 = 1$ and $\dot q^2 = 0$. If we let $d = 2$, then the equation becomes $$f^{c2}\partial_{1}g_{1c} = f^{c2}\partial_{c}g_{11} $$
or $$f^{12}\partial_{1}g_{11} +f^{22}\partial_{1}g_{12}= f^{12}\partial_{1}g_{11} +f^{22}\partial_{2}g_{11} $$

This implies $$\partial_{1}g_{12}= \partial_{2}g_{11} $$ But this is not generally true.

 

[JD_PM]

For example, take the first equation $$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

Suppose that we have only two degrees of freedom so that the indices take on only the values 1 and 2. Suppose at some instant of time $\dot q^1 = 1$ and $\dot q^2 = 0$. If we let $d = 2$, then the equation becomes $$f^{c2}\partial_{1}g_{1c} = f^{c2}\partial_{c}g_{11} $$
or $$f^{12}\partial_{1}g_{11} +f^{22}\partial_{1}g_{12}= f^{12}\partial_{1}g_{11} +f^{22}\partial_{2}g_{11} $$

This implies $$\partial_{1}g_{12}= \partial_{2}g_{11} $$ But this is not generally true.

Thanks for pointing out this example! I am thinking about it.

The last equation is true. But I still don't see how that can be used to show the first two equations. I don't believe the first two equations are valid.

To be honest, I am doubting now why $2 \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$ holds.

My logic was as follows:

Let:

$$A = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b}$$

Then:

$$2A = A + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$$

And for the above equation to be true, we need $\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = A$.

Thus:

$$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

But your example suggests it may not be that simple...

 

[JD_PM]

Now pay attention to the trick which solve the problem: In the second term of (1), the indices (a,b) are dummy indices, so you write that term as the sum of two equal terms with (a,b) \leftrightarrow (b,a) is done in the second term:
2 \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} .

Mmm samalkhaiat said that they are both equal.

Btw, I have edited my previous comment.

 

[TSny]

Actually, to be honest, I still do not see why $2 \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$ holds.

Note that

$2 \partial_{b}g_{ac} \dot q^a \dot q^b = \partial_{b}g_{ac} \dot q^a \dot q^b + \partial_{b}g_{ac} \dot q^a \dot q^b$

Now, interchange the dummy indices $a$ and $b$ in the last term on the right.

My logic was as follows:

Let:

$$A = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b}$$

Then:

$$2A = A + A$$

Must hold and thus I got to the conclusion that:

$$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

I'm not following your steps here in going from $2A=A+A$ to your final conclusion $$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

 

[JD_PM]

Note that

$2 \partial_{b}g_{ac} \dot q^a \dot q^b = \partial_{b}g_{ac} \dot q^a \dot q^b + \partial_{b}g_{ac} \dot q^a \dot q^b$

Now, interchange the dummy indices $a$ and $b$ in the last term on the right.

Oh yes! I see it, thanks.

I'm not following your steps here in going from $2A=A+A$ to your final conclusion $$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

I have edited it.

 

[TSny]

My logic was as follows:

Let:

$$A = \partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b}$$

OK

Then:

$$2A = A + \partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a}$$

And for the above equation to be true, we need $\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = A$.

Yes. And it is correct that $\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = A$.

Thus:

$$f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} = f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

I don't see how you conclude this from the previous steps.

 

[JD_PM]

I don't see how you conclude this from the previous steps.

You do not see them because I was wrong!

Actually I have just seen it!

We know that:

$$f^{cd}\partial_{b}g_{ac} \ \dot{q}^{a}\dot{q}^{b} + f^{cd}\partial_{a}g_{bc} \ \dot{q}^{b}\dot{q}^{a} - f^{cd}\partial_{c}g_{ab} \ \dot{q}^{a}\dot{q}^{b}$$

The key is seeing that $c$ is also a dummy index in the above expression. Then by swapping $a$ by $c$ we get the desired result:

$$f^{cd}\Big(\partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c}$$

Which leads to:

$$2\ddot{q}^{d} + f^{cd}\Big(\partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c} = -f^{cd}\partial_{c} V$$

To get the exact same answer, note that we can swap $c$ by $a$ on the right side ($a$ is summed over). Thus:

$$\ddot{q}^{d} + \frac{1}{2} \ f^{da}\Big( \partial_{c}g_{ab} + \partial_{b}g_{ac} - \partial_{a}g_{bc} \Big) \ \dot{q}^{b}\dot{q}^{c} = -\frac{1}{2} \ f^{da}\partial_{a} V$$

It turned out that my $EQ.2.2$ was wrong!

Thanks TSny! I had so much fun with this one!