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!