Laplace equation derivation, where does the potential go

[Uku]

Homework Statement

Since the potential field is only a function of position, not velocity, Lagrange's equations are as follows:

(Wikipedia, image 1)

Homework Equations

(Wikipedia, image 2)

The Attempt at a Solution

Now, $-\frac{\partial{V}}{\partial{q_{j}}}$

How is speed involved in this derivative of potential by generalized coordinate (upper left, image 1)? Potential only depends on the position, we are assuming a conservative field, and that is what we are doing, how does this term disappear?
Feels like saying $F=-\nabla V=0$.
I mean, I can understand eg. generalized impulse depending on speed: $p_{j}=\frac{\partial{L}}{\partial{\dot q_{j}}}$, but not the potential-evaporating transition:
$\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot q_{j}}}\right)-\frac{\partial{L}}{\partial{q_{j}}}=0$

 

[squigglywolf]

So what is your question exactly? Cant quite follow.

 

[Uku]

My problem is in the sentence:
Since the potential field is only a function of position, not velocity, Lagrange's equations are as follows:
And in:
$\frac{\partial{V}}{\partial{g_{j}}}$ disappearing from the very first statement.
So - where does it go, $g_{j}$ is a generalized coordinate, not generalized velocity..?

 

[I like Serena]

For a conservative field, which is only dependent on position, the lagrangian is defined as: $\mathcal{L} = \mathcal{L}(\dot q_j, q_j, t) = T(\dot q_j, q_j, t) - V(q_j)$.

In particular, this means that:
$${\partial \mathcal{L}\over \partial \dot q_i} = {\partial \over \partial \dot q_i}(T(\dot q_j, q_j, t) - V(q_j))
= {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - {\partial \over \partial \dot q_i}V(q_j)
= {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - 0$$

 

[Uku]

Thanks! That makes perfect sense, but it means that there is an error in:

since here, very literally, $\frac{\partial V}{\partial q_{j}}=0$

 

[I like Serena]

Well, what is the following?
$${\partial \mathcal{L}\over \partial q_i} = {\partial \over \partial q_i}(T(\dot q_j, q_j, t) - V(q_j)) $$

since here, very literally, $\frac{\partial V}{\partial q_{j}}=0$

Uhh
How do you get that?

 

[Uku]

I think I mixed up, the potentials subtract. Thanks!