# Homework Help: Laplace equation derivation, where does the potential go

1. Jan 22, 2012

### Uku

1. The problem statement, all variables and given/known data
Since the potential field is only a function of position, not velocity, Lagrange's equations are as follows:

(Wikipedia, image 1)

2. Relevant equations

(Wikipedia, image 2)

3. 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:
$\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot q_{j}}}\right)-\frac{\partial{L}}{\partial{q_{j}}}=0$

Last edited: Jan 23, 2012
2. Jan 24, 2012

### squigglywolf

So what is your question exactly? Cant quite follow.

3. Jan 24, 2012

### 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..?

4. Jan 24, 2012

### 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:
$${\partial \mathcal{L}\over \partial \dot q_i} = {\partial \over \partial \dot q_i}(T(\dot q_j, q_j, t) - V(q_j)) = {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - {\partial \over \partial \dot q_i}V(q_j) = {\partial \over \partial \dot q_i}T(\dot q_j, q_j, t) - 0$$

5. Jan 24, 2012

### Uku

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

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

6. Jan 24, 2012

### I like Serena

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

Uhh
How do you get that?

7. Jan 26, 2012

### Uku

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