# Why is the potential independent of the generalized velocity?

In Goldstein, Classical Mechanics, Chap. 1.4 we derive Lagrange's equations from D'Alembert's Principle. My question is regarding the last part of the derivation, specifically the part where he introduces the Lagrangian $L$ defined as $T - V$: $$\frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} - Q_j = 0 ,\tag{1.53}$$ $$Q_j = - \frac{\partial V}{\partial q_j}. \tag{1.54}$$ Which when substituting yields: $$\frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial (T-V)}{\partial q_j} = 0\tag{1.55}.$$

The next step is what confuses me. He states that the potential $V$ does not depend on the generalized velocities so he makes the following substitution which "has no effect on differentiation with respect to $\dot{q}_j$": $$\frac{d}{dt} \left(\frac{\partial (T - V)}{\partial \dot{q}_j} \right) - \frac{\partial (T-V)}{\partial q_j} = 0.$$ Which leads to the familiar: $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0.\tag{1.57}$$

It makes sense intuitively. After all, you can vary a particles position without effecting its instantaneous velocity at the end of its path, therefore causing a change in potential without a corresponding change in velocity.

My confusion arises when you know the equations of motion as a priori: $V$ is dependent on position, position is dependent on time (from equations of motions), and velocity is dependent on time. Doesn't this means there exists some relationship between the potential function and velocity?

1. First of all, it should be mentioned that Goldstein rushes to define generalized velocity-dependent potentials on the very next page $$Q_j~=~ - \frac{\partial U}{\partial q^j} + \frac{d}{dt}\left(\frac{\partial U}{\partial v^j}\right). \tag{1.58}$$
2. It seems that the heart of OP's question is equivalent to the oft-asked question How can position $q$ and velocity $v$ be independent variables in the Lagrangian $L(q,v,t)$? This has been answered in e.g. this Phys.SE post.
• Regarding 2: For the simple example of projectile motion, you write the y-coordinate as a function of the velocity as such:$y = y_0 + \frac{\frac{1}{2}v^{2}_0 + 2v_0v + \frac{1}{2}v^{2}}{g}$ Substituting this into the potential V (in this case, mgy) we see that V is dependent on the velocity, which seemingly contradicts ∂U/∂v = 0. – andrew Dec 18 '15 at 22:39
• Here is one attempt to give an explanation: A solution to eoms depends on the choice of initial conditions $(q_0,v_0)$. The Lagrangian $L(q(t),v(t),t)$ (and hence the potential) is a state function, which is only allowed to depend on the state $(q(t),v(t))$ of the system in one instant $t$ at a time. If $L$ depends on more than one instant at a time, it would become non-local. In particular, it is not allowed to depend on initial conditions (which refer to another time). A more complete explanation is given in the linked post. – Qmechanic Dec 18 '15 at 23:00