# Classical Mechanics: Relation between general velocity and general potential function for velocity-dependent potential

How is the general force derived from the general potential for a velocity-dependent potential $$U = U(q_j,\dot{q_j})$$? $$Q_j=-\frac{\partial U}{\partial q_j}+ \frac{\mathrm{d}}{\mathrm{dt}}(\frac{\partial U}{\partial \dot{q_j}}). \tag{1}$$

What I understand

1. In a conservative vector field, the force could be found through the gradient of the scalar potential, i.e. $$F = -\nabla V$$.
2. General force in non-velocity-dependent potentials is given by $$Q_j = -\frac{\partial V}{\partial q_j}$$, in a similar manner as of that in bullet point 1, where $$V$$ is a conservative vector field, $$Q_j$$ is the general force and $$q_j$$ a general coordinate.

My attempt to derive Eqn. (1)

1. Recall that the general potential is a function of general coordinate and the general velocity, $$U = U(q_j,\dot{q_j})$$

2. Apply $$Q_j = -\nabla U$$, and get \begin{aligned} Q_j &= -\nabla U(q_j,\dot{q_j})\\ &=-(\frac{\partial}{\partial q_j}+\frac{\partial}{\partial \dot{q_j}})U(q_j,\dot q_j)\\ &=-\frac{\partial U}{\partial q_j}-\frac{\partial U}{\partial \dot q_j} \end{aligned} \tag{2} where the second term on R.H.S. does not match Eqn. (1)

My question

1. What are the things/concepts that I am missing?
2. How is Eqn.(1) derived?
3. How to intuitively/physically understand the second term on R.H.S. in Eqn.(1)?
• Commented Jan 19, 2022 at 9:02

• 1 & 2: Eq. (1) is not derived. It is a defining property of a generalized velocity-dependent potential.

1. The form of eq. (1) mimics the Euler-Lagrange operator, so that we can bring Lagrange equations $$\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^j}-\frac{\partial T}{\partial q^j}~=~Q_j, \qquad j~\in \{1,\ldots, n\}, \tag{L}$$ in the form of Euler-Lagrange equations $$\frac{d}{dt}\frac{\partial (T-U)}{\partial \dot{q}^j}-\frac{\partial (T-U)}{\partial q^j}~=~0, \qquad j~\in \{1,\ldots, n\}. \tag{EL}$$

Let us start from the general form of Euler-Lagrange equations for a curve (I will discuss all this issue in local coordinates thoug global approaches are possible) $$\mathbb{R} \ni t \mapsto (t, q(t), \dot{q}(t)) \in \mathbb{R}^{2n+1}$$ $$\left.\frac{d}{dt}\left( \frac{\partial T|_R(t, q,\dot{q})}{\partial \dot{q}^k}\right)\right|_{(t,q(t),\dot{q}(t)} - \left.\frac{\partial T|_R(t,q,\dot{q})}{\partial q^k}\right|_{(t,q(t),\dot{q}(t)} = Q_k|_R(t,q(t), \dot{q}(t))\quad \mbox{where}\quad \frac{dq^k}{dt} = \dot{q}^k\:, \quad k=1,\ldots, n\:.\tag{1}$$ Above, $$T|_R$$ is the kinetic energy of the system with respect to some fixed reference frame $$R$$ written as a function of the $$n$$ Lagrangian coordinates $$q^k$$ and their formal time-derivatives $$\dot{q}^k$$ once we have solved'' the possible (ideal) $$c$$ constraints on the configurations of the physical system made of, say $$N$$ material points, so that $$n= 3N-c$$.

The known functions $$Q_k(t,q, \dot{q})$$ are obtained out of the forces acting on the system of $$N$$ points in the reference frame $$R$$. $$Q_k|_R(t,q,\dot{q}) = \sum_{i=1}^N \frac{\partial \vec{x}_i}{\partial q^k} \cdot \vec{F}_i|_R(t, \vec{x}_1,\ldots,\vec{x}_N, \vec{v}_1|_R, \ldots \vec{v}_N|_R)\:,$$ Here, $$\vec{x}_i(t,q^1,\ldots, q^n)$$ is the position vector of the ith material point in the reference frame $$R$$ written as a function of time and the Lagrangian coordinates and $$\vec{v}_i|_R(t,q^1,\ldots, q^n, \dot{q}^1,\ldots, \dot{q}^n)$$ is the velocity of that material point in $$R$$.

It is important to specify the reference frame $$R$$ since, if $$R$$ is not inertial, the $$Q_k|_R$$ also include the contribution of the inertial pseuodoforces.

We are also assuming that we know the functional form of these forces $$\vec{F}_i|_R$$ and thus $$Q_k|_R:= Q_k|_R(t,q, \dot{q})$$ are $$n$$ known functions (differently form the functional form of the reactive forces due to the ideal constraints which are embodied in the Lagrangian formalism).

If the forces $$\vec{F}_i|_R$$ are conservative in $$R$$, i.e., $$\vec{F}_i|_R(\vec{x}_1,\ldots,\vec{x}_N) = -\nabla_{\vec{x}_i} U|_R(\vec{x}_1,\ldots,\vec{x}_N)\:,$$ we immediately have that $$Q_k|_R(t,q) = -\frac{\partial }{\partial q^k} U|_R(t,q^1,\ldots,q^n)\:,$$ where we have adopted the non completely rigorous notation, but very effective $$U|_R(t,q^1,\ldots,q^n) := U|_R(\vec{x}_1(t,q^1,\ldots, q^n), \ldots, \vec{x}_N(t,q^1,\ldots, q^n))\:.$$

Euler-Lagrange's equations specialize to $$\frac{d}{dt}\left( \frac{\partial T|_R(t, q,\dot{q})}{\partial \dot{q}^k}\right) - \frac{\partial T|_R(t,q,\dot{q})}{\partial q^k} = - \frac{\partial U|_R}{\partial q^k} \:.$$

Since, trivially, $$\frac{\partial}{\partial \dot{q}^k} U|_R(t,q^1,\ldots,q^n) =0\:,$$ we can re-write those equations as $$\frac{d}{dt}\left( \frac{\partial T|_R(t, q,\dot{q})- U|_{R}(t,q)}{\partial \dot{q}^k}\right) - \frac{\partial T|_R(t,q,\dot{q})- U|_{R}(t,q)}{\partial q^k} = 0\:,$$ namely $$\frac{d}{dt}\left( \frac{\partial L|_R(t, q,\dot{q})}{\partial \dot{q}^k}\right) - \frac{\partial L|_R(t,q,\dot{q})}{\partial q^k} = 0\:, \tag{2}$$ where we have introduced the Lagrangian referred to $$R$$, $$L|_R(t, q,\dot{q}) := T|_R(t, q,\dot{q}) - U|_R(t, q)\:.\tag{3}$$

It is clear that the procedure can be generalized further encompassing forces, if any, whose functional form may depend also on the velocities like this $$Q_k|_R(t,q, \dot{q}) = \frac{d}{dt}\left( \frac{\partial U|_R(t, q,\dot{q})}{\partial \dot{q}^k}\right) - \frac{\partial U|_R(t,q,\dot{q})}{\partial q^k}\tag{4}$$
for some generalized potential $$U|_R:= U|_R(t,q, \dot{q})\:.$$ Also in that case, starting from (1) one sees immediately that E.L's equations take the form (2) if we have defined a Lagrangian as in (3).

Due to several issues arising from the implementation of this recipe, one eventually sees that a generalized potential can be only linear in the variables $$\dot{q}^k$$. (As a matter of fact, that is due to the structure of the EL equations, without this condition the existence and uniqueness theorem may fail, but I do not want to enter here into the details of this technical issue).

The most general permitted form of a generalize potential is therefore $$U(t,q,\dot{q}) = B(t,q)+ \sum_{k=1}^n A_k(t,q)\dot{q}^k \tag{5}\:.$$

A natural question pops out now. Is this nothing but an elegant mathematical extension of the formalism without physical content?

Nope! There are at least two very important cases.

1. The Lorentz force acting on a charge due to generic external electromagnetic fields. The E.L. equations can be written using a well-known generalized potential.

2. All types of inertial forces when $$R$$ is not inertial and its motion is known with respect to an inertial reference frame $$R_0$$. Also in this case there is a generalized potential giving rise to all inertial pseudoforces.