[Physics] Can potential be velocity dependent


In the lagrangian solution for the equation of motion, there's a seemingly out of place $$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}}$$

term. Potential energy is usually a function of the set of $x_i$ or position only. If $x_i$ can all be rewritten as functions of only $q_i$ and $t$, and $q_i$ can be varied without having to change $\dot{q_i}$. Then we're left with this term being precisely $0$

So at least for conservative forces, this term should equal zero. But where do we find cases where it isn't? Magnetic forces? Is it frictional forces (what is potential for a frictional force anyway? And does Lagrange's equation even work for inelastic systems, considering energy is not conserved?)

Best Answer

If you define the Lagrangian as the difference of the energies, it does seem weird for a potential to be velocity dependant. I've come to terms with situations like that by not using that restrictive definition for the Lagrangian, and instead taking it for what it really is: a function that leads to equations of motion.

By Hamilton's Principle we know that a system with a coordinate $q(t)$ that follows second-order differential equations on $t$ can be equivalently described by the minimization of the functional $$S[q(t)]=\int_{t_1}^{t_2}\mathrm{d}t\,L(q,\dot{q},t),$$ since a necessary condition for writing $\delta S=0$ is itself a second order differential equation for $q(t)$, given by the Euler-Lagrange equation: $$\frac{\partial{L}}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial{L}}{\partial \dot q}=0.$$ After this is defined, we can begin describing interactions.

Suppose our system can be described through a specific Lagrangian, and the intensity of the interactions is parametized by a continuous parameter $\lambda$ (this can be charge, mass, etc.) such that $L(\lambda)$ for $\lambda=0$ describes the motion of the system without any interaction. If we expand $L$ around $\lambda=0$, we get $$L=L(0)+\sum_{n=1}^{\infty}\lambda^n\frac{\partial^nL}{\partial\lambda^n},$$ and defining $L_{\mathrm{free}}=L(0)$, $L_{\mathrm{int}}=\sum_{n=1}^{\infty}\lambda^n\frac{\partial^nL}{\partial\lambda^n}$, we can write the Hamiltonian (i.e. the energy) of our system as $$H=-\{(L_{\mathrm{free}}+L_{\mathrm{int}})-\dot q\frac{\partial}{\partial\dot q}(L_{\mathrm{free}}+L_{\mathrm{int}})\}\\H=(\dot q\frac{\partial}{\partial\dot q}-1)L_{\mathrm{free}}+(\dot q\frac{\partial}{\partial\dot q}-1)L_{\mathrm{int}}.$$ From this we define the kinetic energy $K$ as the first term, and the interaction potential $U$ as the second term. In the specific case where the potential does not depend on the velocity, we see that $U=-L_{int}$, which is what we would use for the usual Lagrangian definition. The usual definition also has $K=L_{\mathrm{free}}$, however, which is only the case in nonrelativistic mechanics, which by solving the differential equation for $L_{\mathrm{free}}$ gives us $L_{\mathrm{free}}\propto\dot q^2$. In the case of frictional forces and inelastic systems, however, what we have is a time dependant Lagrangian, resulting in a time dependant energy and therefore dissipation.