Potential Energy: Dependence on Position, Not Velocity

[observer1]

The form of the Lagrangian is: L = K - U

When cast in terms of generalized coordinates, the kinetic energy (K) can be a function of the rates of generalized coordinates AND the coordinates themselves (velocity and position); a case would be a double pendulum.

However, the potential energy (U) is ONLY a function of the generalized coordinates and NOT the rates: position, NOT velocity.

Yes, I understand that the potential energy is the potential function whose negative gradient delivers the force (from the path independence of the work done) and I can anticipate this will only be function of position and not velocity.

However, could anyone extend a few descriptive words - not equations: I got those - on why it is natural to expect the potential energy NOT to depend on velocities. Because... I am not really content with my explanation and am looking for something... just a few more words so I can rest more assured.

 

[wrobel]

Your question is about the definitions. Actually there is also a generalized potential $V=V(q,\dot q)$, it is determined as follows $Q_j=\frac{d}{dt}\frac{\partial V}{\partial \dot q^j}-\frac{\partial V}{\partial q^j}$. In classical mech. $V$ depends on $\dot q$ linearly: $V=v_i(q)\dot q^i+w(q)$. For example the Lorentz force possesses the generalized potential.

 

[observer1]

In classical mech. $V$ depends on $\dot q$ linearly: $V=v_i(q)\dot q^i+w(q)$. For example the Lorentz force possesses the generalized potential.

OK, this is new to me. If, as you say (and from experience on your responses, I have no reason to doubt you) in classical mechanics, the potential can be a function of velocity, can you give me an example? Springs? dashpots? gravity? mechanisms? I just don't see how. I am a bit shocked by your response because it was not what I was expecting. For if, in classical mechanics, the force must be the negative of the gradient of the potential function, I cannot see how the potential function can be a function of velocities.

 

[wrobel]

I have brought an example, another example is the Coriolis force if you are writing the equations in a non-inertial frame

 

[observer1]

I have brought an example, another example is the Coriolis force if you are writing the equations in a non-inertial frame

Wow... a flash of new insight... Can I say it in my words and ask you to assess my words?

We stand on a rotating merry go round and fire a ball outward and observe the motion of the ball from the rotating frame. It will exhibit unexpected motion due to the Coriolis effect. We relegate this as force and call it the Coriolis Force. Now if we solve the equations of motion of this system using variational methods (Hamilton's principle), then the potential energy is a function of the velocity.

If those words are correct, I am afraid I still do not see it. How can the potential energy be a function of the velocity of the ball if we are observing it in the rotating frame?

 

[wrobel]

Assume we have a particle of mass $m$ which is undergone to the force $\boldsymbol F=\boldsymbol B\times\boldsymbol v,\quad \boldsymbol v=(\dot x,\dot y,\dot z)$. For simplicity suppose that $\boldsymbol B=(B_x,B_y,B_z)=const$.
Then the equations of motion of this particle are the Lagrange equations with the Lagrangian $L=\frac{m}{2}(\dot x^2+\dot y^2+\dot z^2)-V,$ where
$$-2V=(-yB_z+zB_y)\dot x+(-zB_x+xB_z)\dot y+(yB_x-xB_y)\dot z.$$

 

[observer1]

Oh!

Thank you!

 

[vanhees71]

The form of the Lagrangian is: L = K - U

When cast in terms of generalized coordinates, the kinetic energy (K) can be a function of the rates of generalized coordinates AND the coordinates themselves (velocity and position); a case would be a double pendulum.

However, the potential energy (U) is ONLY a function of the generalized coordinates and NOT the rates: position, NOT velocity.

Yes, I understand that the potential energy is the potential function whose negative gradient delivers the force (from the path independence of the work done) and I can anticipate this will only be function of position and not velocity.

However, could anyone extend a few descriptive words - not equations: I got those - on why it is natural to expect the potential energy NOT to depend on velocities. Because... I am not really content with my explanation and am looking for something... just a few more words so I can rest more assured.

This is not always correct. The form $L=K-U$, where $K$ is the kinetic energy and $U$ a potential is only true when the kinetic-energy term is quadratic in the (generalized) velocities. In any case you have to determine the form of energy from Noether's theorem applied to time-translation invariance, leading to the Hamiltonian
$$H=p_k \dot{q}^k-L, \quad p_k=\frac{\partial L}{\partial \dot{q}^k}.$$
In relativistic physics usually the interaction part also depends on $\dot{q}^k$, e.g., the Lagrangian for the motion of a relativsitic charged particle in an external em. field in Cartesian coordinates reads
$$L=-m c^2 \sqrt{1-\vec{v}^2/c^2}-\frac{q}{c} \dot{x}^{\mu} A_{\mu}, \quad (x^{\mu})=(ct,\vec{x}).$$
As you can easily check, in this case $L \neq K-U$.