On slide 16 of this presentation it is stated without proof that given a velocity-dependent potential $U(q,\dot q, t)$, the associated generalized force is $$Q_j = – \frac{\partial U}{\partial q^j} + \frac{d}{dt} \frac{\partial U}{\partial \dot q^j}.\tag{1}$$
I'm trying to prove this for myself using the definition of generalized forces without avail and am looking for some help.
My attempt thus far
Generalized forces are defined as follows.
$$Q_j = \mathbf F \cdot \frac{\partial \mathbf x}{\partial q^j}.\tag{2}$$
Since $$\mathbf F = – \nabla U ,\tag{3}$$ this reduces as follows.
$$\begin{align}
Q_j &= – \nabla U \cdot \frac{\partial \mathbf x}{\partial q^j} \\
&= -\frac{\partial U}{\partial x^k} \frac{\partial x^k}{\partial q^j} \\
&= -\frac{\partial U}{\partial q^j}.
\end{align}\tag{4}$$
So, obviously, I've proven the first term holds, but where does the second come from? Or, perhaps my interpretation that $\mathbf F = – \nabla U$ holds for velocity-dependent potentials is incorrect? Can anyone clarify?
Best Answer
I doubt there is any way through which OP can ascertain the validity of the EL equations through velocity-dependent potentials.
Unless the EL equations (or altenatively, the variational principle) is postulated as an axiom, probably the most natural way to arrive at them is through the argument presented in Classical Mechanics by Goldstein, which is way too long to reproduce here.
Long story short, by analizing systems with holonomic constraints, one can arrive at the fact that the EoMs can be written as $$ 0=\frac{\partial L}{\partial q^i}-\frac{d}{dt}\frac{\partial L}{\partial\dot q^i} $$ where $$ L(q,\dot q,t)=T(q,\dot q)-U(q). $$ Here ordinary conservative forces have been considered, and the coordinates are generalized coordinates that have been adapted to the constraints, then reduced.
From this point on then one can realize that the given an action functional $$ S[\gamma]=\int dt\ L(q(t),\dot q(t),t), $$ the EL equations are equivalent with $$ \frac{\delta S}{\delta q^i(t)}=0, $$ eg. the functional derivative must vanish.
One then proposes this variational principle as fundamental, so we say that a Lagrangian system is one with an action functional of the form $$ S=\int dt\ L(q(t),\dot q(t),t). $$ This doesn't always work, as not all classical mechanical systems are of this form. However pretty much all somewhat "fundamental" systems are!
Now, one can consider the case of an ordinary particle in $\mathbb R^3$, whose Lagrangian is of the form $$ L=\frac{1}{2}m\dot{\mathbf r}^2-U(\mathbf r,\dot{\mathbf r}). $$ Clearly we have $$ \frac{\partial T}{\partial\mathbf r}=0,\ \frac{d}{dt}\frac{\partial T}{\partial\dot{\mathbf r}}=m\ddot{\mathbf r}=\mathbf F, $$ and so we obtain $$ 0=\frac{\partial L}{\partial\mathbf r}-\frac{d}{dt}\frac{\partial L}{\partial\dot{\mathbf r}}=\left(\frac{\partial T}{\partial\mathbf r}-\frac{d}{dt}\frac{\partial T}{\partial\dot{\mathbf r}}\right)-\left(\frac{\partial U}{\partial\mathbf r}-\frac{d}{dt}\frac{\partial U}{\partial\dot{\mathbf r}}\right) \\ =-\mathbf F-\frac{\partial U}{\partial\mathbf r}+\frac{d}{dt}\frac{\partial U}{\partial\dot{\mathbf r}}, $$ and rearranging, we obtain the desired result.
Now, the point is, this is nothing deep basically. All we have derived is that if a Lagrangian has a velocity dependent potential term, then this is how the EoMs look like.
I know of only two kinds of velocity dependent forces in physics (might be more, idk), frictions, and the Lorentz force. Be that as it may, the Lorentz force can be described this way, while most frictious forces cannot be (there are some exceptions, but the Lagrangians do not really have any meaning in those cases, just mathematical curiosities).