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?

  • $\begingroup$ For a velocity-dependent potential it is not true that $\mathbf F=-\boldsymbol{\nabla}U$. A hint is that assume that the Lagrangian is of the form $L=T(\mathbf {\dot r})-U(\mathbf r,\mathbf{\dot r})$. $\endgroup$ Sep 24 '18 at 4:54
  • $\begingroup$ I'm trying to use this as justification for myself as to why the Euler Lagrange equations are valid (as they do here in the presentation—this is the part of the derivation of the EL equations). How can this be justified without referencing the EL equations? $\endgroup$ Sep 24 '18 at 4:55
  • $\begingroup$ @qmechanic is there no other logic to this other than "it's just what makes the Euler Lagrange equations work out?" $\endgroup$ Sep 24 '18 at 4:57
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    $\begingroup$ The equation ${\bf F}=-\nabla U$ only holds for conservative forces. Think about the Lorentz force. Would the work that you have to do to move a point charge from A to B depend on whether or not $\vec{v}$ and $\vec{B}$ were parallel or perpendicular? $\endgroup$ Sep 24 '18 at 5:09
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    $\begingroup$ It is just a generalization so that things 'look' the same as in the velocity/time independent case. Have a look at nhn.ou.edu/~gut/notes/cm/lect_09.pdf. I recommend working out this problem for the Lorentz force and seeing that all is well. $\endgroup$ Sep 24 '18 at 7:10

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).


In my opinion the $^{\prime}$special$^{\prime}$ case of generalized force associated with a velocity-dependent potential $\,U(q,\dot q, t)\,$ \begin{equation} Q_j = - \frac{\:\partial U\hphantom{_j}}{\partial q_j} + \frac{\mathrm d}{\mathrm dt} \left(\frac{\:\partial U\hphantom{_j}}{\partial \dot q_j}\right) \tag{01}\label{eq01} \end{equation} is motivated from the demand to introduce the electromagnetic Lorentz force in the Lagrangian formalism.

Now, we start with one force $\,\mathbf F\,$ \begin{equation} Q_j = \mathbf F \boldsymbol \cdot \frac{\partial \mathbf x\hphantom{_j}}{\partial q_j} \tag{02}\label{eq02} \end{equation} with \begin{equation} \text{generalized coordinate } q_j \equiv \text{ cartesian coordinate } x_j \tag{03}\label{eq03} \end{equation} so that \begin{equation} \mathbf x =\left(q_1,q_2,q_3\right)=\left(x_1,x_2,x_3\right) \tag{04}\label{eq04} \end{equation} Inserting the expression of Lorentz force \begin{equation} \mathbf F = q \left(\mathbf{E}+\mathbf{v}\boldsymbol{\times}\mathbf{B} \right) \tag{05}\label{eq05} \end{equation} in equation \eqref{eq02} we have \begin{equation} Q_j = q \left(\mathbf{E}+\mathbf{v}\boldsymbol{\times}\mathbf{B} \right)_j \tag{06}\label{eq06} \end{equation} Since \begin{equation} \mathbf{E} = -\boldsymbol{\nabla}\phi-\frac{\partial \mathbf{A}}{\partial t}\,,\quad \mathbf{B} = \boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A} \tag{07}\label{eq07} \end{equation} for the first component $\,Q_1\,$ we have \begin{align} \!\!\!\!\!\!\frac{Q_1}{q} & =\left(\mathbf{E}+\mathbf{v}\boldsymbol{\times}\mathbf{B} \right)_1 = E_1+\left(\mathbf{v}\boldsymbol{\times}\mathbf{B} \right)_1= E_1+\mathrm v_2 B_3 -\mathrm v_3 B_2= E_1+\dot{q}_2 B_3 -\dot{q}_3 B_2 \nonumber\\ &=-\frac{\partial\phi\hphantom{_1}}{\partial q_1}-\frac{\partial A_1}{\partial t} +\dot{q}_2\left(\frac{\partial A_2}{\partial q_1}-\frac{\partial A_1}{\partial q_2} \right)-\dot{q}_3\left(\frac{\partial A_1}{\partial q_3}-\frac{\partial A_3}{\partial q_1} \right) \nonumber\\ &=-\frac{\partial\phi\hphantom{_1}}{\partial q_1}+\underbrace{\left(\dot{q}_1\frac{\partial A_1}{\partial q_1}+\dot{q}_2\frac{\partial A_2}{\partial q_1}+\dot{q}_3\frac{\partial A_3}{\partial q_1} \right)}_{\boldsymbol{=}\mathbf{v}\boldsymbol{\cdot}\frac{\partial \mathbf{A}}{\partial q_1}\boldsymbol{=}\frac{\partial (\mathbf{v}\boldsymbol{\cdot}\mathbf{A})}{\partial q_1}}-\underbrace{\left(\frac{\partial A_1}{\partial t}+\dot{q}_1\frac{\partial A_1}{\partial q_1}+\dot{q}_2\frac{\partial A_1}{\partial q_2}+\dot{q}_3\frac{\partial A_1}{\partial q_3} \right)}_{\boldsymbol{=}\frac{\mathrm d A_1}{\mathrm d t}\boldsymbol{=}\frac{\mathrm d }{\mathrm d t}\left[\frac{\partial (\mathbf{v}\boldsymbol{\cdot}\mathbf{A})}{\partial \dot{q}_1}\right]\boldsymbol{=-}\frac{\mathrm d }{\mathrm d t}\left[\frac{\partial (\phi-\mathbf{v}\boldsymbol{\cdot}\mathbf{A})}{\partial \dot{q}_1}\right]} \nonumber\\ &=-\frac{\partial\left(\phi-\mathbf{v}\boldsymbol{\cdot} \mathbf{A}\right)}{\partial q_1}+\frac{\mathrm d }{\mathrm d t}\left[\frac{\partial (\phi-\mathbf{v}\boldsymbol{\cdot}\mathbf{A})}{\partial \dot{q}_1}\right] \tag{08}\label{eq08} \end{align} that is \begin{equation} Q_1 = - \frac{\:\partial U\hphantom{_1}}{\partial q_1} + \frac{\mathrm d\hphantom{t}}{\mathrm dt} \left(\frac{\:\partial U\hphantom{_j}}{\partial \dot q_1}\right) \tag{09}\label{eq09} \end{equation} so \begin{equation} Q_j = - \frac{\:\partial U\hphantom{_1}}{\partial q_j} + \frac{\mathrm d\hphantom{t}}{\mathrm dt} \left(\frac{\:\partial U\hphantom{_j}}{\partial \dot q_j}\right) \tag{10}\label{eq10} \end{equation} where \begin{equation} U(q,\dot q,t) =q\left[\phi-\left(\dot{q}_1 A_1+\dot{q}_2 A_2+\dot{q}_3 A_3\right)\right] = q\left[\phi(q,t)-\mathbf{v}\boldsymbol{\cdot} \mathbf{A}(q,t)\right] \tag{11}\label{eq11} \end{equation}

Note :

This elaboration [to find a velocity-dependent potential $\,U(q,\dot q, t)\,$ for the electromagnetic Lorentz force] is identical to that of the .pdf in the link given by user @Alex Opremcak in his comment

It is just a generalization so that things 'look' the same as in the velocity/time independent case. Have a look at Physics 5153 Classical Mechanics - Velocity Dependent Potentials. I recommend working out this problem for the Lorentz force and seeing that all is well. – Alex Opremcak.


No proof needed. OP's eq. (1) is just the defining relation for a velocity-dependent potential $U$. OP's eq. (3) is not always valid.


Earlier, I thought this could be derived. An earlier answer said it is defined this way, but why was it defined this way is the important thing to find out. To do that, we have to go back to the development of Lagrange's Equations.

In the process of deriving Lagrange's Equations eventually we get to the point where we have:

${\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j}$ $\quad$[Base Eq'n]

Here T is the kinetic energy and $Q_j$ is the generalized force. If the system is conservative then from basic mechanics we know $Q_j$ (a force) can be represented by the negative change in a potential. That is, $Q_j$ can be represented as :

${Q_j = -\frac{\partial V}{\partial q_j} }$

Then we can have:

${\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = -\frac{\partial V}{\partial q_j}} $

${\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial (T-V)}{\partial q_j} = 0} $

And since $V$, our potential, is not a function of velocity (contains no velocity terms) we can write:

${\frac{d}{dt} \Big( \frac{\partial (T-V)}{\partial \dot q_j} \Big) - \frac{\partial (T-V)}{\partial q_j} = 0} $

${\frac{d}{dt} \Big( \frac{\partial L}{\partial \dot q_j} \Big) - \frac{\partial L}{\partial q_j} = 0} $

Suppose now we want to use these equations but we have forces in our system that are velocity dependant. Can we still use the Lagrangian of this form? If we use the recommended prescription for $Q_i$

$ {Q_j = \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot q_j} } $

Then if we put that into the base equation and follow the same procedure:

${\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j}$

${\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot q_j}}$

$ {\frac{d}{dt} \Big( \frac{\partial (T-U)}{\partial \dot q_j} \Big) - \frac{\partial (T-U)}{\partial q_j} = 0} $

${\frac{d}{dt} \Big( \frac{\partial L}{\partial \dot q_j} \Big) - \frac{\partial L}{\partial q_j} = 0} $

So we can use the same Lagrange equations for a velocity dependant force if we can define that force $Q_j$ in the way described in the question. In electrodynamics we can do this. Not easily though, but the process is covered quite well in the previous answer to this question.


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