I have a system of a particle moving under the generalized central potential
$$ V= \frac{1}{r}(1+\dot{r}^2) \tag{1} $$ The general Euler-Lagrange equations for such type of potentials are:
$$ \frac{d}{dt}\frac{\partial{T}}{\partial\dot{q}_\alpha}-\frac{\partial{T}}{\partial{q}_\alpha}+\frac{\partial{V}}{\partial{q}_\alpha}=0 \tag{2} $$ Exists, for this system, some Lagrangian functional such that the equations of motion have this form? $$ \frac{d}{dt}\frac{\partial{L}}{\partial\dot{q}_\alpha}-\frac{\partial{L}}{\partial{q}_\alpha}=0 \tag{3} $$ My intention is not you to solve it. If someone has already faced this problem and knows the answer I would really appreciate it
PD: I've tried some general Lagrangians such as
$$ L_1=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-f(r,\dot{r}) \tag{4} $$ $$ L_2=\frac{1}{2}m(\dot{r}^2g(r)+r^2\dot{\theta}^2)-f(r) \tag{5} $$ The second is a particular case of the first one, assuming that the contribution of the velocity is quadratic. In all this attempts I fail in finding that functions.
Of course, this is not a "do my homework" question since I think this is a nice example of velocity-dependent potential and is not seen very much. Everything I've found is about the electromagnetic field but this one can -maybe- be seen as a generalized gravitational field (with gravitomagnetic corrections).

  • $\begingroup$ May i ask-does the terminology "generalized central potential'' is standard one? $\endgroup$
    – Mass
    Commented Jun 7, 2016 at 12:37
  • 4
    $\begingroup$ Terminology comment to the post (v2): Eq. (2) is a Lagrange-type equation but it is not called an EL equation. That word is reserved for the eq. (3). $\endgroup$
    – Qmechanic
    Commented Jun 7, 2016 at 13:27
  • $\begingroup$ Are you specifically working in a 2d case? How many degrees of freedom should the kinetic term have? I would suggest: Use energy conservation to determine what the equations of motion should be, and use this to determine what general properties $U$ in $L=T-U$ must have. $\endgroup$
    – levitopher
    Commented Jun 7, 2016 at 13:30
  • $\begingroup$ This particular case is 2-d problem, but it would be the same problem if we work with spherical (3d) because the velocity dependance is only in the r coordinate. It only adds an additional term to the kinetic energy. $\endgroup$ Commented Jun 8, 2016 at 7:25
  • $\begingroup$ Ok, i think i figured it out, the terminology ''generalized'' actualy refers to the U in L=T-U instead the usual standard V. $\endgroup$ Commented Jun 8, 2016 at 7:41

1 Answer 1


Yes there is. Notice that the expression

$$ \frac{d}{dt}\frac{\partial{T}}{\partial\dot{q}_\alpha}-\frac{\partial{T}}{\partial{q}_\alpha}+\frac{\partial{V}}{\partial{q}_\alpha}=0 $$

is very close to the Euler-Lagrange equations (ELeq) $$ \frac{d}{dt}\frac{\partial{L}}{\partial\dot{q}_\alpha}-\frac{\partial{L}}{\partial{q}_\alpha}=0 $$

except that the following term is null:


which in your case means (assuming a uni-dimensional problem)

$$\frac{d}{dt}\frac{\partial{V}}{\partial\dot{q}_\alpha} = \frac{d}{dt}(2 \frac{\dot r}{r}) = 0 \rightarrow r(t) = at+b \,\,\,\,\,\, where \,\,\, a,b:constants$$

Which means that your system will follow the ELeq only if its equations of motion are, at most, linear with time.

In other words, when you ask for compliance with ELeq you are restricting the solutions, or motions, to functions of constant or linear dependence with time.


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