I'm currently working through the book Heisenberg's Quantum Mechanics (Razavy, 2010), and am reading the chapter on classical mechanics. I'm interested in part of their derivative of a generalized Lorentz force via a velocity-dependent potential.

I understand the generalized force

$$F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$$

that they derive from a Lagrangian of the form $L = \frac{1}{2}m|\vec v|^2 - V(\vec r,\vec v,t)$.

However, in the next (critical) step of the derivation, the author cites a theorem from Helmholtz saying

...according to Helmholtz, for the existence of the Lagrangian, such a generalized force can be at most a linear function of acceleration, and it must satisfy the Helmholtz identities.

The three Helmholtz identities are then listed as:

$$\frac{\partial F_i}{\partial \dot{v_j}} = \frac{\partial F_j}{\partial \dot{v_i}},$$

$$\frac{\partial F_i}{\partial v_j} + \frac{\partial F_j}{\partial v_i} = \frac{d}{dt}\left(\frac{\partial F_i}{\partial \dot{v_j}} + \frac{\partial F_j}{\partial \dot{v_i}}\right),$$

$$\frac{\partial F_i}{\partial x_j} - \frac{\partial F_j}{\partial x_i} = \frac{1}{2}\frac{d}{dt}\left(\frac{\partial F_i}{\partial v_j} - \frac{\partial F_j}{\partial v_i}\right).$$

I'm trying to understand where this theorem comes from. Razavy cited a 1887 paper by Helmholtz. I was able to find a PDF online, but it is in German, so I could not verify whether or not it proved the theorem. Additionally, I could not find it in any recent literature. I searched online and in Goldstein's Classical Mechanics.

The only similar concept that I can find is in the Inverse problem for Lagrangian mechanics where we have three equations known as Helmholtz conditions. Are these two concepts one in the same? If so, how should I interpret the function $\Phi$ and the matrix $g_{ij}$ that appear in the Helmholtz conditions I found online?

If the cited theorem from Razavy does not relate from the inverse Lagrangian problem, could I have some help finding the right direction?

  • $\begingroup$ I'm still working on it, and I found this article that goes over the Helmholtz conditions with a brief proof has proved pretty helpful! My current thought is that these conditions may be combined with the fact that $F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$ in order to prove the antisymmetry relations that are found in the integration from my other question. $\endgroup$
    – Noah M
    Commented Jul 5, 2019 at 15:41
  • $\begingroup$ However, this doesn't explain why "such a generalized force can be at most a linear function of acceleration". $\endgroup$
    – Noah M
    Commented Jul 5, 2019 at 15:44
  • $\begingroup$ With a linear function of acceleration, the acceleration is aligned with the velocity. With a nonlinear acceleration, the acceleration may constantly change the direction of the velocity. $\endgroup$ Commented Jul 8, 2019 at 7:45

2 Answers 2

  1. We are interested whether a given force $$ {\bf F}~=~{\bf F}({\bf r},{\bf v},{\bf a},t) \tag{1}$$ has a velocity-dependent potential $$U~=~U({\bf r},{\bf v},t),\tag{2}$$ which by definition means that $$ {\bf F}~\stackrel{?}{=}~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}. \tag{3} $$

  2. If we define the potential part of the action as $$ S_p~:=~\int \!dt~U,\tag{4}$$ then the condition (3) can be rewritten with the help of a functional derivative as $$ F_i(t)~\stackrel{(2)+(3)+(4)}{=}~ -\frac{\delta S_p}{\delta x^i(t)}, \qquad i~\in~\{1,\ldots,n\}, \tag{5} $$ where $n$ is the number of spatial dimensions.

  3. It follows from eqs. (2) & (3) that in the affirmative case the force ${\bf F}$ must be an affine function in acceleration ${\bf a}$.

  4. Since functional derivatives commute $$ \frac{\delta}{\delta x^i(t)} \frac{\delta S_p}{\delta x^j(t^{\prime})} ~=~\frac{\delta}{\delta x^j(t^{\prime})} \frac{\delta S_p}{\delta x^i(t)},\tag{6}$$ we derive the following consistency condition (7) for a force with a velocity dependent potential $$ \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} ~\stackrel{(5)+(6)}{=}~[(i,t) \longleftrightarrow (j,t^{\prime})].\tag{7} $$ Eq. (7) is a functional analog of a Maxwell relation, and equivalent to the Helmholtz conditions$^1$

    $$ \begin{align} \frac{\partial F_i(t)}{\partial x^j(t)} ~-~\frac{1}{2}\frac{d}{dt}\frac{\partial F_i(t)}{\partial v^j(t)} ~+~\frac{1}{4}\frac{d^2}{dt^2}\frac{\partial F_i(t)}{\partial a^j(t)}~&=~+[i \longleftrightarrow j], \cr \frac{\partial F_i(t)}{\partial v^j(t)} ~-~\frac{d}{dt}\frac{\partial F_i(t)}{\partial a^j(t)} ~&=~-[i \longleftrightarrow j], \cr \frac{\partial F_i(t)}{\partial a^j(t)}~&=~+[i \longleftrightarrow j] .\end{align}\tag{8} $$

    [The above form (8) of the Helmholtz conditions can be simplified a bit.]

  5. Sketched systematic proof of the Helmholtz conditions (8). The distribution on the LHS of eq. (7) reads $$ \begin{align} \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} &~\stackrel{(1)}{=}~\left[\frac{\partial F_i(t)}{\partial x^k(t)} ~+~ \frac{\partial F_i(t)}{\partial v^k(t)}\frac{d}{dt} ~+~ \frac{\partial F_i(t)}{\partial a^k(t)}\frac{d^2}{dt^2}\right] \frac{\delta x^k(t)}{\delta x^j(t^{\prime})}\cr &~=~\left[\frac{\partial F_i(t)}{\partial x^j(t)} ~+~ \frac{\partial F_i(t)}{\partial v^j(t)}\frac{d}{dt} ~+~ \frac{\partial F_i(t)}{\partial a^j(t)}\frac{d^2}{dt^2}\right]\delta(t\!-\!t^{\prime})\cr &~=~\left[\frac{\partial F_i(t)}{\partial x^j(t)} ~-~ \frac{\partial F_i(t)}{\partial v^j(t)}\frac{d}{dt^{\prime}} ~+~ \frac{\partial F_i(t)}{\partial a^j(t)}\frac{d^2}{dt^{\prime 2}}\right]\delta(t\!-\!t^{\prime}) .\end{align}\tag{9} $$ Let us introduce for later convenience new coordinates $$ t^{\pm}~:=~\frac{t \pm t^{\prime}}{2} \quad\Leftrightarrow\quad \left\{\begin{array}{c} t~=~ t^++t^- \cr t^{\prime}~=~ t^+-t^-\end{array} \right\} \quad\Rightarrow\quad \frac{d}{dt^{\pm}}~=~ \frac{d}{dt} \pm \frac{d}{dt^{\prime}}.\tag{10} $$ If we introduce a testfunction $f\in C^{\infty}_c(\mathbb{R}^2)$ with compact support, there are no boundary terms when we integrate by parts: $$ \begin{align} \iint_{\mathbb{R}^2} \! dt~dt^{\prime}&~f(t^+,t^-)~\frac{\delta F_i(t)}{\delta x^j(t^{\prime})} \cr \stackrel{(9)}{=}~~~~&2\iint_{\mathbb{R^2}} \! dt^+~ dt^-~ f(t^+,t^{-})\left[\frac{\partial F_i(t)}{\partial x^j(t)} - \frac{\partial F_i(t)}{\partial v^j(t)}\frac{d}{dt^{\prime}} + \frac{\partial F_i(t)}{\partial a^j(t)}\frac{d^2}{dt^{\prime 2}} \right] \delta(2t^-) \cr \stackrel{\text{int. by parts}}{=}&2\iint_{\mathbb{R^2}} \! dt^+~ dt^-~ \delta(2t^-)\left[\frac{\partial F_i(t)}{\partial x^j(t)} + \frac{\partial F_i(t)}{\partial v^j(t)}\frac{d}{dt^{\prime}} + \frac{\partial F_i(t)}{\partial a^j(t)}\frac{d^2}{dt^{\prime 2}} \right] f(t^+,t^{-})\cr =~~~~&\int_{\mathbb{R}} \! dt^+~\left[\frac{\partial F_i(t^+)}{\partial x^j(t^+)} + \frac{\partial F_i(t^+)}{\partial v^j(t^+)}\frac{d}{dt^{\prime}} + \frac{\partial F_i(t^+)}{\partial a^j(t^+)}\frac{d^2}{dt^{\prime 2}} \right] f(t^+,0) \cr \stackrel{(10)}{=}~~~&\int_{\mathbb{R}} \! dt^+~\left[\frac{\partial F_i(t^+)}{\partial x^j(t^+)} + \frac{\partial F_i(t^+)}{\partial v^j(t^+)}\frac{1}{2}\left(\frac{d}{dt^+}-\frac{d}{dt^-}\right)\right. \cr &+\left. \frac{\partial F_i(t^+)}{\partial a^j(t^+)}\frac{1}{4}\left(\frac{d}{dt^+}-\frac{d}{dt^-}\right)^2 \right] f(t^+,0)\cr \stackrel{\text{int. by parts}}{=}&\int_{\mathbb{R}} \! dt^+~\left[\left(\frac{\partial F_i(t^+)}{\partial x^j(t^+)}-\frac{1}{2}\frac{d}{dt^+}\frac{\partial F_i(t^+)}{\partial v^j(t^+)}+\frac{1}{4}\frac{d^2}{dt^{+ 2}}\frac{\partial F_i(t^+)}{\partial a^j(t^+)} \right)\right. \cr &+\left.\frac{1}{2}\left(\frac{d}{dt^+}\frac{\partial F_i(t^+)}{\partial a^j(t^+)}- \frac{\partial F_i(t^+)}{\partial v^j(t^+)}\right)\frac{d}{dt^-} + \frac{1}{4}\frac{\partial F_i(t^+)}{\partial a^j(t^+)}\frac{d^2}{dt^{- 2}} \right] f(t^+,0) .\end{align}\tag{11} $$
    Now compare eqs. (7) & (11) to derive the Helmholtz conditions (8). We get 3 conditions because each order of $t^-$-derivatives of the testfunction $f$ along the diagonal $t^-=0$ are independent. There is an additional minus sign in the middle condition (8) because $t^-$ is odd under $t\leftrightarrow t^{\prime}$ exchange. $\Box$

  6. It is in principle straightforward to use the same proof technique to generalize the Helmholtz conditions (8) to the case where the force (1) and potential (2) depend on higher time-derivatives.


$^1$ The other Helmholtz conditions mentioned on the Wikipedia page of the inverse problem for Lagrangian mechanics address a much more difficult problem: Given a set of EOMs, we possibly have to rewrite them before they might have a chance of becoming on the form: functional derivative $\approx 0$. See also this related Phys.SE post.

  • $\begingroup$ Thank you so much for going so in-depth! This was really helpful to me. I'm still trying to parse through everything and follow your steps, but this is fantastic! I especially like how you worked with the action and functional derivatives directly. I need to get better/more acquainted with this kind of math. I appreciate it! $\endgroup$
    – Noah M
    Commented Jul 10, 2019 at 4:20

I found the solution

The criteria that Razavy lists are indeed the Helmholtz conditions for the inverse Lagrangian problem. Their form in the Wikipedia article does not make the connection obvious, but I found this arXiv article that spells them out in just about the same form and proves them.


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