I'm reading "The variational principles of mechanics- Lanczos",

The author mentions a relation between Work-Function $U(q_1,q_2,\cdots,q_n,\dot q_1,\dot q_2,\cdots,\dot q_n)$ and the potential energy $V(q_1,q_2,\cdots,q_n)$

$$V=\sum_{i=0}^n \frac{\partial U}{\partial \dot q_i}\dot q_i-U \tag{1}$$

$q_i$'s are the generalized coordinates

The work function and the generalized force $(Q_j)$ are related as

$$Q_j=\frac{\partial U}{\partial q_j}-\frac{d}{dt}\frac{\partial U}{\partial \dot q_j} \tag{2}$$

Looking at the equation $(1)$ I can only tell that $V$ is the legendre transform of $U$ but I'm not able to prove it, The work function as we can see depends also on $\dot q_i$

And we usually have velocity-independent work functions, in this case equation $(1)$ reduces to $V=-U$, and equation $(2)$ becomes

$$Q_i=-\frac{\partial V}{\partial q_i}$$

Which is the well known equation for conservative forces

I searched the internet but couldn't find anything close to this, Can somebody give me a clue on how to derive this? Any help is appreciated


I) Forget Lanczos's book & terminology for a moment. Recall that

  1. the Lagrangian $$\tag{1}L(q,v,t)~=~T(q,v,t)-U(q,v,t)$$ is usually of the form kinetic term minus potential term, where $U(q,v,t)$ is the generalized (possibly velocity-dependent) potential.

  2. the Lagrangian energy function is usually defined as$^1$ $$\tag{2}h_L(q,v,t)~:=~ v\frac{\partial L(q,v,t)}{\partial v}-L(q,v,t).$$

  3. if the Lagrangian $L$ does not depend explicit on time $t$, then the energy $h$ is conserved on-shell.

II) Now let's define $$\tag{3} h_T(q,v,t)~:=~ v\frac{\partial T(q,v,t)}{\partial v}-T(q,v,t),$$ and $$\tag{4} h_U(q,v,t)~:=~ v\frac{\partial U(q,v,t)}{\partial v}-U(q,v,t).$$ Then $$\tag{5} h_L(q,v,t)~=~h_T(q,v,t)-h_U(q,v,t).$$ Note that (3) is just the kinetic term $h_T=T$ if $T$ is quadratic in the velocities $v$.

III) Now let's return to Lanczos's notation.

  1. What Lanczos calls the work function is minus the above generalized potential $U$.

  2. What Lanczos calls the potential energy is $$\tag{6} V(q,v,t)~:=~-h_U(q,v,t).$$

Note that eq. (6) is not a Legendre transformation of some variables. Lanczos makes this definition (6) so that he gets to say that the total energy (2) is the sum $h_L=T+V$ of the kinetic and the potential energy when $T$ is quadratic in the velocities $v$.


  1. C. Lanczos, The variational principles of mechanics, 1949.


$^1$ The energy function $h_L(q,v,t)$ in the Lagrangian formalism corresponds to the Hamiltonian $H(q,p,t)$ in the Hamiltonian formalism. See also e.g. this Phys.SE post.

| cite | improve this answer | |
  • $\begingroup$ Thank you for answering. isn't the Lagrangian energy function (eq$(2)$) the Hamiltonian? isn't that the lengendre transform of $L$? $\endgroup$ – Courage Dec 29 '15 at 5:06
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Dec 29 '15 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.