# To derive the relation between work function and potential energy

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

## 1 Answer

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 $v$.

References:

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

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$^1$ 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.

• Thank you for answering. isn't the Lagrangian energy function (eq$(2)$) the Hamiltonian? isn't that the lengendre transform of $L$? Dec 29, 2015 at 5:06
• I updated the answer. Dec 29, 2015 at 10:11