How is the Lagrangian function homogeneous in the velocities? I'm working through The Variational Principles of Mechanics by Cornelius Lanczos. In chapter 6, section 10, he says that the Lagrangian function
$$L_1 = L\left(q_1,\dotsc,q_{n+1}; \frac{q_1'}{q'_{n+1}},\dotsc,\frac{q_n'}{q'_{n+1}}\right)q'_{n+1}$$
is homogeneous in the velocities.
The indices go to $n+1$ because he's putting the Lagrangian into parametric form where time becomes a variable $q_{n+1}$ and the variables are all functions of an arbitrary parameter.  That's also why you see the prime notation.  It represents differentiation relative to this parameter.
He says that since the function is homogeneous in the velocities, it can be represented as:
$$L_1 = \Sigma \frac{\partial L_1}{\partial q_i'} q_i'.$$
I don't see how this works because the Lagrangian could include potential energy terms that are a function only of position, not velocity, so it seems like those would be lost if you represent it this way.
I've found one other source ("On the Lagrangian being a homogeneous function of the velocity" by Gaetano Giaquinta) that also says that the Lagrangian can be represented this way, but isn't clear on how.  Can someone explain this to me?
 A: If you define the Lagrangian to be homogeneous in the velocities, then you exclude ex ante any terms not depending on velocities. So a Lagrangian of this kind is per definition a sum of terms $f(q_j, \ddot q_j, \dotsc, t)\dot q_i$. If such a Lagrangian is physically meaningful is of course a completely different question.
A: Reading Lanczos this morning I stumbled on the same section. Since there is no good answer yet I am resurrecting this question for future readers!
This is specifically only true for the extended Lagrangian as he is constructing, hence the confusion. Consider a scaling of velocities and time,
$\{q_1', q_2', \cdots, q_{n+1}'\}\to\{\alpha q_1', \alpha q_2', \cdots, \alpha q_{n+1}'\}$
and its influence in $L$ (not $L_1$ yet!),
$L\Big(\frac{q_1'}{q_{n+1}'},\cdots,\frac{q_n'}{q_{n+1}'}\Big) \to L\Big(\frac{\alpha q_1'}{\alpha q_{n+1}'},\cdots,\frac{\alpha q_n'}{\alpha q_{n+1}'}\Big) = L\Big(\frac{q_1'}{q_{n+1}'},\cdots,\frac{q_n'}{q_{n+1}'}\Big)$
so homogeneity of degree 1 in these variables follows from the reparameterization of time in the action integral
$A = \int_{t_0}^{t_1}L(q_1,\cdots, q_n, \dot{q}_1, \cdots, \dot{q}_n)dt = \int_{\tau_0}^{\tau_1}L\Big(\frac{q_1'}{q_{n+1}'},\cdots,\frac{q_n'}{q_{n+1}'}\Big)q_{n+1}'d\tau \equiv \int_{\tau_0}^{\tau_1}L_1d\tau$
where $q_{n+1}' = \frac{dt}{d\tau}$. Since $\alpha q_{n+1}' = \alpha (q_{n+1}')$, it follows that the extended Lagrangian $L_1$ is homogeneous of degree 1 in the variables $\{q_1', q_2', \cdots, q_{n+1}'\}$, and so by Euler's theorem
$L_1 = \sum_{i=1}^{n+1}\frac{\partial L_1}{\partial q_i'}q_i'$.
Then upon Legendre transform to the extended Hamiltonian, we find
$H_1 = \sum_{i=1}^{n+1}p_iq_i' - L_1 = 0$
as the conjugate momenta are $p_i = \frac{\partial L_1}{\partial q_i'}$.
A little more info. To find the momentum conjugate to time, write down the unextended Legendre transform,
$H = \sum_{i=1}^np_i\dot{q}_i - L$
and change variables to extend time, with $\dot{q}_i = \frac{q_i'}{q_{n+1}'}$,
$H = \sum_{i=1}^np_i\frac{q_i'}{q_{n+1}'} - L\implies Hq_{n+1}'=\sum_{i=1}^np_iq_i' - L_1$
so therefore
$H_1 = \sum_{i=1}^np_iq_i' - Hq_{n+1}' - L_1$
and the momentum conjugate to time is the negative of energy,
$p_{n+1} = -H$.
Hope that helps!
