- #1
Wizard
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- TL;DR
- Confusion about a book "The Variational Principles of Mechanics" by Cornelius Lanczos, equation (610.4) where Lanczos uses the fact that "The function ##L_1## is a homogeneous form of the first order..."
In the book "The Variational Principles of Mechanics" by Cornelius Lanczos, the following statement is made about a lagrangian ##L_1## where time is given as an dependent parameter, and a new parameter ##\tau## is introduced as the independent variable, see (610.3) and (610.4) pg. 186,187 Dover Fourth edition:
$$L_1 = L\left(q_1, \dots q_{n+1}; \frac{q_1'}{q_{n+1}'} , \dots , \frac{q_n'}{q_{n+1}'}\right) q_{n+1}' \text{ where } q_i' = \frac{\mathrm d q_i}{\mathrm d \tau} \text{ and } q_{n+1} = t \text{ (time)}.$$
This fact is used to show that the Hamiltonian, defined in the usual way for this new parametrisation, is identically zero.
I don't see how ##L_1## is a homogeneous form. Can someone explain this? I could understand if Lanczos was referring to the canonical integral, which is indeed a homogeneous form in the ##\dot{q_i}##. The canonical integral in the Hamiltonian formulation is:
$$A = \int_{t_0}^{t^1} \sum_i p_i \dot{q_i} - H((q_i)_i, (p_i)_i) \, \mathrm d t \, .$$
As I understand, the canonical integral for action has exactly the same variation as ##\int_{t_0}^{t_1} L \, \mathrm d t## under variations ##\delta q_i##. This is not an obvious fact to me but it seems this is the case, as shown in the text.
In general the function ##L_1## defined above is certainly not going to be a homogeneous form... i.e. take the lagrangian for planetary motion confined to the plane:
$$L(r,\theta;\dot{r},\dot{\theta}) = \frac{m}{2} \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) + \frac{GMm}{r}$$
where ##M## is the mass of the massive gravitational body (e.g. the Sun). Here, we have
$$L_1\left(r,\theta;\frac{\mathrm d r}{\mathrm d \tau}, \frac{\mathrm d \theta}{\mathrm d \tau}\right) = L\left(r,\theta;\frac{\mathrm d r}{\mathrm d \tau}\left(\frac{\mathrm d t}{\mathrm d \tau}\right)^{-1}, \frac{\mathrm d \theta}{\mathrm d \tau}\left(\frac{\mathrm d t}{\mathrm d \tau}\right)^{-1}\right)\frac{\mathrm d t}{\mathrm d \tau} = \frac{m}{2} \left( \left(\frac{\mathrm d r}{\mathrm d \tau}\right)^2 + r^2 \left(\frac{\mathrm d \theta}{\mathrm d \tau}\right)^2 \right) \left(\frac{\mathrm d t}{\mathrm d \tau}\right)^{-1} + \frac{GMm}{r} \frac{\mathrm d t}{\mathrm d \tau}$$
which is clearly not a homogeneous form in ##\frac{\mathrm d r}{\mathrm d \tau}, \frac{\mathrm d \theta}{\mathrm d \tau}, \frac{\mathrm d t}{\mathrm d \tau}##.
where ##L_1## was defined as
$$L_1 = L\left(q_1, \dots q_{n+1}; \frac{q_1'}{q_{n+1}'} , \dots , \frac{q_n'}{q_{n+1}'}\right) q_{n+1}' \text{ where } q_i' = \frac{\mathrm d q_i}{\mathrm d \tau} \text{ and } q_{n+1} = t \text{ (time)}.$$
This fact is used to show that the Hamiltonian, defined in the usual way for this new parametrisation, is identically zero.
I don't see how ##L_1## is a homogeneous form. Can someone explain this? I could understand if Lanczos was referring to the canonical integral, which is indeed a homogeneous form in the ##\dot{q_i}##. The canonical integral in the Hamiltonian formulation is:
$$A = \int_{t_0}^{t^1} \sum_i p_i \dot{q_i} - H((q_i)_i, (p_i)_i) \, \mathrm d t \, .$$
As I understand, the canonical integral for action has exactly the same variation as ##\int_{t_0}^{t_1} L \, \mathrm d t## under variations ##\delta q_i##. This is not an obvious fact to me but it seems this is the case, as shown in the text.
In general the function ##L_1## defined above is certainly not going to be a homogeneous form... i.e. take the lagrangian for planetary motion confined to the plane:
$$L(r,\theta;\dot{r},\dot{\theta}) = \frac{m}{2} \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) + \frac{GMm}{r}$$
where ##M## is the mass of the massive gravitational body (e.g. the Sun). Here, we have
$$L_1\left(r,\theta;\frac{\mathrm d r}{\mathrm d \tau}, \frac{\mathrm d \theta}{\mathrm d \tau}\right) = L\left(r,\theta;\frac{\mathrm d r}{\mathrm d \tau}\left(\frac{\mathrm d t}{\mathrm d \tau}\right)^{-1}, \frac{\mathrm d \theta}{\mathrm d \tau}\left(\frac{\mathrm d t}{\mathrm d \tau}\right)^{-1}\right)\frac{\mathrm d t}{\mathrm d \tau} = \frac{m}{2} \left( \left(\frac{\mathrm d r}{\mathrm d \tau}\right)^2 + r^2 \left(\frac{\mathrm d \theta}{\mathrm d \tau}\right)^2 \right) \left(\frac{\mathrm d t}{\mathrm d \tau}\right)^{-1} + \frac{GMm}{r} \frac{\mathrm d t}{\mathrm d \tau}$$
which is clearly not a homogeneous form in ##\frac{\mathrm d r}{\mathrm d \tau}, \frac{\mathrm d \theta}{\mathrm d \tau}, \frac{\mathrm d t}{\mathrm d \tau}##.
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