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Lifshitz and Landau's Vol. $1$ explicitly states that

$$ \cfrac{\partial{q_k}}{\partial{p_i}} = 0$$

And seems to imply also that $$ \cfrac{\partial{p_k}}{\partial{q_i}} = 0.$$

I guess that whenever $k \neq i$ the two quantities are independent of each other, but for $k=i$ it seems that they do depend on each other since $p_i \text{ is defined in terms of } \dot{q_i}$.

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David
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    In Hamilton mechanics, the $q_i$s and $p_i$s are a certain type of coordinates of the phase space and as such they clearly fulfill $\frac{\partial q_k}{\partial p_i}=0$, even if $k=i$. It's the same as $\frac{\partial x}{\partial y}=0$ in standard coordinates or $\frac{\partial r}{\partial \phi}=0$ in polar coordinates of $\mathbb{R}^2$. – Johnny Longsom Mar 03 '20 at 13:38

1 Answers1

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There are various ways to see this:

  1. Consider the Hamiltonian formulation. Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Hamiltonian $H(q(t_0),p(t_0),t_0)$ is a function of both the instantaneous position $q^i(t_0)$ and the instantaneous momentum $p_k(t_0)$ at the instant $t_0$. Here $q^i(t_0)$ and $p_k(t_0)$ are independent variables. The point is that since Hamilton's equations are $2n$ coupled first-order ODEs one is entitled to make $2n$ independent choices of initial conditions.

  2. Alternatively, in the Lagrangian formulation with Lagrangian $L(q,v,t)$ the generalized positions $q^i$ and the generalized velocities $v^j$ are independent variables, cf. e.g. this Phys.SE post.

    A Legendre transformation of the Lagrangian wrt. the variables $v^j\leftrightarrow p_k$ yields the Hamiltonian $H(q,p,t)$, where positions $q^i$ and momenta $p_k$ are independent variables.

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