The representation of a Poisson bracket is given by the following equation: $$\tag{1} \{f,g\} = \sum_{s=1}^n \sum_{i=1}^{d=3}\left ( \frac{\partial f}{\partial x_i^{(s)}} \frac{\partial g}{\partial p_i^{(s)}} - \frac{\partial f}{\partial p_i^{(s)}} \frac{\partial g}{\partial x_i^{(s)}}\right),$$ where $n$ is the number of particles, and $d$ is the number of dimensions.

Assume we have an arbitrary Hamiltonian $H$ (possibly explicitly time-dependent). Then according to the Hamilton equation we have: $$\frac{d p_j^{(r)}}{dt} = \{p_j^{(r)}, H\} + \frac{\partial p_j^{(r)}}{\partial t} =\{p_j^{(r)}, H\}.$$

Using the representation given in (1) we can show that $\frac{d p_j^{(r)}}{dt} = -\frac{\partial H}{\partial x_i^{(r)}}$.

In the process of derivation, I get that $$\tag{2} \frac{\partial p_i^{(r)}}{\partial x_i^{(r)}} = 0.$$ I don't understand why the quantity in (2) is zero? Can't the momentum depend on time as well as $x$?

Unfortunately, I don't have any physics background whatsoever, so I will appreciate an intuitive answer or a mathematical proof which does not rely on Lagrangian mechanics.


1 Answer 1


Variables $x_i$ and $p_i$ are independent variables used as arguments for the Hamiltonian $H(x_i, p_i)$. Since they're independent variables, the partial derivative of one w.r.t. the other is identically zero.

  • $\begingroup$ Hi, thanks for the reply. Why the variables $x_i$ and $p_i$ are independent? $\endgroup$
    – MonteNero
    Sep 19, 2022 at 14:17
  • $\begingroup$ By definition, the Hamiltonian is defined as a function with independent variables corresponding to the (generalized) coordinates $x_i$(generalized) momenta $p_i$, and time, $H(x_i, p_i, t)$, see en.wikipedia.org/wiki/Hamiltonian_mechanics. Like the Lagrangian function is defined as a function of generalized coordinates $x_i$ and their time derivatives $\dot{x}_i$, and time $L(\dot{x}_i, x_i, t)$. They are equivalent methods to say that the state of the system is determined by the position and the velocity of its parts, and position and velocity are independent $\endgroup$
    – basics
    Sep 19, 2022 at 14:32
  • $\begingroup$ Position and velocity are independent, because you can be in one place and move in an arbitrary direction with an arbitrary magnitude (feasible for the constraints, if any). Your position doesn't directly determine your velocity, so they can be treated as independent variables $\endgroup$
    – basics
    Sep 19, 2022 at 14:35

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