# Poisson Bracket of a Quantity Involving a Differential

I am working through Warren Siegel's "Fields" and have come across the following exercise on p. 58 involving an action measure and a symmetry generator:

Exercise IA4.1.

For general variables ($$q^m$$,$$p_m$$) and generator $$G$$, show from the definition of the Poisson bracket that $$\delta(dq^m p_m) = -d\left(G - p_m \frac{\partial G}{\partial p_m}\right)$$ and that this vanishes for any coordinate transformation.

As defined, the $$\delta$$ of a quantity $$A$$ means the Poisson bracket $$\{A,G\}$$. Given the definition of a Poisson bracket I'm familiar with,

$$\{A(q,p), B(q,p)\} = \frac{\partial A}{\partial q^m}\frac{\partial B}{\partial p_m} - \frac{\partial A}{\partial p^m}\frac{\partial B}{\partial q_m}$$

I am unsure how to handle the differential appearing in $$\delta(dq^m p_m)$$. What is the correct way to apply a Poisson bracket to such a quantity?

• Guessing $dq^m p_m$ is in fact $\mathrm dq^m p_m$ i.e. the total differential, you can simply open it up as $=\mathrm dq^m\frac{\partial}{\partial q^m}q^mp_m+\mathrm dp_m\frac{\partial}{\partial p_m}q^mp_m$. Partial differentiation of this w.r.t $q^m$ and $p_m$ is straight forward. – acarturk Jan 3 at 16:09

It appears that the question requires a bit more nuance. Rather than defining the $$\delta$$ as the direct application of a Poisson bracket, one can think of $$\delta (d q^m p_m)$$ as

$$\delta (d q^m p_m) = (dq^m + \delta (dq^m))(p_m + \delta p_m) - dq^m p_m$$

Now $$\delta (dq^m) = d (\delta q^m)$$ and one finds (using the Poisson bracket) that

$$\delta q^m = \{q^m, G\} = \frac{\partial G}{\partial p_m}$$ $$\delta p_m = \{p_m, G\} = -\frac{\partial G}{\partial q^m}$$

Thus, omitting terms $$2^{\text{nd}}$$ order in $$\delta$$s,

$$\delta (d q^m p_m) = \delta(dq^m)p_m + dq^m \delta p_m = p_m d \left( \frac{\partial G}{\partial p_m} \right) - dq^m \frac{\partial G}{\partial q^m}$$

whereas the RHS of the question is

$$-d(G - p_m\frac{\partial G}{\partial p_m}) = p_m d\left(\frac{\partial G}{\partial p_m} \right) + dp_m \frac{\partial G}{\partial p_m} - dG = p_m d \left( \frac{\partial G}{\partial p_m} \right) - dq^m \frac{\partial G}{\partial q^m}$$

Thus we see that the LHS and RHS are equal.