# Partial derivatives of canonical momenta in Poisson brackets

I will simply give an example for a general doubt about the Hamiltonian formulation. So, consider the spherical pendulum of length $$l$$ as an example of my perhaps more general question. The Lagrangian is given by

$${\displaystyle L={\frac {1}{2}}ml^{2}\left({\dot {\theta }}^{2}+\sin ^{2}\theta \ {\dot {\phi }}^{2}\right)+mgl\cos \theta .}$$

and the conjugate momenta are $${\displaystyle P_{\theta }={\frac {\partial L}{\partial {\dot {\theta }}}}=ml^{2}\cdot {\dot {\theta }}}$$ and $${\displaystyle P_{\phi }={\frac {\partial L}{\partial {\dot {\phi }}}}=ml^{2}\sin ^{2}\!\theta \cdot {\dot {\phi }}}. \tag{1}$$

Lastly, the Hamiltonian can be calculated via the Legendre transform to be $$H=P_{\theta }{\dot {\theta }}+P_{\phi }{\dot {\phi }}-L = {P_{\theta }^{2} \over 2ml^{2}}+ {P_{\phi }^{2} \over 2ml^{2}\sin ^{2}\theta }-mgl\cos \theta.$$

Now, $$\phi$$ is cyclic and hence, $$P_{\phi}$$ is conserved. I should be able to verify this via the Poisson brackets. In particular, I should find that $$\{P_{\phi}, H\}=0$$

However, when expanding the Poisson bracket, I find myself with the term $$\frac{\partial p_{\phi}}{\partial \theta} \frac{\partial H}{\partial p_{\theta}}$$ which using Eq. (1) doesn't seem to vanish. However, if I simply postulate that the canonical momenta $$p_{\phi},p_{\theta}$$ be independent of the coordinates $$\phi, \theta$$, then this term does indeed vanish.

How does one reconcile this? On the one hand, we define the conjugate momenta and in particular $$p_{\phi}$$ seems to clearly depend on $$\theta$$ but then in the next step, we simply claim to view them as independent.

When evalauating partial derivatives you need to specify what is being kept fixed as well as what is varying. The Poisson bracket is
$$\{F(p,q),G(p,q)\}=\sum_i \left(\frac{\partial F(q,p)}{\partial q_i}\right)_{p_i} \left(\frac{\partial G(q,p)}{\partial p_i}\right)_{q_i}- \left(\frac{\partial F(q,p)}{\partial p_i}\right)_{q_i} \left(\frac{\partial G(q,p)}{\partial q_i}\right)_{p_i}.$$ So yes $$\left(\frac{\partial p_\phi}{\partial \theta}\right)_{p_{\phi}}=0.$$

• Thank you for the answer. I understand your point. However, I think the relevant partial derivative would be keeping $p_{\theta}$ fixed and not $p_{\phi}$, (or would it actually keep all the $p_i$ fixed)? So, I dont see why $\left( \frac{\partial p_{\phi}}{\partial \theta}\right)_{p_{\theta}}$ should necessarily vanish. Indeed, your answer does not address my confusion about the $p_i$ being simultaneously independent but also seemingly dependent on the coordinates $q_i$. Jul 12, 2022 at 7:36
• The independent variables are the $p_i$ and the $q_i$. A partial derivative with respect to any one of them keeps all the other $p$ and $q$ variables fixed. This what you get when you Legendre transform from the $\dot q_i$ to the $p_i$'s. So the appearence of $\theta$ in any of the $p$'s is irrelevent. Jul 12, 2022 at 12:54
• That makes sense. Jul 12, 2022 at 13:10

It's just like in thermodynamics, you need to keep track which variables you are using as coordinates for your function. It's always a good idea to formulate things in a coordinate independent perspective.

$$L$$ is defined on the coordinates $$q_i,v_i$$. For an actual trajectory, $$v_i = \dot q_i$$ but from the point of view of $$L$$ they are all independent variables. For a coordinate independent description, $$L$$ is defined on the tangent bundle of the configuration space, starting from a coordinate system $$q_i$$ of the configuration space, you get a new natural coordinate system of the tangent bundle $$q_i,v_i$$.

When you go to the Hamiltonian formalism, you are doing a change of coordinates $$q_i,p_i$$. $$H$$ is therefore a function of these independent variables. For a coordinate independent description, $$H$$ is defined on the cotangent bundle of the configuration space. However, starting from a coordinate system $$q_i$$ of the configuration space you don't have a natural coordinate system of the cotangent bundle $$q_i,p_i$$, you'll need $$L$$ to define it.

The Poisson bracket is defined on two functions of the cotangent space, so they are functions of $$q_i,p_i$$ which are all treated as independent. There is actually a coordinate independent definition of the Poisson bracket, which I won't detail here, check out Arnold's Mathematical Methods for Classical Mechanics, which motivates the above formula.

Hope this helps.

• YOur third paragraph isn't true. Given coordinates $q_i$ on the configuration space $Q$, you can absolutely lift it to get a natural coordinate system on the cotangent bundle, and in fact you inherit one on every tensor bundle (so there's nothing special about the tangent bundle here). Getting a coordinate system for the cotangent bundle doesn't require the Lagrangian. In fact, if you use the coordinates on $T^*Q$ provided by the Lagrangian (i.e using the Fiber derivative of the Lagrangian $\mathbf{F}L:TQ\to T^*$ and pushing forward charts on $TQ$ to a chart on $T^*Q$) then[...] Jul 12, 2022 at 12:03
• [...] the local representative of $L$ and $H$ are the same function from an open subset of $\Bbb{R}^{2n}\to\Bbb{R}$. Jul 12, 2022 at 12:04

The independent coordinates (on the cotangent bundle of the configuration space) for describing the Hamiltonian are $$(\theta,\phi,p_{\theta},p_{\phi})$$. So, almost by definition, $$\frac{\partial (\text{one of these four guys})}{\partial(\text{one of the three other guys})}=0$$, in particular, $$\frac{\partial p_{\phi}}{\partial\theta}=0$$. The reason you're getting confused is you're overloading the same symbols (namely $$p_{\theta}$$ and $$p_{\phi}$$) to have two different meanings. On the one hand, you write the common but muddled equation, $$p_i=\frac{\partial L}{\partial \dot{q}^i}$$, so this seems to be a function of $$q,\dot{q}$$, and on the other hand, we have to use $$(\theta,\phi,p_{\theta},p_{\phi})$$ as a coordinate system, and it is this double usage which confuses you.

A single Variable Analogue.

Suppose $$I,J\subset\Bbb{R}$$ are open intervals, and $$E:I\to\Bbb{R}$$ is a given smooth function, $$\Phi:I\to J$$ a given diffeomorphism (smooth map with smooth inverse), and define $$H:J\to\Bbb{R}$$ as the mapping $$H:=E\circ\Phi^{-1}$$. For the sake of concreteness, let us take

• $$I=\Bbb{R}$$, $$J=(0,\infty)$$,
• $$E:\Bbb{R}\to\Bbb{R}$$ the function $$E(x)=x+e^x$$,
• $$\Phi:\Bbb{R}\to (0,\infty)$$ the diffeomorphism $$\Phi(x)=e^x$$, so $$\Phi^{-1}:(0,\infty)\to\Bbb{R}$$ is $$\Phi^{-1}(y)=\log(y)$$.
• $$H:(0,\infty)\to\Bbb{R}$$ is then the defined as $$H=E\circ\Phi^{-1}$$, which yields $$H(y)=(\log y) + y$$.

If I now ask you to compute the derivative $$H'(y)$$, you'll immediately tell me that since $$H(y)=\log y + y$$, then $$H'(y)=\frac{1}{y}+1$$, and there's no doubt about it. Notice that there's nothing special about the letters $$x$$ and $$y$$. If I wanted to, I could say $$E(t)=t+e^t$$, and $$H(t)=(\ell\circ\Phi^{-1})(t)=\log t + t$$. This is perfectly correct mathematically.

Now, this same procedure is typically presented in physics texts without introducing a name for the diffeomorphism $$\Phi$$. It is typically presented as saying:

Consider the function $$E(x)=x+e^x$$, and make the change of variables $$y=e^x$$ to get a function $$H(y)$$. This function equals, $$H(y)=\log y + y$$, so $$H'(y)=\frac{1}{y}+1$$.

Notice that there is still an overload of notation in this description! The same symbol $$y$$ is first being used in place of the diffeomorphism $$\Phi$$, and second it is used as the argument of the function $$H$$. It is just that one might be too comfortable with single-variable calculus that they gloss over the notational overload.

In any case, the bottom line is once you make the change of coordinates (i.e composing by $$\Phi^{-1}$$), you should forget about whatever happened before (i.e the $$x$$-coordinates).

The Original Question, in Modified Notation.

What I described above is exactly what we're doing when we change from Lagrangian to Hamiltonian mechanics. We start with an energy function $$E:\Bbb{R}^{2n}\to\Bbb{R}$$ in Lagrangian mechanics (here $$n=2$$), and use a diffeomorphism $$\Phi$$ between open subsets of $$\Bbb{R}^{2n}$$ (this is the Fiber derivative of the Lagrangian, $$\mathbf{F}L:TQ\to T^*Q$$ in fancy differential geometry jargon), and using this we 'change coordinates' to get the Hamiltonian $$H:=E\circ\Phi^{-1}$$.

What follows might be slightly difficult to read initially because I'm going to use non-standard letters to denote the independent variables, but hopefully after this you'll appreciate the difference between the choice of coordinates and the diffeomorphism used.

• The Lagrangian. This is a function $$L:\Bbb{R}^4\to\Bbb{R}$$, given as \begin{align} L(x_1,x_2,y_1,y_2)&=\frac{ml^2}{2}\left(y_1^2+\sin^2(x_1)y_2^2\right)+ mgl\cos(x_1). \end{align} Compared with the standard notation, here I used $$(x_1,x_2,y_1,y_2)=(\theta,\phi,\dot{\theta},\dot{\phi})$$. The choice of letters is of course insignificant, I can also write something like $$L(a,b,c,d)=\frac{ml^2}{2}\left(c^2+\sin^2(a)d^2\right)+ mgl\cos(a)$$.

• The Energy. This is the function $$E:\Bbb{R}^4\to\Bbb{R}$$ defined as \begin{align} E(x_1,x_2,y_1,y_2)&=y_1\frac{\partial L}{\partial y_1}\bigg|_{(x_1,x_2,y_1,y_2)}+ y_2\frac{\partial L}{\partial y_2}\bigg|_{(x_1,x_2,y_1,y_2)} - L(x_1,x_2,y_1,y_2)\\ &=y_1\cdot ml^2y_1 + y_2\cdot ml^2\sin^2(x_1)y_2 -\bigg( \frac{ml^2}{2}\left(y_1^2+\sin^2(x_1)y_2^2\right)+ mgl\cos(x_1)\bigg)\\ &=\frac{ml^2}{2}\left(y_1^2+\sin^2(x_1)y_2^2\right)- mgl\cos(x_1). \end{align}

• The Fiber Derivative. This is the mapping $$\Phi:\Bbb{R}^4\to\Bbb{R}^4$$, defined as \begin{align} \Phi(x_1,x_2,y_1,y_2)&:=\left(x_1,x_2, \frac{\partial L}{\partial y_1}\bigg|_{(x_1,x_2,y_1,y_2)}, \frac{\partial L}{\partial y_2}\bigg|_{(x_1,x_2,y_1,y_2)}\right)\\ &=\bigg(x_1,x_2, ml^2y_1, ml^2\sin^2(x_1)y_2\bigg). \end{align} This map becomes a diffeomorphism if we restrict the domain and target sufficiently, in which case, the inverse map is \begin{align} \Phi^{-1}(\xi_1,\xi_2,\eta_1,\eta_2)&=\bigg(\xi_1,\xi_2,\frac{\eta_1}{ml^2},\frac{\eta_2}{ml^2\sin^2\xi_1}\bigg). \end{align}

• The Hamiltonian. This is the composed mapping $$H:=E\circ \Phi^{-1}$$. So, after some computation, \begin{align} H(\xi_1,\xi_2,\eta_1,\eta_2)&=E(\Phi^{-1}(\xi_1,\xi_2,\eta_1,\eta_2))\\ &=\dots\\ &=\frac{\eta_1^2}{2ml^2}+\frac{\eta_2^2}{2ml^2\sin^2\xi_1}-mgl\cos\xi_1. \end{align}

Now, in the Lagrangian, we have $$\frac{\partial L}{\partial x_2}=0$$, so we expect that for the Hamiltonian side, $$\{\eta_2,H\}=0$$, and indeed, we can verify this is the case: \begin{align} \{\eta_2,H\}&=\left(\frac{\partial \eta_2}{\partial \xi_1}\frac{\partial H}{\partial \eta_1}- \frac{\partial \eta_2}{\partial \eta_1}\frac{\partial H}{\partial \xi_1}\right) + \left(\frac{\partial \eta_2}{\partial \xi_2}\frac{\partial H}{\partial \eta_2}- \frac{\partial \eta_2}{\partial \eta_2}\frac{\partial H}{\partial \xi_2}\right)\\ &=0-0+0-\frac{\partial H}{\partial \xi_2}\\ &=0, \end{align} since the partials of $$\eta_2$$ with respect to the other coordinates $$\xi_1,\xi_2,\eta_1$$ are trivially $$0$$, while $$\frac{\partial\eta_2}{\partial\eta_2}=1$$ and finally since $$H$$ doesn't depend on $$\xi_2$$.

If you want to compare notation, then $$(\xi_1,\xi_2,\eta_1,\eta_2)=(\theta,\phi,p_{\theta},p_{\phi})$$. Anyway, the takeaway is that you need to distinguish between the coordinates used to describe the Lagrangian/Hamiltonian, and the diffeomorphism $$\Phi$$ which allows you to convert between the two formalisms.