# Conservation of angular momentum poisson brackets vs newtonian mechanics [closed]

So I have the following system. For a mass falling due to gravity the given Hamiltonian is

$$H = \frac{1}{2m}\left( P^{2}_{x} + P^{2}_{y} \right) + mgy$$

In Cartesian coordinates then, $$x = v_o t \cos \theta$$ and $$y = v_o t \sin \theta - \frac{1}{2}gt^2$$

From these quantities one can arrive to the fact that the angular momentum with respect to the origin is, $$\vec L = \mathcal{l_z} = \vec r \times \vec p = [(v_o t \cos \theta)\hat x + (v_o t \sin \theta - \frac{1}{2}gt^2) \hat y] \times m[(v_o \cos \theta)\hat x + (v_o \sin \theta - gt) \hat y ]$$ $$= \frac{1}{2}mg v_o t^2 \cos \theta \hat z$$

So then in order to check whether the angular momentum is conserved one would proceed to calculate: $$\frac{ d \mathcal{l_z} }{d t} = \frac{ \partial \mathcal{l_z} }{\partial t} + \left( \frac{\partial \mathcal{l_z} }{\partial x} \frac{\partial H}{\partial P_x} - \frac{\partial \mathcal{l_z} }{\partial P_x} \frac{\partial H}{\partial x}\right) + \left( \frac{\partial \mathcal{l_z} }{\partial y} \frac{\partial H}{\partial P_y} - \frac{\partial \mathcal{l_z} }{\partial P_y} \frac{\partial H}{\partial y}\right) = \tau$$

Where $\tau$ is the torque. ($\vec \tau = - mg v_o t \cos \theta \hat z$). But I have not been able to get the right answer. I am having trouble calculating the poisson brackets with respect to $y$. could someone help?

• You just showed that the angular momentum is time-dependent, and therefore not conserved. – ZachMcDargh Oct 29 '15 at 15:27
• A more explicit way of doing this, would be to use the poissson brackets to show that the result actually equals the torque. – alejandro123 Oct 29 '15 at 15:28

You need to use vectors.

Since $L \neq r \times p$, you need to use $\vec L = \vec r \times \vec p$ instead, where the $\times$ is the vector cross product of vectors, not the scalar multiplication of scalars.

So you have $$\vec L= \left[(v_o t \cos \theta) \hat x+ (v_o t \sin \theta - \frac{1}{2}gt^2)\hat y\right] \times m\left[(v_o \cos \theta)\hat x+ (v_o \sin \theta - gt)\hat y \right].$$

And if you wanted the angular momentum as a function of time you are good by using the above. However if you want to take a Poisson Bracket you have terms like $\partial L_z/\partial x$ in which case you don't want $L_z=L_z(t)$ you instead want $$L_z=L_z(x,y,P_x,P_y)=xP_y-yP_x.$$

Now you can take your partials. And also don't forget that its not just $H$ but again $H=H(x,y,P_x,P_y)$ that you need (which is actually how you wrote it that time).

And finally, you don't have the correct equation of motion, or the Poisson Bracket. You need

$$\frac{ d L_z }{d t} = \frac{ \partial L_z }{\partial t} + \left( \frac{\partial L_z}{\partial x} \frac{\partial H}{\partial P_x} - \frac{\partial L_z}{\partial P_x} \frac{\partial H}{\partial x}\right) + \left( \frac{\partial L_z}{\partial y} \frac{\partial H}{\partial P_y} - \frac{\partial L_z}{\partial P_y} \frac{\partial H}{\partial y}\right)$$

And these are all functions that you take partials as if they were functions. And just like the function $f=f(x)=x^2$ is the same function as $f=f(t)=t^2$ (its the function that takes an input and squares it). So these are all functions that just take their first argument and their second argument and so forth. And you need to take partials that way, just have a function with some inputs in order and assign names to those inputs and see how the functions result depends on the input.

You should be able to do it.

• Sorry, i did not explicitly show that these were vectors. but if you carry out that step it gives you the result i already showed. ill edit my question. thank you. – alejandro123 Oct 29 '15 at 15:06
• The last part is what i'm having trouble with. For example, $dL/dy= -P_x + x(dP_y/dy)$. What would this term be equal to, in simplest terms? – alejandro123 Oct 29 '15 at 15:18
• @alejandro123 you don't need to take $dP_y/dy$. Momenta and positions are not dependent. – ZachMcDargh Oct 29 '15 at 15:34
• @ZachMcDargh They are independent variables in the function $L_z=L_z(x,y,P_x,P_y)$ and so the partial of $P_y$ with respect to $y$ is zero since by definition a partial with respect to $y$ leaves $P_y$ unchanged. So it just gives zero when you do the partial derivative like the derivative of any constant. – Timaeus Oct 29 '15 at 15:38