Momentum conservation laws and explanation

Question

I understand that momentum can be conserved along a certain axis while not along another. If there are forces that are affecting our system in the axis we are interested in that means the momentum along that axis is not conserved since the derivative of momentum is the force. That means if the net force is $0$ then there is no change in the total momentum of the system and if it is not $0$ then there is a change.

How can I identify when the momentum is definitely not conserved and when it is conserved for sure, and how do I identify angular momentum?

Answer 1

If a (mechanical) system is invariant under arbitrary spatial translations in a certain direction (say, the $x$-direction), then the corresponding component of the momentum ($p_x$ in our case) is conserved (i.e. time-independent).

Analogously, if a system is invariant under arbitrary rotations with repect to a certain axis (say, the $z$-axis), then the corresponding angular-momentum component ($L_z$ in this case) is a constant of motion.

Likewise, if the system is time-translation invariant, energy is conserved.

As an example, consider the motion of a particle described by the Lagrangian function $$L(x,y,z,\dot{x},\dot{y},\dot{z})=m(\dot{x}^2+\dot{y}^2+\dot{z}^2)/2-V(z), \tag{1} \label{1}$$ where the potential depends only on the coordinate $z$. The system is obviously invariant under $x\to x+x_0$, $y\to y+y_0$ for arbitrary $x_0, y_0 \in \mathbb{R}$. As a consequence, the momenta $p_x=m\dot{x}$ and $p_y=m \dot{y}$ are conserved. This can be seen explicitly from the equations of motion, $$\begin{align} \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} &= \frac{\partial L}{\partial x} \quad \Rightarrow \quad \frac{d}{dt} (m \dot{x}) = 0, \tag{2} \label{2}\\[5pt] \frac{d}{dt}\frac{\partial L}{\partial \dot{y}} &= \frac{\partial L}{\partial y} \quad \Rightarrow \quad \frac{d}{dt}(m\dot{y})=0. \tag{3} \label{3} \end{align}$$ Because of the $z$-dependence of the potential, $p_z=m\dot{z}$ is not conserved: $$\begin{align}\frac{d}{dt} \frac{\partial L}{\partial \dot{z}}&=\frac{\partial L}{\partial z} \quad \Rightarrow \quad \frac{d}{dt}(m \dot{z}) =-V^\prime(z) \ne 0. \tag{4} \label{4} \end{align}$$ As the Lagrangian \eqref{1} is also invariant under rotations around the $z$-axis by an arbitrary angle $\alpha$, $$ x \to x\cos \alpha + y\sin \alpha , \quad y \to -x\sin \alpha +y \cos \alpha , \quad z \to z, \tag{5} \label{5} $$ the angular momentum $L_z= m (x \dot{y}-y\dot{x})$ is conserved, which can be checked by taking the time-derivative of $L_z$ and using \eqref{2} and \eqref{3}. On the other hand, neither $L_x=m(y \dot{z}-z \dot{y})$ nor $L_y=m(z \dot{x}-x \dot{z})$ are conserved, as the system is not invariant under rotations around the $x$- or $y$-axis.

Finally, as \eqref{1} is invariant under $t \to t+t_0$ for arbitrary $t_0 \in \mathbb{R}$, the energy of the system, given by $$ E= \dot{x} p_x+\dot{y} p_y+\dot{z}p_z-L=m(\dot{x}^2+\dot{y}^2+\dot{z}^2)/2+V(z), \tag{6} \label{6}$$ is also a constant of motion.

Answer 2

See my answer to a similar question here.

Conservation of momentum can be derived from Newton's Laws by considering a system of particles which each obey Newton's Second Law $\left(\sum \vec{F}_i = m_i \vec{a}_i\right)$, and whose interactions are governed by Newton's Third Law $\left(\vec{F}_\text{$i$ on $j$} = -\vec{F}_\text{$j$ on $i$}\right)$. By summing up all of the Newton II equation for the individual particles, the internal forces (between particles) will cancel out and we find this vector equation: $$ \vec{F}_\text{net, external} = M \frac{d V_\text{com}}{dt} $$ where the left-hand side is the sum of all external forces on all of the particles in the system (not due to internal interactions within the system), $M$ is the sum of all of the masses in the system, and, $V_\text{com}$ is the center-of-mass velocity of the system of particles: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt}\left( \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \cdots + m_N \vec{r}_N}{m_1 + m_2 + \cdots + m_N} \right)\,. $$ You can see my derivation of that on pages 6--9 of these notes.

So, to your question. "Conservation of Momentum" occurs when the sum of the external forces on the system is zero: $$ \vec{F}_\text{net, external} = 0 \quad \rightarrow \quad M \frac{d \vec{V}_\text{com}}{dt} = 0 \quad \rightarrow \quad \frac{d \left( M \vec{V}_\text{com} \right)}{dt} = 0 \quad \rightarrow \quad \frac{d \vec{P}_\text{tot}}{dt} = 0\,, $$ where, using the definition of $\vec{V}_\text{com}$, above: $$ \vec{P}_\text{tot} = M \frac{m_1 \vec{v}_1 + \cdots m_N \vec{v}_N}{M} = m_1 \vec{v}_1 + \cdots m_N \vec{v}_N\,. $$ So we see that the sum of the momenta of the particles in the system is a constant, unchanging in time.

But the first equation above, containing $\vec{F}_\text{net, external}$ is a vector equation. That means it is true for any component equations. It could be that $F_\text{net, ext, $x$} = 0$, but $F_\text{net, ext, $y$}$ is nonzero. In that case, we would immediately have conservation of momentum in the $x$ direction: $$ m_1 v_\text{$1$, $x$} + \cdots m_N v_\text{$N$, $x$} = \text{const} $$ but no such equation for the $y$ direction.