# Momentum and energy conservation and preconditions

I'm reading Susskind's Classical Mechanics: The Theroretical Minimum and I also like to restrict my question to classical mechanics:

In chapter 4, momentum conservation is shown for a set of particles. In chapter 5 energy conservation is shown for a body in free fall towards Earth (potential energy decreases, kinetic energy increases). The authors then point out that energy conservation holds here in the free falling body example, but not conservation of momentum, and the reason is that we neglected the Earth. When considering the change of Earth's momentum (which starts to move to the falling body as well as actio = reactio), the sum of Earth's and free falling body's momentum is of course conserved.

This leads me to the question what the preconditions for applying momentum and energy conservation in mechanics are? Seems that momentum conservation needs a closed system to be true, but energy conservation not, because we don't have to consider the change of the Earth's potential energy with respect to the falling body, nor the Earth's kinetic energy towards the small body, to argue that the sum of all energies for body + Earth is constant (like we argue with the momentum...).

what the preconditions for applying momentum and energy conservation in mechanics are?

We write the laws of physics that describe a system in the system’s Lagrangian. If the Lagrangian is not an explicit function of time then energy is conserved. If the Lagrangian is not an explicit function of position then momentum is conserved.

For a particle in free fall near the surface of the earth the Lagrangian would be $$L=\frac{m}{2}\left(\dot x^2+\dot y^2 +\dot z^2 \right)-m g z$$ Note that this Lagrangian is not a function of $$t$$ so energy is conserved. Note that this Lagrangian is not a function of $$x$$ or $$y$$ so the horizontal momentum is conserved, but it is a function of $$z$$ so the vertical momentum is not conserved.

If we include the motion of the Earth then the Lagrangian becomes $$L=\frac{m}{2}\left(\dot x_m^2+\dot y_m^2 +\dot z_m^2 \right)+\frac{M}{2}\left(\dot x_M^2+\dot y_M^2 +\dot z_M^2 \right)+\frac{GM}{\sqrt{(x_m-x_M)^2+(y_m-y_M)^2+(z_m-z_M)^2}}$$ We can rewrite this in terms of the position of the center of mass $$\vec R=\frac{m(x_m,y_m,z_m)+M(x_M,y_M,z_M)}{m+M}$$ and the relative position $$\vec r=(x_m,y_m,z_m)-(x_M,y_M,z_M)$$ so that the Lagrangian simplifies to $$L=\frac{1}{2}(m+M)\dot{\vec R}^2+\frac{1}{2}\frac{m M}{m+M}\dot{\vec r}^2+\frac{GM}{|\vec r|}$$ Now this Lagrangian is not a function of $$t$$ so there is a conserved energy. It is also not a function of $$\vec R$$, the position of the center of mass. So there is a conserved momentum associated with the center of mass of the system.

In general, conserved quantities are associated with some symmetry of the Lagrangian of the system. If you consider a system without that symmetry then you get rid of that conserved quantity.

• Very nice answer, many thanks! Jun 21, 2022 at 17:33

Momentum is conserved in a system if there are no external forces on the system.

Likewise, angular momentum is conserved in a system if there are no external torques on the system.

Energy is conserved in a system if there is no work being done on the system and no heat conducted into the system, or more generally, if these contributions cancel out, i.e. $$dQ+dW=0$$.

Short answer to your question: energy conservation does require both considering the falling body and the earth, as the gravitational potential energy is shared among the earth and the falling body (the same is true for electrostatic forces). To see this, we write down the gravitational potential energy $$V = - G M m/r$$, where $$M$$ is the mass of the earth, $$m$$ is the mass of the falling body and $$r$$ is the distance between the body and the center of earth. Let $$R$$ be the radius of the earth and $$z$$ be the height of the body, then $$r=R+z$$. We will assume $$z\ll R$$, then $$V = m(G M /R^2)z$$ which reduces to the usual form $$V = mgz$$, with $$g=GM/R^2$$.