# Minimum energy required to behave like a turning point?

So I've managed to confuse myself.

We know if the energy equal to the potential energy then point at which the energy exceeds the potential energy it behaves like a turning point (slide 2). Usually in the collision of (say billiard) rigid balls we assume $$V = \infty$$.

I realized since it's behaving like a turning point there is no need to set the potential energy $$V$$ to $$\infty$$. The maximum kinetic energy will suffice. But here's where I'm confused. It s should make intuitively no difference to our potential if our objects $$A$$ and $$B$$ are moving with velocities

$$(\vec v_A ,\vec v_B )$$

or

$$(\vec v_A + \vec c ,\vec v_B + \vec c )$$

But when I write the kinetic energy in the $$1$$st inertial frame I get:

$$E_A = \frac{1}{2} m \vec v_A \cdot \vec v_A$$ and $$E_B = \frac{1}{2} m \vec v_B \cdot \vec v_B$$

So in this frame

$$V \geq \max(\frac{1}{2} m \vec v_A \cdot \vec v_A , \frac{1}{2} m \vec v_B \cdot \vec v_B)$$

where as in the other frame:

$$V \geq \max( \frac{1}{2} m (\vec v_A + \vec c) \cdot (\vec v_A + \vec c) , \frac{1}{2} m (\vec v_B + \vec c) \cdot (\vec v_B + \vec c))$$

Both should require the same potential. Which begs the question: what is the minimum potential energy required for the behavior of a turning point.

• You forgot that in a different reference frame, when the "turn" happens, the kinetic energy isn't zero. Instead, both the ball and wall move together, with the same speed. Jun 23 at 17:34
• @knzhou I was thinking of something akin to the kinetic energy $T$ being $T \propto (\vec v_A + \vec c) \cdot (\vec v_A + \vec c)$ then naively I would expect the potential to be $V> T$ but this is clearly frame dependent. I'm not sure what the transformations law of $V$ should be? Jun 23 at 17:39
• Newton's laws are Galilean invariant so the same thing has to happen. And $V$ doesn't transform in Newtonian mechanics. The answer to your question is just that you're not really keeping track of $T$ properly -- it is the decrease in $T$ that is equal to the increase of $V$, not the absolute value of $T$. Jun 23 at 17:41
• If this still feels confusing, try writing down literally any example and solving the equations, it will fall out pretty quickly. Jun 23 at 17:42
• @knzhou ... still confused. Jun 23 at 17:49

The basic problem you are running into is that different inertial observers do not agree on kinetic energy of objects or systems, but they do agree on changes in the system's kinetic energy. But on the other hand, all observers always agree on potential energies.

At the instant when the potential energy is maximum (the instant when kinetic energy is minimum) the kinetic energy of the system isn't zero. Observers don't agree on what the kinetic energy would be at that moment, but they will all agree on how much it is reduced by. The "easy" way to calculate it is this. The system kinetic energy can be split up into the centre of mass kinetic energy and the "convertible" kinetic energy. The centre of mass kinetic energy is

$$K_{CM} = \frac{1}{2} M_{sys} {v_{CM}}^2$$

where $$M_{sys}$$ is the total system mass and $$v_{CM}$$ is the speed of the centre of mass. For a two-object system in 1D the convertible kinetic energy is

$$K_{conv} = \frac{1}{2} \mu {v_{rel}}^2$$

where $$\mu = m_1 m_2/(m_1 + m_2)$$ is the reduced mass, and $$v_{rel}$$ is the relative speed of the objects.

How does this help? If the system is isolated (no external forces) then the $$K_{CM}$$ never changes, even during the collision. Every observer measures a different $$K_{CM}$$ but it sort of doesn't matter since this is "untouchable" kinetic energy that can't be converted.

On the other hand every observer agrees on $$v_{rel}$$ and so they agree on $$K_{conv}$$. So you just need to make your potential equal to $$K_{conv}$$ to make this work in every reference frame. You don't need to make it equal to the total kinetic energy. At the instant when the objects are moving with the same velocity $$K_{conv} = 0$$. This is the instant of minimum kinetic energy and maximum potential energy.

The issue is not in considering the potential in a different reference frame, the issue is in defining what the "turning point" means in different reference frames. I will use the example (slide 2) of a ball in a uniform gravitational field. In the ground's reference frame the "turning point" is both the point where the velocity of the ball is $$0$$ and also the point where the separation between the ground and the ball is maximized. However, in a reference frame moving downward with a steady velocity $$c$$ the point where the velocity of the ball is $$0$$ is different from the point where the separation between the ground and the ball is maximized. This is not a problem in principle, but requires some thought.

In the ground frame

In this frame the position of the ball is $$y$$, its KE is $$\frac{1}{2}m \dot y^2$$ and its PE is $$mgy$$. So the Lagrangian is $$L=\frac{1}{2}m \dot y^2-m g y$$ which gives the Euler equation $$\ddot y = mg$$ Using initial conditions $$y(0)=0$$ and $$\dot y(0)=v_0$$ we get the turning point ($$\dot y(t)=0$$) is $$y\left( \frac{v_0}{g} \right) = \frac{v_0^2}{2g}$$

In the moving frame using standard coordinates

Here the coordinate is $$Y=y+ct$$. In these coordinates the turning point, $$\dot Y=0$$, is the point where the velocity goes to zero in the moving frame.

With these coordinates the position of the ball is $$Y$$, its KE is $$\frac{1}{2}m (\dot Y+c)^2$$ and its PE is $$mg(Y+ct)$$. So the Lagrangian is $$L=\frac{1}{2}m (\dot Y+c)^2-m g (Y+c t)$$ which gives the Euler equation $$\ddot Y = mg$$ Using initial conditions $$Y(0)=0$$ and $$\dot Y(0)=v_0+c$$ we get the turning point ($$\dot Y(t)=0$$) is $$Y\left( \frac{v_0+c}{g} \right) = \frac{(v_0+c)^2}{2g}$$ Note that this turning point is a different position and, more importantly, occurs at a different time. It has a different physical meaning from the original turning point.

In the moving frame using generalized coordinates

Here the coordinate is $$h$$ which represents the separation between the ground and the ball, even though the ground is moving in this frame. In these coordinates the turning point, $$\dot h=0$$, is the point where the change in the separation between the ground and the ball goes to zero in the moving frame.

With these coordinates the position of the ball is $$h$$, its KE is $$\frac{1}{2}m (\dot h+c)^2$$ and its PE is $$mgh$$. So the Lagrangian is $$L=\frac{1}{2}m (\dot h+c)^2-m g h$$ which gives the Euler equation $$\ddot h = mg$$ Using initial conditions $$h(0)=0$$ and $$\dot h(0)=v_0$$ we get the turning point ($$\dot h(t)=0$$) is $$h\left( \frac{v_0}{g} \right) = \frac{v_0^2}{2g}$$ Note that this turning point occurs at the same time as in the original frame, so it is more in line with the physical interpretation of the turning point in the original frame.

So, since in these coordinates $$t$$ is cyclic we immediately obtain the conserved energy in the moving frame is $$E=\frac{1}{2} m (\dot h^2 - c^2) + m g h$$ So the energy condition for $$\dot h=0$$ is not $$mgh=E$$ as it was in the original frame but rather $$mgh=E+\frac{1}{2}m c^2$$ which does reduce to the original condition for $$c=0$$ as we would expect.