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Motivation: I came across a question while perusing the internet: assuming that mass, momentum, and kinetic energy are conserved in a system, ($mv^0, mv^1$, and $\frac{1}{2}mv^2$, but we can ignore the 1/2 here...), are all further quantities $mv^n$, with $n$ an integer >2, conserved?

My instinct was mathematical induction. But I'm pretty sure worried I have no clue what I'm doing, so I checked a simpler case first.

Given that mass and momentum are conserved, we have $m_0=m_f$ and $m_0v_0=m_fv_f$. Here we're taking $m$ to be the mass of the system, and $v$ to be the velocity of the center of mass.

Now, if $m_0=0$, we already know that $\frac{1}{2}m_0v_0^2=\frac{1}{2}m_fv_0^2=0=\frac{1}{2}m_fv_f^2$ (the last equality is because $m_0=m_f=0$). So without loss of generality, we'll assume that the mass of the system is not $0$.

Now, since we have $m_0v_0=m_fv_f$, squaring both sides gives $$m_0^2v_0^2=m_f^2v_f^2.$$ Since $m_0$ is assumed to be nonzero now, we can divide both sides by it to get $$\frac{m_0^2v_0^2}{m_0}=\frac{m_f^2v_f^2}{m_0}.$$

But we also have $m_0=m_f$, so this becomes

\begin{align} \frac{m_0^2v_0^2}{m_0}&=\frac{m_f^2v_f^2}{m_f}\\ m_0v_0^2&=m_fv_f^2\\ \frac{1}{2}m_0v_0^2&=\frac{1}{2}m_fv_f^2. \end{align}

Which seems to say that kinetic energy is conserved if mass and momentum are conserved.

But I was told that kinetic energy can be changed to thermal or positional (or re-stored in other ways), and it is not necessarily conserved even if momentum is conserved. So where have I gone wrong? And is there any way to repurpose or salvage this inductive argument to answer the italicized question? (or is the answer that, in general, it is not true?)

My thoughts on what may have gone wrong: Does $\frac{1}{2} mv^2$ only give the kinetic energy when all the mass is moving in the same direction? If so, is there any interpretation to $\frac{1}{2} mv^2$ when we are treating $v$ as the velocity of the center of mass?

Qmechanic
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quandle
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    My immediate instinct is the statement is not true in general. To each conserved quantity there must be an underlying symmetry, per Noether's theorem. The usual symmetries are translational symmetry (corresponds to conservation of momentum), time symmetry (corresponds to energy conservation, also mass conservation in SR) and rotational symmetry (corresponds to conservation of angular momentum). There's no symmetry for conserving $mv^3$. – Allure Jan 10 '18 at 11:23
  • Also for your calculation, when you square both sides what you actually need to do is take the vector product. That's why you're getting that KE is always conserved if mass and momentum are conserved, which is not true in general. – Allure Jan 10 '18 at 11:26
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    Comment to the question (v3): This is trivial if there's only 1 free particle. Now repeat with an elastic collision of 2 particles. – Qmechanic Jan 10 '18 at 11:37
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    to elaborate on Qmechanic. Note that for two particles, the total energy is given by $E_{tot}=E_1+E_2=m_1 p_1^2 + m_2 p_2^2$. It is thus not simply given by $m_{tot} p_{tot}^2$ – Crimson Jan 10 '18 at 12:09
  • Possible duplicate of Kinetic energy and momentum conservation in an explosion? – sammy gerbil Jan 10 '18 at 13:17

2 Answers2

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The answer is no. Here's a counterexample.

Start with two objects, each with a mass of $2m$, heading towards each other at the same speed.

    v1        v1
2m ---->    <---- 2m

The collision breaks apart the masses into a total of four pieces, all of equal mass $m$. They have various velocities as shown below:

  v3    v2      v2     v3
<---m  <--m    m-->   m--->

Let's check the stipulations of the problem:

  1. Mass is conserved: $2(2m) = 4m.$
  2. Momentum is conserved: zero before and afterwards.
  3. Kinetic energy is conserved if: $$2\left(\frac{1}{2}\left(2m\right)v_1^2\right) = 2\left(\frac{1}{2}mv_2^2\right) + 2\left(\frac{1}{2}mv_3^2\right)$$ $$2v_1^2 = v_2^2 + v_3^2$$

Let's choose $m = 1$, $v_1 = 5/\sqrt{2}$, $v_2 = 3$, $v_3 = 4$ in whatever units. Before the collision for the quartic conservation law: $$\sum_i m_iv_i^4 = 2\left(2mv_1^4\right) = 4\left(\frac{5}{\sqrt2}\right)^4 = 625$$ After the collision: $$\sum_i m_iv_i^4 = 2\left(mv_2^4\right) + 2\left(mv_3^4\right) = 2\left(3^4\right) + 2\left(4^4\right) = 674$$

So, the higher powers are not necessarily conserved.

A quick note about your attempt: you don't have to prove that kinetic energy is conserved. That's already assumed by the puzzle. You only have to figure out the higher powers.

Mark H
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  • Thanks for the example! And @ the note, yep, I realized that. I attempted the kinetic energy case because it's something where I could see that my method resulted in an absurdity (and hence I was probably overlooking something). – quandle Jan 12 '18 at 15:39
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While the momentum of the CM is the same as the total momentum of the system, the kinetic energy of the CM is not the same as the kinetic energy of the system. For example, two masses $m$ moving apart in the same line each with speed $v$ in the laboratory frame. The momentum of the CM is zero, and the total momentum is zero. However, the CM has zero KE but the total KE of the system is $mv^2$.

Fundamentally this difference arises because momentum is a vector whereas KE is a scalar. Whereas the linear momenta of the two particles cancel out because they are in opposite directions, the fact that the KE of the two masses are in opposite directions does not affect their total KE. KE is independent of direction.

sammy gerbil
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