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This answer to a question about why Newtonian kinetic energy is quadratic in velocity shows that if an inelastic collision's KE loss is invariant under Newtonian boosts it has to quadruple when velocity doubles. A simple calculation shows that the famous $\tfrac12mv^2$ formula implies invariance of this loss. If a mass $m_1$'s velocity changes from $v_1$ to $v_1-\frac{m_2}{m_1+m_2}u$ while a mass $m_2$'s velocity changes from $v_2$ to $v_2+\frac{m_1}{m_1+m_2}u$, the total KE reduction is $\frac{m_1m_2}{m_1+m_2}u\cdot(v_1-v_2-\tfrac12u)$, which is invariant under $v_i\mapsto v_i+w$. However, I know of no other reason to expect such invariance. I'm wondering if we can motivate this without the formula, so we can use the above link's reasoning to then derive the quadratic KE-speed relation.

To be fair, the linked answer also argues that energy conservation in a SUVAT approximation of free fall motivates such a quadratic relation. In fact, it can derive not only proportionality to $mv^2$, but the exact expression including the $\tfrac12$ factor. In theory, we can derive the formula that way, then verify invariance, then point out invariance has the implications the answer mentioned earlier. But those are implications we'd already know at that point. To genuinely start from invariance, we need to know why to expect it. (In particular, an individual body's KE change isn't invariant; even the sign of the change isn't.)

Qmechanic
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J.G.
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2 Answers2

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It's possible to derive the form of kinetic energy using conservation and Galilean relativity by considering an elastic collision [1]. This side-steps the problem of justifying why one should assume that "heat" generated in an inelastic collision is frame invariant. (This assumption is particularly undesirable because it's not true in the special relativistic context).

A benefit of this approach is that it generalizes to the special relativistic context. In that case, one finds that (i) the relativistic energy of a particle is $mc^2$, and (ii) massless energy ("heat") possesses momentum, and that such massless energy & momentum transforms as a four-vector.

[1] P. Goyal, Derivation of Classical Mechanics in an Energetic Framework via Conservation and Relativity, Foundations of Physics 50 1426—1479 (2020). Full text: https://rdcu.be/b73po

Philip
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Indeed this is the weak point of an otherwise interesting-looking argument.

There is no apparent reason to believe loss of $\sum_k E(m_k,v_k)$ after a colission among bodies $k$, $E(m_k,v_k)$ being heat that could be extracted from the colission of the body $k$ with heavy stationary wall, is Galilei-invariant. There is no obvious way to transform the energy loss that happens in a colission (generated heat) to another frame using Galilei transformations.

One way to salvage the argument is to rely more on the experiment instead of this idea of invariance of generated heat. If we define $E_k$ as heat that can be generated by collision with a wall, we can simply stick to this assumption and utilize it: we can measure this heat for bodies of same mass $m$ but different $v$'s and discover quite universal law that $E_k$ is proportional to $v_k^2$.

Knowing $E(m_k,v_k) = cm_kv_k^2$, it is a matter of using algebra and Galilei transformations to velocities and to total energy $\sum_k cm_kv_k^2$ to express the energy loss and then using conservation of momentum to conclude that energy loss in a colission between the bodies is indeed Galilei-invariant.

It seems to me this way of thinking is more natural/physical - we start with physical observations and measurements and then use mathematics to discover new interesting fact (invariance of energy loss).

Ján Lalinský
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