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In Physics, an object of mass $m$ can be said to have a certain amount of kinetic energy $T = T (v)$, which is a function of its speed $v$. Let's presume that we do not know in advance the canonical formula $T = \frac{1}{2} mv^2$.

Intuitively, we can assume the following facts about energy.

  1. $T(0) = 0$.
  2. $T(-v) = T(v)$.
  3. Kinetic energy can be transferred to thermal energy $Q$ during a collision.
  4. $T + Q$ is the same before and after a collision.
  5. $Q$ is invariant under changes in reference frame.

Imagine an collision where two objects of mass $m$ approach each other at a constant velocity $v_0$ and stick together. As a result the kinetic energy they had originally ($T(v_0)$ each originally, so $2T(v_0)$ total) gets converted completely into thermal energy $Q$.

$$Q = 2 T (v_0)$$

If we presume that an observer travelling at a velocity $v$ relative to the center of mass of the collision also would compute the same amount of thermal energy resulting from the collision, then we would find the following expression for $Q$, keeping in mind that the two objects will be travelling together with a velocity $v$ in this new frame of reference after the collision.

$$T(v_0 + v) + T(v_0 - v) = 2 T(v) + Q$$

Substituting the first expression for $Q$ into the second gives the following.

$$T(v_0 + v) + T(v_0 - v) = 2 T(v) + 2 T(v_0) \tag{*}$$

If we plug in $v_0 = v$, this reduces to $T(2v_0) = 4 T(v_0)$, which certainly suggests that $T(v) = Kv^2$, where $K$ is some constant, though this wouldn't be a formal proof.

My question is this: is there any way to prove from $\text{(*)}$ (and possibly the list of five facts at the top of this post) that there exists a constant $K$ such that $T(v) = Kv^2$? In essence, I'm curious if the proportionality between $T$ can follow directly from the five bullet points above, or if more postulates are needed.

The statement $T(2v_0) = 4 T(v_0)$ by itself seems so close to convincing that I'm wondering if I'm missing something obvious in making the full connection.

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$T(v_0 + v) + T(v_0 - v) = 2 T(v) + 2 T(v_0) \tag{*}$

Assuming $\,T\,$ sufficiently smooth, and taking derivatives in $\,v_0, v\,$ respectively:

$$ T'(v_0+v)+T'(v_o-v) = 2 T'(v_0) \\ T'(v_0+v)-T'(v_o-v) = 2 T'(v) \\ $$

Adding the above:

$$ T'(v_0+v) = T'(v_0) + T'(v) $$

Therefore $\,T'\,$ is additive, and by continuity linear, so $\,T\,$ is quadratic.

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