# Proving a function is quadratic?

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.

## 1 Answer

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

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.