# “Rigorous” derivation of kinetic energy

I've always wondered where the formula of (non-relativistic) kinetic energy we learn at high school comes from. This is the "derivation" I came up with: $\Delta W:=\int_{r_0}^{r_1}drF=m\int_{r_0}^{r_1}dr\frac{dv}{dt}=m\int_{v_0}^{v_1}dt v\frac{dv}{dt}=m\int_{v_0}^{v_1}dvv=\frac{1}{2}mv^2$. This is also the derivation you find on various websites. However, isn't that not quite rigorous? Firstly, it is only valid in one dimension and secondly $dt$ is treated like some sort of infinitesimal value (which doesn't exist in standard analysis over the reals). What would be a more "rigorous" derivation of this?

I tried to start with $\Delta W=\int_\gamma d\textbf{r} \cdot \textbf{F}=m\int_{r_0}^{r_1}dt \frac{d\textbf{v}(\textbf{r}(t))}{dt}\cdot \dot{\textbf{r}}$, but I don't know how to continue from here. Is this the correct place to start? How do you "rigorously" derive the formula for the kinetic energy?

• – Qmechanic May 24 '17 at 18:51
• now use the fact that $d{\bf{v}}^2/dt = 2 {\bf{v}} \cdot {d\bf{v}}/dt = 2 {\bf{v}} \cdot \dot{\bf{r}}$, and then integrate. – Run like hell May 24 '17 at 18:52
• en.wikipedia.org/wiki/… – user126422 May 24 '17 at 18:53
• adjoining to Run like hell's comment, also use the definition of work here: physics.stackexchange.com/a/121960/19976 (basically the definition of the line integral of a vector field along a parameterized curve). – joshphysics May 24 '17 at 19:03
• Regarding the infinitesimals, physics does not always use strictly rigorous mathematical tools. These have to be just good enough. The "rigorous" formalization/justification of math tools are usually left to mathematicians. – user126422 May 24 '17 at 19:03


First note that

$$\frac{\dd }{\dd t}(\vect{v}\cdot \vect{v}) =2 \vect{v}\cdot \frac{\dd \vect{v}}{\dd t}$$

Therefore

$$\vect{F}\cdot \vect{v} = \vect{F}\cdot \frac{\dd \vect{r}}{\dd t} = m\vect{v}\cdot \frac{\dd \vect{v}}{\dd t} = m \left({\frac{1}{2}}\frac{\dd }{\dd t}(\vect{v}\cdot \vect{v})\right) = \frac{1}{2}m\frac{\dd v^2}{\dd t}$$

Integrating you have

$$\frac{1}{2}m \Delta v^2 = \int_\gamma\dd {\vect r}\cdot \vect F$$

Two years have gone by since I asked this question, and now after some time at uni here's my take on a "rigorous" (whatever I meant with that at the time) derivation (Since now I don't find the other answer very clear, what's the curve integrated over, what's the force field the particle is moving in?):

Suppose the position of a particle of mass $$m > 0$$ moving between time $$0\in \mathbb{R}$$ and time $$1 \in \mathbb{R}$$ is given by a smooth curve $$x: [0,1]\to\mathbb{R^3}$$. If the particle moves through a force field $$F: \mathbb{R^3} \to \mathbb{R^3}$$, then the work done on the particle is given by the line integral $$\Delta W=\int_xds\cdot F(s).$$

We want to find the work needed to accelerate the object from it's initial velocity $$\dot{x}(0)$$ to it's final velocity $$\dot{x}(1)$$. Suppose there are no external forces, only the inertia of the particle is "trying to counteract" the acceleration. For the sake of the argument, suppose further that $$x$$ is injective. Then the force field of the intertia at point $$y \in x([0,1])$$ is given as follows by Newton's third law: $$F(y):=m\ddot{x}(x^{-1}(y)) \in \mathbb{R^3}.$$ The expression simply assigns every point on the trajectory of the particle the force needed to give the particle it's current acceleration.

The work integral computes as follows: $$\Delta W = \int_x ds\cdot F(s)= \int_0^1 dt \langle F(x(t)) \ | \ \dot{x}(t)\rangle = \int_0^1 dt \langle m\ddot{x}(x^{-1}(x(t))) \ | \ \dot{x}(t)\rangle = m \int_0^1 dt \langle \ddot{x}(t) \ | \ \dot{x}(t)\rangle.$$ One quickly checks that $$\langle\ddot{x}(t) | \dot{x}(t)\rangle = \frac{1}{2}\frac{d}{dt} ||\dot{x}(t)||^2$$. So by defining $$\Delta v^2 := ||\dot{x}(1)||^2-||\dot{x}(0)||^2$$, the work integral becomes $$\Delta W = \frac{1}{2}m \int_0^1 dt \frac{d}{dt}||\dot{x}(t)||^2 = \frac{1}{2}m(||\dot{x}(1)||^2-||\dot{x}(0)||^2) = \frac{1}{2}m\Delta v^2,$$ which is the familiar expression of the kinetic energy.