# "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?

• Commented May 24, 2017 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. Commented May 24, 2017 at 18:52
• en.wikipedia.org/wiki/…
– user126422
Commented May 24, 2017 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). Commented May 24, 2017 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
Commented May 24, 2017 at 19:03

$\newcommand{\dd}{{\rm d}}$ $\newcommand{\vect}[1]{{\bf #1}}$

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.

The answers here are not wrong, but they aren't what the most rigorous derivation would be. This formula for Kinetic Energy actually comes from an approximation of the "true" kinetic energy, the relativistic kinetic energy. Work is only defined because we have kinetic energy. The thing that is "fundamental" here, is the fact that energy/mass as related.

So you would start with the energy-momentum equation, subtracting rest energy to get kinetic energy.

$$EK = \sqrt{(pc)^2 + (mc^2)^2} - mc^2$$

And then expand on $$\frac{p}{2m}$$ in a taylor series. After replacing p=mv for a particle with mass, the first term in the expansion will be the familiar $$\frac{1}{2}mv^2$$