# Lack of rigour in usual derivation of Work-Energy Theorem

The derivation of the Work-Energy theorem usually goes as follows:

You define the work done on a particle under net force $\vec{F}$ as $$W=\int\limits_C \vec{F}\cdot\mathrm{d}\vec{r}$$ And then you use Newton's second law \begin{align} W=\int\limits_C \vec{F}\cdot\mathrm{d}\vec{r}&=\int\limits_C m\vec{a}\cdot\mathrm{d}\vec{r}\\ &=\int\limits_C m\frac{\textrm{d}\vec{v}}{\textrm{d}t}\cdot\mathrm{d}\vec{r}=\int\limits_{t_i}^{t_f} m\frac{\textrm{d}\vec{v}}{\textrm{d}t}\cdot\frac{\mathrm{d}\vec{r}}{\textrm{d}t}\textrm{d}t=\int\limits_{t_i}^{t_f} m\frac{\textrm{d}\vec{r}}{\textrm{d}t}\cdot\frac{\mathrm{d}\vec{v}}{\textrm{d}t}\textrm{d}t \\ &=\int\limits_C m\frac{\textrm{d}\vec{r}}{\textrm{d}t}\cdot\mathrm{d}\vec{v}=\int\limits_C m\vec{v}\cdot\mathrm{d}\vec{v} \end{align} Up until here, I'm convinced. Non the less, the argument $$\textrm{d}(v^2)=\textrm{d}(\vec{v}\cdot\vec{v})=\vec{v}\cdot\textrm{d}\vec{v}+\vec{v}\cdot\textrm{d}\vec{v}=2\vec{v}\cdot\textrm{d}\vec{v}$$is often used. Then the integral resolves to $$W=\int\limits_{v_i}^{v_f}\frac{1}{2}m\textrm{d}(v^2)=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$My problem with that argument is that it is assumed that the exterior derivative works under the dot product just as it does under the normal product of real-valued functions. Can somebody explain to me (either geometrically or formally) why $\textrm{d}(v^2)=\textrm{d}(\vec{v}\cdot\vec{v})=\vec{v}\cdot\textrm{d}\vec{v}+\vec{v}\cdot\textrm{d}\vec{v}=2\vec{v}\cdot\textrm{d}\vec{v}$ is true?

One need not follow these steps. Indeed let $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ be the trajectory of a particle. It's position at time $t$ is $\gamma(t)$, it's velocity is $\gamma'(t)$ and it's acceleration is $\gamma''(t)$. It's easy to see that

$$(\gamma'\cdot \gamma')'(t) = 2\gamma'(t)\cdot \gamma''(t),$$

so the work done by the resultant force because of Newton's Second Law can be written as

$$W = \int_\gamma \mathbf{F} = \int_I \mathbf{F}(\gamma(t))\cdot \gamma'(t)dt = \int_I m\gamma''(t)\cdot \gamma'(t)dt,$$

but as pointed out $\gamma''(t)\cdot\gamma'(t) = (\gamma'\cdot\gamma')'(t)/2$ so that

$$W = \dfrac{m}{2}\int_I(\gamma'\cdot\gamma')'(t)dt,$$

and by virtue of the fundamental theorem of calculus, if $I = [a,b]$ we have

$$W = \dfrac{m}{2}(|\gamma'(b)|^2-|\gamma'(a)|^2),$$

or setting $K(\mathbf{v}) = \dfrac{m}{2}|\mathbf{v}|^2$

$$W = K(\gamma'(b)) - K(\gamma'(a)) = \Delta K.$$

• Why is $\gamma''(t)\cdot\gamma'(t)=(\gamma'\cdot\gamma')'(t)dt$? Apr 20 '15 at 0:37
• @IvánMauricioBurbano Try to find the derivative of $(\gamma \cdot ' \gamma ')$ (hint: use the chain rule) Apr 20 '15 at 0:58
• Work should be a line integral along the path, but I don't see a $\vec F \cdot d\vec r$ or something similar. Aug 2 '15 at 19:55

Formally, we have the standard dot product identity $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{A}\cdot\mathbf{B}=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{B}+\mathbf{A}\cdot \frac{\mathrm{d}\mathbf{B}}{\mathrm{d}t}.$$ Inserting $\mathbf{A}=\mathbf{B}=\mathbf{v}$ gives $$\frac{\mathrm{d}\mathbf{v}^2}{\mathrm{d}t}=2\mathbf{v}\cdot \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}.$$ It then follows that $$\mathrm{d}\mathbf{v}^2\stackrel{\int}=2\mathbf{v}\cdot\mathrm{d}\mathbf{v}$$ by the chain rule.

• And how does one come to the dot product identity?? Apr 20 '15 at 0:30
• @IvánMauricioBurbano 1) Simply take the derivative of $\mathbf{A}\cdot\mathbf{B}:=\sum_{i,j}\delta_{ij}A^i B^j$ or 2) Special case of the Ricci identity $Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ)$. Apr 20 '15 at 0:32
• If you could give me a geometric answer to my previous comment that would be great! Apr 20 '15 at 0:33
• @IvánMauricioBurbano I think the "geometric" answer is the Ricci identity. I'm not sure if you mean differential geometry or simply curve geometry. Apr 20 '15 at 0:37