Gradient Theorem
Contents...
Gradient Theorem | Proof
In this paper I prove the gradient theorem which is all about calculating the net work done (energy used) in any dimensional field. So, in any physical setting $U$ with start point $\mathbb{p}$ and end point $\mathbb{q}$ as vector positions, the net work done along path $\gamma$ defined in terms of time as $\mathbb{r}(t)$ for $t \in (a,b)$ we always have…
\[\begin{equation} \int_{\gamma} \nabla\varphi(\mathbb{r}) \cdot d\mathbb{r} = \varphi(\mathbb{q}) - \varphi(\mathbb{p}) \end{equation}\]this is a generalization of greenʻs theorem. It allows us to calculate really hard line integrals in a fast way. So, letʻs prove it. We begin with the composite function…
\[\begin{equation} \frac{d}{dt}(\varphi \circ \mathbb{r})(t) \end{equation}\]we know how to compute this derivative using the multivariate chain rule, in this case $\mathbb{r}(t) = f(u_0,u_1,…,u_n)$ and $u_i = u_i(t)$ for all $i \in {0,…,n}$. We also know that $\varphi(\mathbb{r}(t)) = \varphi{(f(u_0,…,u_n)})$ as well. We can write now the derivatives of these two functions as…
\[\begin{equation} \mathbb{r}^\prime(t) = \pmatrix{f_{u_0}(t) \\ f_{u_1}(t) \\ \vdots \\ f_{u_n}(t)} \end{equation}\]and
\[\begin{equation} \nabla\varphi(\mathbb{r}(t)) = \pmatrix{\varphi_{u_0}(\mathbb{r}(t)) \\ \varphi_{u_1}(\mathbb{r}(t)) \\ \vdots \\ \varphi_{u_n}(\mathbb{r}(t))} \end{equation}\]By the multivariate chain rule, we can now write out what $(2)$ evaluates to as…
\[\begin{equation} \frac{d}{dt}(\varphi \circ \mathbb{r})(t) = \nabla\varphi(\mathbb{r}(t)) \cdot \mathbb{r}^\prime(t) \end{equation}\]At this point, we can compute the first integral in $(1)$ as…
\[\begin{equation} \int_{\gamma} \nabla(\varphi(\mathbb{r})) \cdot d\mathbb{r} = \int_{\gamma} \nabla(\varphi(\mathbb{r}(t))) \cdot d\mathbb{r}(t) \end{equation}\]to do this part, we used the fact that we parametrized $\mathbb{r}$ with time $t$. Now, we can use the definition of the derivative to get…
\[\begin{equation} \int_{a}^b \nabla(\varphi(\mathbb{r}(t))) \mathbb{r}^\prime(t)dt \\ = \int_{a}^{b} \frac{d}{dt}(\varphi \circ \mathbb{r})(t)dt \\ = \left[(\varphi \circ \mathbb{r})(t)\right]_a^b = \varphi(\mathbb{r}(b)) - \varphi(\mathbb{r}(a)) = \varphi(\mathbb{q}) - \varphi(\mathbb{p}) \end{equation}\]where $\mathbb{r}(b) = \mathbb{q}$ and $\mathbb{r}(a) = \mathbb{p}$ by definition. All of this gives the end result which is the gradient theorem, as desired. $\blacksquare$