PX153 - I10 - conservative fields

what vector fields, $\vec{P}$ , would not show this path dependence?

\vec{P} = \vec{\nabla} V

- vector fields which are the gradient of a potential (gradient of a scalar field) have integrals, $\int_{a}^{b} \vec{P} \cdot d \vec{l}$ , which depend only on the end point, $a$ and $b$
- such fields are called conservative fields

eg: is $\vec{P} = 5 y^{2} \hat{i} + 10 x y \hat{j}$ conservative?

\nabla V = (\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y})

\begin{aligned} \frac{\partial V}{\partial x} & = 5 y^{2} & ⟹ V_{1} & = 5 x y^{2} + f_{1} (y) \\ \frac{\partial V}{\partial y} & = 10 x y & ⟹ V_{2} & = 5 x y^{2} + f_{2} (x) \end{aligned}

- since $V_{1} = V_{2}$ , $\vec{P}$ is conservative, where, $V = 5 x y^{2} + c$

it may be useful to parameterize over some variable, $t$
- eg:

W = \int_{{\vec{l}}_{1}}^{{\vec{l}}_{2}} \vec{F} \cdot d \vec{l}

- the path: $\vec{l} (t) = x (t) \hat{i} + y (t) \hat{j} + z (t) \hat{k}$
- as $t$ varies from $t_{1} \to t_{2}$ , $\vec{l} (t)$ traces the path $C$ :
$$W = \int_{t_{1}}^{t_{2}} \vec F (\vec l (t)) \cdot \frac{d \vec l}{dt} , dt$$

if $\vec{F} = - \vec{\nabla} U$ (conservative field), and $d \vec{l} = \frac{d x}{d t} d t \hat{i} + \frac{d y}{d t} d t \hat{j} + \frac{d z}{d t} d t \hat{k}$ :

\begin{aligned} W & = \int_{t_{1}}^{t_{2}} (\frac{\partial U}{\partial x} \frac{d x}{d t} + \frac{\partial U}{\partial y} \frac{d x}{d t} + \frac{\partial U}{\partial z} \frac{d x}{d t}) d t \\ W & = - \int_{t_{1}}^{t_{2}} \frac{d U}{d t} d t = - [U]_{t_{1}}^{t_{2}} = U (t_{2}) - U (t_{1}) \\ W & = U (x_{2}, y_{2}, z_{2}) - U (x_{1}, y_{1}, z_{1}) \end{aligned}

for conservative fields, the line integral depends only on endpoints, and not the path taken
- eg: electrostatic potential: $Δ U_{a b} = U_{b} - U_{a} = - \int_{a}^{b} \vec{E} \cdot d \vec{l}$
eg: $\vec{F}$ is a vector field: $x y^{2} \hat{i} + 2 \hat{j} + x \hat{k}$ . $L$ is a path parameterized by $x = 2 t, y = \frac{2}{t}, z = 1$ , with $1 \leq t \leq 2$ . find $\int_{L} \vec{F} \cdot d \vec{r}$
- on path $L$ :

\begin{aligned} \vec{F} & = \frac{8}{t} \hat{i} + 2 \hat{j} + 2 t \hat{k} \\ \vec{r} & = 2 t \hat{i} + \frac{2}{t} \hat{j} + \hat{k} \\ d \vec{r} & = 2 d t \hat{i} - \frac{2}{t^{2}} d t \hat{j} + 0 \hat{k} \end{aligned}

\int_{t_{1}}^{t_{2}} \vec{F} \cdot d \vec{r} = \int_{1}^{2} (\frac{16}{t} - \frac{4}{t^{2}}) d t = 16 \ln 2 - 2