Tag Info

New answers tagged

1

Your statement about whether $\vec{F}$ is a conservative field is wrong. $\vec{F}$ is conservative iff $\nabla \times \vec{F} = 0$. In terms of the Cartesian components, the curl is $$ \nabla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial P}{\partial z} \right) \hat{x} + \left( \frac{\partial Q}{\partial z} - \frac{\partial ...


4

Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have ...


2

The work-energy theorem leads us to the following result; \begin{equation} \oint \vec F\cdot d\vec s=0 \end{equation} \begin{equation} \oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n \end{equation} Using the rules of vector calculus there must exist some scalar function such that; \begin{equation} \vec ...


1

Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla =\left\langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, ...


7

On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...



Top 50 recent answers are included