Skip to main content
Physics

Return to Question

Removed from Network Questions by Qmechanic♦
Became Hot Network Question
edited body
Source Link
Jack
Jack
  • 959
  • 3
  • 15

The following quote is from the second volume of Feynman's lectures

The interesting theorem is that if the curl $\textbf{A}$ is zero, then $\textbf{A}$ is always the gradient of something.

Feynman didn't specify anything about that vector field $\textbf{A}$, yet, he asserted that once curl $\textbf{A}=\textbf{0}$, then it follows readily that $\textbf{A}$=grad $\psi$ for some scalar function $\psi$. What I know is that this isn't the case when $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$ or even when $\textbf{A}=\textbf{A}(\textbf{r})$, but $\textbf{A}$ isn't defined onin a simply connected region.

The following quote is from the second volume of Feynman's lectures

The interesting theorem is that if the curl $\textbf{A}$ is zero, then $\textbf{A}$ is always the gradient of something.

Feynman didn't specify anything about that vector field $\textbf{A}$, yet, he asserted that once curl $\textbf{A}=\textbf{0}$, then it follows readily that $\textbf{A}$=grad $\psi$ for some scalar function $\psi$. What I know is that this isn't the case when $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$ or even when $\textbf{A}=\textbf{A}(\textbf{r})$, but $\textbf{A}$ isn't defined on a simply connected region

The following quote is from the second volume of Feynman's lectures

The interesting theorem is that if the curl $\textbf{A}$ is zero, then $\textbf{A}$ is always the gradient of something.

Feynman didn't specify anything about that vector field $\textbf{A}$, yet, he asserted that once curl $\textbf{A}=\textbf{0}$, then it follows readily that $\textbf{A}$=grad $\psi$ for some scalar function $\psi$. What I know is that this isn't the case when $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$ or even when $\textbf{A}=\textbf{A}(\textbf{r})$, but $\textbf{A}$ isn't defined in a simply connected region.

Source Link
Jack
Jack
  • 959
  • 3
  • 15

What theorem Feynman is referring to?

The following quote is from the second volume of Feynman's lectures

The interesting theorem is that if the curl $\textbf{A}$ is zero, then $\textbf{A}$ is always the gradient of something.

Feynman didn't specify anything about that vector field $\textbf{A}$, yet, he asserted that once curl $\textbf{A}=\textbf{0}$, then it follows readily that $\textbf{A}$=grad $\psi$ for some scalar function $\psi$. What I know is that this isn't the case when $\textbf{A}=\textbf{A}(\dot{\textbf{r}})$ or even when $\textbf{A}=\textbf{A}(\textbf{r})$, but $\textbf{A}$ isn't defined on a simply connected region