Checking if $\text{ curl } \vec{F}=\vec{0}$ for $\vec{F}$ & $f$

[evinda]

Hello! (Wave)

Is it true that for a vector field $\vec{F}$, a function $f$ such that $\vec{F}=\nabla{f}$ can exist only if $\text{ curl } \vec{F}=\nabla \times \vec{F}=\vec{0}$ ?

How can we check it? (Thinking)

 

[I like Serena]

Hello! (Wave)

Is it true that for a vector field $\vec{F}$, a function $f$ such that $\vec{F}=\nabla{f}$ can exist only if $\text{ curl } \vec{F}=\nabla \times \vec{F}=\vec{0}$ ?

How can we check it? (Thinking)

Hey evinda! (Smile)

Suppose we have a function $f$ such that $\vec{F}=\nabla{f}$.
Then $\text{ curl } \vec{F}=\nabla \times \vec{F}=\nabla \times (\nabla f)=\vec{0}$.
This is one of the vector calculus identities.

Thus, if there is a point somewhere where $\nabla \times \vec{F} \ne \vec{0}$ then there is no function $f$ such that $\vec{F}=\nabla{f}$. (Nerd)

 

[evinda]

Hey evinda! (Smile)

Suppose we have a function $f$ such that $\vec{F}=\nabla{f}$.
Then $\text{ curl } \vec{F}=\nabla \times \vec{F}=\nabla \times (\nabla f)=\vec{0}$.
This is one of the vector calculus identities.

Thus, if there is a point somewhere where $\nabla \times \vec{F} \ne \vec{0}$ then there is no function $f$ such that $\vec{F}=\nabla{f}$. (Nerd)

I see... Thanks a lot! (Happy)