Simplifiyng the proof of conservative field using rottor properties

[slonopotam]

$$\vec{F}=(\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}})$$
$$\vec{r}=(x,y,z)$$

$$|r|=\sqrt{x^2+y^2+z^2}$$

$$\vec{F}=(\frac{x}{|r|},\frac{y}{|r|},\frac{z}{|r|})$$

so its $$F=\frac{r}{|r|}$$

i need to prove that F is a conservative field
where (x,y,z) differs (0,0,0)
so i need to show that rot f is 0
but for rottor i need a determinant
is there a way to do a rot on simpler way?

 

[jpreed]

A conservative field is also irrotational. Take the curl of the field and show that it is equal to zero. (It is).

 

[tomcue]

Yes, there is a simpler way to prove that \vec{F} is a conservative field using rotor properties. We can rewrite \vec{F} as \vec{F}=\frac{\vec{r}}{|\vec{r}|}, where \vec{r}=(x,y,z) and |\vec{r}|=\sqrt{x^2+y^2+z^2}.

To prove that \vec{F} is conservative, we need to show that the curl of \vec{F} is equal to zero. This can be done by using the identity \nabla \times (\frac{\vec{r}}{|\vec{r}|})=0, which is a property of rotor operators.

Therefore, we can conclude that \vec{F} is a conservative field since its curl is equal to zero. This simplified proof using rotor properties allows us to avoid using determinants and makes the proof more concise and efficient.