Rottor of a vector in a simple way

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Homework Help Overview

The discussion revolves around proving that a given vector field is conservative by showing that its curl is zero. The vector field is defined in three-dimensional space, and the original poster seeks a simpler method to compute the curl, or "rot," of the vector field.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for calculating the curl, including the determinant method in Cartesian coordinates and the use of spherical coordinates. Some express curiosity about the determinant approach and its derivation.

Discussion Status

Participants are exploring different coordinate systems and methods for computing the curl of the vector field. There is a mix of approaches being considered, but no consensus has been reached on a single method. The original poster has not yet indicated a preference for any specific approach.

Contextual Notes

The original poster notes the need to avoid the point (0,0,0) in their calculations, which may influence the methods discussed. There is an emphasis on finding a simpler way to compute the curl, suggesting some constraints in the problem setup.

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[tex]\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}})[/tex]
[tex] \vec{r}=(x,y,z)[/tex]

[tex] |r|=\sqrt{x^2+y^2+z^2}[/tex]

[tex]\vec{F}=(\frac{x}{|r|},\frac{y}{|r|},\frac{z}{|r|})[/tex]

so its [tex]F=\frac{r}{|r|}[/tex]

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?
 
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You have a radial vector. Look up "curl in spherical coordinates" and apply it.
 
The determinant method is usually the simplest way. In Cartesian coordinates, it's

[tex]\text{rot}\textbf{F}=\begin{vmatrix}\hat{x} & \hat{y} & \hat{z} \\ \partial_x & \partial_y & \partial_z \\ F_x & F_y & F_z \end{vmatrix}=\begin{vmatrix}\hat{x} & \hat{y} & \hat{z} \\ \partial_x & \partial_y & \partial_z \\ \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}} \end{vmatrix}[/tex]

But since [itex]\textbf{F}=\frac{\textbf{r}}{r}=\textbf{e}_r[/itex], it is probably best to use spherical coordinates:

[tex]\text{rot}\textbf{F}=\begin{vmatrix}\textbf{e}_{r} & r\textbf{e}_{\theta} & r\sin\theta\textbf{e}_{\phi} \\ \partial_r & \partial_{\theta} & \partial_{\phi} \\ F_r & r F_{\theta} & r\sin\theta F_{\phi} \end{vmatrix}=\begin{vmatrix}\textbf{e}_{r} & r\textbf{e}_{\theta} & r\sin\theta\textbf{e}_{\phi} \\ \partial_r & \partial_{\theta} & \partial_{\phi} \\ 1 & 0 & 0 \end{vmatrix}[/tex]
 
wow i how you came up with this kind of determinant
i am use the the first kind
 
In any 3D Curvilinear coordinate system (u,v,w), the rotation (or 'curl') is given by

[tex]\text{rot}\textbf{F}=\begin{vmatrix}h_u\textbf{e}_{u} & h_v\textbf{e}_{v} & h_w\textbf{e}_{w} \\ \partial_u & \partial_v & \partial_w \\ h_u F_u & h_v F_{v} & h_w F_w \end{vmatrix}[/tex]

where [itex]\textbf{e}_u[/itex], [itex]\textbf{e}_v[/itex], and [itex]\textbf{e}_w[/itex] are unit vectors that point in the direction of increasing [itex]u[/itex], [itex]v[/itex] and [itex]w[/itex] respectively, and [itex]h_u[/itex], [itex]h_v[/itex], and [itex]h_w[/itex] are scale factors given by

[tex]h_u\equiv \left| \frac{\partial \textbf{r}}{\partial u} \right|[/tex]

[tex]h_v\equiv \left| \frac{\partial \textbf{r}}{\partial v} \right|[/tex]

and

[tex]h_w\equiv \left| \frac{\partial \textbf{r}}{\partial w} \right|[/tex]

This is derived in most multivariable calculus textbooks.
 

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