What is the proof for the Vortex vector field equation?
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I'm Struggling to understand why the vortex vector field is given by:
Vortex vector field equation
$vec F(x,y) = (frac{-y}{x^2+y^2}, frac{x}{x^2+y^2})$
If anyone could explain why this is, I would be very grateful.
Thank you.
multivariable-calculus vectors vector-analysis
edited Nov 18 at 1:25
Mark Viola
129k1273170
asked Nov 18 at 0:03
Will Righton
264
add a comment |
up vote
4
down vote
favorite
I'm Struggling to understand why the vortex vector field is given by:
Vortex vector field equation
$vec F(x,y) = (frac{-y}{x^2+y^2}, frac{x}{x^2+y^2})$
If anyone could explain why this is, I would be very grateful.
Thank you.
multivariable-calculus vectors vector-analysis
edited Nov 18 at 1:25
Mark Viola
129k1273170
asked Nov 18 at 0:03
Will Righton
264
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm Struggling to understand why the vortex vector field is given by:
Vortex vector field equation
$vec F(x,y) = (frac{-y}{x^2+y^2}, frac{x}{x^2+y^2})$
If anyone could explain why this is, I would be very grateful.
Thank you.
multivariable-calculus vectors vector-analysis
edited Nov 18 at 1:25
Mark Viola
129k1273170
asked Nov 18 at 0:03
Will Righton
264
I'm Struggling to understand why the vortex vector field is given by:
Vortex vector field equation
$vec F(x,y) = (frac{-y}{x^2+y^2}, frac{x}{x^2+y^2})$
If anyone could explain why this is, I would be very grateful.
Thank you.
multivariable-calculus vectors vector-analysis
multivariable-calculus vectors vector-analysis
edited Nov 18 at 1:25
Mark Viola
129k1273170
asked Nov 18 at 0:03
Will Righton
264
edited Nov 18 at 1:25
Mark Viola
129k1273170
asked Nov 18 at 0:03
Will Righton
264
edited Nov 18 at 1:25
Mark Viola
129k1273170
edited Nov 18 at 1:25
Mark Viola
129k1273170
edited Nov 18 at 1:25
Mark Viola
129k1273170
129k1273170
asked Nov 18 at 0:03
Will Righton
264
asked Nov 18 at 0:03
Will Righton
264
asked Nov 18 at 0:03
Will Righton
264
264
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
For $vec F=-hat x frac {y}{x^2+y^2} +hat y frac{x}{x^2+y^2}$, we see that
$$|vec F|=frac1r$$
where $r=sqrt{x^2+y^2}$ is the polar coordinate for the magnitude of the position vector $vec r=hat r r$.
Moreover, the direction of $vec v$ is the polar unit vector $hat theta$ and is perpendicular to the position vector.
Hence, $vec F$ rotates (circulates) around the origin counterclockwise and its "strength" increases as we move closer to the origin.
A point of interest is that while $nabla times vec F=0$ for all $vec rne0$, the line integral of $vec F$ is not $0$ for any (smooth) contour that encircles the origin.
answered Nov 18 at 0:25
Mark Viola
129k1273170
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
1
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For $vec F=-hat x frac {y}{x^2+y^2} +hat y frac{x}{x^2+y^2}$, we see that
$$|vec F|=frac1r$$
where $r=sqrt{x^2+y^2}$ is the polar coordinate for the magnitude of the position vector $vec r=hat r r$.
Moreover, the direction of $vec v$ is the polar unit vector $hat theta$ and is perpendicular to the position vector.
Hence, $vec F$ rotates (circulates) around the origin counterclockwise and its "strength" increases as we move closer to the origin.
A point of interest is that while $nabla times vec F=0$ for all $vec rne0$, the line integral of $vec F$ is not $0$ for any (smooth) contour that encircles the origin.
answered Nov 18 at 0:25
Mark Viola
129k1273170
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
1
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
add a comment |
up vote
1
down vote
accepted
For $vec F=-hat x frac {y}{x^2+y^2} +hat y frac{x}{x^2+y^2}$, we see that
$$|vec F|=frac1r$$
where $r=sqrt{x^2+y^2}$ is the polar coordinate for the magnitude of the position vector $vec r=hat r r$.
Moreover, the direction of $vec v$ is the polar unit vector $hat theta$ and is perpendicular to the position vector.
Hence, $vec F$ rotates (circulates) around the origin counterclockwise and its "strength" increases as we move closer to the origin.
A point of interest is that while $nabla times vec F=0$ for all $vec rne0$, the line integral of $vec F$ is not $0$ for any (smooth) contour that encircles the origin.
answered Nov 18 at 0:25
Mark Viola
129k1273170
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
1
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For $vec F=-hat x frac {y}{x^2+y^2} +hat y frac{x}{x^2+y^2}$, we see that
$$|vec F|=frac1r$$
where $r=sqrt{x^2+y^2}$ is the polar coordinate for the magnitude of the position vector $vec r=hat r r$.
Moreover, the direction of $vec v$ is the polar unit vector $hat theta$ and is perpendicular to the position vector.
Hence, $vec F$ rotates (circulates) around the origin counterclockwise and its "strength" increases as we move closer to the origin.
A point of interest is that while $nabla times vec F=0$ for all $vec rne0$, the line integral of $vec F$ is not $0$ for any (smooth) contour that encircles the origin.
answered Nov 18 at 0:25
Mark Viola
129k1273170
For $vec F=-hat x frac {y}{x^2+y^2} +hat y frac{x}{x^2+y^2}$, we see that
$$|vec F|=frac1r$$
where $r=sqrt{x^2+y^2}$ is the polar coordinate for the magnitude of the position vector $vec r=hat r r$.
Moreover, the direction of $vec v$ is the polar unit vector $hat theta$ and is perpendicular to the position vector.
Hence, $vec F$ rotates (circulates) around the origin counterclockwise and its "strength" increases as we move closer to the origin.
A point of interest is that while $nabla times vec F=0$ for all $vec rne0$, the line integral of $vec F$ is not $0$ for any (smooth) contour that encircles the origin.
answered Nov 18 at 0:25
Mark Viola
129k1273170
edited Nov 18 at 2:34
answered Nov 18 at 0:25
Mark Viola
129k1273170
answered Nov 18 at 0:25
Mark Viola
129k1273170
answered Nov 18 at 0:25
Mark Viola
129k1273170
129k1273170
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
1
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
add a comment |
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
1
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Would the cowardly down voter care to comment?
– Mark Viola
Nov 18 at 2:22
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
Hi mark, thanks for your reply! Why is the direction of F perpendicular to the position vector r
– Will Righton
Nov 18 at 10:33
1
1
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@WillRighton: Because $vec{F} cdot vec{r} = 0$, since $vec{r} = (x , y)$ in Cartesian coordinates.
– Nominal Animal
Nov 18 at 13:58
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
@willrighton Hi Will. You're welcome. My pleasure. And it appears that your question was answered in another comment.
– Mark Viola
Nov 18 at 14:39
add a comment |
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