Mixing
Calculating grad/curl/div of a vector?
$begingroup$
I'm trying to do some practice for an electromagnetism course, and am trying to calculate the grad, curl and div of a vector $A = (2xy, 3zx, yx^2)$
I know:
Div = $nabla . A$
Curl = $nabla times A$
Grad = $nabla A$
I have worked out the first two, but I seem to be having a mind blank and I'm getting myself all confused with the Grad component if someone could clarify this?
Also if someone could explain when I might need to use the Laplace Operator? What is it used for?
vector-analysis
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to do some practice for an electromagnetism course, and am trying to calculate the grad, curl and div of a vector $A = (2xy, 3zx, yx^2)$
I know:
Div = $nabla . A$
Curl = $nabla times A$
Grad = $nabla A$
I have worked out the first two, but I seem to be having a mind blank and I'm getting myself all confused with the Grad component if someone could clarify this?
Also if someone could explain when I might need to use the Laplace Operator? What is it used for?
vector-analysis
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
$endgroup$
$begingroup$
You can only use the gradient on scalar functions, not on vector fields.
$endgroup$
– dinosaur
Feb 21 '14 at 12:49
$begingroup$
Okay that's why I'm getting confused then! Any chance you could give me an example?
$endgroup$
– Sarah Jayne
Feb 21 '14 at 12:50
$begingroup$
@dinosaur actually you can. The result is a tensor field
$endgroup$
– David H
Feb 21 '14 at 12:50
$begingroup$
@DavidH yes, but I don't think that this is what Sarah had in mind here.
$endgroup$
– dinosaur
Feb 21 '14 at 14:06
$begingroup$
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $nabla f(x,y)=left(frac{partial f}{partial x}, frac{partial f}{partial y}right) = (2x,2y)$.
$endgroup$
– dinosaur
Feb 21 '14 at 14:07
|
show 1 more comment
$begingroup$
I'm trying to do some practice for an electromagnetism course, and am trying to calculate the grad, curl and div of a vector $A = (2xy, 3zx, yx^2)$
I know:
Div = $nabla . A$
Curl = $nabla times A$
Grad = $nabla A$
I have worked out the first two, but I seem to be having a mind blank and I'm getting myself all confused with the Grad component if someone could clarify this?
Also if someone could explain when I might need to use the Laplace Operator? What is it used for?
vector-analysis
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
$endgroup$
I'm trying to do some practice for an electromagnetism course, and am trying to calculate the grad, curl and div of a vector $A = (2xy, 3zx, yx^2)$
I know:
Div = $nabla . A$
Curl = $nabla times A$
Grad = $nabla A$
I have worked out the first two, but I seem to be having a mind blank and I'm getting myself all confused with the Grad component if someone could clarify this?
Also if someone could explain when I might need to use the Laplace Operator? What is it used for?
vector-analysis
vector-analysis
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
edited Feb 21 '14 at 12:43
Sarah Jayne
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
asked Feb 21 '14 at 12:40
Sarah JayneSarah Jayne
333110
333110
$begingroup$
You can only use the gradient on scalar functions, not on vector fields.
$endgroup$
– dinosaur
Feb 21 '14 at 12:49
$begingroup$
Okay that's why I'm getting confused then! Any chance you could give me an example?
$endgroup$
– Sarah Jayne
Feb 21 '14 at 12:50
$begingroup$
@dinosaur actually you can. The result is a tensor field
$endgroup$
– David H
Feb 21 '14 at 12:50
$begingroup$
@DavidH yes, but I don't think that this is what Sarah had in mind here.
$endgroup$
– dinosaur
Feb 21 '14 at 14:06
$begingroup$
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $nabla f(x,y)=left(frac{partial f}{partial x}, frac{partial f}{partial y}right) = (2x,2y)$.
$endgroup$
– dinosaur
Feb 21 '14 at 14:07
|
show 1 more comment
$begingroup$
You can only use the gradient on scalar functions, not on vector fields.
$endgroup$
– dinosaur
Feb 21 '14 at 12:49
$begingroup$
Okay that's why I'm getting confused then! Any chance you could give me an example?
$endgroup$
– Sarah Jayne
Feb 21 '14 at 12:50
$begingroup$
@dinosaur actually you can. The result is a tensor field
$endgroup$
– David H
Feb 21 '14 at 12:50
$begingroup$
@DavidH yes, but I don't think that this is what Sarah had in mind here.
$endgroup$
– dinosaur
Feb 21 '14 at 14:06
$begingroup$
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $nabla f(x,y)=left(frac{partial f}{partial x}, frac{partial f}{partial y}right) = (2x,2y)$.
$endgroup$
– dinosaur
Feb 21 '14 at 14:07
$begingroup$
You can only use the gradient on scalar functions, not on vector fields.
$endgroup$
– dinosaur
Feb 21 '14 at 12:49
$begingroup$
You can only use the gradient on scalar functions, not on vector fields.
$endgroup$
– dinosaur
Feb 21 '14 at 12:49
$begingroup$
Okay that's why I'm getting confused then! Any chance you could give me an example?
$endgroup$
– Sarah Jayne
Feb 21 '14 at 12:50
$begingroup$
Okay that's why I'm getting confused then! Any chance you could give me an example?
$endgroup$
– Sarah Jayne
Feb 21 '14 at 12:50
$begingroup$
@dinosaur actually you can. The result is a tensor field
$endgroup$
– David H
Feb 21 '14 at 12:50
$begingroup$
@dinosaur actually you can. The result is a tensor field
$endgroup$
– David H
Feb 21 '14 at 12:50
$begingroup$
@DavidH yes, but I don't think that this is what Sarah had in mind here.
$endgroup$
– dinosaur
Feb 21 '14 at 14:06
$begingroup$
@DavidH yes, but I don't think that this is what Sarah had in mind here.
$endgroup$
– dinosaur
Feb 21 '14 at 14:06
$begingroup$
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $nabla f(x,y)=left(frac{partial f}{partial x}, frac{partial f}{partial y}right) = (2x,2y)$.
$endgroup$
– dinosaur
Feb 21 '14 at 14:07
$begingroup$
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $nabla f(x,y)=left(frac{partial f}{partial x}, frac{partial f}{partial y}right) = (2x,2y)$.
$endgroup$
– dinosaur
Feb 21 '14 at 14:07
|
show 1 more comment
1 Answer
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$begingroup$
This might help.
You don't take the gradient of a vector field, so $nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $mathbb{R}^n$ and produces a vector field on $mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $nabla f:=langle f_x,f_yrangle$.
As for the Laplacian operator, it is given by $Delta=nablacdot nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$Delta f:=nabla cdot nabla f=nabla cdot langle f_x,f_yrangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
Suppose $mathbf F$ is a given conservative vector field (meaning
$mathbf F=nabla f$ for some scalar function $f$) which is also
divergence-free (meaning $nablacdotmathbf F=0$). How do we find the
$f$ (called a potential function) associated with this vector field
$mathbf F$?
The answer is quickly revealed: $$mathbf F=nabla f implies nabla cdotmathbf F=nablacdot nabla fimplies 0=Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
$endgroup$
add a comment |
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$begingroup$
This might help.
You don't take the gradient of a vector field, so $nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $mathbb{R}^n$ and produces a vector field on $mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $nabla f:=langle f_x,f_yrangle$.
As for the Laplacian operator, it is given by $Delta=nablacdot nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$Delta f:=nabla cdot nabla f=nabla cdot langle f_x,f_yrangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
Suppose $mathbf F$ is a given conservative vector field (meaning
$mathbf F=nabla f$ for some scalar function $f$) which is also
divergence-free (meaning $nablacdotmathbf F=0$). How do we find the
$f$ (called a potential function) associated with this vector field
$mathbf F$?
The answer is quickly revealed: $$mathbf F=nabla f implies nabla cdotmathbf F=nablacdot nabla fimplies 0=Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
$endgroup$
add a comment |
$begingroup$
This might help.
You don't take the gradient of a vector field, so $nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $mathbb{R}^n$ and produces a vector field on $mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $nabla f:=langle f_x,f_yrangle$.
As for the Laplacian operator, it is given by $Delta=nablacdot nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$Delta f:=nabla cdot nabla f=nabla cdot langle f_x,f_yrangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
Suppose $mathbf F$ is a given conservative vector field (meaning
$mathbf F=nabla f$ for some scalar function $f$) which is also
divergence-free (meaning $nablacdotmathbf F=0$). How do we find the
$f$ (called a potential function) associated with this vector field
$mathbf F$?
The answer is quickly revealed: $$mathbf F=nabla f implies nabla cdotmathbf F=nablacdot nabla fimplies 0=Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
$endgroup$
add a comment |
$begingroup$
This might help.
You don't take the gradient of a vector field, so $nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $mathbb{R}^n$ and produces a vector field on $mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $nabla f:=langle f_x,f_yrangle$.
As for the Laplacian operator, it is given by $Delta=nablacdot nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$Delta f:=nabla cdot nabla f=nabla cdot langle f_x,f_yrangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
Suppose $mathbf F$ is a given conservative vector field (meaning
$mathbf F=nabla f$ for some scalar function $f$) which is also
divergence-free (meaning $nablacdotmathbf F=0$). How do we find the
$f$ (called a potential function) associated with this vector field
$mathbf F$?
The answer is quickly revealed: $$mathbf F=nabla f implies nabla cdotmathbf F=nablacdot nabla fimplies 0=Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
$endgroup$
This might help.
You don't take the gradient of a vector field, so $nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $mathbb{R}^n$ and produces a vector field on $mathbb R^n$.
Now if $f(x,y)$ is a scalar function, then $nabla f:=langle f_x,f_yrangle$.
As for the Laplacian operator, it is given by $Delta=nablacdot nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$Delta f:=nabla cdot nabla f=nabla cdot langle f_x,f_yrangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.
The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.
A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:
Suppose $mathbf F$ is a given conservative vector field (meaning
$mathbf F=nabla f$ for some scalar function $f$) which is also
divergence-free (meaning $nablacdotmathbf F=0$). How do we find the
$f$ (called a potential function) associated with this vector field
$mathbf F$?
The answer is quickly revealed: $$mathbf F=nabla f implies nabla cdotmathbf F=nablacdot nabla fimplies 0=Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
answered Nov 20 '14 at 4:17
JohnDJohnD
12k32158
12k32158
add a comment |
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$begingroup$
You can only use the gradient on scalar functions, not on vector fields.
$endgroup$
– dinosaur
Feb 21 '14 at 12:49
$begingroup$
Okay that's why I'm getting confused then! Any chance you could give me an example?
$endgroup$
– Sarah Jayne
Feb 21 '14 at 12:50
$begingroup$
@dinosaur actually you can. The result is a tensor field
$endgroup$
– David H
Feb 21 '14 at 12:50
$begingroup$
@DavidH yes, but I don't think that this is what Sarah had in mind here.
$endgroup$
– dinosaur
Feb 21 '14 at 14:06
$begingroup$
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $nabla f(x,y)=left(frac{partial f}{partial x}, frac{partial f}{partial y}right) = (2x,2y)$.
$endgroup$
– dinosaur
Feb 21 '14 at 14:07