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Gradient of vector field using differential forms
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I have been reading about differential forms from various sources. The usual vector-calculus operations of gradient, divergence, curl are nicely represented by the exterior derivative operator $d$ acting on 0-form, 1-form, and 2-forms respectively. For example, if $f$ is a 0-form (i.e. a scalar field) then its exterior derivative gives the gradient 1-form of $f$: $df=partial_x_1f~dx_1+partial_x_2f~dx_2+partial_x_3f~dx_3$. Using the metric tensor for the $x_1,x_2,x_3$ coordinates I can convert this 1-form into a vector (which is what we engineers usually deal with). So far so good.
But in applications we often have to find the gradient of a vector field, in some convenient coordinate system, say (in my case) prolate spheroidal coordinates $(xi,eta,phi)$. These are related to Cartesian coordinates $(x,y,z)$ by:
$$x=dsqrt(xi^2-1)(1-eta^2)cosphi\ y=dsqrt(xi^2-1)(1-eta^2)sinphi\ z=dxieta$$
in which $d>0$ is a constant. The range of spheroidal coordinates are: $xigeq 1,~-1leqetaleq1,~0leqphileq2pi$.
I have a vector field $vecu=u_1(xi,eta,phi)vece_xi+u_2(xi,eta,phi)vece_eta+u_3(xi,eta,phi)vece_phi$, in which $vece$ are unit coordinate vectors. I need to find its gradient $nablavecu$ (which is a second-order tensor) in spheroidal coordinates. How do I do that using differential forms? Using the metric tensor for the spheroidal coordinates, I can write $vecu$ as a 1-form, but what next?
vectors vector-analysis coordinate-systems differential-forms
asked Jul 25 at 6:11
Deep
364210
 |Â
show 2 more comments
up vote
1
down vote
favorite
I have been reading about differential forms from various sources. The usual vector-calculus operations of gradient, divergence, curl are nicely represented by the exterior derivative operator $d$ acting on 0-form, 1-form, and 2-forms respectively. For example, if $f$ is a 0-form (i.e. a scalar field) then its exterior derivative gives the gradient 1-form of $f$: $df=partial_x_1f~dx_1+partial_x_2f~dx_2+partial_x_3f~dx_3$. Using the metric tensor for the $x_1,x_2,x_3$ coordinates I can convert this 1-form into a vector (which is what we engineers usually deal with). So far so good.
But in applications we often have to find the gradient of a vector field, in some convenient coordinate system, say (in my case) prolate spheroidal coordinates $(xi,eta,phi)$. These are related to Cartesian coordinates $(x,y,z)$ by:
$$x=dsqrt(xi^2-1)(1-eta^2)cosphi\ y=dsqrt(xi^2-1)(1-eta^2)sinphi\ z=dxieta$$
in which $d>0$ is a constant. The range of spheroidal coordinates are: $xigeq 1,~-1leqetaleq1,~0leqphileq2pi$.
I have a vector field $vecu=u_1(xi,eta,phi)vece_xi+u_2(xi,eta,phi)vece_eta+u_3(xi,eta,phi)vece_phi$, in which $vece$ are unit coordinate vectors. I need to find its gradient $nablavecu$ (which is a second-order tensor) in spheroidal coordinates. How do I do that using differential forms? Using the metric tensor for the spheroidal coordinates, I can write $vecu$ as a 1-form, but what next?
vectors vector-analysis coordinate-systems differential-forms
asked Jul 25 at 6:11
Deep
364210
How do you define the gradient of a vector field? Is it just the tensor $(partial_i u_j)$?
– md2perpe
Jul 25 at 7:47
What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field.
– md2perpe
Jul 25 at 8:02
@md2perpe In Cartesian coordinates gradient of a vector field is indeed $partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here).
– Deep
Jul 25 at 8:45
Is it for the viscosity term of the integral form of Navier-Stokes? $$nu oint_partial Omega partial_i vec u , n^i , dS$$
– md2perpe
Jul 25 at 10:58
@md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates.
– Deep
Jul 25 at 15:06
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been reading about differential forms from various sources. The usual vector-calculus operations of gradient, divergence, curl are nicely represented by the exterior derivative operator $d$ acting on 0-form, 1-form, and 2-forms respectively. For example, if $f$ is a 0-form (i.e. a scalar field) then its exterior derivative gives the gradient 1-form of $f$: $df=partial_x_1f~dx_1+partial_x_2f~dx_2+partial_x_3f~dx_3$. Using the metric tensor for the $x_1,x_2,x_3$ coordinates I can convert this 1-form into a vector (which is what we engineers usually deal with). So far so good.
But in applications we often have to find the gradient of a vector field, in some convenient coordinate system, say (in my case) prolate spheroidal coordinates $(xi,eta,phi)$. These are related to Cartesian coordinates $(x,y,z)$ by:
$$x=dsqrt(xi^2-1)(1-eta^2)cosphi\ y=dsqrt(xi^2-1)(1-eta^2)sinphi\ z=dxieta$$
in which $d>0$ is a constant. The range of spheroidal coordinates are: $xigeq 1,~-1leqetaleq1,~0leqphileq2pi$.
I have a vector field $vecu=u_1(xi,eta,phi)vece_xi+u_2(xi,eta,phi)vece_eta+u_3(xi,eta,phi)vece_phi$, in which $vece$ are unit coordinate vectors. I need to find its gradient $nablavecu$ (which is a second-order tensor) in spheroidal coordinates. How do I do that using differential forms? Using the metric tensor for the spheroidal coordinates, I can write $vecu$ as a 1-form, but what next?
vectors vector-analysis coordinate-systems differential-forms
asked Jul 25 at 6:11
Deep
364210
I have been reading about differential forms from various sources. The usual vector-calculus operations of gradient, divergence, curl are nicely represented by the exterior derivative operator $d$ acting on 0-form, 1-form, and 2-forms respectively. For example, if $f$ is a 0-form (i.e. a scalar field) then its exterior derivative gives the gradient 1-form of $f$: $df=partial_x_1f~dx_1+partial_x_2f~dx_2+partial_x_3f~dx_3$. Using the metric tensor for the $x_1,x_2,x_3$ coordinates I can convert this 1-form into a vector (which is what we engineers usually deal with). So far so good.
But in applications we often have to find the gradient of a vector field, in some convenient coordinate system, say (in my case) prolate spheroidal coordinates $(xi,eta,phi)$. These are related to Cartesian coordinates $(x,y,z)$ by:
$$x=dsqrt(xi^2-1)(1-eta^2)cosphi\ y=dsqrt(xi^2-1)(1-eta^2)sinphi\ z=dxieta$$
in which $d>0$ is a constant. The range of spheroidal coordinates are: $xigeq 1,~-1leqetaleq1,~0leqphileq2pi$.
I have a vector field $vecu=u_1(xi,eta,phi)vece_xi+u_2(xi,eta,phi)vece_eta+u_3(xi,eta,phi)vece_phi$, in which $vece$ are unit coordinate vectors. I need to find its gradient $nablavecu$ (which is a second-order tensor) in spheroidal coordinates. How do I do that using differential forms? Using the metric tensor for the spheroidal coordinates, I can write $vecu$ as a 1-form, but what next?
vectors vector-analysis coordinate-systems differential-forms
asked Jul 25 at 6:11
Deep
364210
asked Jul 25 at 6:11
Deep
364210
asked Jul 25 at 6:11
Deep
364210
asked Jul 25 at 6:11
Deep
364210
364210
How do you define the gradient of a vector field? Is it just the tensor $(partial_i u_j)$?
– md2perpe
Jul 25 at 7:47
What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field.
– md2perpe
Jul 25 at 8:02
@md2perpe In Cartesian coordinates gradient of a vector field is indeed $partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here).
– Deep
Jul 25 at 8:45
Is it for the viscosity term of the integral form of Navier-Stokes? $$nu oint_partial Omega partial_i vec u , n^i , dS$$
– md2perpe
Jul 25 at 10:58
@md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates.
– Deep
Jul 25 at 15:06
 |Â
show 2 more comments
How do you define the gradient of a vector field? Is it just the tensor $(partial_i u_j)$?
– md2perpe
Jul 25 at 7:47
What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field.
– md2perpe
Jul 25 at 8:02
@md2perpe In Cartesian coordinates gradient of a vector field is indeed $partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here).
– Deep
Jul 25 at 8:45
Is it for the viscosity term of the integral form of Navier-Stokes? $$nu oint_partial Omega partial_i vec u , n^i , dS$$
– md2perpe
Jul 25 at 10:58
@md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates.
– Deep
Jul 25 at 15:06
How do you define the gradient of a vector field? Is it just the tensor $(partial_i u_j)$?
– md2perpe
Jul 25 at 7:47
How do you define the gradient of a vector field? Is it just the tensor $(partial_i u_j)$?
– md2perpe
Jul 25 at 7:47
What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field.
– md2perpe
Jul 25 at 8:02
What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field.
– md2perpe
Jul 25 at 8:02
@md2perpe In Cartesian coordinates gradient of a vector field is indeed $partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here).
– Deep
Jul 25 at 8:45
@md2perpe In Cartesian coordinates gradient of a vector field is indeed $partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here).
– Deep
Jul 25 at 8:45
Is it for the viscosity term of the integral form of Navier-Stokes? $$nu oint_partial Omega partial_i vec u , n^i , dS$$
– md2perpe
Jul 25 at 10:58
Is it for the viscosity term of the integral form of Navier-Stokes? $$nu oint_partial Omega partial_i vec u , n^i , dS$$
– md2perpe
Jul 25 at 10:58
@md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates.
– Deep
Jul 25 at 15:06
@md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates.
– Deep
Jul 25 at 15:06
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If $vec n$ is a vector field on $mathbb R^3$, its "gradient" is actually its total covariant derivative. This makes sense on an arbitrary Riemannian manifold and is usually denoted by $nabla vec n$. If you want to compute it for $mathbb R^3$ in different coordinates, you'll have to first compute the Christoffel symbols of the metric in those coordinates, and then $nablavec n$ is the matrix-valued function whose components are given in any coordinate chart by
$$
n^i _;j = partial_j xi^i + sum_k Gamma_jk^i xi^k.
$$
For details, check out any differential geometry book that treats Riemannian metrics. (You can try my Introduction to Riemannian Manifolds, but there are plenty of other good choices.)
answered Jul 27 at 21:21
Jack Lee
25.2k44262
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
1
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $vec n$ is a vector field on $mathbb R^3$, its "gradient" is actually its total covariant derivative. This makes sense on an arbitrary Riemannian manifold and is usually denoted by $nabla vec n$. If you want to compute it for $mathbb R^3$ in different coordinates, you'll have to first compute the Christoffel symbols of the metric in those coordinates, and then $nablavec n$ is the matrix-valued function whose components are given in any coordinate chart by
$$
n^i _;j = partial_j xi^i + sum_k Gamma_jk^i xi^k.
$$
For details, check out any differential geometry book that treats Riemannian metrics. (You can try my Introduction to Riemannian Manifolds, but there are plenty of other good choices.)
answered Jul 27 at 21:21
Jack Lee
25.2k44262
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
1
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
add a comment |Â
up vote
2
down vote
accepted
If $vec n$ is a vector field on $mathbb R^3$, its "gradient" is actually its total covariant derivative. This makes sense on an arbitrary Riemannian manifold and is usually denoted by $nabla vec n$. If you want to compute it for $mathbb R^3$ in different coordinates, you'll have to first compute the Christoffel symbols of the metric in those coordinates, and then $nablavec n$ is the matrix-valued function whose components are given in any coordinate chart by
$$
n^i _;j = partial_j xi^i + sum_k Gamma_jk^i xi^k.
$$
For details, check out any differential geometry book that treats Riemannian metrics. (You can try my Introduction to Riemannian Manifolds, but there are plenty of other good choices.)
answered Jul 27 at 21:21
Jack Lee
25.2k44262
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
1
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $vec n$ is a vector field on $mathbb R^3$, its "gradient" is actually its total covariant derivative. This makes sense on an arbitrary Riemannian manifold and is usually denoted by $nabla vec n$. If you want to compute it for $mathbb R^3$ in different coordinates, you'll have to first compute the Christoffel symbols of the metric in those coordinates, and then $nablavec n$ is the matrix-valued function whose components are given in any coordinate chart by
$$
n^i _;j = partial_j xi^i + sum_k Gamma_jk^i xi^k.
$$
For details, check out any differential geometry book that treats Riemannian metrics. (You can try my Introduction to Riemannian Manifolds, but there are plenty of other good choices.)
answered Jul 27 at 21:21
Jack Lee
25.2k44262
If $vec n$ is a vector field on $mathbb R^3$, its "gradient" is actually its total covariant derivative. This makes sense on an arbitrary Riemannian manifold and is usually denoted by $nabla vec n$. If you want to compute it for $mathbb R^3$ in different coordinates, you'll have to first compute the Christoffel symbols of the metric in those coordinates, and then $nablavec n$ is the matrix-valued function whose components are given in any coordinate chart by
$$
n^i _;j = partial_j xi^i + sum_k Gamma_jk^i xi^k.
$$
For details, check out any differential geometry book that treats Riemannian metrics. (You can try my Introduction to Riemannian Manifolds, but there are plenty of other good choices.)
answered Jul 27 at 21:21
Jack Lee
25.2k44262
answered Jul 27 at 21:21
Jack Lee
25.2k44262
answered Jul 27 at 21:21
Jack Lee
25.2k44262
answered Jul 27 at 21:21
Jack Lee
25.2k44262
25.2k44262
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
1
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
add a comment |Â
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
1
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
+1 I guess there is no straightforward way of doing it using just differential forms.
– Deep
Jul 28 at 4:23
1
1
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
@Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one.
– Jack Lee
Jul 30 at 22:07
add a comment |Â
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How do you define the gradient of a vector field? Is it just the tensor $(partial_i u_j)$?
– md2perpe
Jul 25 at 7:47
What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field.
– md2perpe
Jul 25 at 8:02
@md2perpe In Cartesian coordinates gradient of a vector field is indeed $partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here).
– Deep
Jul 25 at 8:45
Is it for the viscosity term of the integral form of Navier-Stokes? $$nu oint_partial Omega partial_i vec u , n^i , dS$$
– md2perpe
Jul 25 at 10:58
@md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates.
– Deep
Jul 25 at 15:06