Grad in polar coordinates
$begingroup$
As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...
Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.
I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$
Help would be appreciated....
vector-analysis polar-coordinates spherical-coordinates
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
$endgroup$
add a comment |
$begingroup$
As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...
Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.
I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$
Help would be appreciated....
vector-analysis polar-coordinates spherical-coordinates
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
$endgroup$
$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04
add a comment |
$begingroup$
As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...
Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.
I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$
Help would be appreciated....
vector-analysis polar-coordinates spherical-coordinates
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
$endgroup$
As part of my lectures, it is noted that $nabla sin theta propto frac{1}{r^2}$ and $nabla phi propto frac{1}{r^2}$ where we are working in spherical polar coordinates with $theta$ as the polar angle and $phi$ as the azimuthal (so the physics convention)...
Now I can’t seem to see why this is true. I’ve tried $$nabla sin theta = frac{partial}{partial r} (sin theta) + frac{partial}{partial theta} (sin theta) + frac{partial}{partial phi} (sin theta)$$ but I can’t see how a $frac{1}{r^2}$ is going to come out of this.
I’ve also tried to work with grad in spherical polars but I still can’t seem to get the $frac{1}{r^2}$, likewise for $nabla phi$
Help would be appreciated....
vector-analysis polar-coordinates spherical-coordinates
vector-analysis polar-coordinates spherical-coordinates
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 5 at 11:38
PhysicsMathsLovePhysicsMathsLove
1,208414
1,208414
$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04
add a comment |
$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04
$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04
$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.
You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.
answered Jan 5 at 11:49
TreborTrebor
85415
$endgroup$
add a comment |
$begingroup$
The thing is that
$$
nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
$$
only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,
$$
nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
$$
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.
You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.
answered Jan 5 at 11:49
TreborTrebor
85415
$endgroup$
add a comment |
$begingroup$
The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.
You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.
answered Jan 5 at 11:49
TreborTrebor
85415
$endgroup$
add a comment |
$begingroup$
The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.
You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.
answered Jan 5 at 11:49
TreborTrebor
85415
$endgroup$
The $mathrm{grad}$ operator is of the form $partial_xmathbf{i}+partial_ymathbf{j}+partial_zmathbf{k}$ in Catesian coordinates. In other coordinate systems, the operator has different forms. See this wikipedia article for a list of them.
You can easily see the reason for this when you write out the conversion formula from Cartesian coordinates to polar/spherical coordinates, and differentiate it. You get extra factors on the polar coordinates side, because the relationship is no longer linear. To systematically address this problem, we can introduce connections. Furthermore, this can be generalized to curved spaces.
answered Jan 5 at 11:49
TreborTrebor
85415
answered Jan 5 at 11:49
TreborTrebor
85415
answered Jan 5 at 11:49
TreborTrebor
85415
answered Jan 5 at 11:49
TreborTrebor
85415
85415
add a comment |
add a comment |
$begingroup$
The thing is that
$$
nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
$$
only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,
$$
nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
$$
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
add a comment |
$begingroup$
The thing is that
$$
nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
$$
only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,
$$
nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
$$
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
add a comment |
$begingroup$
The thing is that
$$
nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
$$
only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,
$$
nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
$$
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
$endgroup$
The thing is that
$$
nabla f = frac{partial f}{partial u_1}hat{bf e}_1 + frac{partial f}{partial u_2}hat{bf e}_2 + frac{partial f}{partial u_3}hat{bf e}_3
$$
only works when $u$ are cartesian coordinates. In spherical coordinates is not as trivial,
$$
nabla f = frac{partial f}{partial r}hat{bf e}_r + frac{1}{r}frac{partial f}{partial theta} hat{bf e}_theta + frac{1}{rsintheta}frac{partial f}{partial phi}hat{bf e}_phi
$$
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
edited Jan 5 at 12:26
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
answered Jan 5 at 11:49
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
add a comment |
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
Ok but I still don’t see how I’ll get $frac{1}{r^2}$? All the derivatives apart from the $theta$ derivative will evaluate to 0, and I’ll just have $cos theta hat{e_{theta}}$
$endgroup$
– PhysicsMathsLove
Jan 5 at 11:57
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
$begingroup$
@PhysicsMathsLove That may be a problem with your notes. You see that dimensionally it does not make sense $nabla$ has units of $L^{-1}$ and $sintheta$ is dimensionless, so $nabla sintheta$ must have dimensions $L^{-1}$, so it should be something like $1/r$
$endgroup$
– caverac
Jan 5 at 12:21
add a comment |
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$begingroup$
Are you sure that your lecturer was talking about the gradient, and not the divergence, curl (with some included unit vector) or laplacian? For the gradient the magnitudes are proportional to $1/r$, for the Laplacian $1/r^2$, and for the others it depends on the unit vector, but should be $1/r$ also.
$endgroup$
– Eddy
Jan 5 at 20:04