divergence in polar coordinates
$begingroup$
For a vector field $X$, the divergence in coordinates is given by $nablacdot X=sum_nfrac{X^i}{partial x^i}$. In polar coordinates, the metric is $begin{bmatrix}1 & 0\ 0 & r^2end{bmatrix}$, and so $frac{1}{sqrt{g(frac{partial}{partial r},frac{partial}{partial r})}}frac{partial}{partial r}=frac{partial}{partial r}$ and $frac{1}{sqrt{g(frac{partial}{partialtheta},frac{partial}{partialtheta})}}frac{partial}{partialtheta}=frac{1}{r}frac{partial}{partialtheta}$ are unit vectors. Then for $X=X_{r}frac{partial}{partial r}+X_{theta}frac{partial}{rpartialtheta}$, $nablacdot X=frac{partial X_r}{partial r}+frac{partial}{partialtheta}frac{X_{theta}}{r}=frac{partial X_r}{partial r}+frac{1}{r}frac{partial X_{theta}}{partialtheta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
differential-geometry vector-spaces
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
$endgroup$
add a comment |
$begingroup$
For a vector field $X$, the divergence in coordinates is given by $nablacdot X=sum_nfrac{X^i}{partial x^i}$. In polar coordinates, the metric is $begin{bmatrix}1 & 0\ 0 & r^2end{bmatrix}$, and so $frac{1}{sqrt{g(frac{partial}{partial r},frac{partial}{partial r})}}frac{partial}{partial r}=frac{partial}{partial r}$ and $frac{1}{sqrt{g(frac{partial}{partialtheta},frac{partial}{partialtheta})}}frac{partial}{partialtheta}=frac{1}{r}frac{partial}{partialtheta}$ are unit vectors. Then for $X=X_{r}frac{partial}{partial r}+X_{theta}frac{partial}{rpartialtheta}$, $nablacdot X=frac{partial X_r}{partial r}+frac{partial}{partialtheta}frac{X_{theta}}{r}=frac{partial X_r}{partial r}+frac{1}{r}frac{partial X_{theta}}{partialtheta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
differential-geometry vector-spaces
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
$endgroup$
add a comment |
$begingroup$
For a vector field $X$, the divergence in coordinates is given by $nablacdot X=sum_nfrac{X^i}{partial x^i}$. In polar coordinates, the metric is $begin{bmatrix}1 & 0\ 0 & r^2end{bmatrix}$, and so $frac{1}{sqrt{g(frac{partial}{partial r},frac{partial}{partial r})}}frac{partial}{partial r}=frac{partial}{partial r}$ and $frac{1}{sqrt{g(frac{partial}{partialtheta},frac{partial}{partialtheta})}}frac{partial}{partialtheta}=frac{1}{r}frac{partial}{partialtheta}$ are unit vectors. Then for $X=X_{r}frac{partial}{partial r}+X_{theta}frac{partial}{rpartialtheta}$, $nablacdot X=frac{partial X_r}{partial r}+frac{partial}{partialtheta}frac{X_{theta}}{r}=frac{partial X_r}{partial r}+frac{1}{r}frac{partial X_{theta}}{partialtheta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
differential-geometry vector-spaces
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
$endgroup$
For a vector field $X$, the divergence in coordinates is given by $nablacdot X=sum_nfrac{X^i}{partial x^i}$. In polar coordinates, the metric is $begin{bmatrix}1 & 0\ 0 & r^2end{bmatrix}$, and so $frac{1}{sqrt{g(frac{partial}{partial r},frac{partial}{partial r})}}frac{partial}{partial r}=frac{partial}{partial r}$ and $frac{1}{sqrt{g(frac{partial}{partialtheta},frac{partial}{partialtheta})}}frac{partial}{partialtheta}=frac{1}{r}frac{partial}{partialtheta}$ are unit vectors. Then for $X=X_{r}frac{partial}{partial r}+X_{theta}frac{partial}{rpartialtheta}$, $nablacdot X=frac{partial X_r}{partial r}+frac{partial}{partialtheta}frac{X_{theta}}{r}=frac{partial X_r}{partial r}+frac{1}{r}frac{partial X_{theta}}{partialtheta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
differential-geometry vector-spaces
differential-geometry vector-spaces
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
edited Dec 18 '18 at 2:31
user124910
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
asked Dec 18 '18 at 0:27
user124910user124910
1,300816
1,300816
add a comment |
add a comment |
1 Answer
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$DeclareMathOperatordiv{div}$The formula for $nablacdot X$ is incorrect. The notation with the 'usual' dot product is misleading. Properly it is for a diagonal metric:
$$div F = frac 1rhofrac{partial(rho F^i)}{partial x^i}$$
where $rho=sqrt{det g}$ is the coefficient of the differential volume element $dV=rho, dx^1wedgeldots wedge dx^n$, meaning $rho$ is also the Jacobian determinant, and where $F^i$ are the components of $F$ with respect to an unnormalized basis.
In polar coordinates we have $rho=sqrt{det g}=r$, and:
$$div X = frac 1r frac{partial(r X^r)}{partial r}
+ frac 1rfrac{partial(r X^theta)}{partial theta}$$
In the usual normalized coordinates $X=hat X^{r}frac{partial}{partial r} + hat X^{theta}frac 1rfrac{partial}{partialtheta}$ this becomes:
$$div X = frac 1r frac{partial(r hat X^{r})}{partial r}
+ frac 1rfrac{partial hat X^{theta}}{partial theta}$$
which agrees with the usual formula given in calculus books.
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
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1 Answer
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$begingroup$
$DeclareMathOperatordiv{div}$The formula for $nablacdot X$ is incorrect. The notation with the 'usual' dot product is misleading. Properly it is for a diagonal metric:
$$div F = frac 1rhofrac{partial(rho F^i)}{partial x^i}$$
where $rho=sqrt{det g}$ is the coefficient of the differential volume element $dV=rho, dx^1wedgeldots wedge dx^n$, meaning $rho$ is also the Jacobian determinant, and where $F^i$ are the components of $F$ with respect to an unnormalized basis.
In polar coordinates we have $rho=sqrt{det g}=r$, and:
$$div X = frac 1r frac{partial(r X^r)}{partial r}
+ frac 1rfrac{partial(r X^theta)}{partial theta}$$
In the usual normalized coordinates $X=hat X^{r}frac{partial}{partial r} + hat X^{theta}frac 1rfrac{partial}{partialtheta}$ this becomes:
$$div X = frac 1r frac{partial(r hat X^{r})}{partial r}
+ frac 1rfrac{partial hat X^{theta}}{partial theta}$$
which agrees with the usual formula given in calculus books.
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperatordiv{div}$The formula for $nablacdot X$ is incorrect. The notation with the 'usual' dot product is misleading. Properly it is for a diagonal metric:
$$div F = frac 1rhofrac{partial(rho F^i)}{partial x^i}$$
where $rho=sqrt{det g}$ is the coefficient of the differential volume element $dV=rho, dx^1wedgeldots wedge dx^n$, meaning $rho$ is also the Jacobian determinant, and where $F^i$ are the components of $F$ with respect to an unnormalized basis.
In polar coordinates we have $rho=sqrt{det g}=r$, and:
$$div X = frac 1r frac{partial(r X^r)}{partial r}
+ frac 1rfrac{partial(r X^theta)}{partial theta}$$
In the usual normalized coordinates $X=hat X^{r}frac{partial}{partial r} + hat X^{theta}frac 1rfrac{partial}{partialtheta}$ this becomes:
$$div X = frac 1r frac{partial(r hat X^{r})}{partial r}
+ frac 1rfrac{partial hat X^{theta}}{partial theta}$$
which agrees with the usual formula given in calculus books.
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperatordiv{div}$The formula for $nablacdot X$ is incorrect. The notation with the 'usual' dot product is misleading. Properly it is for a diagonal metric:
$$div F = frac 1rhofrac{partial(rho F^i)}{partial x^i}$$
where $rho=sqrt{det g}$ is the coefficient of the differential volume element $dV=rho, dx^1wedgeldots wedge dx^n$, meaning $rho$ is also the Jacobian determinant, and where $F^i$ are the components of $F$ with respect to an unnormalized basis.
In polar coordinates we have $rho=sqrt{det g}=r$, and:
$$div X = frac 1r frac{partial(r X^r)}{partial r}
+ frac 1rfrac{partial(r X^theta)}{partial theta}$$
In the usual normalized coordinates $X=hat X^{r}frac{partial}{partial r} + hat X^{theta}frac 1rfrac{partial}{partialtheta}$ this becomes:
$$div X = frac 1r frac{partial(r hat X^{r})}{partial r}
+ frac 1rfrac{partial hat X^{theta}}{partial theta}$$
which agrees with the usual formula given in calculus books.
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
$endgroup$
$DeclareMathOperatordiv{div}$The formula for $nablacdot X$ is incorrect. The notation with the 'usual' dot product is misleading. Properly it is for a diagonal metric:
$$div F = frac 1rhofrac{partial(rho F^i)}{partial x^i}$$
where $rho=sqrt{det g}$ is the coefficient of the differential volume element $dV=rho, dx^1wedgeldots wedge dx^n$, meaning $rho$ is also the Jacobian determinant, and where $F^i$ are the components of $F$ with respect to an unnormalized basis.
In polar coordinates we have $rho=sqrt{det g}=r$, and:
$$div X = frac 1r frac{partial(r X^r)}{partial r}
+ frac 1rfrac{partial(r X^theta)}{partial theta}$$
In the usual normalized coordinates $X=hat X^{r}frac{partial}{partial r} + hat X^{theta}frac 1rfrac{partial}{partialtheta}$ this becomes:
$$div X = frac 1r frac{partial(r hat X^{r})}{partial r}
+ frac 1rfrac{partial hat X^{theta}}{partial theta}$$
which agrees with the usual formula given in calculus books.
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
edited Jan 16 at 8:59
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
answered Dec 31 '18 at 20:10
I like SerenaI like Serena
4,1671722
4,1671722
add a comment |
add a comment |
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