Formula for a Line Integral in Curvilinear Coordinates.
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I am familiar with the formula for a path integral given a parametrisation $textbf{x}(t)$, $t in l subseteq mathbb{R}$ of a curve $C$, and given some scalar function $f(x,y,z)$, $$int_C f ds = int_l f(textbf{x}(t)) Big | Big |frac{d textbf{x}}{d t} Big | Big| dt $$
Now, my lecture notes say if we are in some curvilinear coordinates $(xi_1, xi_2, xi_3)$, we describe the curve in curvilinear coordinates by $boldsymbol{xi}(u), u in l' subseteq mathbb{R}$, we have some transformation $boldsymbol{phi}$ such that the curve $C$ is parametrised by $boldsymbol{phi}(boldsymbol{xi}(u))$, and we have some scalar function $g$ = $g(boldsymbol{xi})$, $$int_C g(boldsymbol{xi}) ds = int_{l'} g(boldsymbol{xi}) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du $$
My understanding is that usually when we have a scalar function $f(x,y,z)$, for every point in 3D space, there is a (unique) corresponding scalar value. But in the curvilinear case, when we consider scalar functions, do we consider functions that directly create a correspondence between points in parameter space $(xi_1, xi_2, xi_3)$ to scalar values? Or do we first convert these points into cartesian coordinates? Why is the formula not $$int_C g(boldsymbol{phi}(boldsymbol{xi})) ds = int_{l'} g(boldsymbol{phi}(boldsymbol{xi})) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du$$
Which seems to be equivalent to the first formula. When we integrate a scalar field over some curve, don't we want the scalar field to be evaluated over that curve?
vector-spaces vector-analysis line-integrals scalar-fields curvilinear-coordinates
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add a comment |
$begingroup$
I am familiar with the formula for a path integral given a parametrisation $textbf{x}(t)$, $t in l subseteq mathbb{R}$ of a curve $C$, and given some scalar function $f(x,y,z)$, $$int_C f ds = int_l f(textbf{x}(t)) Big | Big |frac{d textbf{x}}{d t} Big | Big| dt $$
Now, my lecture notes say if we are in some curvilinear coordinates $(xi_1, xi_2, xi_3)$, we describe the curve in curvilinear coordinates by $boldsymbol{xi}(u), u in l' subseteq mathbb{R}$, we have some transformation $boldsymbol{phi}$ such that the curve $C$ is parametrised by $boldsymbol{phi}(boldsymbol{xi}(u))$, and we have some scalar function $g$ = $g(boldsymbol{xi})$, $$int_C g(boldsymbol{xi}) ds = int_{l'} g(boldsymbol{xi}) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du $$
My understanding is that usually when we have a scalar function $f(x,y,z)$, for every point in 3D space, there is a (unique) corresponding scalar value. But in the curvilinear case, when we consider scalar functions, do we consider functions that directly create a correspondence between points in parameter space $(xi_1, xi_2, xi_3)$ to scalar values? Or do we first convert these points into cartesian coordinates? Why is the formula not $$int_C g(boldsymbol{phi}(boldsymbol{xi})) ds = int_{l'} g(boldsymbol{phi}(boldsymbol{xi})) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du$$
Which seems to be equivalent to the first formula. When we integrate a scalar field over some curve, don't we want the scalar field to be evaluated over that curve?
vector-spaces vector-analysis line-integrals scalar-fields curvilinear-coordinates
$endgroup$
add a comment |
$begingroup$
I am familiar with the formula for a path integral given a parametrisation $textbf{x}(t)$, $t in l subseteq mathbb{R}$ of a curve $C$, and given some scalar function $f(x,y,z)$, $$int_C f ds = int_l f(textbf{x}(t)) Big | Big |frac{d textbf{x}}{d t} Big | Big| dt $$
Now, my lecture notes say if we are in some curvilinear coordinates $(xi_1, xi_2, xi_3)$, we describe the curve in curvilinear coordinates by $boldsymbol{xi}(u), u in l' subseteq mathbb{R}$, we have some transformation $boldsymbol{phi}$ such that the curve $C$ is parametrised by $boldsymbol{phi}(boldsymbol{xi}(u))$, and we have some scalar function $g$ = $g(boldsymbol{xi})$, $$int_C g(boldsymbol{xi}) ds = int_{l'} g(boldsymbol{xi}) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du $$
My understanding is that usually when we have a scalar function $f(x,y,z)$, for every point in 3D space, there is a (unique) corresponding scalar value. But in the curvilinear case, when we consider scalar functions, do we consider functions that directly create a correspondence between points in parameter space $(xi_1, xi_2, xi_3)$ to scalar values? Or do we first convert these points into cartesian coordinates? Why is the formula not $$int_C g(boldsymbol{phi}(boldsymbol{xi})) ds = int_{l'} g(boldsymbol{phi}(boldsymbol{xi})) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du$$
Which seems to be equivalent to the first formula. When we integrate a scalar field over some curve, don't we want the scalar field to be evaluated over that curve?
vector-spaces vector-analysis line-integrals scalar-fields curvilinear-coordinates
$endgroup$
I am familiar with the formula for a path integral given a parametrisation $textbf{x}(t)$, $t in l subseteq mathbb{R}$ of a curve $C$, and given some scalar function $f(x,y,z)$, $$int_C f ds = int_l f(textbf{x}(t)) Big | Big |frac{d textbf{x}}{d t} Big | Big| dt $$
Now, my lecture notes say if we are in some curvilinear coordinates $(xi_1, xi_2, xi_3)$, we describe the curve in curvilinear coordinates by $boldsymbol{xi}(u), u in l' subseteq mathbb{R}$, we have some transformation $boldsymbol{phi}$ such that the curve $C$ is parametrised by $boldsymbol{phi}(boldsymbol{xi}(u))$, and we have some scalar function $g$ = $g(boldsymbol{xi})$, $$int_C g(boldsymbol{xi}) ds = int_{l'} g(boldsymbol{xi}) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du $$
My understanding is that usually when we have a scalar function $f(x,y,z)$, for every point in 3D space, there is a (unique) corresponding scalar value. But in the curvilinear case, when we consider scalar functions, do we consider functions that directly create a correspondence between points in parameter space $(xi_1, xi_2, xi_3)$ to scalar values? Or do we first convert these points into cartesian coordinates? Why is the formula not $$int_C g(boldsymbol{phi}(boldsymbol{xi})) ds = int_{l'} g(boldsymbol{phi}(boldsymbol{xi})) Big | Big |frac{d boldsymbol{phi}(boldsymbol{xi})(u)}{d u} Big | Big| du$$
Which seems to be equivalent to the first formula. When we integrate a scalar field over some curve, don't we want the scalar field to be evaluated over that curve?
vector-spaces vector-analysis line-integrals scalar-fields curvilinear-coordinates
vector-spaces vector-analysis line-integrals scalar-fields curvilinear-coordinates
edited Jan 9 at 13:44
asked May 23 '18 at 10:42
user445909
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