Mixing
Why is the circulation the line integral of the vector field over closed path bounding the region?
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The circulation of vector field around some region is the line integral of the field over the closed curve surrounding it and that integral is integrating the field dotted with the vector of the derivative of the vector valued function that represents the curve (the parametrization of the curve) with respect to the parameter(t) right??
But since the parametrization of the curve determines the value of the derivative of the vector valued function at each point.
Then the parametrization of the curve should change the value of the circulation for the same curve??
calculus derivatives vector-fields
asked Jan 26 at 18:11
gogo okagogo oka
11
$endgroup$
add a comment |
$begingroup$
The circulation of vector field around some region is the line integral of the field over the closed curve surrounding it and that integral is integrating the field dotted with the vector of the derivative of the vector valued function that represents the curve (the parametrization of the curve) with respect to the parameter(t) right??
But since the parametrization of the curve determines the value of the derivative of the vector valued function at each point.
Then the parametrization of the curve should change the value of the circulation for the same curve??
calculus derivatives vector-fields
asked Jan 26 at 18:11
gogo okagogo oka
11
$endgroup$
add a comment |
$begingroup$
The circulation of vector field around some region is the line integral of the field over the closed curve surrounding it and that integral is integrating the field dotted with the vector of the derivative of the vector valued function that represents the curve (the parametrization of the curve) with respect to the parameter(t) right??
But since the parametrization of the curve determines the value of the derivative of the vector valued function at each point.
Then the parametrization of the curve should change the value of the circulation for the same curve??
calculus derivatives vector-fields
asked Jan 26 at 18:11
gogo okagogo oka
11
$endgroup$
The circulation of vector field around some region is the line integral of the field over the closed curve surrounding it and that integral is integrating the field dotted with the vector of the derivative of the vector valued function that represents the curve (the parametrization of the curve) with respect to the parameter(t) right??
But since the parametrization of the curve determines the value of the derivative of the vector valued function at each point.
Then the parametrization of the curve should change the value of the circulation for the same curve??
calculus derivatives vector-fields
calculus derivatives vector-fields
asked Jan 26 at 18:11
gogo okagogo oka
11
asked Jan 26 at 18:11
gogo okagogo oka
11
asked Jan 26 at 18:11
gogo okagogo oka
11
asked Jan 26 at 18:11
gogo okagogo oka
11
asked Jan 26 at 18:11
gogo okagogo oka
11
11
add a comment |
add a comment |
1 Answer
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No, it actually does not change the circulation. The circulation can only change its sign if you reverse the orientation of your parametrization. If you take two different parametrizations $gamma_1(t)$ and $gamma_2(s)$ linked by some smooth change $s=s(t)$ preserving orientation, you have
$$
intlimits_{t_1}^{t_2}F(gamma_1(t))cdotgamma_1'(t),dt=intlimits_{s_1}^{s_2}F(gamma_1(t(s))cdotgamma_1'(t(s))frac{dt}{ds},ds
$$
$$
=intlimits_{s_1}^{s_2}F(gamma_2(s))cdotgamma_2'(s),ds
$$
You can easily check that if the change of parameter changes orientation you get a minus sign.
answered Jan 26 at 18:27
GReyesGReyes
2,23315
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it actually does not change the circulation. The circulation can only change its sign if you reverse the orientation of your parametrization. If you take two different parametrizations $gamma_1(t)$ and $gamma_2(s)$ linked by some smooth change $s=s(t)$ preserving orientation, you have
$$
intlimits_{t_1}^{t_2}F(gamma_1(t))cdotgamma_1'(t),dt=intlimits_{s_1}^{s_2}F(gamma_1(t(s))cdotgamma_1'(t(s))frac{dt}{ds},ds
$$
$$
=intlimits_{s_1}^{s_2}F(gamma_2(s))cdotgamma_2'(s),ds
$$
You can easily check that if the change of parameter changes orientation you get a minus sign.
answered Jan 26 at 18:27
GReyesGReyes
2,23315
$endgroup$
add a comment |
$begingroup$
No, it actually does not change the circulation. The circulation can only change its sign if you reverse the orientation of your parametrization. If you take two different parametrizations $gamma_1(t)$ and $gamma_2(s)$ linked by some smooth change $s=s(t)$ preserving orientation, you have
$$
intlimits_{t_1}^{t_2}F(gamma_1(t))cdotgamma_1'(t),dt=intlimits_{s_1}^{s_2}F(gamma_1(t(s))cdotgamma_1'(t(s))frac{dt}{ds},ds
$$
$$
=intlimits_{s_1}^{s_2}F(gamma_2(s))cdotgamma_2'(s),ds
$$
You can easily check that if the change of parameter changes orientation you get a minus sign.
answered Jan 26 at 18:27
GReyesGReyes
2,23315
$endgroup$
add a comment |
$begingroup$
No, it actually does not change the circulation. The circulation can only change its sign if you reverse the orientation of your parametrization. If you take two different parametrizations $gamma_1(t)$ and $gamma_2(s)$ linked by some smooth change $s=s(t)$ preserving orientation, you have
$$
intlimits_{t_1}^{t_2}F(gamma_1(t))cdotgamma_1'(t),dt=intlimits_{s_1}^{s_2}F(gamma_1(t(s))cdotgamma_1'(t(s))frac{dt}{ds},ds
$$
$$
=intlimits_{s_1}^{s_2}F(gamma_2(s))cdotgamma_2'(s),ds
$$
You can easily check that if the change of parameter changes orientation you get a minus sign.
answered Jan 26 at 18:27
GReyesGReyes
2,23315
$endgroup$
No, it actually does not change the circulation. The circulation can only change its sign if you reverse the orientation of your parametrization. If you take two different parametrizations $gamma_1(t)$ and $gamma_2(s)$ linked by some smooth change $s=s(t)$ preserving orientation, you have
$$
intlimits_{t_1}^{t_2}F(gamma_1(t))cdotgamma_1'(t),dt=intlimits_{s_1}^{s_2}F(gamma_1(t(s))cdotgamma_1'(t(s))frac{dt}{ds},ds
$$
$$
=intlimits_{s_1}^{s_2}F(gamma_2(s))cdotgamma_2'(s),ds
$$
You can easily check that if the change of parameter changes orientation you get a minus sign.
answered Jan 26 at 18:27
GReyesGReyes
2,23315
answered Jan 26 at 18:27
GReyesGReyes
2,23315
answered Jan 26 at 18:27
GReyesGReyes
2,23315
answered Jan 26 at 18:27
GReyesGReyes
2,23315
2,23315
add a comment |
add a comment |
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