Scalar product of vectors with different basis?
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Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
$$ê_ycdotê_rho$$
Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?
Or how does it work?
vectors vector-analysis change-of-basis
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Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
$$ê_ycdotê_rho$$
Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?
Or how does it work?
vectors vector-analysis change-of-basis
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
$$ê_ycdotê_rho$$
Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?
Or how does it work?
vectors vector-analysis change-of-basis
Let's say I have this scalar product with the basis in both Cartesian and cylindrical coordinates.
$$ê_ycdotê_rho$$
Do I need to convert one of them to the other basis, e.g. convert $ê_rho$ to $cosphiê_x+sinphiê_y$ and then perform the scalar product $(0,1,0)cdot(cosphi, sinphi,0)=(0,sinphi,0)$?
Or how does it work?
vectors vector-analysis change-of-basis
asked Aug 28 at 13:14
Defindun
296
296
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1 Answer
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The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is
$overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$
where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is
$overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$
where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
add a comment |Â
up vote
0
down vote
accepted
The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is
$overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$
where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is
$overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$
where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.
The scalar product must be a scalar, so it is just $sin phi$. You can see this by converting one vector from cylindrical to cartesian co-ordinates or by using the co-ordinate free definition of the scalar product which is
$overrightarrowa cdot overrightarrowb = |overrightarrowa||overrightarrowb| cos theta$
where $theta$ is the angle between vectors $overrightarrowa$ and $overrightarrowb$. In you example $|overrightarrowa|=|overrightarrowb=1$ and $theta = fracpi2-phi$.
answered Aug 28 at 14:14
gandalf61
6,061522
6,061522
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
add a comment |Â
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, of course, it's just sin(phi). Scalar product! Okay, thank you, so to convert is the answer (or use the coordinate free definition of scalar product). If I was to convert to cartesian to cylindrical the answer would always be the same I guess?
â Defindun
Aug 28 at 15:06
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
Yes, but the formula for scalar product in cylindrical co-ordinates becomes more complex. It is $(r_1, theta_1, z_1) cdot (r_2, theta_2, z_2) = r_1r_2 cos(theta_1-theta_2) + z_1z_2$
â gandalf61
Aug 28 at 15:34
add a comment |Â
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