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Why did they define $hattheta=frachatztimeshatrsin theta$ and $hattheta=hatphitimes hatr?$
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I am seeing this approach first time in an Electrodynamic book. They have used straightforward formula.
From the figure, I could able to prove $$x=rsin theta cos phi,$$ $$y=rsin theta sin phi$$ and $$z=rcostheta$$
I can see that $hatr=fracxhati+yhatj+zhatkr.$ I can deduce the given expression of $hatr$. But, I am not able to deduce the other two expresions. Why did they define $hattheta=frachatztimeshatrsin theta$ and $hattheta=hatphitimes hatr?$ How $theta$ related to $hattheta$? How $phi$ related to $hatphi$?
calculus vector-analysis mathematical-physics
asked yesterday
N. Maneesh
2,4271823
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I am seeing this approach first time in an Electrodynamic book. They have used straightforward formula.
From the figure, I could able to prove $$x=rsin theta cos phi,$$ $$y=rsin theta sin phi$$ and $$z=rcostheta$$
I can see that $hatr=fracxhati+yhatj+zhatkr.$ I can deduce the given expression of $hatr$. But, I am not able to deduce the other two expresions. Why did they define $hattheta=frachatztimeshatrsin theta$ and $hattheta=hatphitimes hatr?$ How $theta$ related to $hattheta$? How $phi$ related to $hatphi$?
calculus vector-analysis mathematical-physics
asked yesterday
N. Maneesh
2,4271823
These are the unit vectors along the directions the point moves if you make an infinitesimal change in $phi$ or $theta$ respectively while holding the other spherical coordinates fixed. (More formally, if the point is $vec r = xhat x+yhat y+zhat z$, then $hatphi = dfracpartialvec r/partialphi$, and similarly for $hattheta$.)
– Rahul
yesterday
How $hatphi$ related to the angle $phi$ in the figure?
– N. Maneesh
yesterday
$hatphi$ is the unit vector along the direction the point moves if you make an infinitesimal change in the angle $phi$ while holding the other spherical coordinates $r$ and $theta$ fixed.
– Rahul
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am seeing this approach first time in an Electrodynamic book. They have used straightforward formula.
From the figure, I could able to prove $$x=rsin theta cos phi,$$ $$y=rsin theta sin phi$$ and $$z=rcostheta$$
I can see that $hatr=fracxhati+yhatj+zhatkr.$ I can deduce the given expression of $hatr$. But, I am not able to deduce the other two expresions. Why did they define $hattheta=frachatztimeshatrsin theta$ and $hattheta=hatphitimes hatr?$ How $theta$ related to $hattheta$? How $phi$ related to $hatphi$?
calculus vector-analysis mathematical-physics
asked yesterday
N. Maneesh
2,4271823
I am seeing this approach first time in an Electrodynamic book. They have used straightforward formula.
From the figure, I could able to prove $$x=rsin theta cos phi,$$ $$y=rsin theta sin phi$$ and $$z=rcostheta$$
I can see that $hatr=fracxhati+yhatj+zhatkr.$ I can deduce the given expression of $hatr$. But, I am not able to deduce the other two expresions. Why did they define $hattheta=frachatztimeshatrsin theta$ and $hattheta=hatphitimes hatr?$ How $theta$ related to $hattheta$? How $phi$ related to $hatphi$?
calculus vector-analysis mathematical-physics
asked yesterday
N. Maneesh
2,4271823
edited yesterday
asked yesterday
N. Maneesh
2,4271823
asked yesterday
N. Maneesh
2,4271823
asked yesterday
N. Maneesh
2,4271823
2,4271823
These are the unit vectors along the directions the point moves if you make an infinitesimal change in $phi$ or $theta$ respectively while holding the other spherical coordinates fixed. (More formally, if the point is $vec r = xhat x+yhat y+zhat z$, then $hatphi = dfracpartialvec r/partialphi$, and similarly for $hattheta$.)
– Rahul
yesterday
How $hatphi$ related to the angle $phi$ in the figure?
– N. Maneesh
yesterday
$hatphi$ is the unit vector along the direction the point moves if you make an infinitesimal change in the angle $phi$ while holding the other spherical coordinates $r$ and $theta$ fixed.
– Rahul
yesterday
add a comment |Â
These are the unit vectors along the directions the point moves if you make an infinitesimal change in $phi$ or $theta$ respectively while holding the other spherical coordinates fixed. (More formally, if the point is $vec r = xhat x+yhat y+zhat z$, then $hatphi = dfracpartialvec r/partialphi$, and similarly for $hattheta$.)
– Rahul
yesterday
How $hatphi$ related to the angle $phi$ in the figure?
– N. Maneesh
yesterday
$hatphi$ is the unit vector along the direction the point moves if you make an infinitesimal change in the angle $phi$ while holding the other spherical coordinates $r$ and $theta$ fixed.
– Rahul
yesterday
These are the unit vectors along the directions the point moves if you make an infinitesimal change in $phi$ or $theta$ respectively while holding the other spherical coordinates fixed. (More formally, if the point is $vec r = xhat x+yhat y+zhat z$, then $hatphi = dfracpartialvec r/partialphi$, and similarly for $hattheta$.)
– Rahul
yesterday
These are the unit vectors along the directions the point moves if you make an infinitesimal change in $phi$ or $theta$ respectively while holding the other spherical coordinates fixed. (More formally, if the point is $vec r = xhat x+yhat y+zhat z$, then $hatphi = dfracpartialvec r/partialphi$, and similarly for $hattheta$.)
– Rahul
yesterday
How $hatphi$ related to the angle $phi$ in the figure?
– N. Maneesh
yesterday
How $hatphi$ related to the angle $phi$ in the figure?
– N. Maneesh
yesterday
$hatphi$ is the unit vector along the direction the point moves if you make an infinitesimal change in the angle $phi$ while holding the other spherical coordinates $r$ and $theta$ fixed.
– Rahul
yesterday
$hatphi$ is the unit vector along the direction the point moves if you make an infinitesimal change in the angle $phi$ while holding the other spherical coordinates $r$ and $theta$ fixed.
– Rahul
yesterday
add a comment |Â
1 Answer
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We define them as the covariant basis vectors in terms of the spherical coordinates $$q^1 = r, qquad q^2 = theta qquad q^3 = phi$$
The new basis vectors are tangential to a coordinate line.
In other words a line where one $q^alpha$ is hold variable and the rest of the $q^i$ is hold constant.
The new basis vectors $vecg_alpha$ are derived through
$$left(vecg_alpharight)^i=fracpartial x^ipartial q^alpha frac1partial vecx / partial q^alpha $$
where $x^i$ are the Cartesian coordinates in spherical coordinates.
Using this we can derive $hattheta, hatphi, hatr$ which have the property that they are perpendicular to one another.
The same property or vectors can be derived by the way your book defined it.
Since the cross product of two vectors is perpendicular two those two vectors, $hattheta$ should be clear.
Now, can you see the idea behind $frachatz times hatrtextsintheta$?
answered yesterday
P-A
6319
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
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
We define them as the covariant basis vectors in terms of the spherical coordinates $$q^1 = r, qquad q^2 = theta qquad q^3 = phi$$
The new basis vectors are tangential to a coordinate line.
In other words a line where one $q^alpha$ is hold variable and the rest of the $q^i$ is hold constant.
The new basis vectors $vecg_alpha$ are derived through
$$left(vecg_alpharight)^i=fracpartial x^ipartial q^alpha frac1partial vecx / partial q^alpha $$
where $x^i$ are the Cartesian coordinates in spherical coordinates.
Using this we can derive $hattheta, hatphi, hatr$ which have the property that they are perpendicular to one another.
The same property or vectors can be derived by the way your book defined it.
Since the cross product of two vectors is perpendicular two those two vectors, $hattheta$ should be clear.
Now, can you see the idea behind $frachatz times hatrtextsintheta$?
answered yesterday
P-A
6319
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
add a comment |Â
up vote
0
down vote
We define them as the covariant basis vectors in terms of the spherical coordinates $$q^1 = r, qquad q^2 = theta qquad q^3 = phi$$
The new basis vectors are tangential to a coordinate line.
In other words a line where one $q^alpha$ is hold variable and the rest of the $q^i$ is hold constant.
The new basis vectors $vecg_alpha$ are derived through
$$left(vecg_alpharight)^i=fracpartial x^ipartial q^alpha frac1partial vecx / partial q^alpha $$
where $x^i$ are the Cartesian coordinates in spherical coordinates.
Using this we can derive $hattheta, hatphi, hatr$ which have the property that they are perpendicular to one another.
The same property or vectors can be derived by the way your book defined it.
Since the cross product of two vectors is perpendicular two those two vectors, $hattheta$ should be clear.
Now, can you see the idea behind $frachatz times hatrtextsintheta$?
answered yesterday
P-A
6319
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We define them as the covariant basis vectors in terms of the spherical coordinates $$q^1 = r, qquad q^2 = theta qquad q^3 = phi$$
The new basis vectors are tangential to a coordinate line.
In other words a line where one $q^alpha$ is hold variable and the rest of the $q^i$ is hold constant.
The new basis vectors $vecg_alpha$ are derived through
$$left(vecg_alpharight)^i=fracpartial x^ipartial q^alpha frac1partial vecx / partial q^alpha $$
where $x^i$ are the Cartesian coordinates in spherical coordinates.
Using this we can derive $hattheta, hatphi, hatr$ which have the property that they are perpendicular to one another.
The same property or vectors can be derived by the way your book defined it.
Since the cross product of two vectors is perpendicular two those two vectors, $hattheta$ should be clear.
Now, can you see the idea behind $frachatz times hatrtextsintheta$?
answered yesterday
P-A
6319
We define them as the covariant basis vectors in terms of the spherical coordinates $$q^1 = r, qquad q^2 = theta qquad q^3 = phi$$
The new basis vectors are tangential to a coordinate line.
In other words a line where one $q^alpha$ is hold variable and the rest of the $q^i$ is hold constant.
The new basis vectors $vecg_alpha$ are derived through
$$left(vecg_alpharight)^i=fracpartial x^ipartial q^alpha frac1partial vecx / partial q^alpha $$
where $x^i$ are the Cartesian coordinates in spherical coordinates.
Using this we can derive $hattheta, hatphi, hatr$ which have the property that they are perpendicular to one another.
The same property or vectors can be derived by the way your book defined it.
Since the cross product of two vectors is perpendicular two those two vectors, $hattheta$ should be clear.
Now, can you see the idea behind $frachatz times hatrtextsintheta$?
answered yesterday
P-A
6319
edited 20 hours ago
answered yesterday
P-A
6319
answered yesterday
P-A
6319
answered yesterday
P-A
6319
6319
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
add a comment |Â
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The new basis vectors are tangential to a coordinate line. I don't understand this statement. can you explain bit?
– N. Maneesh
yesterday
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
The basis vectors are the normalized tangent vectors to the coordinate lines. So like rahul said $fracpartial vecr / partial q^alpha$. Is there an expression that is unclear?
– P-A
20 hours ago
add a comment |Â
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These are the unit vectors along the directions the point moves if you make an infinitesimal change in $phi$ or $theta$ respectively while holding the other spherical coordinates fixed. (More formally, if the point is $vec r = xhat x+yhat y+zhat z$, then $hatphi = dfracpartialvec r/partialphi$, and similarly for $hattheta$.)
– Rahul
yesterday
How $hatphi$ related to the angle $phi$ in the figure?
– N. Maneesh
yesterday
$hatphi$ is the unit vector along the direction the point moves if you make an infinitesimal change in the angle $phi$ while holding the other spherical coordinates $r$ and $theta$ fixed.
– Rahul
yesterday