Basis vectors as a differentiation of position vector
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Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$
position vector $vec {R}=r hat{r}$.
That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$
But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$
Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$
Where am I wrong? Why is r appearing?
vector-spaces tensors
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Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$
position vector $vec {R}=r hat{r}$.
That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$
But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$
Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$
Where am I wrong? Why is r appearing?
vector-spaces tensors
$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26
Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29
add a comment |
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0
down vote
favorite
up vote
0
down vote
favorite
Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$
position vector $vec {R}=r hat{r}$.
That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$
But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$
Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$
Where am I wrong? Why is r appearing?
vector-spaces tensors
Basis vectors are defined as $vec {E_i}=vec{E_i}(x_1,x_2,x_3)=frac{partialvec{r}}{partial x^i}$ i=1,2,3.
In spherical coordinate system $x^1=r, x^2=theta, x^3=phi$
position vector $vec {R}=r hat{r}$.
That means $hat {r}=frac{partialvec{R}}{partial r}=hat {r}$
But $hat {theta}=frac{partialvec{R}}{partial theta}=rhat {theta}$
Similarly $hat {phi}=frac{partialvec{R}}{partial phi}=rhat {phi}$
Where am I wrong? Why is r appearing?
vector-spaces tensors
vector-spaces tensors
asked Nov 26 at 9:14
Asit Srivastava
256
256
$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26
Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29
add a comment |
$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26
Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29
$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26
$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26
Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29
Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29
add a comment |
2 Answers
2
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0
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You're missing the scale factors
$$
hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
$$
As an example
$$
{bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
$$
So
$$
frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
$$
The scale factor is just
$$
h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
$$
So the unit vector is
$$
hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
$$
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
add a comment |
up vote
0
down vote
In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing
$frac{partialvec{R}}{partial x}=hat{x}$ etc.
However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So
$frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$
$Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$
and similarly for $phi$:
$frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$
$Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You're missing the scale factors
$$
hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
$$
As an example
$$
{bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
$$
So
$$
frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
$$
The scale factor is just
$$
h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
$$
So the unit vector is
$$
hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
$$
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
add a comment |
up vote
0
down vote
accepted
You're missing the scale factors
$$
hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
$$
As an example
$$
{bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
$$
So
$$
frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
$$
The scale factor is just
$$
h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
$$
So the unit vector is
$$
hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
$$
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You're missing the scale factors
$$
hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
$$
As an example
$$
{bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
$$
So
$$
frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
$$
The scale factor is just
$$
h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
$$
So the unit vector is
$$
hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
$$
You're missing the scale factors
$$
hat{e}_alpha = frac{1}{h^alpha}frac{partial {bf x}}{partial u^alpha}
$$
As an example
$$
{bf x} = pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta}
$$
So
$$
frac{partial {bf x}}{partial theta} = frac{partial}{partial theta}pmatrix{rsinthetacosphi \ rsinthetasinphi \ r costheta} = pmatrix{rcosthetasinphi \ rcosthetacosphi \ -rsintheta}
$$
The scale factor is just
$$
h^2 = [r^2cos^2theta(sin^2phi + cos^2phi) + r^2sin^2theta]^{1/2} = r
$$
So the unit vector is
$$
hat{e}_2 = pmatrix{costhetasinphi \ costhetacosphi \ -sintheta}
$$
answered Nov 26 at 9:29
caverac
12.4k21027
12.4k21027
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
add a comment |
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
But what about the way of defining basis vectors? As mentioned above it doesn't mention scale factors.
– Asit Srivastava
Nov 26 at 11:07
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
@AsitSrivastava The vectors are defined through the derivatives as you said, but they are not unitary. You need to scale them first, hence the scale factors
– caverac
Nov 26 at 11:18
add a comment |
up vote
0
down vote
In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing
$frac{partialvec{R}}{partial x}=hat{x}$ etc.
However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So
$frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$
$Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$
and similarly for $phi$:
$frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$
$Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$
add a comment |
up vote
0
down vote
In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing
$frac{partialvec{R}}{partial x}=hat{x}$ etc.
However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So
$frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$
$Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$
and similarly for $phi$:
$frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$
$Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$
add a comment |
up vote
0
down vote
up vote
0
down vote
In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing
$frac{partialvec{R}}{partial x}=hat{x}$ etc.
However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So
$frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$
$Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$
and similarly for $phi$:
$frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$
$Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$
In Cartesian co-ordinates the unit vectors parallel to the co-ordinate axes $hat{x}, hat{y}, hat{z}$ do not change with position. So if we have $vec{R}=xhat{x}+yhat{y}+zhat{z}$ then we are used to seeing
$frac{partialvec{R}}{partial x}=hat{x}$ etc.
However, in other co-ordinate systems this is not true. In particular, in spherical co-ordinates the direction of the $hat{r}$ vector changes depending on $theta$ and $phi$ - so we should more properly write $hat{r}$ as $hat{r}(theta, phi)$. So
$frac{partialvec{R}}{partial theta}=rfrac{partialvec{r}}{partial theta}=r hat{theta}$
$Rightarrow hat{theta} =frac{1}{r}frac{partialvec{R}}{partial theta}$
and similarly for $phi$:
$frac{partialvec{R}}{partial phi}=rfrac{partialvec{r}}{partial phi}=rsin theta hat{phi}$
$Rightarrow hat{phi} =frac{1}{r sin theta}frac{partialvec{R}}{partial phi}$
answered Nov 26 at 10:16
gandalf61
7,365623
7,365623
add a comment |
add a comment |
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$hat{r} = sin (theta) cos (phi) hat{x} + sin (theta) sin(phi) hat{y} + cos (theta) hat{z}$
– user42298
Nov 26 at 9:26
Differentiating this gives what we need. But we are differentiating position vector and it doesn't give the answer.
– Asit Srivastava
Nov 26 at 9:29