Why should the position vector be noted as $Rhat{R}$ in spherical polar coordinates?
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Why should the position vector be noted as $Rhat{R}$ in spherical polar coordinates? Now i did the calculation like this: $vec R = R sintheta cosphi hat{i} + R sintheta sinphi hat{j} + R costheta hat{k}$ so now i am manipulating the unit vectors. As :- $$hat{R}= frac{frac{partial vec{R}}{partial R}}{left|frac{partial vec{R}}{partial R}right|}=sintheta cosphi hat{i} + sintheta sinphi hat{j} + costheta hat{k}$$ by doing similiar calculations i found $hat{theta}=costheta cosphi hat{i} + costheta sinphi hat{j} -sinthetahat{k}$. Similarly I found $hat{phi}= cosphi hat{i} + sinphihat{j}$ now position vector can be written as $vec R= [vec R. hat{R}]hat{theta} + [vec R. hat{theta}]hat{theta} + [vec{R},hat{phi}] hat{phi}$. Which gives me $vec{R} = Rhat{R} + Rsintheta hat{phi}$ not $Rhat{R}$ now where i am misunderstanding or miscalculating ?
vectors vector-analysis spherical-coordinates
edited Jan 14 at 16:13
mechanodroid
28.7k62548
asked Jan 13 at 1:19
user187604user187604
285111
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add a comment |
$begingroup$
Why should the position vector be noted as $Rhat{R}$ in spherical polar coordinates? Now i did the calculation like this: $vec R = R sintheta cosphi hat{i} + R sintheta sinphi hat{j} + R costheta hat{k}$ so now i am manipulating the unit vectors. As :- $$hat{R}= frac{frac{partial vec{R}}{partial R}}{left|frac{partial vec{R}}{partial R}right|}=sintheta cosphi hat{i} + sintheta sinphi hat{j} + costheta hat{k}$$ by doing similiar calculations i found $hat{theta}=costheta cosphi hat{i} + costheta sinphi hat{j} -sinthetahat{k}$. Similarly I found $hat{phi}= cosphi hat{i} + sinphihat{j}$ now position vector can be written as $vec R= [vec R. hat{R}]hat{theta} + [vec R. hat{theta}]hat{theta} + [vec{R},hat{phi}] hat{phi}$. Which gives me $vec{R} = Rhat{R} + Rsintheta hat{phi}$ not $Rhat{R}$ now where i am misunderstanding or miscalculating ?
vectors vector-analysis spherical-coordinates
edited Jan 14 at 16:13
mechanodroid
28.7k62548
asked Jan 13 at 1:19
user187604user187604
285111
$endgroup$
add a comment |
$begingroup$
Why should the position vector be noted as $Rhat{R}$ in spherical polar coordinates? Now i did the calculation like this: $vec R = R sintheta cosphi hat{i} + R sintheta sinphi hat{j} + R costheta hat{k}$ so now i am manipulating the unit vectors. As :- $$hat{R}= frac{frac{partial vec{R}}{partial R}}{left|frac{partial vec{R}}{partial R}right|}=sintheta cosphi hat{i} + sintheta sinphi hat{j} + costheta hat{k}$$ by doing similiar calculations i found $hat{theta}=costheta cosphi hat{i} + costheta sinphi hat{j} -sinthetahat{k}$. Similarly I found $hat{phi}= cosphi hat{i} + sinphihat{j}$ now position vector can be written as $vec R= [vec R. hat{R}]hat{theta} + [vec R. hat{theta}]hat{theta} + [vec{R},hat{phi}] hat{phi}$. Which gives me $vec{R} = Rhat{R} + Rsintheta hat{phi}$ not $Rhat{R}$ now where i am misunderstanding or miscalculating ?
vectors vector-analysis spherical-coordinates
edited Jan 14 at 16:13
mechanodroid
28.7k62548
asked Jan 13 at 1:19
user187604user187604
285111
$endgroup$
Why should the position vector be noted as $Rhat{R}$ in spherical polar coordinates? Now i did the calculation like this: $vec R = R sintheta cosphi hat{i} + R sintheta sinphi hat{j} + R costheta hat{k}$ so now i am manipulating the unit vectors. As :- $$hat{R}= frac{frac{partial vec{R}}{partial R}}{left|frac{partial vec{R}}{partial R}right|}=sintheta cosphi hat{i} + sintheta sinphi hat{j} + costheta hat{k}$$ by doing similiar calculations i found $hat{theta}=costheta cosphi hat{i} + costheta sinphi hat{j} -sinthetahat{k}$. Similarly I found $hat{phi}= cosphi hat{i} + sinphihat{j}$ now position vector can be written as $vec R= [vec R. hat{R}]hat{theta} + [vec R. hat{theta}]hat{theta} + [vec{R},hat{phi}] hat{phi}$. Which gives me $vec{R} = Rhat{R} + Rsintheta hat{phi}$ not $Rhat{R}$ now where i am misunderstanding or miscalculating ?
vectors vector-analysis spherical-coordinates
vectors vector-analysis spherical-coordinates
edited Jan 14 at 16:13
mechanodroid
28.7k62548
asked Jan 13 at 1:19
user187604user187604
285111
edited Jan 14 at 16:13
mechanodroid
28.7k62548
asked Jan 13 at 1:19
user187604user187604
285111
edited Jan 14 at 16:13
mechanodroid
28.7k62548
edited Jan 14 at 16:13
mechanodroid
28.7k62548
edited Jan 14 at 16:13
mechanodroid
28.7k62548
28.7k62548
asked Jan 13 at 1:19
user187604user187604
285111
asked Jan 13 at 1:19
user187604user187604
285111
asked Jan 13 at 1:19
user187604user187604
285111
285111
add a comment |
add a comment |
1 Answer
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$begingroup$
You made a mistake when calculating $hat{phi}$. We have
$$frac{dhat{R}}{dphi} = -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}$$
so
$$hatphi = frac{frac{dhat{R}}{dphi}}{left|frac{dhat{R}}{dphi}right|} = frac{ -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}}{Rsintheta}= - sinphihat{i} + cosphihat{j}$$
Now we have
$$leftlangle vec{R}, hatphirightrangle = -Rsinthetacosphisinphi + Rsinthetasinphicosphi = 0$$
which gives the correct result $vec{R} = Rhat{R}$.
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
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add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
You made a mistake when calculating $hat{phi}$. We have
$$frac{dhat{R}}{dphi} = -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}$$
so
$$hatphi = frac{frac{dhat{R}}{dphi}}{left|frac{dhat{R}}{dphi}right|} = frac{ -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}}{Rsintheta}= - sinphihat{i} + cosphihat{j}$$
Now we have
$$leftlangle vec{R}, hatphirightrangle = -Rsinthetacosphisinphi + Rsinthetasinphicosphi = 0$$
which gives the correct result $vec{R} = Rhat{R}$.
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
$endgroup$
add a comment |
$begingroup$
You made a mistake when calculating $hat{phi}$. We have
$$frac{dhat{R}}{dphi} = -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}$$
so
$$hatphi = frac{frac{dhat{R}}{dphi}}{left|frac{dhat{R}}{dphi}right|} = frac{ -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}}{Rsintheta}= - sinphihat{i} + cosphihat{j}$$
Now we have
$$leftlangle vec{R}, hatphirightrangle = -Rsinthetacosphisinphi + Rsinthetasinphicosphi = 0$$
which gives the correct result $vec{R} = Rhat{R}$.
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
$endgroup$
add a comment |
$begingroup$
You made a mistake when calculating $hat{phi}$. We have
$$frac{dhat{R}}{dphi} = -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}$$
so
$$hatphi = frac{frac{dhat{R}}{dphi}}{left|frac{dhat{R}}{dphi}right|} = frac{ -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}}{Rsintheta}= - sinphihat{i} + cosphihat{j}$$
Now we have
$$leftlangle vec{R}, hatphirightrangle = -Rsinthetacosphisinphi + Rsinthetasinphicosphi = 0$$
which gives the correct result $vec{R} = Rhat{R}$.
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
$endgroup$
You made a mistake when calculating $hat{phi}$. We have
$$frac{dhat{R}}{dphi} = -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}$$
so
$$hatphi = frac{frac{dhat{R}}{dphi}}{left|frac{dhat{R}}{dphi}right|} = frac{ -Rsinthetasinphi hat{i} + Rsinthetacosphi hat{j}}{Rsintheta}= - sinphihat{i} + cosphihat{j}$$
Now we have
$$leftlangle vec{R}, hatphirightrangle = -Rsinthetacosphisinphi + Rsinthetasinphicosphi = 0$$
which gives the correct result $vec{R} = Rhat{R}$.
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
answered Jan 14 at 16:11
mechanodroidmechanodroid
28.7k62548
28.7k62548
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