proving the laplacian of a vector in cylindrical coordnates
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I am proving the following identity for the laplacian of a vector $vecv=<v_r,v_theta,v_z>$ in cylindrical coordinates:
$$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2+frac1rfracpartial v_rpartial r-frac2r^2fracpartial v_thetapartial theta -fracv_rr^2right )vece_r \ + left (fracpartial^2 v_thetapartial r^2+frac1r^2fracpartial^2 v_thetapartial theta^2+fracpartial^2 v_thetapartial z^2+frac1rfracpartial v_thetapartial r+frac2r^2fracpartial v_rpartial theta-fracv_thetar^2 right )vece_theta \ left( fracpartial^2 v_zpartial r^2+frac1r^2fracpartial^2 v_zpartial theta^2+frac1rfracpartial v_zpartial r+fracpartial^2 v_zpartial z^2 right)vece_z$$ I am able to derive the following identity for the Laplacian operator in cylindrical coordinates $$nabla^2=fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 $$. So to prove the desired identity, $$nabla^2 vecv=left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r+v_theta vece_theta+v_zvece_z) \
= left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_thetavece_theta)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_zvece_z)$$. And upon distributing the vector components to the operator I finally get $$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1rfracpartial v_rpartial r+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2-fracv_rr^2 right)vece_r \
+left( fracpartial^2 v_thetapartial r^2+frac1rfracpartial v_thetapartial r+frac1r^2fracpartial^2 v_thetapartial theta^2-fracv_thetar^2+fracpartial^2 v_thetapartial z^2 right)vece_theta \
left( fracpartial^2 v_zpartial r^2+frac1rfracpartial v_zpartial r+frac1r^2fracpartial^2 v_zpartial theta^2+fracpartial^2 v_zpartial z^2 right)vece_z$$ which is not the same with the identity. I am confused how the $-frac2r^2fracpartial v_thetapartial theta$ and $frac2r^2fracpartial v_rpartial theta$ appeared in the $vece_r$ and $vece_theta$ components, respectively. Where did I go wrong? Need help...thanks
vector-analysis
asked Feb 24 '15 at 18:41
james25
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I am proving the following identity for the laplacian of a vector $vecv=<v_r,v_theta,v_z>$ in cylindrical coordinates:
$$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2+frac1rfracpartial v_rpartial r-frac2r^2fracpartial v_thetapartial theta -fracv_rr^2right )vece_r \ + left (fracpartial^2 v_thetapartial r^2+frac1r^2fracpartial^2 v_thetapartial theta^2+fracpartial^2 v_thetapartial z^2+frac1rfracpartial v_thetapartial r+frac2r^2fracpartial v_rpartial theta-fracv_thetar^2 right )vece_theta \ left( fracpartial^2 v_zpartial r^2+frac1r^2fracpartial^2 v_zpartial theta^2+frac1rfracpartial v_zpartial r+fracpartial^2 v_zpartial z^2 right)vece_z$$ I am able to derive the following identity for the Laplacian operator in cylindrical coordinates $$nabla^2=fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 $$. So to prove the desired identity, $$nabla^2 vecv=left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r+v_theta vece_theta+v_zvece_z) \
= left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_thetavece_theta)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_zvece_z)$$. And upon distributing the vector components to the operator I finally get $$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1rfracpartial v_rpartial r+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2-fracv_rr^2 right)vece_r \
+left( fracpartial^2 v_thetapartial r^2+frac1rfracpartial v_thetapartial r+frac1r^2fracpartial^2 v_thetapartial theta^2-fracv_thetar^2+fracpartial^2 v_thetapartial z^2 right)vece_theta \
left( fracpartial^2 v_zpartial r^2+frac1rfracpartial v_zpartial r+frac1r^2fracpartial^2 v_zpartial theta^2+fracpartial^2 v_zpartial z^2 right)vece_z$$ which is not the same with the identity. I am confused how the $-frac2r^2fracpartial v_thetapartial theta$ and $frac2r^2fracpartial v_rpartial theta$ appeared in the $vece_r$ and $vece_theta$ components, respectively. Where did I go wrong? Need help...thanks
vector-analysis
asked Feb 24 '15 at 18:41
james25
7818
2
The vectors $e_r, e_z$ and $e_theta$ are not constant when $r, theta, z$ vary. (This is a distinctive trait of Cartesian coordinates). You should differentiate those vectors as well. mathworld.wolfram.com/CylindricalCoordinates.html
– Giuseppe Negro
Oct 5 '15 at 16:23
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I am proving the following identity for the laplacian of a vector $vecv=<v_r,v_theta,v_z>$ in cylindrical coordinates:
$$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2+frac1rfracpartial v_rpartial r-frac2r^2fracpartial v_thetapartial theta -fracv_rr^2right )vece_r \ + left (fracpartial^2 v_thetapartial r^2+frac1r^2fracpartial^2 v_thetapartial theta^2+fracpartial^2 v_thetapartial z^2+frac1rfracpartial v_thetapartial r+frac2r^2fracpartial v_rpartial theta-fracv_thetar^2 right )vece_theta \ left( fracpartial^2 v_zpartial r^2+frac1r^2fracpartial^2 v_zpartial theta^2+frac1rfracpartial v_zpartial r+fracpartial^2 v_zpartial z^2 right)vece_z$$ I am able to derive the following identity for the Laplacian operator in cylindrical coordinates $$nabla^2=fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 $$. So to prove the desired identity, $$nabla^2 vecv=left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r+v_theta vece_theta+v_zvece_z) \
= left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_thetavece_theta)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_zvece_z)$$. And upon distributing the vector components to the operator I finally get $$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1rfracpartial v_rpartial r+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2-fracv_rr^2 right)vece_r \
+left( fracpartial^2 v_thetapartial r^2+frac1rfracpartial v_thetapartial r+frac1r^2fracpartial^2 v_thetapartial theta^2-fracv_thetar^2+fracpartial^2 v_thetapartial z^2 right)vece_theta \
left( fracpartial^2 v_zpartial r^2+frac1rfracpartial v_zpartial r+frac1r^2fracpartial^2 v_zpartial theta^2+fracpartial^2 v_zpartial z^2 right)vece_z$$ which is not the same with the identity. I am confused how the $-frac2r^2fracpartial v_thetapartial theta$ and $frac2r^2fracpartial v_rpartial theta$ appeared in the $vece_r$ and $vece_theta$ components, respectively. Where did I go wrong? Need help...thanks
vector-analysis
asked Feb 24 '15 at 18:41
james25
7818
I am proving the following identity for the laplacian of a vector $vecv=<v_r,v_theta,v_z>$ in cylindrical coordinates:
$$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2+frac1rfracpartial v_rpartial r-frac2r^2fracpartial v_thetapartial theta -fracv_rr^2right )vece_r \ + left (fracpartial^2 v_thetapartial r^2+frac1r^2fracpartial^2 v_thetapartial theta^2+fracpartial^2 v_thetapartial z^2+frac1rfracpartial v_thetapartial r+frac2r^2fracpartial v_rpartial theta-fracv_thetar^2 right )vece_theta \ left( fracpartial^2 v_zpartial r^2+frac1r^2fracpartial^2 v_zpartial theta^2+frac1rfracpartial v_zpartial r+fracpartial^2 v_zpartial z^2 right)vece_z$$ I am able to derive the following identity for the Laplacian operator in cylindrical coordinates $$nabla^2=fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 $$. So to prove the desired identity, $$nabla^2 vecv=left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r+v_theta vece_theta+v_zvece_z) \
= left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_rvece_r)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_thetavece_theta)+ left(fracpartial^2partial r^2+frac1rfracpartialpartial r+frac1r^2fracpartial^2partial theta^2+fracpartial^2z^2 right)(v_zvece_z)$$. And upon distributing the vector components to the operator I finally get $$nabla^2 vecv=left( fracpartial^2 v_rpartial r^2+frac1rfracpartial v_rpartial r+frac1r^2fracpartial^2 v_rpartial theta^2+fracpartial^2 v_rpartial z^2-fracv_rr^2 right)vece_r \
+left( fracpartial^2 v_thetapartial r^2+frac1rfracpartial v_thetapartial r+frac1r^2fracpartial^2 v_thetapartial theta^2-fracv_thetar^2+fracpartial^2 v_thetapartial z^2 right)vece_theta \
left( fracpartial^2 v_zpartial r^2+frac1rfracpartial v_zpartial r+frac1r^2fracpartial^2 v_zpartial theta^2+fracpartial^2 v_zpartial z^2 right)vece_z$$ which is not the same with the identity. I am confused how the $-frac2r^2fracpartial v_thetapartial theta$ and $frac2r^2fracpartial v_rpartial theta$ appeared in the $vece_r$ and $vece_theta$ components, respectively. Where did I go wrong? Need help...thanks
vector-analysis
asked Feb 24 '15 at 18:41
james25
7818
asked Feb 24 '15 at 18:41
james25
7818
asked Feb 24 '15 at 18:41
james25
7818
asked Feb 24 '15 at 18:41
james25
7818
7818
2
The vectors $e_r, e_z$ and $e_theta$ are not constant when $r, theta, z$ vary. (This is a distinctive trait of Cartesian coordinates). You should differentiate those vectors as well. mathworld.wolfram.com/CylindricalCoordinates.html
– Giuseppe Negro
Oct 5 '15 at 16:23
add a comment |Â
2
The vectors $e_r, e_z$ and $e_theta$ are not constant when $r, theta, z$ vary. (This is a distinctive trait of Cartesian coordinates). You should differentiate those vectors as well. mathworld.wolfram.com/CylindricalCoordinates.html
– Giuseppe Negro
Oct 5 '15 at 16:23
2
2
The vectors $e_r, e_z$ and $e_theta$ are not constant when $r, theta, z$ vary. (This is a distinctive trait of Cartesian coordinates). You should differentiate those vectors as well. mathworld.wolfram.com/CylindricalCoordinates.html
– Giuseppe Negro
Oct 5 '15 at 16:23
The vectors $e_r, e_z$ and $e_theta$ are not constant when $r, theta, z$ vary. (This is a distinctive trait of Cartesian coordinates). You should differentiate those vectors as well. mathworld.wolfram.com/CylindricalCoordinates.html
– Giuseppe Negro
Oct 5 '15 at 16:23
add a comment |Â
1 Answer
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The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
answered Nov 11 '17 at 16:17
katiousa
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
answered Nov 11 '17 at 16:17
katiousa
1
add a comment |Â
up vote
0
down vote
The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
answered Nov 11 '17 at 16:17
katiousa
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
answered Nov 11 '17 at 16:17
katiousa
1
The product rule for second order differentiation is $(fg)'' = f''g + 2f'g'+ fg''$. You simply omitted the middle value.
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
answered Nov 11 '17 at 16:17
katiousa
1
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
edited Nov 11 '17 at 16:53
TheSimpliFire
9,79461952
9,79461952
answered Nov 11 '17 at 16:17
katiousa
1
answered Nov 11 '17 at 16:17
katiousa
1
answered Nov 11 '17 at 16:17
katiousa
1
1
add a comment |Â
add a comment |Â
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The vectors $e_r, e_z$ and $e_theta$ are not constant when $r, theta, z$ vary. (This is a distinctive trait of Cartesian coordinates). You should differentiate those vectors as well. mathworld.wolfram.com/CylindricalCoordinates.html
– Giuseppe Negro
Oct 5 '15 at 16:23