Motion equation given the following function, is this a potential function?

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I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.



I'm considering the vector field



$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$



($k$ positive constant). I've checked the condition



$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.



Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.



The other question is I have the potential function



$$
U(x,y,z) = k sqrtx^2 + y^2
$$



and I clearly have



$$
nabla U = F
$$



How do I derive the motion equation? My intuition is that since



$$
F = ma = m fracd^2xdt^2
$$



I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?










share|cite|improve this question





















  • You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
    – Botond
    Sep 3 at 10:42











  • Your vector field is undefined at the origin, so it shouldn’t be a surprise that none of the derivatives exist there, either.
    – amd
    Sep 3 at 21:17










  • It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
    – user8469759
    Sep 4 at 10:30














up vote
0
down vote

favorite












I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.



I'm considering the vector field



$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$



($k$ positive constant). I've checked the condition



$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.



Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.



The other question is I have the potential function



$$
U(x,y,z) = k sqrtx^2 + y^2
$$



and I clearly have



$$
nabla U = F
$$



How do I derive the motion equation? My intuition is that since



$$
F = ma = m fracd^2xdt^2
$$



I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?










share|cite|improve this question





















  • You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
    – Botond
    Sep 3 at 10:42











  • Your vector field is undefined at the origin, so it shouldn’t be a surprise that none of the derivatives exist there, either.
    – amd
    Sep 3 at 21:17










  • It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
    – user8469759
    Sep 4 at 10:30












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.



I'm considering the vector field



$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$



($k$ positive constant). I've checked the condition



$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.



Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.



The other question is I have the potential function



$$
U(x,y,z) = k sqrtx^2 + y^2
$$



and I clearly have



$$
nabla U = F
$$



How do I derive the motion equation? My intuition is that since



$$
F = ma = m fracd^2xdt^2
$$



I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?










share|cite|improve this question













I'm a bit rusty in this stuff.
I'm trying to design an energy function that would allow me to derive a motion equation.



I'm considering the vector field



$$
F(x,y,z) = kleft( fracxsqrtx^2+y^2,fracysqrtx^2+y^2,0 right)^T
$$



($k$ positive constant). I've checked the condition



$$
curl(F) = 0
$$
for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.



Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.



The other question is I have the potential function



$$
U(x,y,z) = k sqrtx^2 + y^2
$$



and I clearly have



$$
nabla U = F
$$



How do I derive the motion equation? My intuition is that since



$$
F = ma = m fracd^2xdt^2
$$



I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?







calculus physics vector-fields






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share|cite|improve this question











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asked Sep 3 at 10:01









user8469759

1,1801515




1,1801515











  • You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
    – Botond
    Sep 3 at 10:42











  • Your vector field is undefined at the origin, so it shouldn’t be a surprise that none of the derivatives exist there, either.
    – amd
    Sep 3 at 21:17










  • It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
    – user8469759
    Sep 4 at 10:30
















  • You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
    – Botond
    Sep 3 at 10:42











  • Your vector field is undefined at the origin, so it shouldn’t be a surprise that none of the derivatives exist there, either.
    – amd
    Sep 3 at 21:17










  • It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
    – user8469759
    Sep 4 at 10:30















You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
– Botond
Sep 3 at 10:42





You are in 3D, so the motion equation is $-textgrad(U)=underlineF=munderlinea=munderlineddotr$: $$k ,textgrad(r)+munderlineddotr=0$$.
– Botond
Sep 3 at 10:42













Your vector field is undefined at the origin, so it shouldn’t be a surprise that none of the derivatives exist there, either.
– amd
Sep 3 at 21:17




Your vector field is undefined at the origin, so it shouldn’t be a surprise that none of the derivatives exist there, either.
– amd
Sep 3 at 21:17












It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
– user8469759
Sep 4 at 10:30




It's not a surprise and I was indeed puzzled, but the field is conservative anyway right?
– user8469759
Sep 4 at 10:30










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.






share|cite|improve this answer






















  • But is my function $U$ a valid potential?
    – user8469759
    Sep 3 at 13:02










  • Sure, why shouldn't it be?
    – denklo
    Sep 3 at 13:08










  • Thank you, If I post a more complicated question do you think you can give me some insight on that one?
    – user8469759
    Sep 3 at 13:54










  • @user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
    – denklo
    Sep 3 at 14:04










  • Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
    – user8469759
    Sep 3 at 14:33











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.






share|cite|improve this answer






















  • But is my function $U$ a valid potential?
    – user8469759
    Sep 3 at 13:02










  • Sure, why shouldn't it be?
    – denklo
    Sep 3 at 13:08










  • Thank you, If I post a more complicated question do you think you can give me some insight on that one?
    – user8469759
    Sep 3 at 13:54










  • @user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
    – denklo
    Sep 3 at 14:04










  • Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
    – user8469759
    Sep 3 at 14:33















up vote
2
down vote



accepted










You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.






share|cite|improve this answer






















  • But is my function $U$ a valid potential?
    – user8469759
    Sep 3 at 13:02










  • Sure, why shouldn't it be?
    – denklo
    Sep 3 at 13:08










  • Thank you, If I post a more complicated question do you think you can give me some insight on that one?
    – user8469759
    Sep 3 at 13:54










  • @user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
    – denklo
    Sep 3 at 14:04










  • Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
    – user8469759
    Sep 3 at 14:33













up vote
2
down vote



accepted







up vote
2
down vote



accepted






You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.






share|cite|improve this answer














You should use the cylindric coordinates $phi$, $r$, $z$, where
$$[x,y,z]= [rcos(phi),rsin(phi),z].$$
We then have for the kinetic energy
$$
T = fracm2left(r^2dotphi^2 + dotr^2 + dotz^2right)
$$
and for the potential simply
$$
U = k r.
$$
Now apply Euler-Lagrange:
$$
L = T - V
$$
and
$$
d_tpartial_dotqL = partial_q L$$
where $qin lbracephi,r,zrbrace$. This will give you your equations of motion:
$$
ddotz = 0,
$$
conservation of momentum in $z$-direction,
$$
rddotphi + 2dotrdotphi = 0 = dotJ,
$$
conservation of angluar momentum $J = mr^2dotphi$, and finally
$$
mddotr = mrdotphi^2-k = fracJ^2mr^3 - k.
$$
$J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 3 at 13:00

























answered Sep 3 at 12:17









denklo

2335




2335











  • But is my function $U$ a valid potential?
    – user8469759
    Sep 3 at 13:02










  • Sure, why shouldn't it be?
    – denklo
    Sep 3 at 13:08










  • Thank you, If I post a more complicated question do you think you can give me some insight on that one?
    – user8469759
    Sep 3 at 13:54










  • @user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
    – denklo
    Sep 3 at 14:04










  • Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
    – user8469759
    Sep 3 at 14:33

















  • But is my function $U$ a valid potential?
    – user8469759
    Sep 3 at 13:02










  • Sure, why shouldn't it be?
    – denklo
    Sep 3 at 13:08










  • Thank you, If I post a more complicated question do you think you can give me some insight on that one?
    – user8469759
    Sep 3 at 13:54










  • @user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
    – denklo
    Sep 3 at 14:04










  • Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
    – user8469759
    Sep 3 at 14:33
















But is my function $U$ a valid potential?
– user8469759
Sep 3 at 13:02




But is my function $U$ a valid potential?
– user8469759
Sep 3 at 13:02












Sure, why shouldn't it be?
– denklo
Sep 3 at 13:08




Sure, why shouldn't it be?
– denklo
Sep 3 at 13:08












Thank you, If I post a more complicated question do you think you can give me some insight on that one?
– user8469759
Sep 3 at 13:54




Thank you, If I post a more complicated question do you think you can give me some insight on that one?
– user8469759
Sep 3 at 13:54












@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
– denklo
Sep 3 at 14:04




@user8469759 feel free to post whatever. If i can't answer, maybe someone else may do so ;).
– denklo
Sep 3 at 14:04












Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
– user8469759
Sep 3 at 14:33





Posted math.stackexchange.com/questions/2903927/…. If you have time please have a look.
– user8469759
Sep 3 at 14:33


















 

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