Vector convergence in cartesian coordinates

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Yasmin Tamimi
Yasmin Tamimi 2014-9-25
编辑： Yasmin Tamimi 2014-9-27
Hey everyone,
How can I help my vector in cartesian coordinates converge to its resting point. The vector flips from +x-axis to the negative but instead of converging to it it keeps rotating around it!
Thanks in advance.
  2 个评论
Geoff Hayes
Geoff Hayes 2014-9-25
Yasmine - you might want to provide some context concerning what problem you are trying to solve. As well, posting some of the code may allow others to get an idea of what is happening and be in a better position to provide/offer some guidance.
Matt J
Matt J 2014-9-26
Yasmine Tamimi Commented:
Ok,, so I have a magnetization vector that according to some torques flips from +x-axis to -x-axis.. the problem I'm facing is that it doesn't converge to -x-axis it keeps rotating.. So, what shall I add to help it converge?
Steps = 20000; % Arrays size
time = 100e-9; % Total time
dt = (time/(Steps)); % Time step
Tswitch = 0:dt:time; % in seconds
for t = 1:Steps %Euler's Method
dtheta(t) = dt*((C1*G0)*(hMaz(t) + alpha*hMpol(t))) ; %3.1239
dphi(t) = dt*(((C1*G0)/sin(theta(t)))*(alpha*hMaz(t)- hMpol(t))); %1.57
theta(t+1) = theta(t) + dtheta(t) ; %3.1239
phi(t+1) = phi(t) + dphi(t); %1.57
Mi(t+1) = sin(theta(t+1))*cos(phi(t+1));
Mj(t+1) = sin(theta(t+1))*sin(phi(t+1));
Mk(t+1) = cos(theta(t+1));
u(t+1) = (EAi*Mi(t+1))+(EAj*Mj(t+1))+(EAk*Mk(t+1)); % dot prod. of M and EA.
Anis(t+1) = acos(u(t+1));
epoli(t+1) = cos(theta(t+1))*cos(phi(t+1));
epolj(t+1) = cos(theta(t+1))*sin(phi(t+1));
epolk(t+1) = -sin(theta(t+1));
eazi(t+1) = -sin(phi(t+1));
eazj(t+1) = cos(phi(t+1));
eazk(t+1) = 0 ;
ddu_acos(t+1) = -1/sqrt(1-(u(t+1)^2));
du_dMpol(t+1) = (EAi*cos(theta(t+1))*cos(phi(t+1))+EAj*cos(theta(t+1))*sin(phi(t+1))- EAk*sin(theta(t+1)));
du_dMaz(t+1) = sin(theta(t+1))*(EAj*cos(phi(t+1)) - EAi*sin(phi(t+1)));
dd_Anis(t+1) = Ku*Vol*sin(2*Anis(t+1));
g(t+1) = (-4 + ((1+P)^3)*(3+u(t+1))/(4*(P^1.5))) ;
G(t+1) = 1/g(t+1) ;
Ffi(t+1) = EAj*cos(theta(t+1)) - EAk*sin(theta(t+1))*sin(phi(t+1)) ;
Ffj(t+1) = -EAi*cos(theta(t+1)) + EAk*sin(theta(t+1))*cos(phi(t+1)) ;
Ffk(t+1) = EAi*sin(theta(t+1))*sin(phi(t+1))- EAj*sin(theta(t+1))*cos(phi(t+1)) ;
hMpol(t+1) = - (dd_Anis(t+1) * ddu_acos(t+1) * du_dMpol(t+1))/C2 - ((Is*P_hbar/(2*P_Q*g(t+1)*C2))*((epoli(t+1)*Ffi(t+1))+(epolj(t+1)*Ffj(t+1))+(epolk(t+1)*Ffk(t+1))))- Ms*sin(2*theta(t+1))*((Nx-Nz)+ (Ny-Nx)*(sin(phi(t+1))^2))+((hx*cos(theta(t+1))*cos(phi(t+1))+hy*cos(theta(t+1))*sin(phi(t+1))-hz*sin(theta(t+1))));
hMaz(t+1) = - (dd_Anis(t+1) * ddu_acos(t+1) * du_dMaz(t+1))/(C2*sin(theta(t+1)))- ((Is*P_hbar/(2*P_Q*g(t+1)*C2))*((eazi(t+1)*Ffi(t+1))+(eazj(t+1)*Ffj(t+1))))- Ms*(Ny-Nx)*sin(theta(t+1))*sin(2*phi(t+1))+(hy*cos(phi(t+1)) - hx*sin(phi(t+1))) ;
end

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回答（1 个）

Youssef Khmou
Youssef Khmou 2014-9-25
There are 24 undefined variables C1;G0;hMaz;hMpol;Alpha;theta;phi;EAj;EAi;EAk;Ku;Vol;P;C2;Is;P_hbar;P_Q;Ms;Nx;Nz;Ny;hx;hy;hz;
I tired using random values but the vector hMaz gives NaN values, because we need precise values of these variables, because i think this is inside quantum box right?
  13 个评论
显示 11更早的评论
Youssef Khmou
Youssef Khmou 2014-9-26
Converging M means that the magnetization is zero? what about the volume ?
Yasmin Tamimi
Yasmin Tamimi 2014-9-27
编辑：Yasmin Tamimi 2014-9-27
Vector M is position vector initially its M=[Mi Mj Mk]= [1 0 0] and the end result must be [-1 0 0]. The volume stays constant bcz it is the area multiplied by the thickness of the material.

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