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NbodyOne.m
The program is used in the introduction set of notes to look at the five body
problem with symmetric initial conditions. It calls the program 'rstTOijk.m' to
carry out some coordinate transformations described in those notes, and the
program 'nBodyWpar.m' for the integration.
%--------------------------------------------------------------------------%----------------------NOTES-----------------------------------------------%--------------------------------------------------------------------------
%Dependencies:%This program calls 'nBodyWpar.m' in the integration.%It calls 'rstTOijk.m' when transforming coordinates of the initial
%conditions.
%Use:%Just type 'NbodyOne' from the matlab command line with the programand%all it's dependencies in the search path.
%The values of the masses and the initial conditions can easily betinkered%with if you want to experiment with the system. More drastic changes%would be needed to adjust the number of bodies (because of the waythe%initial conditions are entered) but this program could be used as a
modle%for others with other numbers of bodies.
%Purpose:%N body simulation of 5 bodies with symmetric initial data.%This program and it's results are described in the introduction setof%notes for the N-Body problem. The special component of this code isthe%way that the intial conditions are specified (essentially inspherical%coordinates.
%Output:%The output is a plot of the trajectories in R^3.
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N=5; %number of bodiesG=1; %Gravational ConstantMass=[1.0 1.0 1.0 1.0 1.0]; %their masses
%NOTE: The masses are currently all equal. I the problem describedin the%notes, the output of this program with the present masses iscompaired to%the output when the value of the first mass is changed to 2.
%Coordinates given in special system have to be converted toRectangular.
PI=3.1415926535898;c=PI/180;
%Initial positions in r alpha and betar01=1.0;r02=1.0;r03=1.0;r04=1.0;r05=1.0;ar1=0.0*c;ar2=72.0*c;ar3=144.0*c;ar4=216.0*c;ar5=288.0*c;br1=0.0;br2=0.0;br3=0.0;br4=0.0;br5=0.0;
%initial velocities in v alpha and beta.
v01=0.6;v02=0.6;v03=0.6;v04=0.6;v05=0.6;av1=90.0*c;av2=90.0*c;av3=90.0*c;
av4=90.0*c;av5=90.0*c;bv1=0.0;bv2=0.0;bv3=0.0;bv4=0.0;bv5=0.0;
Rec=0;%convert positions to cartesean frame
Rec(1:3,1)=[r01*cos(ar1)*cos(br1);r01*sin(ar1)*cos(br1);r01*sin(br1)];
Rec(4:6,1)=[r02*cos(ar2)*cos(br2);r02*sin(ar2)*cos(br2);r01*sin(br2)];
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Rec(7:9,1)=[r03*cos(ar3)*cos(br3);r03*sin(ar3)*cos(br3);r03*sin(br3)];Rec(10:12,1)=[r04*cos(ar4)*cos(br4);r04*sin(ar4)*cos(br4);r04*sin(br4)];Rec(13:15,1)=[r05*cos(ar5)*cos(br5);r05*sin(ar5)*cos(br5);r05*sin(br5)];
A=0;%convert spherical velocities to rst frameA(1:3,1)=[v01*cos(av1)*cos(bv1);v01*sin(av1)*cos(bv1);v01*sin(bv1)];A(4:6,1)=[v02*cos(av2)*cos(bv2);v02*sin(av2)*cos(bv2);v02*sin(bv2)];A(7:9,1)=[v03*cos(av3)*cos(bv3);v01*sin(av3)*cos(bv3);v03*sin(bv3)];A(10:12,1)=[v04*cos(av4)*cos(bv4);v04*sin(av4)*cos(bv4);v04*sin(bv4)];A(13:15,1)=[v05*cos(av5)*cos(bv5);v05*sin(av5)*cos(bv5);v05*sin(bv5)];
%convert rst velocities to ijk. Each velociety has a differentmatrix.
Rec(16:18,1)= rstTOijk(Rec(1:3,1))*A(1:3,1);Rec(19:21,1)= rstTOijk(Rec(4:6,1))*A(4:6,1);Rec(22:24,1)= rstTOijk(Rec(7:9,1))*A(7:9,1);Rec(25:27,1)= rstTOijk(Rec(10:12,1))*A(10:12,1);Rec(28:30,1)= rstTOijk(Rec(13:15,1))*A(13:15,1);
%Integrate the Systemtspan = [0 10];y0=Rec
options=odeset('RelTol',1e-10,'AbsTol',1e-10);[t,y] = ode113('nBodyWpar',tspan,y0,options,flag,N,G,Mass);
%--------------Below Should not be modified----------------------------%------------The program should be fit to the problem------------------%------------data by adjusting the above and 'Nbody'-------------------%-----------The only exception is the plotting-------------------------
A=size(y) %vector containing (rows,columns)timeMax=A(1,1) %number of rows of y is number of timesteps
M=0; %initialize Mfor i=1:N
M=M+Mass(1,i);end
%The next section computes the 10 classical integrals of motion.
%First the components of the velociety of the center
%of mass. Then the components of the position of the%center of mass (which are not constant, but linear in
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%time.) Then the appropriate difference which is a%constant.
%Next the three components of angular momentum are computed.
%And finally energy.
%Linear Momentum
%first integral; component one of V_cmI1=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*y(i,3*N+(3*j-2));end
I1(i)=S/M;
end
%second integral; component two of V_cmI2=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*y(i,3*N+(3*j-1));end
I2(i)=S/M;end
%third integral; component three of V_cmI3=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*y(i,3*N+(3*j));end
I3(i)=S/M;end
%Center of Mass as a function of time; component one of R_cm(t)CM1=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*y(i,3*j-2);end
CM1(i)=S/M;end
%Center of Mass as a function of time; component two of R_cm(t)CM2=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*y(i,3*j-1);end
CM2(i)=S/M;
end
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%Center of Mass as a function of time; component three of R_cm(t)CM3=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*y(i,3*j);
endCM3(i)=S/M;end
%Fourth integral: component one of R_cm(t)-[V_cm(t)]*tI4=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=CM1(i)-I1(i)*t(i);end
I4(i)=S;end
%Fifth integral: component two of R_cm(t)-[V_cm(t)]*tI5=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=CM2(i)-I2(i)*t(i);end
I5(i)=S;end
%Sixth integral: component three of R_cm(t)-[V_cm(t)]*tI6=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=CM3(i)-I3(i)*t(i);end
I6(i)=S;end
%Angular Momentum
%seventh integral; component one of hI7=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*[y(i,3*j-1)*y(i,3*N+3*j)-y(i,3*j)*y(i,3*N+3*j-1)];
endI7(i)=S/M;
end
%eighth integral; component two of hI8=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*[y(i,3*j)*y(i,3*N+3*j-2)-y(i,3*j-2)*y(i,3*N+3*j)];
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endI8(i)=S/M;
end
%nineth integral; component three of hI9=0;
for i=1:timeMaxS=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*[y(i,3*j-2)*y(i,3*N+3*j-1)-y(i,3*j-1)*y(i,3*N+3*j-2)];end
I9(i)=S/M;end
%Energy: (T-U)(t)
%Potential Energy: U(t)U=0;
for i=1:timeMaxS=0; %Initialize Sfor j=1:N
s=0;for k=1:N
Rjk=(y(i,3*j-2)-y(i,3*k-2))^2+(y(i,3*j-1)-y(i,3*k-1))^2+(y(i,3*j)-y(i,3*k))^2;
if k~=js=s+Mass(1,k)/[sqrt(Rjk)];
elses=s+0;
endendS=S+Mass(1,j)*s;
endU(i)=(G/2)*S;
end
%Kenetic Energy; T(t)T=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*[(y(i,3*N+(3*j-2)))^2+(y(i,3*N+(3*j-
1)))^2+(y(i,3*N+(3*j)))^2];endT(i)=S/2;
end
%Tenth integral; Energy E(t)E=0;for i=1:timeMax
E(i)=T(i)-U(i);end
%Aux Quanties
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%I or moment of ineritaI=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*[(y(i,3*j-2))^2+(y(i,3*j-1))^2+(y(i,3*j))^2];
endI(i)=S;end
%d/dt(I)Idot=0;for i=1:timeMax
S=0; %Initialize Sfor j=1:N
S=S+Mass(1,j)*[y(i,3*N+(3*j-2))*y(i,3*j-2)+y(i,3*N+(3*j-1))*y(i,3*j-1)+y(i,3*N+(3*j))*y(i,3*j)];
end
Idot(i)=S*2;end
%d^2/(dt)^2(I)Idotdot=0;for i=1:timeMax
Idotdot(i)=2*T(i)+2*E(i);end
%Change in Energy. If integration is perfect then this is alwayszero.%Its deviation from zero measures the solutions deviation from thereal%one.
%deltaEdeltaE=0;deltaE(1)=0;for i=2:timeMax
deltaE(i)=E(i)-E(1);end
%plotting
%Three dimensional plotting
%Plot all bodies in physical space%plot3(y(:,1), y(:,2), y(:,3))%plot3(y(:,4),y(:,5),y(:,6))plot3(y(:,1), y(:,2), y(:,3),'r', y(:,4), y(:,5), y(:,6), 'b',y(:,7),y(:,8), y(:,9), 'g', y(:,10), y(:,11), y(:,12), 'k', y(:,13),y(:,14),y(:,15), 'y')grid onxlabel('x axis'), ylabel('y axis'), zlabel('z axis')
%Plotting the integrals
%plot(t(1:timeMax),I5(1:timeMax))
