Items tagged with subs subs Tagged Items Feed

a1:= f(x) :
> T1 :=simplify((taylor(a1,x=alpha,N+3))):
> E1:=subs([seq(((D@@i)(f))(alpha) = 0,i=1..m-1),f(alpha)=0,x=e[n]+alpha],T1):
> g1 :=(convert(simplify(series((E1,e[n]=0,N))),polynom));

 

Hello,

I would like to plot coupled elliptic parametric curves.

I have a problem in the for loop in my code. The index i of the phi variable is not incremented.

Here my code:

for i to 4
do
u[i]:=t->L/2*cos(w*t);
v[i]:=t->H*sin(w*t+phi[i]);
end do;

The result looks like this :

u[1](t) -> - L cos(w t)
v[1](t) -> H sin(w t + phi[i])

u[2](t) -> - L cos(w t)
v[2](t) -> H sin(w t + phi[i])

u[3](t) -> - L cos(w t)
u[3](t) -> H sin(w t + phi[i])

u[4](t)  -> - L cos(w t)
u[4](t)  -> H sin(w t + phi[i])

And i would have prefered :

u[1](t) -> - L cos(w t)
v[1](t) -> H sin(w t + phi[1])

u[2](t) -> - L cos(w t)
v[2](t) -> H sin(w t + phi[2])

u[3](t) -> - L cos(w t)
v[3](t) -> H sin(w t + phi[3])

u[4](t)  -> - L cos(w t)
v[4](t)  -> H sin(w t + phi[4])

Do you have ideas how I can obtain iteration on my phi variable ?

Thank you for your help.

To check one of the MythBusters TV episodes, i.e., the fall of a mannequin (80 Kg) from a plane at an altitude of 1200 m in 31 s, with Maple13 (Windows Vista) I solved the following differential equation with initial conditions:

> de:= m*(D@@2)(x)(t) = m*g - k*(D(x)(t))^2:

> ini:= (x(0) = 0, D(x)(0) = 0):


> X:= unapply(rhs(expand(dsolve({de,ini}))),t);

   
     X := proc (t) options operator, arrow; -m^(1/2)*g^(1/2)*t/k^(1/2)-m*ln(2)/k+m*ln(exp(2*g^(1/2)*k^(1/2)*t/m^(1/2))+1)/k end proc

1) Posing: V0 = sqrt(m * g / k),  T = sqrt(m/(g*k)) one has  V0*T = m/k. I want to have:

   > Xxb:= t -> V0*(- t + T * ln((exp(2*t/T) + 1)/2) ); # m.

 How to obtain this equivalent equation  Xxb(t), without retyping the equation of  X(t) ? With subs , convert or simplify  applied to  X(t), Maple 13 gives error messages.

2) Taking the derivative of  Xxb(t), one find :

     V := proc (t) options operator, arrow; V0*(exp(2*t/T)-1)/(exp(2*t/T)+1) end proc.

How to transform it to V = V0 tanh(t/T) ?

Again, subs , convert or simplify  seem not working, even with  assume(t, real), assume(T, real) ! I know simplify is a difficult task, but Maple should recognize a tanh !

Thank you in advance for any suggestion.

( Note that with m = 80 Kg, g = 9.81 m/s^2 and k = 0.267482 Kg/m, which correspond to a speed limit of V0 =195 km/h, on find t(1200m) = 26 s, instead of 31 s ).

i am new on maple while usig maple i have to substitute a general value in an equation but could not do it like

> restart;
> taylor(f(x), x = gamma);

i typed this and want to replace x-gamma with e[n] which is easy and i did it but whn i want to substitute D(f)(gamma),(D@@2)(f)(gamma) and so on by (D@@k)(f)(gamma)=k!*c[k]*f(gamma) then i could not understand how can i do 2nd part 
pls help in this i m very thanlful to you

 

after following a example , got error

 

                             2                  
               1   / d      \    1        2     2
               - m |--- x(t)|  - - m omega  x(t)
               2   \ dt     /    2              
Error, (in Mechanics:-LagrangeEqs) invalid input: subs received subst1, which is not valid for its 1st argument
Error, invalid input: Mechanics:-GeneralSol expects its 1st argument, eqs, to be of type list, but received eqs
Error, invalid input: rhs received sol, which is not valid for its 1st argument, expr
L;

 

Mechanics := module()
export SetVariables, LagrangeEqs, GeneralSol;
option package;
local subst1, subst2, varN, t;

SetVariables = proc( vars:: list, time )
local i;
t := time;
varN := nops( vars );
subst1 := {};
subst2 := {};
for i from 1 to var N do
subst1 := subst1 union
{vars[i](t) = q[i], diff(vars[i](t), t) = v[i]};
subst2 := subst2 union
{q[i] = vars[i](t), v[i] = diff(vars[i](t), t)};
end do;
print( subst1 );
print( subst2 );
NULL;
end proc;

LagrangeEqs := proc (L)
local i, l1, term1, term2;
l1 := subs(subst1, L):
for i to varN do
term1 := [seq(diff(subs(subst2, diff(l1, v[i])), t), i = 1..varN)]:
term2 := [seq(subs(subst2, diff(l1, q[i])), i = 1..varN)]:
end do;
[ seq(simplify(term1[i]-term2[i]) = 0, i = 1..varN) ];
end proc;

RayleighEqs := proc(L, R)
local i, l1, r1, term1, term2, term3;
l1 := subs( subst1, L ):
r1 := subs( subst1, R ):
for i from 1 to varN do
term1:=[seq(diff(subs(subst2, diff(l1, v[i])), t), i=1..varN)]:
term2:=[seq(subs(subst2, diff(l1, q[i])), i=1..varN)]:
term3:=[seq(subs(subst2, diff(r1, v[i])), i=1..varN)]:
end do:
[ seq(simplify(term1[i]-term2[i]+term3[i]), i=1..varN) ];
end proc;

LagrEqsII := proc( L, Q::list )
local i, l1, term1, term2;
l1 := subs(subst1, L):
for i to varN do
term1 := [seq(diff(subs(subst2, diff(l1, v[i])), t), i = 1 .. varN)]:
term2 := [seq(subs(subst2, diff(l1, q[i])), i = 1 .. varN)]:
end do;
[seq(simplify(term1[i]-term2[i]) = Q[i], i = 1 .. varN)];
end proc;

LagrEqsIII := proc (L, R, Q::list)
local i, l1, r1, term1, term2, term3;
l1 := subs(subst1, L):
r1 := subs(subst1, R):
for i to varN do
term1 := [seq(diff(subs(subst2, diff(l1, v[i])), t), i = 1 .. varN)]:
term2 := [seq(subs(subst2, diff(l1, q[i])), i = 1 .. varN)]:
term3 := [seq(subs(subst2, diff(r1, v[i])), i = 1 .. varN)]:
end do;
[seq(simplify(term1[i]-term2[i]+term3[i]) = Q[i], i = 1 .. varN)];
end proc;

GeneralSol := proc (eqs::list)
local i, initconds, eqs2;
initconds := NULL:
eqs2 := eqs[][]:
for i to varN do
initconds:=VarNames[i](0)=q[i], (D(VarNames[i]))(0)=v[i], initconds:
end do;
dsolve({initconds, eqs2});
end proc;


end module;

with(LibraryTools):
LibLocation := cat("c:\\Temp");
Save(Mechanics, LibLocation);
with(FileTools):
march('list',"c:\\Temp\\Mechanics.lib");
save(Mechanics, "c:\\Temp\\Mechanics.m");
read "c:\\Temp\\Mechanics.m";

 

with(Mechanics):
SetVariables([x], t);
L := (1/2)*m*diff(x(t), t)^2 - (1/2)*m*omega^2 * x(t)^2;
eqs := LagrangeEqs(L);
sol := GeneralSol( eqs );
X := unapply( rhs(sol), t );

 

 

dsol := dsolve([eq2=diff(s(t),t), eq3=diff(s(t),t), eq4=diff(s(t),t)], [a(t),b(t),c(t)]);
eval(subs(t(t)=t,subs(_C1(t) = _C1, subs(_C2(t) = _C2, subs(_C3(t) = _C3, sol1)))));

 

after subs t(t) = t , it still  have t(t)

in harmonic ocillator version, why it has error after it dsolve?

with(Physics[Vectors]):
with(SumTools):
Setup(mathematicalnotation = true):

Qs is Qss*cos(m*t+phi);

H := 1/(8*Pi*c^2)*Summation(Commutator(diff(Qs*cos(w*t+phi),t)*diff(conjugate(Qs*cos(w*t+phi)),t)+w^2*Qs*cos(w*t+phi)*conjugate(Qs*cos(w*t+phi))),ks=1..3)-1/c*(Summation(conjugate(Qs*cos(w*t+phi)),ks=1..3));

or

H := 1/(8*Pi*c^2)*Summation(Commutator(diff(Qs,t)*diff(conjugate(Qs),t)+w^2*Qs*conjugate(Qs)),ks=1..3);

#Qs = p, Ps := Pss*sin(m*t+phi) Ps = qs

i use

H := 1/(8*Pi*c^2)*Summation(Commutator(diff(Qs*cos(w*t+phi),t)*diff(conjugate(Qs*cos(w*t+phi)),t)+w^2*Qs*cos(w*t+phi)*conjugate(Qs*cos(w*t+phi))),ks=1..3)-1/c*(Summation(conjugate(Qs*cos(w*t+phi)),ks=1..3));

 HJ := subs(Qs= diff(f(q,P,t), q), H);

H:=subs( f(q,P,t) = f1(q) + f2(t), HJ);
sol:=dsolve({rhs(H)=E,lhs(H)=E});
S:=rhs(sol[1][1]+sol[1][2]);
p:=diff(S,q);
Q:=diff(S,E);

Hello to you all,

I have this DE

but when I try to change the variable

I get this

>algsubs(r(t) = 1/u(t), diff(diff(r(t), t), t)-k/r(t) = -K/r(t)^2);

Why is it not done for the rest of the terms.  Is there a more easy way to do it.  Take into account that I cannot use dsolve because it is in an course in integral.

 

Thank in advance for your help.

 

--------------------------------------
Mario Lemelin
Maple 2015 Ubuntu 14.04 - 64 bits
Maple 2015 Win 10 Pro - 64 bits messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987

I have y"+y=sinx

If I want to substitute y=sinx, how should I do it?

I tried subs(y=sin(x),f(y))(ofc after defining f:=y as above).

The most troubling thing is that it does not automatically apply double derivative.

https://drive.google.com/file/d/0Bxs_ao6uuBDUVkE1b2pmcnljcnc/view?usp=sharing
https://drive.google.com/file/d/0Bxs_ao6uuBDUbE56R3Z1c0tLRkE/view?usp=sharing

restart;
changering := proc(Equation1, f3,g3)
g1 := (x,y)-> f3;
f1 := (x,y)-> g3;
h:=subs(g=g1, Equation1);
h:=subs(f=f1, h);
h:=subs(0=0, h);
return h;
end proc:
Eq1 := f(x,g(x,y)) + f(x,y);
h2 := changering(Eq1,x+y, x+y);
h2;

g1 := (x,y)-> x+y;
f1 := (x,y)-> x+y;
h:=subs(g=g1, Eq1);
h:=subs(f=f1, h);
h:=subs(0=0, h);

 

Hey Guys, 

I was wondering how big an algebriac equation has to be before maple crashes. (i cant think of a better word) . I am trying to derive an epression that involves solving 9 equations that all pretty big. about 16 terms. I am substituting in each equation into the next and I get to the last one and it just says evaluating. I left it up overnight and it was still there evaluating. Any help would be appreciated. 

Hello All,

I have 6 equations that are similair in size to the one listed below. I am trying to find a single equation in terms of only 1 of the variables (s11). After finding this equation I want to solve it for that single variable. When I start plugging all these equations into eachother I get to about the last one then maple seems to get stuck evaluating. I left it on overnight thinking that if I gave it time it would eventually solve, but this didn't seem to work. My questions is, If I gave it enough time would it solve this? Is there another way to do this? Any help you guys could offer would be a great help. Thanks!

 

Sample Equation

restart:

Xsubs:=X=x*exp(epsilon*alpha[1]);

Ysubs:=Y=y*exp(epsilon*alpha[2]);

psubs := P = p*exp(epsilon*alpha[3])

where P(X,Y) and p(x,y). epsilon, alpha[i], i=1,2,3 are parameters.

how to changeover from p(x,y) to P(X,Y) in the following equation?

Eq:= diff(p,y$2)+diff(p,x)=0;

Cheers!

 

Sorry for the bad format, maple input is not working.

In particular,from an expression, something like

 Int(x*f(x), x = 0 .. 1)+ Int(x, x = 0 .. 1/3)+3* Int(x*f(x), x = 0 .. 2/3)

I want to extract 

x*f(x)+x+3*x*f(x)

That is all the integrands occuring in the expression. GetIntegrand helps to find the integrand but only for one integral, how do I do it for an expression like above.

 

I tried to apply following approach

subs(sin(c::anything)=c, sin(x)+f);


and the output was

f

I was expecting to get only 'x+f' for output.

EDIT 1: 

Thank you so much for your help. You guys answered exactly what I asked for. I needed this to do something different but I should have been more careful. I realized, I could not get that part now. This is what I actually need to do, if

A:=-(5/8)*x*(Int(f(x), x = 0 .. x))+(5/24)*(Int(f(x), x = 1/3 .. x))-(1/4)*x*(Int(x*f(x), x = 1/3 .. x));

then I wish to get a function, g(xi),


g(xi)= -(5/8)x*f(xi) ;   if 0<= xi <= x
                                                  (5/24)*f(xi)-(1/4)*x*xi*f(xi) ; if 1/3<= xi<= x

Notice, all the terms after integral should use variable 'xi' and terms occurring before integral remains as it is.

I am trying to simplify equation 18 using equations 8 and 9. It should look a little like equation 21, but instead I get the results in equations 19 and 20.  I tried using different substituions, but algsubs gets the closest answer. A few terms are going to zero after the substitution.

When I substitute Z(X) then Zbar(X) terms vanish, and visa versa.


Initialize the metric and tetrad

 

restart; with(Physics); with(Tetrads)

0, "%1 is not a command in the %2 package", Tetrads, Physics

(1.1)

X = [zetabar, zeta, v, u]

X = [zetabar, zeta, v, u]

(1.2)

ds2 := Physics:-`*`(Physics:-`*`(2, dzeta), dzetabar)+Physics:-`*`(Physics:-`*`(2, du), dv)+Physics:-`*`(Physics:-`*`(2, H(zetabar, zeta, v, u)), (du+Physics:-`*`(Ybar(zetabar, zeta, v, u), dzeta)+Physics:-`*`(Y(zetabar, zeta, v, u), dzetabar)-Physics:-`*`(Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), dv))^2)

2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2

(1.3)

PDEtools:-declare(ds2)

Ybar(zetabar, zeta, v, u)*`will now be displayed as`*Ybar

(1.4)

NULL

vierbien = Matrix([[1, 0, -Ybar(zetabar, zeta, v, u), 0], [0, 1, -Y(zetabar, zeta, v, u), 0], [Physics:-`*`(H(zetabar, zeta, v, u), Y(zetabar, zeta, v, u)), Physics:-`*`(H(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), 1-Physics:-`*`(Physics:-`*`(H(zetabar, zeta, v, u), Y(zetabar, zeta, v, u)), Ybar(zetabar, zeta, v, u)), H(zetabar, zeta, v, u)], [Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u), -Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), 1]])

vierbien = (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -Ybar(zetabar, Zeta, v, u), (1, 4) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = -Y(zetabar, Zeta, v, u), (2, 4) = 0, (3, 1) = H(zetabar, Zeta, v, u)*Y(zetabar, Zeta, v, u), (3, 2) = H(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (3, 3) = 1-H(zetabar, Zeta, v, u)*Y(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (3, 4) = H(zetabar, Zeta, v, u), (4, 1) = Y(zetabar, Zeta, v, u), (4, 2) = Ybar(zetabar, Zeta, v, u), (4, 3) = -Y(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (4, 4) = 1}))

(1.5)

``

NULL

Setup(tetrad = rhs(vierbien = Matrix(%id = 18446744078408794830)), metric = ds2, mathematicalnotation = true, automaticsimplification = true, coordinatesystems = (X = [zetabar, zeta, v, u]), signature = "+++-")

[automaticsimplification = true, coordinatesystems = {X}, mathematicalnotation = true, metric = {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}, signature = `+ + + -`, tetrad = {(1, 1) = 1, (1, 3) = -Ybar(X), (2, 2) = 1, (2, 3) = -Y(X), (3, 1) = H(X)*Y(X), (3, 2) = H(X)*Ybar(X), (3, 3) = 1-H(X)*Y(X)*Ybar(X), (3, 4) = H(X), (4, 1) = Y(X), (4, 2) = Ybar(X), (4, 3) = -Y(X)*Ybar(X), (4, 4) = 1}]

(1.6)

``

Verification of Tetrad

 

I will try to verify the tetrad from (Kerr and Schild (1965)). However, the tetrad given in the paper seems to have the third tetrad with the wrong sign. I changed the sign and get the correct verification,

    e_[]

`&efr;`[a, mu] = (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -Ybar(X), (1, 4) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = -Y(X), (2, 4) = 0, (3, 1) = H(X)*Y(X), (3, 2) = H(X)*Ybar(X), (3, 3) = 1-H(X)*Y(X)*Ybar(X), (3, 4) = H(X), (4, 1) = Y(X), (4, 2) = Ybar(X), (4, 3) = -Y(X)*Ybar(X), (4, 4) = 1}))

(2.1)

g_[]

g[mu, nu] = (Matrix(4, 4, {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 1) = 1+2*H(X)*Y(X)*Ybar(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 1) = -2*H(X)*Y(X)^2*Ybar(X), (3, 2) = -2*H(X)*Ybar(X)^2*Y(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 1) = 2*H(X)*Y(X), (4, 2) = 2*H(X)*Ybar(X), (4, 3) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}))

(2.2)

Physics:-`*`(e_[a, mu], e_[a, nu]) = g_[mu, nu]

Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[`~a`, nu] = Physics:-g_[mu, nu]

(2.3)

TensorArray(Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[`~a`, nu] = Physics:-g_[mu, nu])

Matrix(4, 4, {(1, 1) = 2*H(X)*Y(X)^2 = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X) = 2*H(X)*Y(X), (2, 1) = 1+2*H(X)*Y(X)*Ybar(X) = 1+2*H(X)*Y(X)*Ybar(X), (2, 2) = 2*H(X)*Ybar(X)^2 = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X) = 2*H(X)*Ybar(X), (3, 1) = -2*H(X)*Y(X)^2*Ybar(X) = -2*H(X)*Y(X)^2*Ybar(X), (3, 2) = -2*H(X)*Ybar(X)^2*Y(X) = -2*H(X)*Ybar(X)^2*Y(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2 = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X) = 1-2*H(X)*Y(X)*Ybar(X), (4, 1) = 2*H(X)*Y(X) = 2*H(X)*Y(X), (4, 2) = 2*H(X)*Ybar(X) = 2*H(X)*Ybar(X), (4, 3) = 1-2*H(X)*Y(X)*Ybar(X) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X) = 2*H(X)})

(2.4)

Physics:-`*`(e_[a, mu], e_[b, mu]) = eta_[a, b]

Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[b, `~mu`] = Physics:-Tetrads:-eta_[a, b]

(2.5)

NULL

TensorArray(Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[b, `~mu`] = Physics:-Tetrads:-eta_[a, b])

Matrix(4, 4, {(1, 1) = 0 = 0, (1, 2) = 1 = 1, (1, 3) = 0 = 0, (1, 4) = 0 = 0, (2, 1) = 1 = 1, (2, 2) = 0 = 0, (2, 3) = 0 = 0, (2, 4) = 0 = 0, (3, 1) = 0 = 0, (3, 2) = 0 = 0, (3, 3) = 0 = 0, (3, 4) = 1 = 1, (4, 1) = 0 = 0, (4, 2) = 0 = 0, (4, 3) = 1 = 1, (4, 4) = 0 = 0})

(2.6)

``

gamma_[4, 2, 1]

diff(Y(X), zeta)-(diff(Y(X), u))*Ybar(X)

(2.7)

SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(D_[nu](e_[4, mu]), e_[2, mu]), e_[1, `~nu`]))

diff(Y(X), zeta)-(diff(Y(X), u))*Ybar(X)

(2.8)

NULL

``

For equation 2.8 we get the following:

SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(Riemann[`~sigma`, rho, mu, nu], e_[4, `~rho`]), e_[4, `~nu`]))

(-Physics:-Riemann[`~sigma`, 4, 4, mu]*Y(X)^2+(Physics:-Riemann[`~sigma`, 4, 1, mu]+Physics:-Riemann[`~sigma`, 1, 4, mu])*Y(X)-Physics:-Riemann[`~sigma`, 1, 1, mu])*Ybar(X)^2+((Physics:-Riemann[`~sigma`, 2, 4, mu]+Physics:-Riemann[`~sigma`, 4, 2, mu])*Y(X)^2+(-Physics:-Riemann[`~sigma`, 2, 1, mu]+Physics:-Riemann[`~sigma`, 3, 4, mu]+Physics:-Riemann[`~sigma`, 4, 3, mu]-Physics:-Riemann[`~sigma`, 1, 2, mu])*Y(X)-Physics:-Riemann[`~sigma`, 3, 1, mu]-Physics:-Riemann[`~sigma`, 1, 3, mu])*Ybar(X)-Physics:-Riemann[`~sigma`, 2, 2, mu]*Y(X)^2+(-Physics:-Riemann[`~sigma`, 2, 3, mu]-Physics:-Riemann[`~sigma`, 3, 2, mu])*Y(X)-Physics:-Riemann[`~sigma`, 3, 3, mu]

(1)

 

Now we replicate eqn 2.16. These are the conditions for e[4,mu] to be geodesic and shear-free. The outputs are eqn 3.5.

 

gamma_[4, 1, 1] = 0

diff(Ybar(X), zeta)-(diff(Ybar(X), u))*Ybar(X) = 0

(2)

gamma_[4, 2, 2] = 0

diff(Y(X), zetabar)-(diff(Y(X), u))*Y(X) = 0

(3)

gamma_[1, 4, 4] = 0

(diff(Ybar(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Ybar(X), zeta))-Ybar(X)*(diff(Ybar(X), zetabar))-(diff(Ybar(X), v)) = 0

(4)

gamma_[2, 4, 4] = 0

(diff(Y(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Y(X), zeta))-(diff(Y(X), zetabar))*Ybar(X)-(diff(Y(X), v)) = 0

(5)

gamma_[3, 4, 4] = 0

0 = 0

(6)

gamma_[4, 4, 4] = 0

0 = 0

(7)

shearconditions := {diff(Y(X), zetabar)-(diff(Y(X), u))*Y(X) = 0, diff(Ybar(X), zeta)-(diff(Ybar(X), u))*Ybar(X) = 0, (diff(Y(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Y(X), zeta))-(diff(Y(X), zetabar))*Ybar(X)-(diff(Y(X), v)) = 0, (diff(Ybar(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Ybar(X), zeta))-Ybar(X)*(diff(Ybar(X), zetabar))-(diff(Ybar(X), v)) = 0}:

 

Now we can define the rotation coefficients associated with rotation and expansion z = theta - i omega

 

gamma_[2, 4, 1] = Z(X)

-(diff(Y(X), zeta))+(diff(Y(X), u))*Ybar(X) = Z(X)

(8)

gamma_[1, 4, 2] = Zbar(X)

-(diff(Ybar(X), zetabar))+(diff(Ybar(X), u))*Y(X) = Zbar(X)

(9)

PDEtools:-declare(Z(X), Zbar(X))

Zbar(zetabar, zeta, v, u)*`will now be displayed as`*Zbar

(10)

Zdefinitions := {-(diff(Y(X), zeta))+(diff(Y(X), u))*Ybar(X) = Z(X), -(diff(Ybar(X), zetabar))+(diff(Ybar(X), u))*Y(X) = Zbar(X)}

{-(diff(Y(X), zeta))+(diff(Y(X), u))*Ybar(X) = Z(X), -(diff(Ybar(X), zetabar))+(diff(Ybar(X), u))*Y(X) = Zbar(X)}

(11)

We now show that the tetrad vectors are propogated parallel along each curve of the congruence of null geodesics which have e[4,~mu] as tangents.

 

   

We now use the tetrad form of the Ricci tensor. In order to use this in Maple we need to create a Ricci Tensor Tetrad function.

 

RicciT := proc (a, b) options operator, arrow; SumOverRepeatedIndices(Ricci[mu, nu]*e_[a, `~mu`]*e_[b, `~nu`]) end proc

proc (a, b) options operator, arrow; Physics:-SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(Physics:-Ricci[mu, nu], Physics:-Tetrads:-e_[a, `~mu`]), Physics:-Tetrads:-e_[b, `~nu`])) end proc

(12)

SlashD := proc (f, a) options operator, arrow; SumOverRepeatedIndices(D_[b](f)*e_[a, `~b`]) end proc

proc (f, a) options operator, arrow; Physics:-SumOverRepeatedIndices(Physics:-`*`(Physics:-D_[b](f), Physics:-Tetrads:-e_[a, `~b`])) end proc

(13)

SlashD(f(X), 1)

diff(f(X), zeta)-Ybar(X)*(diff(f(X), u))

(14)

SlashD(f(X), 2)

diff(f(X), zetabar)-Y(X)*(diff(f(X), u))

(15)

SlashD(f(X), 3)

(1+H(X)*Y(X)*Ybar(X))*(diff(f(X), u))-H(X)*((diff(f(X), zeta))*Y(X)+Ybar(X)*(diff(f(X), zetabar))+diff(f(X), v))

(16)

SlashD(f(X), 4)

-Y(X)*Ybar(X)*(diff(f(X), u))+(diff(f(X), zeta))*Y(X)+Ybar(X)*(diff(f(X), zetabar))+diff(f(X), v)

(17)

NULL

The geodesic and shear free condition given by Lemma 1 in (Goldberg and Sachs (1962)). Kerr uses the fourth tetrad instead of the third so we need to modify the Ricci tensor conditions. The equations (2) - (5) enforce the first Lemma.

 

   

 

Notice that none of the previous Ricci conditions can be used to solve for H.  We can use the remaining field equations to find the partial differential equations necessary to derive the metric.

 

  simplify(RicciT(1, 2), shearconditions) = 0

H(X)*(diff(diff(Y(X), zeta), zetabar))*Ybar(X)-H(X)*Ybar(X)*Y(X)*(diff(diff(Ybar(X), u), zetabar))-H(X)*Ybar(X)^2*(diff(diff(Y(X), u), zetabar))-H(X)*Y(X)^2*(diff(diff(Ybar(X), u), zeta))-2*H(X)*Y(X)*Ybar(X)*(diff(diff(Y(X), u), zeta))+H(X)*Y(X)^2*Ybar(X)*(diff(diff(Ybar(X), u), u))-H(X)*Y(X)*(diff(diff(Ybar(X), u), v))+H(X)*Y(X)*Ybar(X)^2*(diff(diff(Y(X), u), u))-H(X)*(diff(diff(Y(X), u), v))*Ybar(X)+H(X)*(diff(Ybar(X), zetabar))^2+(-3*H(X)*Y(X)*(diff(Ybar(X), u))-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Ybar(X), zetabar))+H(X)*(diff(Y(X), zeta))^2+(-4*H(X)*(diff(Y(X), u))*Ybar(X)-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Y(X), zeta))+2*H(X)*Y(X)^2*(diff(Ybar(X), u))^2-Y(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Ybar(X), u))+2*(H(X)*(diff(Y(X), u))*Ybar(X)+(1/2)*(diff(H(X), u))*Y(X)*Ybar(X)-(1/2)*(diff(H(X), zeta))*Y(X)-(1/2)*(diff(H(X), zetabar))*Ybar(X)-(1/2)*(diff(H(X), v)))*(diff(Y(X), u))*Ybar(X) = 0

(18)

-(diff(H(X), zetabar))*Ybar(X)*Z(X)-Y(X)*(diff(H(X), zeta))*Z(X)-H(X)*(diff(Y(X), zeta))^2+Z(X)*((diff(H(X), u))*Y(X)*Ybar(X)+2*H(X)*Z(X)-(diff(H(X), v))) = 0

-(diff(H(X), zetabar))*Ybar(X)*Z(X)-(diff(H(X), zeta))*Y(X)*Z(X)-H(X)*(diff(Y(X), zeta))^2-Z(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)-2*H(X)*Z(X)+diff(H(X), v)) = 0

(19)

Zbar(X)*(-(diff(H(X), v))-(diff(H(X), zetabar))*Ybar(X)-(diff(H(X), zeta))*Y(X)+(diff(H(X), u))*Y(X)*Ybar(X)+H(X)*(diff(Ybar(X), zetabar)+2*Zbar(X))) = 0

-Zbar(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)-H(X)*(diff(Ybar(X), zetabar))-2*H(X)*Zbar(X)+diff(H(X), v)) = 0

(20)

Physics:-`*`(SlashD(H(X), 4), Z(X)+Zbar(X)) = Physics:-`*`(H(X), SlashD(Z(X), 4)+SlashD(Zbar(X), 4))

-(-(diff(H(X), v))-(diff(H(X), zeta))*Y(X)+Ybar(X)*((diff(H(X), u))*Y(X)-(diff(H(X), zetabar))))*(Z(X)+Zbar(X)) = H(X)*(-Y(X)*Ybar(X)*(diff(Z(X), u))+Ybar(X)*(diff(Z(X), zetabar))+Y(X)*(diff(Z(X), zeta))+diff(Z(X), v)-Y(X)*Ybar(X)*(diff(Zbar(X), u))+Ybar(X)*(diff(Zbar(X), zetabar))+Y(X)*(diff(Zbar(X), zeta))+diff(Zbar(X), v))

(21)

``

NULL

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