Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hello every one,

 

I am trying to solve a backward induction problem (game theoretic problem) in maple. Lets say we have S rounds. I start from bottom (S-th round) and solve a parametric equation. Then I put the solution in the upper level (S-1). Then, update the functions at level and solve the equation pertatining to this level (S-1). Again, I put the solution ofround S-1 into the equations of S-2 and update the equations and parametrically solve the equation belonging to level S-2 . This process repeats till the first level.

 

The issue is that the solution gradually becomes larger and larger. I guess that's why Maple is not able to solve it. 

 

Have any of you guys faced to a similar problem. Any suggestion?

 

I was thinking of asking maple to reduce its precision in computations. I mean there may not be necessary to store a number with 100 digits precision! However, I don't know how to do that (I don't know the command). Any suggestion?

 

Thank you in advance.

 

Ahmadreza

Using Maple 18, I solved for minimum and maximum price. Instead of using fsolve I wanna use procedure programming structure in order to get the same results. How can I do it?

min_sol := fsolve([bc_cond, slope_cond, x[G, 1] = w[aggr, 1]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_min := subs(min_sol, p); max_sol := fsolve([bc_cond, slope_cond, x[G, 2] = w[aggr, 2]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_max := subs(max_sol, p);
{p = 0.3857139820, x[G, 1] = 127.8000000, x[G, 2] = 38.99045418}
0.3857139820
{p = 0.8841007104, x[G, 1] = 44.30160890, x[G, 2] = 164.2000000}
0.8841007104

 

Hello,

please explain how to write a code to calculate and output the actual area using integration for y=X^3 over range (0,2) using left-hand rule and 200 subdivisions?

 

Thank you 

I currently have a function quadsum(n) that determines the [x,y] solutions of the above equation for an integer n. :

quadsum:= proc(n::nonnegint)
local
k:= 0, mylist:= table(),
x:= isqrt(iquo(n,2)), y:= x, x2:= x^2, y2:= y^2;
if 2*x2 <> n then x:= x+1; x2:= x2+2*x-1; y:= x; y2:= x2; end if;
while x2 <= n do
y:= isqrt(n-x2); y2:= y^2;
if x2+y2 = n then k:= k+1; mylist[k]:= [x,y] end if;
x:= x+1; x2:= x2+2*x-1;
end do;
convert(mylist, list)
end proc:

How would I alter this so that I get [x,y] for n= (5^a).(13^b).(17^c)(29^d) for non-negative integers a,b,c,d?

please is there any one can help me to find a solution of a sytem of 3 non linear equations each with 3 variable and with more than 30 unknown coefficients

this is the system

solve({EEE_x(x, y, z) = 0, EEE_y(x, y, z) = 0, EEE_z(x, y, z) = 0}, {x, y, z})

where x,y,r are the unknowns

and the three equations are simply the partial derivative with respect to x,y and z repectively

EEE_x(x,y,z):=(&DifferentialD;)/(&DifferentialD; x) EE(x,y,z)

EEE_y(x,y,z):=(&DifferentialD;)/(&DifferentialD; y) EE(x,y,z)

EEE_z(x,y,z):=(&DifferentialD;)/(&DifferentialD;z)EE(x,y,z)

the main equation is EE where (it has 3 variables and more than 30 qunknowns coefficients

(x, y, z) ->

1
----------------------------------------------------------------
2
/ 2 2 2\
hh \ii + jj x + ll z + mm y + 100. y + nn y z + oo x + pp z /

/ 2 2 2 2 3 2
\p z y + q z y + l z x + g z x + o z y + n z x + m y x

2 2 2 3 2 2 2
+ j y x + k y x + i z y + d z y + f z x + h z y

2 2 4 3 2 3
+ e y x + u z y x + v z y x + a + b x + c x + r x + s z

2 2 4 3 4 \
+ t z + bb z + cc y + dd y + ee y + ff y + gg z + aa x/

 

1.  a procedure quadsumstats whose input is an integer n. This procedure should return a list of length 

n whose kth  entry is the number of solutions to
x^2 + y^2 = k 
for
1 <= k and k <= n

I am sort of confused as to how to construct that list of length n and how to obtain integer solutions to the equation in maple.

2.

a procedure firstCount(k) that finds the first integer
n
with
k
representations as
"x^2+y^2= n." What does it mean for an integer to have k representations?

 

 

 

 

This is a question I have also submitted to the technical support, I am worried that it is a bit too technical for them, however:)

I am debugging a C program which links against the OpenMaple API library (under Linux and with Maple 17 and 18). I am using valgrind memcheck, because I am experiencing strange behavior which could be due to writes beyond allocated blocks of memory.  

The first thing which jumps to my eye, are many errors of the types

Use of uninitialised value of size (4/8/16)

Invalid read of size (4/8/16)

Conditional jump or move depends on uninitialised value(s)

The same errors are also printed when I use the examples that ship with Maple. For instance, I compile "simple.c" with

gcc  -Wl,--no-as-needed -lmaple -lmaplec -lrt -L /usr/lib -L $MAPLEDIR/bin.X86_64_LINUX -I $MAPLEDIR/extern/include -o simple simple.c

and run valgrind as

valgrind --tool=memcheck --error-limit=no --log-file=memcheck.log ./simple

memcheck.txt 

Some, but not all, of the errors occur in __intel_sse2_strcpy or __intel_sse2_strlen. Furthermore, according to valgrind there are definite memory leaks. which appear in the library. 

Practically this makes it hard for me to identify my potential own errors. I am a bit surprised to see so many warnings because I tend to fix my own programs until memcheck does not print these anymore (before I give it away at least). The question is: Can I consider these errors as safe to ignore? How would I distinguish real errors which may appear in my application?

If I have the following system of first order diff eq's:

x'(t)=2x(t)+3y(t)

y(t)=-3x(t)-2y(t)

then can I consider the coefficient matrix A=<<2,-3>,<3,-2>> and compute the eigenvalues of A and infer as follows:

if the eigenvalues are of the same sign- eq point is a node

if they are of opposite signs- eq point is a saddle

if they are pure imaginary- eq point is a center

if they are complex conjugates- eq. point is a spiral

I've been given these conditions but my text says for a linear system of the form x'=Ax, the eigenvalues of A can be used to identify the nature of the eq. point. I am confused as to whether this applies to the given system as well; I have obtained 5 different trajectories and drawn the phase diagram for the system

I'm trying to solve a system of four pdes and I know that the Newton method won't converge.

Are there other numerical methods that I can use?

Any help would be greatly appreciated!

Thanks,

Eve

Hi,

I'm an industrial engineering student who's running into problems with solving a simple system of non-linear equations.

Why do I get this error when issuing the solve command? I'm pretty sure I'm only passing two lists into the function?

Thanks a lot in advance,

Joshua

 

Hi!

Is there any way to remove the empty space that comes under images when printing the project, while using document mode?

Regards

Nicolai

I have a matrix A for which the basis of the left null space using NullSpace() is the empty set {}  while the column space is {e1,e2,e3}. By definition, we need every vector in col space . every vector in basis of left null space =0 but how would I show that in this case? Can I determine another basis for the left null space?

I've got the following matrix :

A:=[<a,a-1,-b>|<a-1,a,-b>,<b,b,2a-1>] where <> are the column elements of A, a is  a real number defined on [0,1] and b^2=2a(1-a) 

a) to show A is an orthogonal matrix, I understand that I need A.Transpose(A)=Identity(3*3) but is there a way in which I can let a take a random real numbered value between 0 and 1? The rand() only returns an integer within a range. Directly multiplying A and Transpose(A) will return an expression in a, so what's the right approach?

b) from a) we can infer that A is a matrix that describes a rotation in e1,e2,e3 where these are the standard bases vectors in R3. How can I determine the rotation axis? The hint I've been given says I need to consider the Eigenvalues and eigen vectors but I don't quite understand how.

I wish to use closed Newton-Cotes with n=2, also known as Simpson's Rule to numerically integrate an improper integral.

 

If it matters the integrand is (cos(2x))/(x^1/3), integrating between x=0..1

I've tried a few different (but similar) code but to no avail. Here is some stuff I've tried:

 

1.

 

with(Student[NumericalAnalysis]):

with(Student[Calculus 1]):

Simp1 := ApproximateInt(cos(2*x)/x^(1/3), x = 0 .. 1, method = newtoncotes[2]);

 

This gives me an output message that says "Float(infinity)".

 

2.

with(Student[NumericalAnalysis]):

with(Student[Calculus 1]):

Simp2 := int(exp(-x)/sqrt(1-x), x = 0 .. 1);

 

This doesn't have Simpson's rule as an option.

 

I think I'm on the right track with my first try, since I guess it wasn't tecnically an error message, but I'm not sure how to alter the code accordingly to get a numerical value instead. Thanks for any help.

 

 

 

 

 

 

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

NULL


Download Deriving_the_Kerr_Metric.mw

First 181 182 183 184 185 186 187 Last Page 183 of 2097