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When I do simplify(LegendreP(n, 1, cos(t))), Maple gives me -sqrt(1-cos(t))*sqrt(cos(t)+1). Isn't it the same thing as -sin(t)? How can I have Maple further convert/simplify it to -sin(t)? (I tried simplify(%, trig) but it didn't work). I am new on Maple. Thanks in advance for anyone's help!

 

I want to simplify an expression via simplify();

However since the expression is very long (length(expression)=2713899) Maple is taking too long...So far one hour.

I have already simplified several expressions of a similar kind with lengths of up to 400.000 which worked fine within minutes.

Is there any way to speed up the process. Am I making a mistake?

Thanks in advance

Hi,

I m trying to simplify an expression involving square roots in Maple. But instead of giving the expected output it is not simplifying it. Please find the attached maple file for reference. I calculated manually and the result is 1. Please help me out for this. Thanx in advance.

Regards

Sunit

restart

temp := (1/2)*(a*r*t-b*p)*sqrt(p+v*sqrt(a*r))*sqrt(b)/(sqrt(a*b*r*t)*sqrt(a*r*t*(p+v*sqrt(a*r))))+(1/2)*(a*r*v*t-v*b*p+p*t*sqrt(a*r)-b*v^2*sqrt(a*r))*sqrt(b)/(sqrt(a*b*r*t*(p+v*sqrt(a*r)))*sqrt(t*(p+v*sqrt(a*r))))+(1/2)*b*(p+v*sqrt(a*r))/(a*r*t)

(1/2)*(a*r*t-b*p)*(p+v*(a*r)^(1/2))^(1/2)*b^(1/2)/((a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2))+(1/2)*(a*r*v*t-v*b*p+p*t*(a*r)^(1/2)-b*v^2*(a*r)^(1/2))*b^(1/2)/((a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2))+(1/2)*b*(p+v*(a*r)^(1/2))/(a*r*t)

(1)

simplify(temp)

-(1/2)*(-(p+v*(a*r)^(1/2))^(1/2)*b^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a^2*r^2*t^2+(p+v*(a*r)^(1/2))^(1/2)*b^(3/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*p-b^(1/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a^2*r^2*t^2*v+b^(3/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*v*p-b^(1/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t^2*p*(a*r)^(1/2)+b^(3/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*v^2*(a*r)^(1/2)-b*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*p-b*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*v*(a*r)^(1/2))/((a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t)

(2)

``


Download question.mw

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

(1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

(2)

``


Download consistency_expmle_4.mw

.

 

Its pretty obvious that i would like the to get the answer -exp(-n(cos(phi)+sin(phi))) from the 'simplify' command. What am i doing wrong? the problem

When solving a nonlinear differential equation on some variable x, but using some other parameter w, I am finding on Maple some complicated solution, which I would like to simplify by making evident what is the x dependence, and where I can compact complicated functions of the parameter w alone into new constants. How can I do that automatically?

 

For example, to have

 

(sinh(w) + ln(w))*x 

 

to be automatically called

 

c*x

 

Thank you in advance.

Hello.

given this expression

T:=unapply((1/6930)*exp(-(1/7938)*(X[4]-933)^2)*exp(-(1/6050)*(X[2]-805)^2)/((1+exp((1/50)*X[4]-(1/50)*X[2]))*Pi),X[2]);

U := unapply(sum(T(X[2]), X[4] = 0 .. 3600), X[2]):

I want to display U, but not all 3600 terms. is there anyway to simplify/reduce this sum?

kind of like geo series a+ar+ar^2+ar^3+...+ar^(n-1)=sum(ar^k,k=0..n-1) can be reduced to a*(1-r^n)/(1-r)

 

hello
i have a problem that you could help me
i have an expression that i want convert it to an expression according to the expression q[d] with maple
i have bellow expressions
B[d]:=(-d*w[1]+w[3])/(((-2*w[1]+w[3])^2)-((-d*w[1]+w[3])^2))
B[o]:=(-2*w[1]+w[3])/(((-2*w[1]+w[3])^2)-((-d*w[1]+w[3])^2))
A[o]:=w[1]*(alpha[o]-c[o]-t[o])+2*w[2]*e[o]
A[d]:=w[1]*(alpha[d]-c[d]-t[d])+2*w[2]*e[d]
q[o]:=B[o]*A[o]-B[d]*A[d]
q[d]:=B[o]*A[d]-B[d]*A[o]
i want simplify expression U[d] such as this one
U[d]:=w[1]*(q[d]*(alpha[d]-q[d]-d*q[o]-c[d]-t[d])-C)+w[2]*(e[d]*q[d]+e[o]*q[o])+w[3]*((1/2)*(q[d]+q[o])^2)
I'm looking to simplify U[d] according to the expression q[d]
please please help me

I am trying to simplify sums of a few LaguerreL polinomials of different n using the identities in the function advisor such as recurrsion relations. How does one go about in using the FunctionAdvisor identities when trying to simplify expressions containing orthogonal polynomials? 

 

 

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_[]

`𝔢`[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

Hello all,

 

I'm experiencing an aggravating issue when substituting numerical values into a symbolic expression.  I'm using Maple 14.  I believe the issue may be related to floating point precision.  Either way, it's ruining my life.  If anyone has a solution or a work-around, I'd be most grateful.  The following code reproducibly produces the anomolay on my machine:

 

vx := Vector( 4, symbol=x ):

# A numerator and a denominator. They're equal.
num := (x[1]*x[3]+x[1]*x[4]+x[2]*x[3]+x[2]*x[4])^2:
den := ((x[1]+x[2])^2*(x[3]+x[4])^2):

# Prints true...
evalb(num=den);

# This equals zero
y := 0.1 - 0.1*num/den;

# Prints 0...
simplify(y);

# Prints 0...
simplify(subs(x[3]=1,x[4]=3,0.1-0.1*num/den));

# Prints 0...
simplify(subs(x[3]=1,x[4]=3,y));

# Prints 0.1 x 10^-10 !?!?
simplify(subs(x[3]=1,x[4]=2,y));

 

Hi all.

I try to get the real part from the complex expression. But it turns out to not be the simplest result:

A:=I*sin(k*Pi*(x-h*cos(theta))/a)*sin(l*Pi*(y-h*sin(theta))/b)*exp(-I*k[0]*h)*sin(k*Pi*x/a)*sin(l*Pi*y/b)

convert(exp(-I*k[0]*h), sin);

simplify(Re(A));

Maple results in:

Re(sin(k*Pi*(-x+h*cos(theta))/a)*sin(l*Pi*(-y+h*sin(theta))/b)*exp(-I*k[0]*h)*sin(k*Pi*x/a)*sin(l*Pi*y/b))

while the simplified result should be:

sin(k*Pi*(x-h*cos(theta))/a)*sin(l*Pi*(y-h*sin(theta))/b)*sin(k*Pi*x/a)*sin(l*Pi*y/b)*sin(k[0]*h)

 

I wander how to get the simplifyed result in maple. Thanks

there is a solution of equation,so the equation can be divided by the solution,but because the equation is complex,it can't be simplify by the soution,can anyone give me some help?thanks a lot.


I am trying to do a substitution as shown in the attached document. I know variants of this question have been asked before but dont quiet get what to do. It is problem with algsubs and how it handles denominators I think. Can get substiturion to work for simple fractions but more complicated ones fail. Would appreciate any guidance here.

restart 

``

``

CR := proc (a, b, c, d) options operator, arrow; (a-c)*(b-d)/((a-d)*(b-c)) end proc

proc (a, b, c, d) options operator, arrow; (a-c)*(b-d)/((a-d)*(b-c)) end proc

(1)

eqns := CR(a, b, c, d)

(a-c)*(b-d)/((a-d)*(b-c))

(2)

e1 := CR(b, a, c, d)

(b-c)*(a-d)/((b-d)*(a-c))

(3)

simplify(e1, {(a-c)*(b-d)/((a-d)*(b-c)) = lambda})

(a*b-a*c-b*d+c*d)/(a*b-a*d-b*c+c*d)

(4)

e1

(b-c)*(a-d)/((b-d)*(a-c))

(5)

``

lambda

lambda

(6)

applyrule((a-c)*(b-d)/((a-d)*(b-c)) = lambda, e1)

(b-c)*(a-d)/((b-d)*(a-c))

(7)

alias(lambda = (a-c)*(b-d)/((a-d)*(b-c)))

lambda

(8)

e1

(b-c)*(a-d)/((b-d)*(a-c))

(9)

``

NULL

``

f := a/b

a/b

(10)

``

f := algsubs(a/b = alpha, f)

alpha

(11)

f

alpha

(12)

algsubs((a-c)*(b-d)/((a-d)*(b-c)) = lambda, e1)

Error, (in algsubs) cannot compute degree of pattern in a

 

``

 

Download UHG5_substitution.mw

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