Solving Time Delay Integral?

Hi,

When trying to integrate this guy:

fD= @(r) (1/(1-rs/r)).*((1-((b^2)./(r.^2)).*(1-(rs/r))).^(-1/2)-1);

I get the error:

Error using  / 
Matrix dimensions must agree.

Error in @(r)(1/(1-rs/r)).*((1-((b^2)./(r.^2)).*(1-(rs/r))).^(-1/2)-1)

Error in integralCalc/iterateScalarValued (line 315)
                fx = FUN(t);

Error in integralCalc/vadapt (line 133)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 84)
        [q,errbnd] = vadapt(@AToInfInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in Bolometric_Flux (line 42)
    t=integral(fD,R,Inf);*

This is the full code. Matlab can solve the commented out integral (f) but for some reason it won't work for (fD).

for k=1:length(a);
    b=(R/(sqrt(1-u)))*sin(a(k));
    %f= @(r) (1./r.^2).*(1/(b(k).^2)-(1./r.^2).*(1-(rs./r))).^(-1/2); %light bending integrand
    fD= @(r) (1/(1-rs/r)).*((1-((b^2)./(r.^2)).*(1-(rs/r))).^(-1/2)-1); %time delay integrand
    t=integral(fD,R,Inf);
    %triangle_phi(i)=2*pi*h*triangle_t(i);
end

R,u and rs are defined in the code.

Answer 1

[Rebecca Krosnick]

How is rs defined? If it is not a scalar, then the error is caused by

1/(1-rs/r)

You cannot divide the scalar "1" by a matrix. You instead need to use element-wise division like you have elsewhere in your code. Specifically, the code would instead be:

1./(1-rs/r)

Otherwise, to troubleshoot the issue, break down your definition of fD into variables. Define a variable for each matrix, and then compare the dimensions of the matrices to ensure they are appropriate for each /, ./, and .* operation.