CosmoCoffee

Dear all,

I want to figure out the ISW part (i.e. the ([tex]\dot{\Phi} + \dot{\Psi}[/tex])) of the CMB source terms. The CMB source term is in subroutine output in equations.f90:


t4 = 1.D0/adotoa
t92 = k**2

sources(1) = (4.D0/3.D0*EV%Kf(1)*expmmu(j)*sigma+2.D0/3.D0*(-sigma-t4*etak)*expmmu(j))*k+ &
(3.D0/8.D0*ypol(2)+pig/16.D0+clxg/4.D0)*vis(j)
sources(1) = sources(1)-t4*expmmu(j)*dgrho/3.D0+((11.D0/10.D0*sigma- &
3.D0/8.D0*EV%Kf(2)*ypol(3)+vb+ 3.D0/40.D0*qg-9.D0/80.D0*EV%Kf(2)*y(9))*dvis(j)+(5.D0/3.D0*grho+ &
gpres)*sigma*expmmu(j)+(-2.D0*adotoa*etak*expmmu(j)+21.D0/10.D0*etak*vis(j))/ &
EV%Kf(1)+(vbdot-3.D0/8.D0*EV%Kf(2)*ypolprime(3)+3.D0/40.D0*qgdot-21.D0/ &
5.D0*sigma*adotoa-9.D0/80.D0*EV%Kf(2)*yprime(9))*vis(j))/k+(((-9.D0/160.D0*pigdot- &
27.D0/80.D0*ypolprime(2))*opac(j)-21.D0/10.D0*dgpi -27.D0/80.D0*dopac(j)*ypol(2) &
-9.D0/160.D0*dopac(j)*pig)*vis(j) - diff_rhopi*expmmu(j)+((-27.D0/80.D0*ypol(2)-9.D0/ &
160.D0*pig)*opac(j)+3.D0/16.D0*pigdot+9.D0/8.D0*ypolprime(2))*dvis(j)+9.D0/ &
8.D0*ddvis(j)*ypol(2)+3.D0/16.D0*ddvis(j)*pig)/t92


Are the terms multiplied by expmmu(j) the ([tex]\dot{\Phi} + \dot{\Psi}[/tex])?
Many thanks!


Cheers
Gong-Bo Zhao

This seems to be a popular question at the moment!

Probably not. If you want to get just that term best bet would be start with the source maple file and modify it to output just the source for that term.

If someone sorts it out perhaps they could send me the code and I'll put in in equations.f90 as a comment for future reference.

Thanks, Antony! Since I am not familiar with covariant gauge theory, I want to get some sense in comparison with CMBFAST4.5.1 where the ISW term is just the s1 in foutput. And in your thesis(page 108), you said 'For testing purposes and convenience of other users we include an option to use the synchronous gauge equations originally used by CMBFAST.' Can you tell me where the option is?


Thanks,
Gongbo

That was removed ages ago - it was too much effort maintaining two versions (the result was essentially identical). The CAMB notes give some hints on the equivalence between the code and normal synchronous gauge notation.

Sorry, the expmmu(j) terms are just exactly what you'd expect, so this is easy.