[mesa-users] Total Energy Integral

[Robert FARMER]

Hi
If you search for total_energy_integral in star/private then you find it in
star/private/profile_getval.f90.
Then you find that total_energy_integral is computed in
get_total_energy_integral().
That function is in star/private/star_utils.f90, that then calls
eval_section_total_energy_integrals (in the same file) and the result we
want is sum_total.

Maybe you can see in this function what MESA does that you do differently?

Rob

On Mon, Mar 9, 2015 at 1:22 PM, Michael Medford <michaelmedford at gmail.com>
wrote:

> Hello everyone,
>
> Over the last few weeks I have been investigating the profile column
> variable labelled total_energy_integral. The description of this variable
> is "sum from surface inwards of dm*(v^2/2 - G m / r + E) (ergs)". I was
> confused as to whether this sum includes the necessary volume elements to
> properly sum the spherical shells within this integral. So I set about
> calculating my own energy integral using the necessary data from each cell
> within the profile. Here is how I calculated my integral:
>
> -Routines to shift each of the mass, radii and density terms to the center
> of each cell by taking averages of the outer and inner boundaries
> -Define "dr" as the radii width of each shell
> -Defining an "energy element" = (-Gm/r + E) * (4*pi*r^2*density*dr) where
> all of the terms are in cgs and the velocity of all of the cells are 0
> -Sum from the surface inwards of this energy element
>
> Attached you will see the results of my variable versus the output from
> the profile's total_energy_integral. As the energy integral claims to have
> a dm multiplied into each term, I did not multiply the
> total_energy_integral times anything, since this should be equivalent to
> (4*pi*r^2*density*dr).
>
> Does anyone know why these two values would be different, and by such
> large orders of magnitude? What am I missing about the
> total_energy_integral and what it is recording? Or am I manually summing
> the integral incorrectly?
>
> Thanks so much for the help.
> ~Michael
>
> ----
> Michael Medford
> Northwestern University
>
>
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