Changes between Version 1 and Version 2 of u/erica/ResRadFeedback

Timestamp:

01/28/16 11:57:44 (9 years ago)

Author:

Erica Kaminski

Comment:

—

u/erica/ResRadFeedback

@@ -1 @@
-== Choosing formula for accretion luminosity ==
 
 These plots (described in [https://astrobear.pas.rochester.edu/trac/blog/erica01262016 blog post]), 
@@ -17 +16 @@
 
 
-which describes the luminosity due to gas falling in from a distance r away (instead of infinity). However, as these plots show, using this equation for different values of the kinetic energy at r (ke(r)), i.e. ke(r)=1/2 mv^2^(r), does not produce wildly differing accretion energies until you move to the far left on the x-axis. In fact, in all but the most extreme cases, the degree of error is negligible (<1%), even at a distance of 1/100,000 of a parsec (i.e. much less than a typical cell size). At that distance, it doesn't matter whether the gas parcel is starting from rest, moving slowly (less than freefall speed), or moving fast (up to 10x freefall speed), the accretion energy that parcel will release at the stellar surface ''is the same''.  This is a statement that ''most'' of the energy gained from gravitational infall occurs in the final legs of the journey. Therefore under most circumstances, using the simpler equation (the first one listed) should be a *great* approximation. 
+which describes the luminosity due to gas falling in from a distance r away (instead of infinity). However, as these plots show, using this equation for different values of the kinetic energy at r (ke(r)), i.e. ke(r)=1/2 mv^2^(r), does not produce wildly differing accretion energies until you move to the far left on the x-axis. In fact, in all but the most extreme cases, the degree of error is negligible (<1%), even at a distance of 1/100,000 of a parsec (i.e. much less than a typical cell size). At that distance, it doesn't matter whether the gas parcel is starting from rest, moving slowly (less than freefall speed), or moving fast (up to 10x freefall speed), the accretion energy that parcel will release at the stellar surface ''is the same''.  This is a statement that ''most'' of the energy gained from gravitational infall occurs in the final legs of the journey. Therefore under most circumstances, using the simpler equation (the first one listed) should be a *great* approximation, and therefore is the way AstroBEAR computes accretion luminosity. 
 
 To get a handle of the resolution where this approximation may break down, I made a few tables of error for some different scenarios. In what follows recall that the kinetic energy at r which would have been acquired from freefall alone is just GM/r.