Choosing formula for accretion luminosity

These plots (described in ​blog post),

give an indication of how good or bad of an approximation it is to use the equation,

L= \frac{G\dot{m} M}{R}

for the expected accretion luminosity of cells surrounding a sink. Given gas in the accretion volume has not traveled from infinity in free-fall to the sink surface, one might be inclined to instead use,

L= \frac{1}{2}\dot{m} v^2(r) - \frac{GM \dot{m}}{r}+\frac{GM\dot{m}}{R}

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)=½ mv²(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.

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.

First, let's consider ke(r) > GM/r. This corresponds to the upper plot. In this plot, the accretion energy (per unit mass) for a freely falling particle from infinity, GM/R, is normalized to 1 and lies exactly on top of the x-axis. This plot shows that particles falling in from r with ke(r) > GM/r, would produce stronger accretion energy than those which would have fallen in from freefall alone. As the speed increases, they would produce greater and greater accretion energies. I am next going to solve the following equations for r given various ke(r)>GM/r at the .01% error level and the 30% error level,

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = 1.01 (\frac{GM}{R})

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = 1.3 (\frac{GM}{R})

Note the RHS is our approximation, i.e. the accretion energy from infinity, whereas the LHS of these equations give the assumed more 'realistic' accretion energy at the surface of the star, ke®, due to gas haven fallen in from a distance r away. Here is a table of error and r for various ke(r)>GM/r,

ke(r)	Error (%)	Distance (pc)
10*freefall	30	6.8*10-7
10*freefall	.01	.00002
100*freefall	30	7.4*10-6
100*freefall	.01	.0002
1000*freefall	30	7.5*10-5
1000*freefall	.01	.002

This shows that you get the greatest error closer into the surface of the star, and even when you are going very fast starting from a distance r = .002 pc you still only have a .01% error. This means that for any cells larger than this r, it virtually doesn't matter how fast the material is going (it can even be going 1000x freefall speed!!) — it is going to end up with nearly the same accretion luminosity as if it traveled from infinity. Therefore, the approximation seems good over a wide range of potential speeds and down to very small distances away from the sink.

Now, what if a gas parcel started from rest, a distance r away from the star surface? Now we are solving,

-\frac{G*M}{r}+\frac{G*M}{R} = .99 (\frac{GM}{R})

-\frac{G*M}{r}+\frac{G*M}{R} = .70 (\frac{GM}{R})

for a .01 and 30% error. This gives the following distances,

ke(r)	Error (%)	Distance (pc)
0	.01	2.2*10-6
0	30	7.5*10-8

Again, the closer into the star we get, the worse the approximation gets — but that a cell size of r~ 2.2*10^-6 shouldn't produce much deviation (<1%), assuming the material is moving slowly.

Lastly, what if the parcel was moving slower than freefall? In particular, what if ke(r) was ½, 1/5, 1/10 the freefall kinetic energy at r… at what distance would we see a 0.01% error? What about a 30% error?

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = .99 (\frac{GM}{R})

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = .70 (\frac{GM}{R})

ke(r)	Error (%)	Distance (pc)
½*freefall	.01	1.1*10-6
½*freefall	30	3.8*10-8
1/5*freefall	.01	1.8*10-6
1/5*freefall	30	6*10-8
1/10*freefall	.01	2*10-6
1/10*freefall	30	6.8*10-8

Again, we can expect gas falling in from something like 10^-6 pc away from the sink to have enough time to gain a significant fraction of the accretion luminosity from infinity. Therefore, unless we are really modeling distances much smaller than this — we have shown that really for a wide range of starting kinetic energies, the infinity luminosity should do just fine. As a caveat, the sink radius is so much less than a cell size (it is a sub-grid model, no way of telling its size other than it is small compared to a grid zone).. this implies that these are all upper limits to the resolution requirements. So basically, it makes sense to use L = G M m-dot/R.