These plots (described in blog post),

give an indication of how good or bad using the equation,

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

is for the accretion luminosity. Given gas in the box has not traveled from infinity in free-fall to the sink surface, one might be inclined to instead use,

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

However, as these plots show, using the previous equation for different values of the kinetic energy at r, ke(r) = ½ mv²(r), does not produce wildly differing accretion energies at the stellar surface. Therefore using the simpler equation (the first one listed) is a fine approximation. Instead of reading off the error from these plots, I made a few tables for the different curves. This gives us a sense of how small a zone-size would have to be before we would start to expect deviations. Note that the kinetic energy at r which would have been acquired from freefall alone is just GM/r.

First, let's consider cases where the kinetic energy was greater than GM/r. This corresponds to the lower 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 starting at r with kinetic energies > GM/r, would produce stronger accretion energy than those which would have fallen in from freefall alone. As the speed increases, the produce greater and greater accretion energies. As the distance to the star decreases, they I am next going to solve the following equations for r given various ke(r),

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 LHS of these equations give the (specific) accretion energy at the surface of the star, ke®, from gas that fell from a distance r away, i.e.,

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

Here is a table of error and r for various ke(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 you still only have a .01% error. This means that for any cells larger than this r, wouldn't expect great deviation, even in the extreme 1000*freefall case.

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.

Lastly, what if the parcel was moving slower than freefall? Now ke(r) will be a fraction of the freefall energy in the table below. In particular, what if ke(r) was ½, 1/5, 1/10 the freefall kinetic energy… at what distance, r, 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

This all shows us that the majority of energy gains from a stellar source occurs very close to the source, and thus, using the equation of accretion luminosity for a particle starting from rest at infinity is a fine approximation.