Sure. Sectional density (Sρ) in physics is the mass of the bullet divided by its cross-sectional area. Ballistics uses a shortcut convention in dividing pounds mass by the square of diameter to escape making the area calculation, so it simplifies some calculations, like BC, where the ratios of two Sρs is all that matters. You can divide the ballistic Sρ by ¼π to get the actual physical sectional density if you need it, but in most ballistics calculations, like the ones below, it's about ratios of Sρs, so it is only a matter of being consistent with which one you use.
So here's the value: As you know, the force pushing a bullet down a barrel may be found by dividing the gas pressure behind it by the bullet's cross-sectional area. Thus (not counting rifling's reduction of the area, which is small), a 0.451" diameter bullet at 15,000 psi experiences 2396 lbs of force trying to accelerate it, while a 0.357 diameter bullet at that same pressure experiences 1501 lbs of force on its base. So if the two bullets have the same weight, that pressure will accelerate the 0.451" bullet 2396/1501=1.596 times faster. Or, if the 0.357" bullet is 1.596 times lighter, the same force will accelerate them both at the same rate. So, if both bullets are, say 200 grains (0.02857 lbs), their Sρs by ballistic convention are:
.357200 = 0.2242
.451200 = 0.1405
When you divide the former by the latter, you get:
0.2242/0.1405 = 1.596
That's the same multiplier we got before, but it let us bypass calculating the area of the bullet cross-section and multiplying it by the force, so the ratio of the Sρs was a shortcut to that result. You get a number showing how much higher pressure will have to be to accelerate the 0.357" bullet of that same weight.
To get the same acceleration from the same pressure at both bullet diameters, you can lighten the 0.357" bullet by 200/1.596 = 125.3 grains. Thus, at the same pressure, a 125.3-grain 0.357" bullet will have the same acceleration as the 200-grain 0.451" bullet does.
Another way to look at it is a bullet's Sρ is proportional to how much pressure it needs for a given rate of acceleration. This means, with a lower Sρ you don't need as much pressure to impart a given rate of acceleration.
That last fact brings up an additional problem. If you try to use a slow powder with a low Sρ bullet, it can accelerate so easily that it scoots down the bore faster than the powder can make gas to keep up with it. This means, the lower the Sρ, the faster the powder burn rate has to be to reach the desired peak pressure. And this is why, with a low Sρ bullet, you often can get better performance with a fast powder than you might expect.