# Discrete quantities and calculus

Why we can apply calculus in cases where discrete quantities take place?

Suppose we have a box that has two partitions, namely A and B (look at the figure below). Suppose we know the rate that particles pass from A to B. That is we know the rate $$r=\frac{dN}{dt}$$. Then we calculate the number of particles that passed from A to B from $$t_i$$ to $$t_f$$ as following:

$$N = \int_{t_i}^{t_f}rdt$$

Suppose that the rate is constant and equal to $$2$$ particles per second (or just $$2$$ per second) and the interval is equal to $$2.6$$ s. Then the amount of particles that went from A to B is $$5.2$$. But how is it possible the amount of particles to be a real number and not an integer? I just used that example for particles but it can generalized to charge, photons etc.

I was trying to understand what the instantaneous rate of transfer of a discrete quantity means but I just can't. I thought one could measure that rate like we measure the power of some engine ($$\frac{dE}{dt})$$. But the main difference is that $$dE$$ is an infinitesimal of a continuous quantity.

What is the justification that allows us to use calculus in case of discrete quantities?

• Calculus is used for continuous functions and your function is not continuous in time. Jul 10 at 23:15
• The rate of particles can be non integer, multiply it by time again gives back an integer Jul 11 at 2:06
• When you wanted to avoid calculus you need to fiddle around with a lot of expressions involving discrete sums and differences instead of integrals and derivatives. Calculus is a lot more convenient. For systems with discrete numbers $N$ it is typically applied when $N$ is very large, say, $10836.6$. This is clearly $\approx 10836$ so we can safely apply calculus for problems with large $N$. Jul 11 at 4:27

In your example, a uniform rate of $$2$$ particles per second means that in a random interval of $$2.6$$ seconds there could be either $$5$$ or $$6$$ particles passing from A to B. The result of $$5.2$$ particles that you get from a continuous approximation means that over a large number of $$2.6$$ second intervals the average number of particles passing from A to B in each interval will be $$5.2$$ - but the result in any individual interval will be either $$5$$ or $$6$$.
It is like saying that the average family has $$2.1$$ children - but no individual family actually has $$2.1$$ children.