Squiggle Math II: The Dawn of Differentiation

We are bounded in a nutshell of Infinite Space: Worksheet # 12, Problem #1: Squiggle Math II: The Dawn of Differentiation

1. In this worksheet we’re going to do some “order of magnitude differentiation”. Let’s start off with a simple example that is completely unrelated to stellar structure:

(a) The velocity of a particle is $v = \alpha t^2$ where $\alpha$ is a constant, and we want to find the scaling of position with time. First write down the equation in the form of a differential equation for x, the position. Next, we are going to say that $dx \approx \Deltax \sim x$ and $dt \approx \Delta t \sim t$ . In English: “$dx$ is approximately the change in x, which scales as x.” Now it should be easy to show the scaling of x with t. What is the form of this scaling relationship?

(b) So you are probably saying to yourself, “This doesn’t feel right mathematically. How can you treat differential quantities with such disdain?!” But this is a simple differential equation, so you can actually integrate it. What do you get? How does it compare to your scaling relationship?

a. This system of mathematics fairly similar to the “squiggle math” we had worked with on a previous occasion, in the sense we are establishing the simple relationships and the way different aspects scale with one another in an equation. This is the case of a standard velocity equation with a constant: $$v = \alpha t^2,$$ and if we were to describe velocity in terms of the differentiated form of position, we see that:   $$\frac{dx}{dt} = \alpha t^2 ,$$ and as such becomes a description of the relationship once we “integrate” to find the scaling relationship of the true position, not the change in it: $$\frac{x}{t} \sim \alpha t^2 ,$$ therefore:   $$x \sim \alpha t^3,$$ and equivalently: $$x \propto t^3,$$ which is the scaling relationship we were looking for.  

b. However, we could to this calculation precisely, finding the correct proportionality and scaling factors of the equation we solved a minute ago: $$\frac{dx}{dt} = \alpha t^2 ,$$ we separate the variables: $$dx = \alpha t^2 dt,$$ and integrate using simple integration limits: $$\int^x _0 dx = \int^t _ \alpha t^2 dt$$ $$x = \alpha \frac{1}{3} t^3 ,$$ and now we see that the scaling relationship of the previous problem is perfectly sensible, and with the correct integration we may find the precise scaling factors which make the equation precise:   $$x = \frac{a}{3} t^3$$ $$x \propto t^3.$$