Blog #17 - Orders of Magnitude

Orders of Magnitude and Scaling

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

(a) The velocity of a particle is v = αt2, where α 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 ≈ ∆x ∼ x and dt ≈ ∆t ∼ 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?

We begin with the information we are given. \[v = \alpha t^2 \]From there we know the differential version of velocity is simply the change in distance over the change in time. \[v = \frac{dx}{dt} = \alpha t^2 \] Which gives us \[\frac{\Delta x}{\Delta t} \approx \frac{x}{t} \approx \alpha t^2\]Now we simply solve for x to give the scaling relation \[x = \alpha t^3\]

(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?

To prove through integration that the scaling relation holds, we have to show the integral of velocity is the position scaling relation that we expect. We start with \[\int v = x\]Thus we have the integral for velocity as \[\int v = \int \alpha t^2 dt\]Which leads to \[\int \alpha t^2 dt = \frac{1}{3} \alpha t^3 + C_1\]Which gives us the same scaling relation as part a). \[\frac{1}{3} \alpha t^3 \approx \alpha t^3\]