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I'm taking a module on stars and the solar system; I've attached notes from our first lecture- hydrostatic equilibrium. I'm confused about the notation $\mathrm{d}$ for $\mathrm{d}A, \, \mathrm{d}r$, and $\mathrm{d}m$. I understand that $\mathrm{d}$ is the derivative operator, signifying an infinitesimal change in the variable it's next to. I'm unsure, however, how to visually think of this "change". Why is mass a function, and why is it changing? With regard to what is it changing? Also, I understand that in this example, "$\mathrm{d}r$" is simply the thickness of the planet/ star's shell. How does that relate to "a change in radius $r$", which I assume is measured from the center of the planet all the way until the shell starts?

Thank you so much for any help!!

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M. A.
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canihavealmondmilk
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1 Answers1

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Physicists tend to be informal about the $d$ notation, using it initially to represent a relatively small change in a variable, and then eventually dividing one small change by another to produce, out of a hat, a derivative. Mathematicians reach the same end point with a more formal notation in which a small change in a variable is represented by a $\delta$, and the $d$ notation is reserved for the true derivative when we divide one $\delta$ quantity by another $\delta$ quantity and then take the limit as $\delta$ goes to zero (physicists take the limit step as read).

Using the $\delta$ notation, we would say $\delta m$ is the mass of the small cylinder in the diagram, with cross sectional area $\delta A$ and length $\delta r$ (because it lies in the shell between radius $r$ and radius $r+\delta r$). If we call the volume of this small cylinder $\delta V$ then we have

$\delta V = \delta r \cdot \delta A$

But the density of material in the shell between radius $r$ and radius $r+\delta r$ is $\rho(r)$. Here we use the fact that $\delta r$ is small compared with $r$ to make the simplifying assumption that density is approximately constant across the shell. So

$\delta m = \rho(r) \cdot \delta V = \rho(r) \cdot \delta r \cdot \delta A$

It is somewhat confusing that $r$ is used both to represent the generic radius of the planet and also the specific radius of the shell.

gandalf61
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