Why is $dm$ and $\Delta m$ used this way in physics textbooks?

Question

I would assume that $\Delta m$ would mean a finite change in mass and $dm$ to be an infinitesimal change in mass. But many physics textbooks use it to denote a small but finite mass and an infinitesimal mass element respectively while writing a differential equation. Why? And how does this make sense?

Similarly, $dU$, which should mean the infinitesimal change in the potential energy function is treated as the "small" potential energy of mass $dm$. Same for $dF$ etc.

One e.g would be to look at the derivation of the gravitational potential energy of a point mass and a shell. The potential energy of the ring is changed from $U_i$ to $dU$ and the mass of the ring is taken as $dM$. Doesn't $d$ mean an infinitesimal change in some quantity?

In rocket, $\Delta m$ refers to mass change due to gas propulsion. In calculating C.G. or moment of inertia, $dm$ means finite mass. — Ng Chung Tak, Feb 21 '17 at 17:21
Possible duplicates: http://physics.stackexchange.com/q/70376/2451 , http://physics.stackexchange.com/q/92925/2451 and links therein. — Qmechanic, Feb 21 '17 at 18:32
Possible duplicate of Difference between $\Delta$, $d$ and $\delta$ — Kyle Kanos, Feb 21 '17 at 18:50
No that is not what I'm asking. I know the difference between $\delta m$ and $dm$. I just want to know why $dm$ is regarded as a very small mass rather than an infinitesimal change in mass, and so on for $dU$ etc. — xasthor, Feb 22 '17 at 01:30

score 2 · Answer 1 · answered Feb 21 '17 at 17:15

You must always consider the specific notation definitions which an author may use. There is not a unique use of Δm. The notation dm is closer to being a universal notation for an infinitesimal m element, although m might be defined as something other than mass. Pay attention to context and book definitions.

score 0 · Answer 2 · answered Feb 21 '17 at 18:00

The way they use the $dm$ (or $dU$ or $dF$) notations is to represent an infinitesimal element.

You seem to be getting it confused with rates of change, given by $\frac{dm}{dt}$ for example.

When it is a ratio, then you are determining the change in $m$ as a function of $t$. $dm$ still represents an infinitesimal mass, but $\frac{dm}{dt}$ can be used to find the change in mass relative to time at any time (t). The two infinitesimals themselves don't represent a change, but by comparing the ratio of the two infinitesimals as a parameter changes, you can determine the rate of change with respect to another variable.

Why is $dm$ and $\Delta m$ used this way in physics textbooks?

2 Answers2