# How to treat differentials?

In my Calculus class, my math teacher said that differentials such as $dx$ are not numbers, and should not be treated as such.

In my physics class, it seems like we treat differentials exactly like numbers, and my physics teacher even said that they are in essence very small numbers.

Can someone give me an explanation which satisfies both classes, or do I just have to accept that the differentials are treated differently in different courses? For example, if the linear density of a solid rod is $d$, in Physics class we would say that the mass of a very small part of the rod $dx$, is $d*dx$, so my physics teacher would say $dm=d*dx$.

P.S. I took Calculus 2 so please try to keep the answers around that level.

P.S.S. Feel free to edit the tags if you think it is appropriate.

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One is a mathematician, the other isn't. Seems like an obvious choice to me. –  Git Gud Jan 9 '14 at 0:12
If you're a math major: Don't treat them as numbers, because it makes a difference. If you're an engineering/physics major: go right ahead (but only in those applied classes)--it probably won't matter for the topics in that field, and most people in that field tend to do so anyway. ;) –  apnorton Jan 9 '14 at 0:12
@GitGud My physics textbook seems to take the same approach as the physics teacher. –  Ovi Jan 9 '14 at 0:14
It is a physics book, hence.... –  Git Gud Jan 9 '14 at 0:15
$dx$ aren't vectors in $\Bbb{R}^n$, instead they are covectors or linear functionals i.e. linear maps $\Bbb{R}^n\to\Bbb{R}$.. and at the end $dx$ are the gradient of the coordinated functions $x$'s –  janmarqz Jan 9 '14 at 0:18

The definition of the derivative is that $$\lim_{\Delta x\rightarrow 0 }\frac{\Delta f}{\Delta x} = \frac{df}{dx}.$$ In other words, for a fixed $x_{0}$, you can set $\Delta x = x-x_{0}$ and $\Delta f=f(x)-f(x_{0})$ and, for $\Delta x \ne 0$, you'll have some function $o$ such that $$\frac{\Delta f}{\Delta x} = \frac{df}{dx} + o(\Delta x),$$ where $\lim_{\Delta x\rightarrow 0}o(\Delta x)=0$. So, just think of what the Physicists are doing as an approximate where the error terms will, hopefully, disappear in the limit, and leave you with a differential equation. If you're careful it works just about as easily, but without leaving you in doubt about what you're doing. The good Physicists have an excellent ability to think in these terms, and they know what they're doing.

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I'm in the process of mathematical physics training and this is incredibly spot on to what I have been taught! –  DanZimm Jan 9 '14 at 3:30

I sympathize with the physicist and with the mathematician in different ways, but for the moment I lean toward the physicist's point of view.

Here is one of my own efforts toward explaining "$dx$": What is $dx$ in integration?

In some obvious ways $dx$ is treated like a number. One writes the chain rule as $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$, as if $du$ is a number that cancels. One says that if $u=\sin x$ then since $\dfrac{du}{dx}=\cos x$, one concludes that $du=\cos x \, dx$, just multiplying both sides by $dx$.

Mathematicians tend to warn people against naively thinking that intuitive arguments are valid, not suspecting that few of their students have advanced to the point of being able to see the plausibility of those intuitive arguments. "I realize it will naturally seem irresistably plausible to you people that [INSERT ARGUMENT HERE THAT IS INCOMPREHENSIBLE TO THE STUDENTS], and you've been thinking that since you were in first grade, but what you should learn from this course is what's wrong with thinking that." There is a fallacy in the thinking of mathematicians who say that.

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