# Product of two differentials (infinitesimals)

In a physics derivation today my lecturer ended up with two differentials multiplied together, like this term in a longer summation:

$p^2 q^2 \mathrm{d} p ~ \mathrm{d} q$

And then he said this term becomes zero because the two differentials are multiplied together, and can be removed. Is there a calculus justification for this? It doesn't make sense to me, but we got the right result...

• It depends on what it's appearing next to. By itself it can be interpreted as the area of an infinitessimally small rectangle, so it will be negligible if added to a term like $$\frac{pq\:\text dp}{10^{10^{10^{10}}}}$$ which is the area of a rectangle with only one infinitessimal side, and the other side finite. Besides, physicians do a lot of "not really correct but somehow giving the correct answer anyway" shortcuts when doing calculus, like reducing fractions by differentials. – Arthur Oct 4 '12 at 7:26

## 1 Answer

Often when physicists write $\mathrm{d}p$ on its own, they view it as some very small quantity representing some change in $p.$ For example, if $\dfrac{\mathrm{d}p}{\mathrm{d}t} =5$ then they might write $\mathrm{d}p = 5\mathrm{d}t$ and interpret that as "a small change in time causes a small change in $p$ five times greater". Also, when they ignore terms with multiple differentials, they are using their intutition that when we are already dealing with very small quantities, the terms that are products of such small quantities are an entire order of magnitude smaller and thus, negligible.

They do this to avoid repeating the same steps of formalism to reach the same result more rigorously, which would be to use $\Delta t = t-t_0$ to represent some change in time, $\Delta p = p(t) - p(t_0)$ to represent the corresponding change in $p,$ do all their working like this, and then finally taking the limits like $\dfrac{\mathrm{d}p}{\mathrm{d}t} = \displaystyle\lim_{\Delta t \to 0} \frac{ \Delta p}{\Delta t}$ so all the $\Delta$'s become $\mathrm{d}$'s.

So now suppose we were doing it properly and we have gotten to this:

$$\Delta y = 52 \Delta p + \pi \Delta q + p^2 q^2 \Delta p \Delta q.$$

Divide through by $\Delta t$:

$$\frac{\Delta y}{\Delta t} = 52 \frac{\Delta p}{\Delta t} + \pi \frac{\Delta q}{\Delta t} + p^2 q^2 \frac{\Delta p}{\Delta t} \Delta q.$$

Now when we take the limit as $\Delta t \to 0$ we get $$\dfrac{\mathrm{d}y}{\mathrm{d}t} = 52\dfrac{\mathrm{d}p}{\mathrm{d}t} +\pi \dfrac{\mathrm{d}q}{\mathrm{d}t}.$$

The phenomenon you observed occurs - the final result makes it appear that only the terms with a single differential mattered while terms with two or more differentials vanish. That is because in this limit, all $\Delta$ terms go to $0$ but the ones written above as ratios tend towards derivatives, so in the third term:

$$p^2 q^2 \frac{\Delta p}{\Delta t} \Delta q\to p^2q^2 p'(t) \cdot 0 =0.$$

So the reason you can ignore them is because when you do the background formalism with limits, the terms with two or more differentials will go to $0$ in the end anyway.

• Thanks, it seemed awkward having to write "dp dq is zero" in my upcoming assignment. Now I know it is a valid thing to do, and your example is a great explanation! :) – Thomas Oct 4 '12 at 9:00
• Sweet, great explanation! – Asad Saeeduddin Dec 15 '16 at 14:29