This is something that has been bothering me and I hope the title kind of makes sense. It may be a stupid question but please be gentle. My question is, let's say we have current:


And I understand that if we want to find the total charge over some time, for example, you do:

$$q=\int I \; dt$$

And what this does, by the definition of an integral, is it sums every product of the infinitesimally small $dt$ and the value of $I$ at that $t$.

However, how come you can't have an infinitesimally small current? Meaning, why can't this exist:


Is this a calculus question? Or is this just by the definition of current? I'm really confused.

Thanks in advance.

  • 1
    $\begingroup$ An infinitesimally small current can exist, however it would be similar to the second time derivative of the charge. Your notation, however, doesn't work. But, to be honest, I don't know the correct notation with infinitesimals, because this is NOT how you define it in modern calculus/real analysis. "Infinitesimals" don't exist there - you'd have to do what is called "nonstandard analysis" to get to them. In usual analysis, you work with limits instead. Still, part of the notation remains... $\endgroup$ – Martin Sep 2 '14 at 16:11
  • $\begingroup$ I'm not quite sure what you are asking, but this old question has many discussions of differentials and integrals at different levels. $\endgroup$ – ACuriousMind Sep 2 '14 at 16:11
  • $\begingroup$ @Martin I know this is far-fetched but can you ever manipulate this equation to be something like $t=\frac{dq}{dI}$? Why/why not? $\endgroup$ – Shahar Sep 2 '14 at 16:19
  • $\begingroup$ @Shahar: $t=dq/dI$ doesn't make sense because $I\equiv dq/dt$ (it's the definition). You could say $t=\int dq/I$, but that's not the same thing as what you wrote (and many mathematicians will cringe when they see that). $\endgroup$ – Kyle Kanos Sep 2 '14 at 17:05
  • $\begingroup$ Try drawing a diagram or graph. $\endgroup$ – akrasia Sep 2 '14 at 18:41

Meaning, why can't this exist:


Is this a calculus question? Or is this just by the definition of current?

It can exist, and I would consider this a calculus question.

When Newton and Leibniz originally invented calculus, they conceived of a derivative as a (usually finite) ratio of two infinitesimal numbers. Infinitesimal means smaller in absolute value than any positive real number, but greater than zero. This was the way people did calculus for about 150 years. During the 19th century, mathematicians decided to redo the foundations of calculus using limits rather than infinitesimals. Ca. 1960 it was shown that there was actually nothing wrong with the infinitesimal approach. There is a nice online book by Keisler http://www.math.wisc.edu/~keisler/calc.html that does freshman calc using an updated version of the infinitesimal approach. There is nothing particularly hard or scary about any of this, and modern historical scholarship has shown that Leibniz basically had the whole system figured out correctly in many ways, even by modern standards of rigor: Blaszczyk, Katz, and Sherry, Ten Misconceptions from the History of Analysis and Their Debunking, http://arxiv.org/abs/1202.4153

In the Leibniz notation, d just means "a little bit of..." (where "little" means "infinitesimal"). Your equation $dI=dq/dt$ makes perfect sense. It simply says that a little bit of charge flowed over a little bit of time, and the result was a little bit of current.

There is a notational issue, which is that although we often think of the derivative $q'(t)$ as being the same as the ratio of the infinitesimals $dq/dt$, actually if you want to be consistent you have to define the derivative as the standard part of that ratio. The standard part of a number is the closest real number to that number. In your equation, if division of the numbers $dq$ and $dt$ gives an infinitesimally small result, then the standard part of $dq/dt$ is zero, so the derivative $q'(t)$ is zero. So there is a distinction between a number and its standard part. Leibniz was in fact careful about this distinction and had a special notation to represent it, but later writers felt that it was too cumbersome and stopped writing it. When infinitesimals were rehabilitated in the 1960s, they had to reinvent this distinction, and they invented the term "standard part."

There are certain grammatical rules for manipulating infinitesimals, but your equation doesn't violate any of them. For example, it would be grammatically incorrect to write

$$ du=1/dv .$$

This is because if you take a finite number divided by an infinitesimal number, you are guaranteed to get an infinite number, but a notation like $du$ is used to represent an infinitesimal number, not an infinite one.

Your equation $dI=dq/dt$ doesn't violate these rules because it's perfectly possible to divide an infinitesimal by another infinitesimal and get a result that is infinitesimal rather than finite. For example, suppose that $dx$ is a nonzero infinitesimal. Then we have as an identity $dx=dx^2/dx$.


The current by definition is the flow of charge. Also, time and charge can be quantified infinetsimally at least when dealing with macrospcopic system. As such, a small amount of charge during a small time interval is what current it. As the system gets smaller, quantum level, current is quantized based on ballistic charge trasport.

So, dealing with current, if you want an infinitesimal current, it will be the second derivative if the current flow. This can be used to talk about charge drift-diffusion for example, as used to describe curvature is in calculus

  • $\begingroup$ "This can be regarded as curvature"? What's that? $\endgroup$ – innisfree Sep 2 '14 at 20:33
  • $\begingroup$ See Edit. I calculus, the second derivative is used to describe curvature. In transport theory, it is used to describe diffusion process. $\endgroup$ – mcodesmart Sep 2 '14 at 20:36
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    $\begingroup$ No, this is just wrong. A second derivative is a real number, not an infinitesimal. $\endgroup$ – user4552 Oct 11 '14 at 0:30

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