I am reading Griffiths’ Introduction to Electrodynamics wherein the author defines a line integral as $\int^b_a\mathbf{v}\cdot d \mathbf{l}$. However, this definition in terms of the dot product seems problematic. Permitting this concept of a differential displacement vector, then, since $ \mathbf{v}\cdot d \mathbf{l} = $$|\mathbf{v}||d\mathbf{l}|$cos$\theta$, this definition is equivalent to $\int_a^b|\mathbf{v}||d\mathbf{l}|$cos$\theta$. But $|d\mathbf{l}|$ is the magnitude of an infinitesimal, a number which I believe is supposed to be infinitely small, so small that multiplying it by any normal finite number we might encounter in physics will also be an infinitesimal; therefore the value of the line integral should also be infinitesimal. But the value of all our line integrals being infinitesimal seems like a very wrong consequence. Am I missing something? Is there a better way to understand a line integral?

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    $\begingroup$ I am curious whether you have this same objection for ordinary integrals. $\endgroup$
    – d_b
    Commented Jun 25 at 17:17
  • $\begingroup$ Recall that integrals are defined as limits of sums. Since $|\Delta x| = |-\Delta x|$, we have $$\int_a^b \; f(x) \,|\mathrm{d}x| = \int_b^a \; f(x) \,|\mathrm{d}x| \text{.}$$ That is, the $|\mathrm{d}x|$ is indifferent to the direction along the interval of integration. $\endgroup$ Commented Jul 4 at 16:15

2 Answers 2


I don't have your textbook, so I don't know if he's made infinitesimals rigorous in some way or if it's just supposed to be a heuristic. Either way, the answer to your objection is that your integral is accumulating an infinite number of these infinitesimals, and so it can have some non-infinitesimal value.

You should think of the vector-valued infinitesimal $d\mathbf{l}$ and the scalar-valued infinitesimal $|d \mathbf{l}|$ as both being "infinitesimal" objects, the latter being like one from one-variable calculus, and so the paradox/miracle that you are encountering is really already one of one-variable calculus: in the expression $$ \int_0^1 x^2 \ dx = \frac{1}{3}, $$ we have this object $x^2 \ dx$ that "evaluates to zero" at any given point (the "area of the line segment above the point") but accumulates up to more than that (a limit of sums of areas of increasingly thin rectangles). Nothing has changed here in the vector-valued case, except that you're computing a dot product before you add up infinitely many infinitesimals.


Since we can't measure anything to infinite precision anyway, in physics it is never necessary to use this concept of "infinitely small." You can safely assume it merely means "very very small," say $10^{-10}$ meters, and then you never have to explicitly reason about infinitesimals anywhere.

With this shift in perspective the answer is easy to explain in finite terms: you can get a normal number from numbers as small as $10^{-10}$ if you add up $10^{10}$ of them, and that's how integration works. If you were trying to numerically integrate this integral using a Riemann sum with step size $10^{-10}$, for example, this is literally what you would have to do!

(The only change is you have to say that instead of holding exactly various identities hold up to an error of order $10^{-10}$. But, again, in physics this is not a problem since nothing can be measured to infinite precision anyway.)

This also has nothing to do with vectors and dot products, you could make exactly the same objection to the ordinary one-variable integral from ordinary calculus (what does that $dx$ actually mean, anyway?) and the answer is the same. You could even make the same objection to the ordinary derivative (isn't $\frac{df}{dx}$ trying to divide an infinitesimal by another infinitesimal?) and the answer is the same (we are just dividing a very very small number by another very very small number).

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    $\begingroup$ It would be enormously clarifying, I think, to teach students calculus this way - never talking about limits or infinity but only ever working with some very large number $N = 10^{10}$ and defining the derivative as $N \left( f \left( x + \frac{1}{N} \right) - f(x) \right)$ and so on, and writing everything down with explicit error terms. I wonder if anyone has done this. $\endgroup$ Commented Jun 25 at 0:39
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    $\begingroup$ @QiachuYuan Technically, to define an error you need to know the correct value, so you need to define the limit. In fact what you suggest is done in most textbooks on real analysis, as the $\epsilon$-$N$ and $\epsilon$-$\delta$ formulations of limit are nothing but estimating the error of successive approximations. The problem however, as many students can attest, is that explicitly calculating even the simplest limits in this way is extremely tedious. People still do it when they have to, though, like in numerical analysis, where the approximation is the actual object of interest. $\endgroup$
    – mlk
    Commented Jun 25 at 9:00
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    $\begingroup$ A perfect response for physics students. Similarly, it took me years to realize that (apparently) physicists did not really think of Dirac's delta as having to do with paradoxical things, but just as (an idealization of) a very narrow, very tall thing. The seeming mathematical imperative to really "go to the limit" is tangential to the physical idea... :) $\endgroup$ Commented Jun 25 at 17:22
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    $\begingroup$ I had always thought that manipulation of objects such as $df$ and $dx$ etc. can be made simpler for physics students to understand (and visualize) if one wrote them explicitly using finite differences $\delta f$ and $\delta x$, and then stipulating that any equality involving them is valid up to order $\sim\delta x$, i.e. ignoring $(\delta x)^2$ and above. E.g. if you have $f=f(x)$, and you write $df(x)/dx = g(x)$, a standard manipulation is to write $df=g\,dx$. What does that mean? Well, you can write this as $\delta f = g\cdot\delta x + O(\delta x^2)$. $\endgroup$
    – printf
    Commented Jun 26 at 5:39

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