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I'm a bit confused with some notation I encounter in physics calculus. Consider this: enter image description here

Taken from here. Integration operates on functions, correct? What does it mean to integrate $\frac{d{\bf p}}{dt} dt$? Could someone point me to some reading about the manipulation of Leibniz notation?

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  • $\begingroup$ Unless I'm missing something, the answer is in Wikipedia. $F = \frac{dp}{dt}$ relates force to momentum. $\endgroup$
    – HDE 226868
    Commented Aug 18, 2014 at 20:53
  • $\begingroup$ Ok, but is that the net force or the change in momentum that a single force imparts? How do we factor that change in? I apologize if I'm not making much sense; I'm struggling with some intuition here. $\endgroup$ Commented Aug 18, 2014 at 20:57
  • $\begingroup$ Comment to the question (v2): It seems the question essentially is How to interpret vector-valued integration formulas? $\endgroup$
    – Qmechanic
    Commented Aug 18, 2014 at 20:57
  • $\begingroup$ @RustyDoornobs: I would think it's the total change in momentum - note the integral. But if QMechanic is right, and the question is about vectors, then this would be wrong. $\endgroup$
    – HDE 226868
    Commented Aug 18, 2014 at 20:58
  • $\begingroup$ OK, I believe one source of confusion for me was not realizing that NET FORCE = dp/dt, not just a single force. Generally, I see a lot of manipulation with differentials in physics. Is there something I could read to better understand this? I think that there is a weakness in my calculus intuition in this area. $\endgroup$ Commented Aug 18, 2014 at 21:01

3 Answers 3

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While in some contexts,

$$ \frac{d\textbf{p}}{dt}dt = d\textbf{p} $$

is correct and is mathematically rigorous, there is a straightforward way to derive impulse as the change in momentum. Consider a function $f(x)$ where $f:\mathbb{R}^m \to \mathbb{R}^n$ and $f \in C^1$. Suppose its derivative is $g = f'(x)$. We can consider integrating $g(x)$ over some interval $I = [x_i, x_f] \subset \mathbb{R}^m$ such that

$$ \int_I g(t)dt = \int_{x_1}^{x_2}g(t)dt = f(x_2) - f(x_1), $$

by the fundamental theorem of calculus. We can apply this result to momentum to get your result. Consider that the momentum $\textbf{p}(t)$ of a particle is simply a function $\textbf{p}: \mathbb{R} \to \mathbb{R}^3$. The impulse is defined as the net external force $\textbf{F}_{ext}(t)$ that a particle experiences over some time interval $I = [t_1, t_2]$, or

$$\textbf{J} \equiv \int_I \textbf{F}(t)dt,$$

where the "ext" has been dropped for brevity. (It is still understood that it represents the net external force.) By definition, $\textbf{F}(t) \equiv \textbf{p}'(t),$ so we can apply the fundamental theorem of calculus to obtain the desired result:

$$\int_{t_1}^{t_2} \textbf{F}(t)dt = \textbf{p}(t_2) - \textbf{p}(t_1) = \Delta \textbf{p}.$$

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  • $\begingroup$ Yes, I understand from the second line and below in the image. My confusion lies in the logic made to go from the first line to the second line. $\endgroup$ Commented Aug 18, 2014 at 21:09
  • $\begingroup$ Edited to show how the fundamental theorem of calculus is used to derive the result. $\endgroup$
    – Ultima
    Commented Aug 18, 2014 at 21:20
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    $\begingroup$ No, $(dp/dt)dt=dp$ is not an abuse of notation. In this notation, $dp$ and $dt$ can be interpreted as infinitesimal numbers. That's how Leibniz interpreted them, and that interpretation has rigorous logical support in non-standard analysis: en.wikipedia.org/wiki/Non-standard_analysis $\endgroup$
    – user4552
    Commented Aug 19, 2014 at 1:09
  • $\begingroup$ One can also write it like $dp = (\partial p/\partial t) \, dt$, which is true in a well-defined way when viewed as 1-form, using only standard calculus. $\endgroup$ Commented Aug 19, 2014 at 18:53
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What does it mean to integrate $\frac{d\mathbf p}{dt}dt$?

First, and in scalar form, recall from elementary calculus that

$$\int_{x_1}^{x_2} dx = x_2 - x_1 $$

Second, recall that

$$f(x + dx) = f(x) + f'(x)dx$$

where

$$f'(x) = \frac{df(x)}{dx} $$

Denoting the differential of $f$ as

$$df = f(x + dx) - f(x)$$

we have

$$df = f'(x)dx$$

Since

$$\int_{x_1}^{x_2} f'(x)dx = f(x_2) - f(x_1) = f_2 - f_1$$

it follows that

$$\int_{f_1}^{f_2} df = f_2 - f_1 = f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(x)dx$$

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Integration operates on functions, correct?

No. Integration dates back to ca. 1670. The notion of a function gradually evolved and didn't get put into its modern form until ca. 1830. Let's look at your expression

$$ \int_{p_1}^{p_2} dp .$$

This is Leibniz's notation, and what he means by it is a sum of infinitely many terms $dp$. Each of these terms is an infinitesimal change in the variable $p$, which doesn't have to be a function of anything.

There is a lot of confusion about the way physicists and engineers use differentials, because there was a period ca. 1880-1960 when it was believed that they were logically suspect and the only correct way to understand calculus was in terms of limits. Non-standard analysis cleared up this confusion and showed that infinitesimals were not logically problematic. A nice book at the freshman calc level that explains the modern way of looking at this is Keisler, https://www.math.wisc.edu/~keisler/calc.html , which is free online.

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  • $\begingroup$ So, what does it mean when we integrate dy/dx, for example? The following I understand: integrating dx results in one, big delta x. That makes sense intuitively to me since we are adding up many infinitesimal changes to determine the total change. However, often I hear that we integrate the 'derivative'. If the derivative is dy/dx, how do we integrate that? We just tack on a dx so the dx's cancel out? There is my confusion. By the way, this is the second time I've been linked to that book. I'll definitely give it a good look. Thanks for your reply. $\endgroup$ Commented Aug 19, 2014 at 1:53

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