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This question relates to this post.

From what I know in calculus and standard analysis, strictly speaking, there is no meaning of $dx$. It only makes sense when combining with another $d$, e.g. $df/dx$ as derivative or integration, e.g. $\int f(x) dx$. The $dx$ in derivative or integration has a definite meaning inside the definition of derivative or integration, respectively.

However, in differential form, it is given as

$\omega = \frac{ \omega_{\mu_1,\cdots, \mu_r}}{r!} dx^{\mu_1} \wedge\cdots\wedge dx^{\mu_r} $

where $dx$ appears explicitly. What does $dx$ mean in differential form? Physicist usually say it is infinitesimal. However, infinitesimal does not mean anything in standard analysis, or I am completely mistaken?

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2 Answers 2

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$dx_1$ is a differential 1-form (aka a covector field) which associates to each point in space a linear map from $\mathbb{R}^n \to \mathbb{R}$. The action of this linear map is to take a vector and spit out its component in the $x_1$ direction. In other words, it is the covector $\begin{bmatrix} 1&0&0&...&0 \end{bmatrix}$.

The covector fields $dx_i$ span all covector fields in the algebra of these guys over real valued functions, just because you can write

$$\begin{bmatrix} f_1(x)&f_2(x)&f_3(x)&...&f_n(x) \end{bmatrix}$$ as $$f_1(x)dx_1+f_2(x)dx_2 + ...+ f_n(x)dx_n$$

The way you integrate a differential one form $\omega$ along a curve is pretty simple. Given a curve $\gamma: [0,1] \to \mathbb{R}^n$, partition it into $k$ pieces. Then you can form the sum

$$\sum_{i=1}^k \omega\big|_{\gamma(\frac{i}{k})}(\gamma(\frac{i}{k}) - \gamma(\frac{i-1}{k}))$$

The limit of this sum as $k \to \infty$ is the integral of the one form. Alternatively you could have plugged in actual tangent vectors at each point, instead of the approximate tangent vectors I used, but I think it is somewhat easier to conceptualize what is going on this way.

Why are these reasonable things to look at? Why would anyone ever think of integrating such a thing? Answer: because the derivative of a function $f:\mathbb{R}^n \to \mathbb{R}$ IS a covector field, and we certainly want to be able to integrate a derivative over a curve, and have a fundamental theorem of calculus. In fact, $dx_1$ is just the derivative of the coordinate function $f(x_1,x_2,...,x_n) = x_1$. If you think about how you would sensibly integrate the derivative of a function, you will probably recover my notion of integration above.

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  • $\begingroup$ Little typo at the beggining: "[...] and spit out its component in the $x_1$ direction. [...]" instead ;-) $\endgroup$ Feb 5, 2014 at 15:04
  • $\begingroup$ So is $dx_i$ a collection of points (field) for the covectors in the differential form? $\endgroup$
    – user26143
    Feb 5, 2014 at 15:11
  • $\begingroup$ @user26143 Here is an example $xydx +dy$ is a family of covectors. At the point $(x,y)$ in the plane, this covector is the linear map whose matrix is given by $\begin{bmatrix} xy & 1 \end{bmatrix}$. $\endgroup$ Feb 5, 2014 at 15:15
  • $\begingroup$ Sorry, how did you get $[x^2\,\,\,\, 1]$? $\endgroup$
    – user26143
    Feb 5, 2014 at 15:16
  • $\begingroup$ Oops, typo. haha $\endgroup$ Feb 5, 2014 at 15:18
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My first answer just answered the question directly, but I would like to take a little bit of time to explore this circle of ideas.

Let $f(x,y) = x^2y$.

The derivative of a function gives the best linear approximation to the function at a point. This remains true in higher dimensions. The derivative of the function $f$ above is

$df = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\end{bmatrix} = \begin{bmatrix} 2xy & x^2\end{bmatrix}$.

The conceptual meaning of the derivative is this:

$$f(x+\Delta x, y+\Delta y) \approx f(x,y) + df(\begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix}) = x^2y + \begin{bmatrix} 2xy & x^2\end{bmatrix}\begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix} = x^2y+2xy\Delta x + x^2\Delta y$$

In other words, at each point $(x,y)$ the derivative is a linear map which takes a small change $\begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix}$ away from the point $(x,y)$ and returns the approximate change in $f$ resulting from that.

Now say someone told me that a certain function $g$ with $g(0,0)=0$ had derivative $\begin{bmatrix} y\cos(xy) & x\cos(xy)\end{bmatrix}$, and I wanted to figure out what $g$ was. In this case I could probably just solve the differential equations $\frac{\partial g}{\partial x} = y\cos(xy)$ and $\frac{\partial g}{\partial y} = y\cos(xy)$ by inspection, but this would not always be possible.

Let us stick to the somewhat easier problem of approximating $g(1,1)$. Here is my idea for doing that: I will pick a path from $(0,0)$ (whose value I know) to $(1,1)$. I will split that path up into millions of vector changes. Then I will use what I know about the derivative to approximate the change in $g$ over each of those small changes and add them up. This should give me a pretty reasonable approximation.

In this case, I can see that I can pick the path $\gamma:[0,1] \to \mathbb{R}^2$ given by $\gamma(t) = (t,t)$. Splitting this into $k$ pieces, I have the following approximations:

$$g(\frac{1}{k},\frac{1}{k}) \approx g(0,0) + dg|_{(0,0)}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right)$$.

So then

$$g(\frac{2}{k},\frac{2}{k}) \approx g(\frac{1}{k},\frac{1}{k}) + dg|_{(\frac{1}{k},\frac{1}{k})}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right) \approx g(0,0) + dg|_{(0,0)}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right) + dg|_{(\frac{1}{k},\frac{1}{k})}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right)$$.

Continuing on in this way , we will see that

$$g(1,1) \approx g(0,0) + \sum_{i=0}^k dg\big|_{\frac{i}{k}}\left(\begin{bmatrix} \frac{1}{k} \\ \frac{1}{k}\end{bmatrix}\right)$$

It makes sense to give some name to this process. We define the limit of the sum above to be the integral of the covector field $dg$ along the path $\gamma$. Refer to my other post for the general definition, instead of just a particular example like this.

So far we have defined the integral only for derivatives of functions, and we have defined it exactly in such a way that the following fundamental theorem of calculus holds:

$$g(P_1) - g(P_0) = \int_\gamma dg$$ for any path $\gamma$ from $P_0$ to $P_1$. But the definition of the integral never used the fact that we were integrating the derivative of a function: it only mattered that we are integrating a covector field (i.e. a gadget which eats change vectors and spits out numbers). So we can use exactly the same definition to give the integral of a general covector field $\begin{bmatrix} f(x,y) & g(x,y)\end{bmatrix}$, which may or may not be the differential of a function.

(There are certainly covector fields which are not derivatives of functions. For example, $\begin{bmatrix} x & x\end{bmatrix}$ could not be the differential of a function, for if it were we would have $\frac{\partial f}{\partial x} = x$ and $\frac{\partial f}{\partial y} = x$. But then the mixed partials of $f$ would not be equal, contradicting Clairout's theorem.)

$dx$ is the constant covector field $\begin{bmatrix} 1 & 0\end{bmatrix}$, and $dy$ is the constant covector field $\begin{bmatrix} 0 & 1\end{bmatrix}$. So we can write any covector field $\begin{bmatrix} f(x,y) & g(x,y)\end{bmatrix}$ as $f(x,y)dx + g(x,y)dy$. Integrating this thing along a curve is PRECISELY what you defined as a line integral in your first multivariable calculus course.

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    $\begingroup$ Thanks a lot for this detailed explanation. $\endgroup$
    – user26143
    Feb 5, 2014 at 15:57
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    $\begingroup$ Very nice explanation $\endgroup$
    – Karl
    Apr 5, 2016 at 6:13

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