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?


2 Answers 2


$dx_1$ 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.

  • $\begingroup$ 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

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.

  • 1
    $\begingroup$ Thanks a lot for this detailed explanation. $\endgroup$
    – user26143
    Feb 5, 2014 at 15:57
  • 1
    $\begingroup$ Very nice explanation $\endgroup$
    – Karl
    Apr 5, 2016 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.