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I am trying to understand section 3 of Here, titled "what is a derivative". At equation (3.2) Hestenes defines the derivation in relation to the integral, as follows:

$$ \partial A =\lim_{d\omega\to0} \frac{1}{d\omega} \oint_{\partial \mathcal{M}} d\sigma A \tag{1} $$

where $d\omega$ is a volume element. Hestenes suggests that this is a very good way to think of a derivative (perhaps even the best way).

I am trying to apply this definition to the 1D case, but I am struggling to do so. Specifically, my goal is to start from (1) and obtain (2):

$$ \frac{d A[x]}{d x} = \lim_{dx\to 0} \frac{A[x+dx]-A[x]}{dx} \tag{2} $$

My thoughts and assumptions are as follows.

  1. Since we are dealing with the 1D case, should I be using $Cl_1(\mathbb{R})$; that is, the Clifford algebra of dimension 1 over the reals with basis element $\{\hat{\mathbf{x}}_1\}$?
  2. Hestene claims $d\omega$ is m-vector-valued differential; that is, it is a pseudo-scalar from the tangent space of $\mathcal{M}$ evaluated at point $x \in \mathcal{M}$. In our 1D case, $d\omega=Idx=\hat{\mathbf{x}}_1 dx$ where I is the unit pseudoscalar of $Cl_1(\mathbb{R})$
  3. $A$ is a function of $x$. Thus, I write $A[x]$.
  4. Hestenes claims that $\partial=\partial_x$ is the derivated with respect to a vector $x$. In 1D, therefore $\partial_x=\partial/\partial x$.
  5. Hestenes claims that $d\sigma$ is a (m-1)-valued pseudoscalar also in the tangent space of $\partial \mathcal{M}$ evaluated at point $x$. I am not sure how to downgrade $\mathcal{M}$ to $\partial \mathcal{M}$ such that it is $0$-dimensional? Am I supposed to consider $d\sigma$ as a pseudoscalar of $Cl_0(\mathbb{R})$? If so then is the answer just $d\sigma=dx$?
  6. Finally, Hestenes claims (starting from equation 3.2) that one needs $d\omega \wedge \partial =0$ in order to get to the geometric product. In 1D, why is $\hat{\mathbf{x}}_1dx \wedge \partial=0$? Is $\partial$ assumed in the tangent space of $\mathcal{M}$ and thus parallel to $\hat{\mathbf{x}}_1$?
  7. What becomes of the counter integral in 1D... does it collapses to a simple definite integral? I hope I dont have to integrate from a to b then from b to a to get back to the original point and thus to complete the "contour". If so then the integrals would simply cancel each other: $\oint_R f(x)dx = \int_a^b f(x)dx + \int_b^a f(x)dx=0$... that can't be good :(
  8. Since the left-most term of (1) is a derivative of A and the right-most term contains $A$ and not $A'$, then I feel the contour integral in 1D must collapse to a non-integral in order to avoid raising A to its anti-derivative.
  9. What is $\partial \mathcal{M}$ for a 1D manifold $\mathcal{M}$ - I am assuming it is simply an interval $[x,x+h]$, where h is an infinitesimal element?
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    $\begingroup$ Isn't the right 1d analog for (1) $$ \lim_{\varepsilon \rightarrow 0} \dfrac{1}{\varepsilon}\int_{x-\varepsilon/2}^{x+\varepsilon/2}f(z)dz=F'(x) $$, rather than (2)? It seems like you want to integrate over the boundary of a "volume" element, the boundary of $(x-\varepsilon/2,x+\varepsilon/2)$, and then take the volume $\varepsilon$ to zero. $\endgroup$
    – user762914
    Jun 26, 2020 at 21:32
  • $\begingroup$ @Renard You may be correct, but I do not understand how the definite integral is a 1D version of a contour integral? My intuition regarding the contour integral is that the concept of returning to the starting point is important... but of course, in 1D returning to the starting point always nullifies the result because the integral is anti-symmetric. How is the 1D version of the contour integral the definite integral? $\endgroup$
    – Anon21
    Jun 26, 2020 at 21:52
  • $\begingroup$ @Renard I guess that the point of the contour integral is to create a finite region within the integrated space; thus it is equivalent to the definite integral in 1D. $\endgroup$
    – Anon21
    Jun 26, 2020 at 22:05

1 Answer 1

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The boundary of a 1D manifold is a 0D manifold: its two endpoints. A 0-dimensional integral is just a finite sum.

I'll call the basis vector $\mathbf e_1$, and the variable position vector $\mathbf x=x\mathbf e_1$. The manifold is $M=\{x\mathbf e_1\mid a\leq x\leq b\}\cong[a,b]$, and its boundary is $\partial M=\{a\mathbf e_1,b\mathbf e_1\}\cong\{a,b\}$. Actually, $M$ and $\partial M$ should also include information about orientation. The integrals are

$$\int_Md^1\mathbf x\,A(\mathbf x)=\int_a^b\mathbf e_1dx\,A(x)=\mathbf e_1\int_a^bA(x)\,dx,$$

$$\oint_{\partial M}d^0\mathbf x\,A(\mathbf x)=({}^-1)A(a)+({}^+1)A(b)=A(b)-A(a).$$

The derivative at a point $\mathbf y$ is defined thus (with different notation; I hope it's clear):

$$\partial A(\mathbf y)=\lim_{|M|\to0\\M\to\{\mathbf y\}}\frac{1}{\left(\int_M\,d^1\mathbf x\right)}\oint_{\partial M}d^0\mathbf x\,A(\mathbf x)$$

$$=\lim_{a\to y^-\\b\to y^+}\frac{1}{(b-a)\mathbf e_1}\big(A(b)-A(a)\big)$$

$$=\mathbf e_1^{-1}\lim_{a\to y^-\\b\to y^+}\frac{A(b)-A(a)}{b-a}.$$

This is not exactly the same as the usual definition, though we could take the limit "along a certain path in interval space", such as keeping one endpoint fixed: $a=y,\,b\to y^+$.


More generally, suppose $M$ is a 1D curve in a higher-dimensional space, parametrized as $\mathbf x=f(t)$ with endpoints $\mathbf x_1=f(t_1)$ and $\mathbf x_2=f(t_2)$. The integrals are

$$\int_Md^1\mathbf x\,A(\mathbf x)=\int_{t_1}^{t_2}\frac{d\mathbf x}{dt}A(\mathbf x)\,dt,$$

$$\oint_{\partial M}d^0\mathbf x\,A(\mathbf x)=({}^-1)A(\mathbf x_1)+({}^+1)A(\mathbf x_2)=A(\mathbf x_2)-A(\mathbf x_1).$$

The derivative on $M$ is defined in terms of integrals over sub-curves $M'\subset M$ containing the given point $\mathbf y=f(t_0)\in M$:

$$\partial A(\mathbf y)=\lim_{|M'|\to0\\M'\to\{\mathbf y\}}\frac{1}{\left(\int_{M'}\,d^1\mathbf x\right)}\oint_{\partial M'}d^0\mathbf x\,A(\mathbf x)$$

$$=\lim_{t_1\to t_0^-\\t_2\to t_0^+}\frac{1}{(\mathbf x_2-\mathbf x_1)}\big(A(\mathbf x_2)-A(\mathbf x_1)\big)$$

$$=\lim_{t_1\to t_0^-\\t_2\to t_0^+}\frac{1}{\left(\frac{\mathbf x_2-\mathbf x_1}{t_2-t_1}\right)}\left(\frac{A(\mathbf x_2)-A(\mathbf x_1)}{t_2-t_1}\right)$$

$$=\frac{1}{f'(t_0)}\lim_{t_1\to t_0^-\\t_2\to t_0^+}\frac{A\big(f(t_2)\big)-A\big(f(t_1)\big)}{t_2-t_1}.$$

If $t$ is arclength, then $f'(t_0)=\frac{1}{f'(t_0)}$ is the unit tangent vector to $M$ at $\mathbf y$.

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  • $\begingroup$ Actually the limit $$\lim_{(a,b)\to(y,y)\in\mathbb R^2\\ \quad\;\,a\leq y\leq b}\frac{A(b)-A(a)}{b-a}$$ is equivalent to the usual definition of the derivative. But there's also the strong derivative (see the linked questions there), which doesn't have the restriction $a\leq y\leq b$, and is not equivalent. $\endgroup$
    – mr_e_man
    Jul 9, 2020 at 21:23

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