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My question is, that given a vector field only numerically discrete in space, is there a way to estimate its vector potential?

Theoretically, I see this which requires the vector field over all of $\mathbb{R}^3$. Further, I also see Will Jagy's answer using Poincare's lemma which I like much more. But in this case I still need to know the entire vector field from $(0,0,0)$ to the point $(x,y,z)$ where I want to estimate the vector potential because of $t\in(0,1)$. Better than $\mathbb{R}^3$ but still requiring too much information.

Is there any way to use only some local information from a vector field and estimate the local vector potential? I don't need to know the entire vector potential and I don't have the entire vector field anyway. It seems to me it should be possible because from what I understand, the curl is defined as a limit and seems to be a local property. So can I reverse it locally? Any insight or references would be appreciated.

Just to clarify, I only have discrete (in space) numerical measurements of the vector field in a very small finite corner of $\mathbb{R}^3$ away from the origin. And I am hoping to use them to estimate the local vector potential numerically in the same corner.

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  • $\begingroup$ My first guess: numerically approximate the integral in the Poincare lemma, using sums in place of integrals, since the integral is a limit of sums. $\endgroup$
    – Ben McKay
    Commented Jan 15, 2014 at 9:00
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    $\begingroup$ This is not really a research level question. $\endgroup$
    – Ben McKay
    Commented Jan 15, 2014 at 9:00
  • $\begingroup$ @BenMcKay A matter of opinion surely. This question really did come up in my current original research and I can't seem to find anything where this problem has been discussed before. If you know of any, please let me know. $\endgroup$
    – Fixed Point
    Commented Jan 15, 2014 at 9:43
  • $\begingroup$ @BenMcKay Sorry for the confusion. The question isn't about how to compute a (1D Riemann) integral. Those are easy. The question is rather how to "invert" the curl operator, and even that, locally. The question is, what is the integral that I should be evaluating, i.e. I don't have field values from $[0,5100]$. I only have values in $[5000,5100]$. Is there another formulation of the curl inverse which requires only local values? $\endgroup$
    – Fixed Point
    Commented Jan 15, 2014 at 10:02

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Let's put the origin in your corner of ${\mathbb R}^3$, and say you have measurements of the field $\vec{F}$ at lattice points $\epsilon [i,j,k]$ for integers $0 \le i \le m$, $0 \le j \le n$, $0 \le k \le p$. One version of the vector potential at $(x,y,z)$ is $$ {\vec V}(x,y,z) = \left[ \int_0^z F_2(x,y,t)\ dt, \int_0^x F_3(s,y,0)\ ds - \int_0^z F_1(x,y,t)\ dt, 0\right]$$ The integrals (when $x,y,z$ are lattice points) can then be approximated by sums over lattice points by the usual numerical methods.

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  • $\begingroup$ Thanks Robert, but the field is a physical (magnetic) field so I can't move the origin. The origin is fixed in physical space and suppose my measurements are at some lattice in let's say $(x,y,z)\in[5000,5100]^3$. I don't have any measurements in $[0,5000]^3$. Any way for me to estimate the vector potential in a subset of $[5000,5100]^3$? If I have the necessary info needed for an integral then of course numerical integrating won't be a problem. $\endgroup$
    – Fixed Point
    Commented Jan 15, 2014 at 9:51
  • $\begingroup$ @FixedPoint You can surely relabel your coordinates so that $(x,y,z)\in[0,100]^3$. Alternatively, just integrate starting from $(5000,5000,5000)$ instead of $(0,0,0)$. $\endgroup$
    – j.c.
    Commented Jan 15, 2014 at 10:48
  • $\begingroup$ Aha, the translation invariance of the curl. Perfect, exactly what I was looking for. Thanks everyone. $\endgroup$
    – Fixed Point
    Commented Jan 18, 2014 at 1:02

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