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February 28

Conservative field in high dimension

Given a bounded, continuously differentiable vector field on a bounded simply connected domain in $\mathbb {R} ^{n}$ (where n>3); How can I prove that this field is conservative? Differentiating the field gives the Hessian matrix of the potential (if it exists). Clearly it is a necessary condition that the Hessian is symmetric, and for n ≤ 3 a symmetric Hessian is equivalent to zero curl, so in that case it is also a sufficient condition. However, curl is not defined for n>3, and I can't find any citeable reference that will tell me if a symmetric derivative matrix implies a conservative field. Any help will be appreciated. PeR (talk) 09:07, 28 February 2014 (UTC)[reply]

This can be formulated in the language of differential forms. Instead of your vector field

F=(F_{1},\dots ,F_{n})

, you look at the 1-form

v_{F}=F_{1}dx_{1}+F_{2}dx_{2}+\dots +F_{n}dx_{n}

. By Poincaré's lemma, if F is defined on all of

\mathbb {R} ^{n}

,

v_{F}

is exact if and only if it is closed. Closed means

dv_{F}=0

, which is (look at exterior derivative) another way of writing

\partial _{x_{i}}F_{j}=\partial _{x_{j}}F_{i}

. Exact means there is a 0-form (which is the same as a function) U such that

v_{F}=dU

, which is the same as

F=\nabla U

.

So a "symmetric derivative matrix" implies that the field is conservative, provided the domain on which it is defined is contractible (although simply connected is enough).

I hope that was not too confusing. There are elementary treatments of the fact (and they are not difficult), but all citable references I have on my desk at the moment are in German. For example, H. Heuser, Lehrbuch der Analysis 2, 8th edition, Stuttgart: Teubner 1993, Satz 182.2. Basically, the proof is the same as for two or three dimensions. —Kusma (t·c) 09:53, 28 February 2014 (UTC)[reply]

Great! Thank you so much for the explanation. PeR (talk) 08:09, 1 March 2014 (UTC)[reply]

Think you can find most of it in Introduction to smooth manifolds by John M. Lee YohanN7 (talk) 03:52, 3 March 2014 (UTC)[reply]

Sum of Reciprocal (prime) powers.

Let D be the set of integers > 1. Let E be the set of Prime Numbers (2,3,5,7,11...).

W = (sum over s in D, sum over t in D, (1/(s^t))
X = (sum over s in D, sum over t in E, (1/(s^t))
Y = (sum over s in E, sum over t in D, (1/(s^t))
Z = (sum over s in E, sum over t in E, (1/(s^t))

Any ideas on how to prove any of these being finite or infinite. Note that W>Y>Z and W>X>Z.Naraht (talk) 19:14, 28 February 2014 (UTC)[reply]

Using the sum of a geometric series,

W=\sum _{n\geq 2}\sum _{k\geq 2}{\frac {1}{n^{k}}}=\sum _{n\geq 2}{\frac {1}{n^{2}(1-{\frac {1}{n}})}},

and that sum converges. So all of your sums converge. —Kusma (t·c) 20:06, 28 February 2014 (UTC)[reply]

But isn't

\sum _{n\geq 2}{\frac {1}{n^{2}(1-{\frac {1}{n}})}}

=1?

— Preceding unsigned comment added by Naraht (talk • contribs) 23:21, 1 March 2014‎

Indeed it is, since

{\frac {1}{n^{2}(1-{\frac {1}{n}})}}={\frac {1}{n(n-1)}}={\frac {1}{n-1}}-{\frac {1}{n}}

. So

W=1

. —Kusma (t·c) 06:44, 2 March 2014 (UTC)[reply]

Any information on the values for X, Y, or Z or which is greater, X or Y?Naraht (talk) 04:48, 3 March 2014 (UTC)[reply]

I don't know anything about these values. For Y, the closest thing I can find is the "integral" section in Prime zeta function. —Kusma (t·c) 19:54, 4 March 2014 (UTC)[reply]

Thank you.Naraht (talk) 20:04, 4 March 2014 (UTC)[reply]