BR Maths Corner-1

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Re: BR Maths Corner-1

Postby vina » 10 Aug 2008 07:11

Thread clean up please ..AniB. What does mathematics have anything with your post.

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Re: BR Maths Corner-1

Postby vsunder » 10 Aug 2008 07:57

Sorry times up. People on this forum are generally becoming extremely arrogant.
The lemmas are stated precisely as in my original papers. If the referees had no problems with the statement
then what is the issue.

The Swiss Cheese Lemma appears here:

Swiss Cheese Lemma

My institutional affiliation appears very clearly as I said in my post.

The lemma 2 that makes no sense appears here. If you feel it does not make sense write to the editorial board.
Also check Google to see how this second paper has influenced many people. Type the key words Nodal and my last name and it will spit tons of stuff.

Covering lemma

You can read the proofs now of the problems I posed. I have stated original problems that I discovered.
I dont need to be told what to read and do in mathematics. I had already made my name at 22 with the Kolmogorov inequality for Legendre polynomials. Here it is:

Legendre Polynomials

Here is a survey though in Spanish that tells you exactly what Kolmogorov did for Fourier series and I did for Legendre series. See what the author says on page 3, 20 odd years after my article.
Survey

Amber you need to be polite to people, you are very arrogant and have no mathematical research to your credit.
Googling further will lead you to many directions that people have taken my original ideas. As I said the circle of ideas in Lemma 2 have sparked a lot of activity. Instead you have chosen to ridicule me. J. Differential geometry is a top journal in the world and extremely difficult to get a paper published just take a look at the editorial board,
Yau, Donaldson are Fields medallists and Hamilton had a hand in the Poincare conjecture. As far as the Swiss cheese lemma, Freeman Dyson himself was excited since he had done work on stability of matter. Now you insult me. I hope you do realize the arrogance of your post. Look at my post I just posed the problem I did not taunt anyone.
Forum admins: I must say the forum has degenerated into a cesspool of name calling arrogance by petty foggers
and sundry others. I have decided to stop coming here. It was nice in the old days and I am happy I learnt many things from people.

Sagun

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Re: BR Maths Corner-1

Postby Rahul M » 10 Aug 2008 10:09

vsunder ji, plz check AOL.
will delete this post after you acknowledge and will clean up the thread.

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Re: BR Maths Corner-1

Postby AniB » 10 Aug 2008 11:12

Sorry to sully the thread (said the drunken soldier to Archimedes, before he killed him). Was merely trying to reach Amber G. Further discussion could be conducted elsewhere.

Actually the problem is interesting. Humans have distinct circadian rhythms that govern mood, awareness, and physiology. Oscillating genes like the homeobox (Hox) cause spatial body patterning.

There are also interesting periodicities in the exosphere/heliosphere. NASA’s “Living around a Star" program heads in this direction. Subtle changes in solar behavior can cause mini-ice ages like in the 1600s.

All that is needed is a coupling mechanism. I could propose one. (No not a physical force or 'cosmos' rays)

You are surely intrigued by dissipative systems, Brusselators, Bendixon limit cyles and n-body problems?

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Re: BR Maths Corner-1

Postby Rahul M » 10 Aug 2008 11:17

AniB, could you repost that astrology related post to the nukkad thread ?? I'm sure Amber G will notice it there.

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Re: BR Maths Corner-1

Postby ArmenT » 10 Aug 2008 11:58

Re: Carpet cutting - Would the answer be 2? I'm envisioning a staircase like pattern cut.

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Re: BR Maths Corner-1

Postby Satya_anveshi » 10 Aug 2008 22:26

Dear Posters,

I don't think I have any *qualification (especially in maths)* other than being a student to make this post and address to gurus. However, Can you all "gurus" please get off your taller than mountain egos?

1) don't offend fellow postors especially if the poster has put a personal stake into the post; Little bit of humility in seeking clarification will go a long way in setting the tone of the debate

2) don't take offence to another poster if the said postor turns out to be not deserving the high pedestal he claims he is holding (like what happened in case of Amber G's arrogant posts to VSunder).

Threads like these are a nice opportunity to learn from gurus but situations like above deprive me (and others) of this opportunity.

VSunder ji, If I may request you to please reconsider your decision for the benefit of people like me and I am sure there are many others who feel the same.

This forum is a very dynamic place and focus of the forum keeps changing based on India centric events. So, keeping option open to posts and contribute to analysis is one way of giving back to community.

Again there is nothing new in my posts that you gurus don't know of but sometimes students in class need to tell teachers to get back to topic in the interest of learning.

Regards,

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Re: BR Maths Corner-1

Postby SK Mody » 11 Aug 2008 03:05

vsunder wrote:Here is the first lemma:

Take a square it works for a cube in any dimension. Now play the following game. Lets take a square for example.
Bisect each side you get now 4 congruent mini squares. Call this Step 1. Each congruent mini square can be bisected further to get now 16 mini squares in step 2. Proceed so on. One gets at step n, 4^n squares.

Now remove N DISJOINT squares in any way, that is remove a step 10 square, a step 25 square etc. The remaining
figure is a big square with holes like Swiss cheese Prove you can cover the remaining figure; the Swiss cheese looking part by exactly C N squares. C is a universal constant that only depends on the dimension. I forget the value I found for C for dimension 2.

By cover I mean nothing must penetrate the holes where squares were taken out.


Not sure I understand the problem as stated. Do you mean as n-> infinity? Otherwise for example:
for n=1 and N=1, C=3
for n=2 N=2, C=2.5 (smallest value of C)
for n=3 N=3, C=3.33.. (smallest value of C)

Also, are you saying remove at most one square from every level (which is what I have assumed above)?
Certainly seems like a very interesting type of problem, but some ambiguities in your post.
Thanks.
Last edited by SK Mody on 11 Aug 2008 03:37, edited 2 times in total.

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Re: BR Maths Corner-1

Postby SK Mody » 11 Aug 2008 03:18

I would like make a request to members.

To problem posters:
Please give sufficient time for solution. I think even if a person visits the thread every other day, he should have a chance to have a go at some of the problems. I suggest the following time limits.
Problems considered easy: at least 3 days.
Problems considered moderately difficult: at least 1 week.
Problems considered very difficult (and I think we can put the problems such as posted by vsunder in that category): at least 3 weeks.

You can give more time but these should be the minimum. If you think this is too much, please remember that what matters is how much you learn by trying to solve a problem and not who gives the fastest solution. In fact I think it would be even better not to post a solution unless someone explicitly asks for it. That is the policy I intend to follow if I post a problem. Once in a while a problem poster can remind people of his problem by posting a link to it.

To problem solvers:
Please post solutions in tiny font.
Thanks.

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Re: BR Maths Corner-1

Postby ramana » 11 Aug 2008 08:59

AmberG, Vsunder is respected mathematician. his work was appreciated by the greats like S Chandrasekhar himself. I understand your strong point is physics. If one didn't understand the problem one can ask for clarification. No need to consider it humorous etc. I am sure you didnt mean to insult. So please edit your posts. Thanks, ramana

Vsunder, please don't go off. We do need you. ramana

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Re: BR Maths Corner-1

Postby Multatuli » 11 Aug 2008 13:49

Some info on V. S. Sunder :

http://www.imsc.res.in/~vijay/

I have to agree with Amber G, the problem should have been formulated more precisely, very sloppy formulation.

V Sunder, why are you so annoyed and insulted ? Can anyone diminish you as a mathematician through his/her comments ?

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Re: BR Maths Corner-1

Postby SK Mody » 11 Aug 2008 19:06

Amber G wrote:
There is a carpet 9(units) x 16 (units), and you want to cover 12 x 12 square. In how many (minimum) pieces you have to cut the original carpet (and resew them) to make it fit 12 x 12.?


Solution 1:

Two Pieces. Use a staircase cut.
Start 4 units down from top left and cut 3 units right. Stop.
Cut 4 units down. Stop.
Cut 3 units right. Stop.
cut 4 units down. Stop.
Finally cut 3 units right. You end up 4 units up from bottom right..
Now slide the upper right piece 3 units right and 4 units down - you get your 12 square carpet.


Solution 2 (Non-constructive)

1. Clearly it can be done using four pieces. Cut four units off the top 16. Cut the remaining pieces into 3x4 pieces and attach them to the
right.
2. Therefore the answer must be <= 4.
3. It is given that the answer is surprising.
4. 4 is not surprising and and 3 is not surprising. 0 or 1 are clearly impossible.
5. Therefore the answer must be two.
QED.


I should get a toffee.

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Re: BR Maths Corner-1

Postby ramana » 12 Aug 2008 20:49

x-posted..

Hyderabad is going to host the next International Congress of Mathematicians?This is HUGE...it is held once every 4 years and is the biggest math conference in the world...and it hands out the Fields Medal which is considered the Nobel Prize in Mathematics...

http://en.wikipedia.org/wiki/Internatio ... ematicians

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Re: BR Maths Corner-1

Postby Multatuli » 12 Aug 2008 22:24

a, b, c are positive and a + b + c = 1

Show that ( 1/a - 1 ) * ( 1/b - 1 ) * ( 1/c - 1 ) is equal to or greater then 8.

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 00:33

Multatuli wrote:a, b, c are positive and a + b + c = 1

Show that ( 1/a - 1 ) * ( 1/b - 1 ) * ( 1/c - 1 ) is equal to or greater then 8.



Given that a+b+c = 1
The minimum value of 1/a + 1/b + 1/c is when a = b = c = 1/3. (This can be shown by first noting that the minimum in such a case must be a local turning point. Then show that a=b=c=1/3 is the only turning point.)

Therefore 1/a + 1/b + 1/c >= 3 + 3 + 3 = 9
=> (bc + ac + ab)/abc >= 9
=> bc + ac + ab >= 9abc

Adding a + b + c - 1 (=0) to the left hand side:
=> 1 - a - b - c + bc + ac + ab - abc >= 8abc
=> (1-a)(1-b)(1-c) >= 8abc
=> (1/a - 1)(1/b - 1)(1/c - 1) >= 8

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 01:12

There is a box containing lots of milk chocolates, coconut toffees and bittersweets (there may be different quantities of each). I choose 3 with eyes closed. Show that the probability that I will get at least two of one type is greater than or equal to 7/9. (Assume that there are enough sweets in the box that taking a few out does not change the proportion significantly).

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Re: BR Maths Corner-1

Postby Nandu » 13 Aug 2008 01:42

SK Mody wrote:There is a box containing lots of milk chocolates, coconut toffees and bittersweets (there may be different quantities of each). I choose 3 with eyes closed. Show that the probability that I will get at least two of one type is greater than or equal to 7/9. (Assume that there are enough sweets in the box that taking a few out does not change the proportion significantly).


It is variant on Multatali's poser, isn't it?

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 02:14

Yup, it is.

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Re: BR Maths Corner-1

Postby Neshant » 13 Aug 2008 09:47

Bear with me as math is not my strong point. I may be talking nonsense here but I'm gonna school you guys on what i think is the solution :

Number of ways of picking 2 out of 3 candies being the same for a single item in the set :
C(3,2) = 3! / 2!(3-2)!
C(3,2) = 3

Total number of ways of picking 2 of the same kind of candies for all items in the set :
3 * 3 = 9

Number of ways to pick 3 out of 3 being of the same kind of candy :

C(3,3) = 3! / 3!(3-3)!
C(3,3) = 1

Total number of ways of picking two or more of the same kind of candies : (3 + 3 + 1)/9 = 7/9

Is that the solution or am I dreaming?
Last edited by Rahul M on 13 Aug 2008 09:50, edited 1 time in total.
Reason: Please post solutions in small font.

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Re: BR Maths Corner-1

Postby Satya_anveshi » 13 Aug 2008 10:03

Multatuli wrote:Some info on V. S. Sunder :

http://www.imsc.res.in/~vijay/



You might want to read his posts carefully.

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 10:21

Neshant wrote:Bear with me as math is not my strong point. I may be talking nonsense here but I'm gonna school you guys on what i think is the solution :

Number of ways of picking 2 out of 3 candies being the same for a single item in the set :
C(3,2) = 3! / 2!(3-2)!
C(3,2) = 3

Total number of ways of picking 2 of the same kind of candies for all items in the set :
3 * 3 = 9

Number of ways to pick 3 out of 3 being of the same kind of candy :

C(3,3) = 3! / 3!(3-3)!
C(3,3) = 1

Total number of ways of picking two or more of the same kind of candies : (3 + 3 + 1)/9 = 7/9

Is that the solution or am I dreaming?


Yes and no. Your answer is equivalent to assuming that the quantity of each of the sweets is the same. But there may be different amounts of each sweet.

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 10:30

another easier solution to;
Show that ( 1/a - 1 ) * ( 1/b - 1 ) * ( 1/c - 1 ) is equal to or greater then 8.



(1/a - 1)(1/b - 1)(1/c - 1) = (1 - a)(1 - b)(1 - c)/abc
= (b + c)(a + c)(a + b)/abc

Now a - 2sqrt(a)sqrt(b) + b = ( sqrt(a) - sqrt(b) )^2 >= 0
Therefore a + b >= 2sqrt(a)sqrt(b)
Same for (a + c) and (a + b)

So we get (b + c)(a + c)(a + b)/abc >= 8sqrt(b)sqrt(c)sqrt(a)sqrt(c)sqrt(a)sqrt(b)/abc
= 8abc/abc
= 8

Arithmetic-mean, Geometric Mean, Harmonic mean inequality can be used in a number of ways here.

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Re: BR Maths Corner-1

Postby sugriva » 13 Aug 2008 14:58

SK Mody wrote:There is a box containing lots of milk chocolates, coconut toffees and bittersweets (there may be different quantities of each). I choose 3 with eyes closed. Show that the probability that I will get at least two of one type is greater than or equal to 7/9. (Assume that there are enough sweets in the box that taking a few out does not change the proportion significantly).



This is equal to 1 - p where p = probability of picking up exactly one of each. Any other way of picking up will have at least two of a single type.
Now p is A/B where
A = number of ways of filling up 3 slots, each with a different kind of sweet. If X,Y,Z are the quantities of the three types. this is equal to
X * Y * Z
and
B = number of ways of selecting 3 sweets among all candies = (X + Y + Z) choose 3.
Therefore A/B = XYZ/C(X+Y+Z,3) This is equal to 6XYZ/((X+Y+Z)(X+Y+Z - 1)(X+Y+Z -2)) . This quantity is maximum when X = Y = Z
which makes A/B almost equal to 6X^3/27X^3 = 2/9
Therefore required probability is 1 - 2/9 = 7/9

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 22:58

Good Sol.

Those who are interested check out RM>AM>GM>HM. It is a good exercise to prove.
Also take a look at this Mean page on wiki.

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Re: BR Maths Corner-1

Postby SK Mody » 13 Aug 2008 23:12

Hope that vsunder will be back. It will take some of us from being freshers to post grads.

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Re: BR Maths Corner-1

Postby shiv » 15 Aug 2008 08:36

vsunder wrote:Forum admins: I must say the forum has degenerated into a cesspool of name calling arrogance by petty foggers
and sundry others. I have decided to stop coming here. It was nice in the old days and I am happy I learnt many things from people.

Sagun


Sagun, I admit that the post by AmberG (quoted below) was unnecessarily aggressive and egotistic. Since I know you personally I would like to apologize for this idiocy. Postor AmberG has not spared even a few microseconds to think that anyone else might know any math, which is sad for a person who wanted to test the math ability of forum members on the nuclear discussion thread a few weeks ago.

Amber G. wrote:
vsunder wrote:Many years ago in 1985 I proved the following lemma that I used in proving a major theorem about
the stability of matter when I was at the Institute in Princeton as a post-doc. Can anyone reprove it? I have another lemma that I proved lets see if you guys can prove that too. Here is the first lemma:

Take a square it works for a cube in any dimension. Now play the following game. Lets take a square for example.
Bisect each side you get now 4 congruent mini squares. Call this Step 1. Each congruent mini square can be bisected further to get now 16 mini squares in step 2. Proceed so on. One gets at step n, 4^n squares.

Now remove N DISJOINT squares in any way, that is remove a step 10 square, a step 25 square etc. The remaining
figure is a big square with holes like Swiss cheese Prove you can cover the remaining figure; the Swiss cheese looking part by exactly C N squares. C is a universal constant that only depends on the dimension. I forget the value I found for C for dimension 2.

By cover I mean nothing must penetrate the holes where squares were taken out.


It's not April 1, but I will bite. If this is not a joke, could one please write it in a way so it makes sense mathematically by being precise in definition.
For example: What exactly the sentence mean "Take a square it works for a cube in any dimension." ?? If rest of the part is particular example in D2, then state so, if not, then were exactly, say 4 in 4^n comes from?
Also what is "N" is is different than "n"? for example, when n=1, we have only one choice (3 squares) making C=3 (Assuming N and n are same) while for n=2, one can easily get an example where you need only 2 squares to cover.

In short it makes no sense.

If this was a joke, can the spam be removed.

(I am not even commenting on the rest of the post.. as it makes even less sense to me)
Thanks in advance.

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Re: BR Maths Corner-1

Postby Satya_anveshi » 15 Aug 2008 09:14

But I am amused by the astounding absense of Amber G after that post.

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Re: BR Maths Corner-1

Postby shiv » 15 Aug 2008 20:43

Multatuli wrote:Some info on V. S. Sunder :

http://www.imsc.res.in/~vijay/

I have to agree with Amber G, the problem should have been formulated more precisely, very sloppy formulation.

V Sunder, why are you so annoyed and insulted ? Can anyone diminish you as a mathematician through his/her comments ?



The post by VSunder has been signed "Sagun" and a link from the same post bears the name of the author of the paper as "Sagun Chanillo". Who is this V.S. Sunder you have Googled? I don't suppose you actually read the post or followed the link.


I'm no mathematician but I can read. I must comment on your sloppy reading.

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Re: BR Maths Corner-1

Postby Vriksh » 15 Aug 2008 21:24

Here is a math question I have been trying to solve for a particular problem perhaps a guru here can shed some light.

How can I find out if 2 closed paths in a 3-Dimensional space are entangled?

Here the word entanglement means that I cannot separate the 2 loops without breaking the loops.

Consider 2 elements of a chain
Image

Are there any mathematical formulation that will give an boolean answer (True/False) for 2 closed paths.

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Re: BR Maths Corner-1

Postby sugriva » 15 Aug 2008 22:22

Vriksh wrote:Here is a math question I have been trying to solve for a particular problem perhaps a guru here can shed some light.
How can I find out if 2 closed paths in a 3-Dimensional space are entangled?
Here the word entanglement means that I cannot separate the 2 loops without breaking the loops.
Are there any mathematical formulation that will give an boolean answer (True/False) for 2 closed paths.


I don't know but you can ask this guy ;-)

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Re: BR Maths Corner-1

Postby SK Mody » 15 Aug 2008 23:06

Vriksh wrote:Here is a math question I have been trying to solve for a particular problem perhaps a guru here can shed some light.
How can I find out if 2 closed paths in a 3-Dimensional space are entangled?
Are there any mathematical formulation that will give an boolean answer (True/False) for 2 closed paths.


There is a concept of winding number but more relevant is "linking number".
Check this page: linking number.

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Re: BR Maths Corner-1

Postby Vriksh » 16 Aug 2008 00:59

SK Mody thanks!

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Re: BR Maths Corner-1

Postby vsunder » 17 Aug 2008 07:29

I have received several e-mails telling me I should post. It is true though that I increasingly have
little to post. I have promised an admin that I will make a token post. But then a token
post with no content is bad. So here is some food for thought. I scanned the comments above.

I have some comments about the Wiki link that Mody provides. It misleads a little bit
and does not make a very fundamental connection. First the winding number and linking number
are two different ideas. I have used the winding number in my own work. Now what is the best
way to understand the linking number. I will explain and if I am in a very generous mood actually write
a PDF file put it on my web page where you can download it. Well this is the best way to understand the
linking number which the Wiki does not do so shame on them. Well most of you must have heard of the Biot-Savart Law in magnetostatics and also Ampere's Law in magnetostatics. So pretend one closed loop is carrying a unit
current. Write down by Biot-Savart the magnetic field generated by this current. Now compute the
circulation around the second loop of this magnetic field and apply Ampere's law. Immediately you will
get the formula of the linking number as on the Wiki page. It will very quickly give you a physical idea
why the linking numbers on the Wiki link of Mody are as they are in the link for different type of links. This is a classic example of Physics and Math coming together and a beautiful example of topological notions that arise in
Mathematics that actually are Physics in disguise.

Now there are particular configurations of entangled loops that have linking number zero. So just because the linking number is zero does not mean no entanglement so beware. The example is very simple but I am unable to draw it but the picture is in this book by the way a good cheap book by Flanders

Differential Forms

Naively the example consists of one closed loop which is a simple circle. Now the second loop consists
of two pieces of string going through the circle and carrying current in opposite directions.
When you see the picture in Flanders and keep the current stuff in your mind will see right away that
the linking number has to be nothing but zero. So the linking number is not a fantastic way to measure entanglement. The mathematician VFR Jones found various algebraic invariants to attach to knots and got a Fields medal in 1986 or so in the process. These are more refined quantities to "measure" entanglements.

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Re: BR Maths Corner-1

Postby ramana » 17 Aug 2008 11:24

Thanks Vsunder.

BTW I finally got the book by Chandrasekhar ji on Stochastic problems in Physics and Astronomy so I can understand you.

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Re: BR Maths Corner-1

Postby SK Mody » 18 Aug 2008 01:54

So pretend one closed loop is carrying a unit
current. Write down by Biot-Savart the magnetic field generated by this current. Now compute the
circulation around the second loop of this magnetic field and apply Ampere's law. Immediately you will
get the formula of the linking number as on the Wiki page.


Thanks. Regarding linking no = 0 does not imply no entanglement - for the special case of 2 loops it does imply no entanglement (correct me if I'm wrong). After Vriksh asked I was trying to think of some way to use the idea of winding no to get a formula determining whether or not the loops were entangled, when I stumbled on the linking no. on wiki. The Biot-savart law did occur to me.

Regarding your swiss cheese lemma and other links, the only one that is accessible is not in english.

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Re: BR Maths Corner-1

Postby Vriksh » 18 Aug 2008 03:09

Alright I tried to code the linking number criteria for 2 closed paths that exist on a cubic lattice and have failed miserably to get the linking number perhaps due to bug in code.

From wiki link

http://en.wikipedia.org/wiki/Linking_coefficient

There is the gauss linking integral formulation, however I think I made a mistake in coding, very possibly since my understanding of the mathematical notation is suspect.

Here is my attempt at the pseudocode for gauss linking integral
Notation:
p1 = path 1
p2 = path 2

r(l_k) = coordinates of k node on path l
dr(l_k) = r(l_k)-r(l_k+1)

LINKNUM=0
for k=0; k<node(p1); k++ {
for m=0; m<node(p2); m++
LINKNUM += [r(1_k)-r(2_m)] . [dr(1_k) x dr(1_m)]/|r(1_k)-r(2_m)|^3
}
}
return (LINKNUM/4pi)

Can some gurus tell me if this is correct?

SK Mody
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Re: BR Maths Corner-1

Postby SK Mody » 18 Aug 2008 13:58

[dr(1_k) x dr(1_m)]


Don't know if you actually mistyped it here or in the actual program. [dr(1_k) x dr(2_m)].

SK Mody
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Posts: 251
Joined: 15 Mar 2002 12:31

Re: BR Maths Corner-1

Postby SK Mody » 18 Aug 2008 14:08

Also there may be another problem with your code. Even assuming that your curves are actually polygons defined by the vertices at the nodes, I'm not sure of the accuracy of the calculation. If two nodes are spaced far apart, you may need to split that straight line segment into a further set of nodes spaced more closely. JMT.

vsunder
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Location: Ulan Bator, Mongolia

Re: BR Maths Corner-1

Postby vsunder » 18 Aug 2008 21:17

Mr. Mody: What I said is this: There is a configuration of TWO and just two entangled loops
that have zero linking number. Please read the book by Flanders I linked you will see the picture
of this example in the book. So I said beware because the Wiki article seems not to point
out this situation. Hence my further remark that the linking number is an imperfect tool
to measure entanglement. If you find the linking number is zero it means nothing could be or need not be entangled.


In fact once you see the picture in Flanders you will be able to construct more complicated entanglements
with zero linking number keeping the analogy with currents and Biot-Savart in mind. Some biologists
are interested in this because of protein folding and DNA. In fact Siddhartha Gadgil of the math. deptt.
at IISc Bangalore has a paper on this aspect you can print it out from his web page, it might have appeared
already in a journal.

Also there is a confusion it seems between winding number and linking number. They are two different objects.

SK Mody
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Posts: 251
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Re: BR Maths Corner-1

Postby SK Mody » 18 Aug 2008 22:11

vsunder wrote:What I said is this: There is a configuration of TWO and just two entangled loops
that have zero linking number.


If the linking number is the circulation of the magnetic field around the loop, and the linking number is zero, the net current through the loop is zero. How is that possible for only two entangled loops? Sorry if I'm missing something.

I would have to order that book from dover publications or amazon.com to read it. Can't find it that easily in this part of the world.

Yes winding number is a different concept, but I feel it is somewhat related.
Thanks.

Added later: OK, I see your point. For the benefit of others who may be interested, check this page:
Linking number
See the third diagram from the top.
Last edited by SK Mody on 18 Aug 2008 22:47, edited 1 time in total.


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