Re: Mathematical methods for the discrete space-time.
Jason Resch skrev: I am not sure how related this is to what you ask in your original post, but as for a model (and candidate TOE) of physics which is discrete, there is a theory known as Hiem Theory ( http://en.wikipedia.org/wiki/Heim_Theory ) which posits there are six discrete dimensions. Interestingly, the theory is able to predict the masses of many subatomic particles entirely from some force constants, something which even the standard model is unable to explain. I have now looked at Heim Theory, but it does not look enough serious to me. Every theory that compute the masses of the elementary particles from nothing, must be wrong. Because in different possible universa the masses of the elementary particles are different. Besides, the Heim Theory could not explain the quarks. But from the Heim Theory article I followed a link to Difference operator ( http://en.wikipedia.org/wiki/Difference_operator ), and that article was much more interesting, because there you could find the extended Leibniz rule. And from that article I found a link to Umbral calculus ( http://en.wikipedia.org/wiki/Umbral_calculus ), that look like exactly what I am looking for. The Umbral calculus seems to be a good candidate for a tool for handling discrete space-time! -- Torgny Tholerus --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Mathematical methods for the discrete space-time.
On Mon, Nov 24, 2008 at 9:30 AM, Torgny Tholerus [EMAIL PROTECTED] wrote: Jason Resch skrev: I am not sure how related this is to what you ask in your original post, but as for a model (and candidate TOE) of physics which is discrete, there is a theory known as Hiem Theory ( http://en.wikipedia.org/wiki/Heim_Theory ) which posits there are six discrete dimensions. Interestingly, the theory is able to predict the masses of many subatomic particles entirely from some force constants, something which even the standard model is unable to explain. I have now looked at Heim Theory, but it does not look enough serious to me. Every theory that compute the masses of the elementary particles from nothing, must be wrong. Because in different possible universa the masses of the elementary particles are different. Besides, the Heim Theory could not explain the quarks. Well the masses are not derived out of no where, the article states: The predicted masses were claimed to have been derived by Heim using only 4 parameters - h (Planck's Constanthttp://en.wikipedia.org/wiki/Planck%27s_Constant), G (Gravitational constanthttp://en.wikipedia.org/wiki/Gravitational_constant ), vacuum permittivity http://en.wikipedia.org/wiki/Vacuum_permittivity and permeabilityhttp://en.wikipedia.org/wiki/Permeability_(electromagnetism) . But from the Heim Theory article I followed a link to Difference operator ( http://en.wikipedia.org/wiki/Difference_operator ), and that article was much more interesting, because there you could find the extended Leibniz rule. And from that article I found a link to Umbral calculus ( http://en.wikipedia.org/wiki/Umbral_calculus ), that look like exactly what I am looking for. The Umbral calculus seems to be a good candidate for a tool for handling discrete space-time! Great! I'm glad it helped. Jason --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Mathematical methods for the discrete space-time.
Torgny Tholerus skrev: What I want to know is what result you will get if you start from the axiom that *everything in universe is finite*. One important function in Quantum Theory is the harmonic oscillator. So I want to know: What is the corresponding function in discrete mathematics? In continuous mathematics you have the harmonic oscillator defined by the differential equation D^2(f) + k^2*f = 0, which will have one of its solutions as: f(t) = exp(i*k*t) = cos(k*t) + i*sin(k*t), where i is sqrt(-1). In discrete mathematics you have the corresponding oscillator defined by the difference equation D^2(f) + k^2*f = 0, which will have one of its solutions as: f(t) = (1 + i*k)^t = dcos(k*t) + i*dsin(k*t), where dcos() och dsin() are the corresponding discrete functions of the continuous functions cos() and sin(). So what is dcos() and dsin()? If you do Taylor expansion of the continuos function you get: exp(i*k*t) = Sum((i*k*t)^n/n!) = Sum((-1)^m*k^(2*m)*t^(2*m)/(2*m)!) + i*Sum((-1)^m*k^(2*m+1)*t^(2*m+1)/(2*m+1)!) And if you do binominal expansion of the discrete function you get: (1 + i*k)^t = Sum(t!/((t-n)!*n!)*(i*k)^n) = Sum((-1)^m*k^(2*m)*(t!/(t-2*m)!)/(2*m)!) + i*Sum((-1)^m*k^(2*m+1)*(t!/(t-2*m-1)!)/(2*m+1)!) When you compare these two expession, you see a remarkable resemblance! If you replace t^n in the upper expression with t!/(t-n)! you will then get exactly the lower expression! This suggest the general rule: If the Taylor expansion of a continuous function f(x) is: f(x) = Sum(a(n)*x^n) = Sum(a(n)*Prod(n;x)), then the corresponding discrete funtion f(x) is: f(x) = Sum(a(n)*x!/(x-n)!) = Sum(a(n)*Prod(n;x-m)), where Prod(n;x-m) = x*(x-1)*(x-2)* ... *(x-n+2)*(x-n+1) is a finite product. I have no strict proof of this general rule. But this rule is such a beautifil result, that it simply *must* be true! -- Torgny --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Mathematical methods for the discrete space-time.
On 13 Nov 2008, at 14:21, Torgny Tholerus wrote: Bruno Marchal skrev: I have to think. I think that to retrieve a Leibniz rule in discrete mathematics, you have to introduce an operator and some non commutativity rule. This can be already found in the book by Knuth on numerical mathematics. This has been exploited by Kauffman and one of its collaborator, and they have published a book which I have ordered already two times ... without success. It is a very interesting matter. Dirac quantum relativistic wave equation can almost be retrieved form discrete analysis on complex or quaternion. It is worth investigating more. Look at Kauffman page (accessible from my url), and download his paper on discrete mathematics. I will look closer at the Kauffman paper on Non-commutative Calculus and Discrete Physics. It seems interesting, but not quite what I am looking for. Kauffman only gets the ordinary Leibniz rule, not the extended rule I have found. Ah? What I want to know is what result you will get if you start from the axiom that *everything in universe is finite*. Like with comp + occam. Look I think I will concentrate on the MGA thread for a period. Meanwhile I will ask one of my student, who has a craving for discrete math, to take a look on your finite calculus, and he will contact you in case he find it interesting. Sorry but I have not so much time those days. Best, Bruno For this you will need a function calculus. A function is then a mapping from a (finite) set of values to this set of values. Because this value set is finite, you can then map the values on the numbers 0,1,2,3, ... , N-1. So a function calculus can be made starting from a set of values consisting of the numbers 0,1,2,3, ... , N-1, where N is a very large number, but not too large. N should be a number of the order of a googol, ie 10^100. Because the size of our universe is 10^60 Planck units, and our universe has existed for 10^60 Planck times. As the arithmetic, we can count modulo N, ie (N-1) + 1 = 0. This makes it possible for the calculus to describe our reality. A function can then be represented as an ordered set of N numbers, namely: f = [f(0), f(1), f(2), f(3), ... , f(N-1)]. This means that S(f) becomes: S(f) = [f(1), f(2), f(3), ... , f(N-1), f(0)]. The sum or the product of two functions is obtained by adding or multiplying each element, namely: f*g = [f(0)*g(0), f(1)*g(1), f(2)*g(2), ... , f(N-1)*g(N-1)]. and to apply a function f on a function g then becomes: f(g) = [f(g(0)), f(g(1)), f(g(2)), ... , f(g(N-1))]. Exercise: Show that the extended Leibniz rule in the discrete mathematics: D(f*g) = f*D(g) + D(f)*g + D(f)*D(g), is correct! -- Torgny Tholerus http://iridia.ulb.ac.be/~marchal/ --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Mathematical methods for the discrete space-time.
Torgny Tholerus skrev: Exercise: Show that the extended Leibniz rule in the discrete mathematics: D(f*g) = f*D(g) + D(f)*g + D(f)*D(g), is correct! Another way to see both form of the Leibniz rule is in the graphical set theory, where you represent the sets by circles on a paper. Here I will represent the union of the sets A and B with A + B, and the intersection as A*B. Then you can represent the D operator as the border of the circle. Then you will have: D(A*B) = A*D(B) + D(A)*B, ie the Leibniz rule, ie the border of the area of the intersection is the union of the border of B inside A, and the border of A inside B. I can not show this figure in this message, but you can draw two circles on a paper before you, and you will then see what I mean. Now the interesting thing is what will happen if the circles have *thick* borders: Then the set A is represented by two circles inside each other, and the border will then be the area between the two circles. The set A will then be the interior of the inner circle, and the outside of A will be the outside of the outer circle. What will you then get if you look at the border of the intersection of A and B? This time you will get: D(A*B) = A*D(B) + D(A)*B + D(A)*D(B), ie the extended Leibniz rule. The extra term then comes from the two small squares you get where the two borders cross each other. (Do draw this figure om the paper before you, and you will understand.) This picture with the circles with thick borders is a way to represent intiutionistic logic. The interior of the inner circle is the objects that represent A (such as red), and the outside of the outer circle represent not-A (such as not red). Inside the border you will have all that is neither A nor not-A (such as red-orange, where you don't know if it is red or not...) -- Torgny Tholerus --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Mathematical methods for the discrete space-time.
I have to think. I think that to retrieve a Leibniz rule in discrete mathematics, you have to introduce an operator and some non commutativity rule. This can be already found in the book by Knuth on numerical mathematics. This has been exploited by Kauffman and one of its collaborator, and they have published a book which I have ordered already two times ... without success. It is a very interesting matter. Dirac quantum relativistic wave equation can almost be retrieved form discrete analysis on complex or quaternion. It is worth investigating more. Look at Kauffman page (accessible from my url), and download his paper on discrete mathematics. There are also interesting relations with knots, and even with the way lambda calculus could be used to provide semantics for the Fourth and fifth arithmetical hypostases, but to be sure I have failed to exploit this. If this were true, the background comp physical reality would be described by a sort of number theoretical quantum topology. That would explain also the role of exceptional (and monstruous) finite simple groups. You are perhaps on a right track, but in a incredibly complex labyrinth ... to be honest ... Bruno Le 12-nov.-08, à 18:44, Torgny Tholerus a écrit : When you are going to do exact mathematical computations for the discrete space-time, then the continuous mathematics is not enough, because then you will only get an approximation of the reality. So there is a need for developing a special calculus for a discrete mathematics. One difference between continuous and discrete mathematics is the rule for how to derívate the product of two functions. In continuous mathematics the rule says: D(f*g) = f*D(g) + D(f)*g. But in the discrete mathematics the corresponding rule says: D(f*g) = f*D(g) + D(f)*g + D(f)*D(g). In discrete mathematics you have difference equations of type: x(n+2) = x(n+1) + x(1), x(0) = 0, x(1) = 1, which then will give the number sequence 0,1,1,2,3,5,8,13,21,34,55,... etc. For a general difference equation you have: Sum(a(i)*x(n+i)) = 0, plus a number of starting conditions. If you then introduce the step operator S with the effect: S(x(n)) = x(n+1), then you can express the difference equation as: Sum((a(i)*S^i)(x(n)) = 0. You will then get a polynom in S. If the roots (the eigenvalues) to this polynom are e(i), you will then get: Sum(a(i)*S^i) = Prod(S - e(i)) = 0. This will give you the equations S - e(i) = 0, or more complete: (S - e(i))(x(n)) = S(x(n)) - e(i)*x(n) = x(n+1) - e(i)*x(n) = 0, which have the solutions x(n) = x(0)*e(i)^n. The general solution to this difference equation will then be a linear combination of these solutions, such as: x(n) = Sum(k(i)*e(i)^n), where k(i) are arbitrary constants. To get the integer solutions you can then build the eigenfunctions: x(j,n) = Sum(k(i,j)*e(i)^n) = delta(j,n), for n the grade of the difference equation. With the S-operator it is then very easy to define the difference- or derivation-operator D as: D = S-1, so D(x(n)) = x(n+1) - x(n). What do you think, is this a good starting point for handling the mathematics of the discrete space-time? -- Torgny Tholerus http://iridia.ulb.ac.be/~marchal/ --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Re: Mathematical methods for the discrete space-time.
Bruno Marchal skrev: I have to think. I think that to retrieve a Leibniz rule in discrete mathematics, you have to introduce an operator and some non commutativity rule. This can be already found in the book by Knuth on numerical mathematics. This has been exploited by Kauffman and one of its collaborator, and they have published a book which I have ordered already two times ... without success. It is a very interesting matter. Dirac quantum relativistic wave equation can almost be retrieved form discrete analysis on complex or quaternion. It is worth investigating more. Look at Kauffman page (accessible from my url), and download his paper on discrete mathematics. I will look closer at the Kauffman paper on Non-commutative Calculus and Discrete Physics. It seems interesting, but not quite what I am looking for. Kauffman only gets the ordinary Leibniz rule, not the extended rule I have found. What I want to know is what result you will get if you start from the axiom that *everything in universe is finite*. For this you will need a function calculus. A function is then a mapping from a (finite) set of values to this set of values. Because this value set is finite, you can then map the values on the numbers 0,1,2,3, ... , N-1. So a function calculus can be made starting from a set of values consisting of the numbers 0,1,2,3, ... , N-1, where N is a very large number, but not too large. N should be a number of the order of a googol, ie 10^100. Because the size of our universe is 10^60 Planck units, and our universe has existed for 10^60 Planck times. As the arithmetic, we can count modulo N, ie (N-1) + 1 = 0. This makes it possible for the calculus to describe our reality. A function can then be represented as an ordered set of N numbers, namely: f = [f(0), f(1), f(2), f(3), ... , f(N-1)]. This means that S(f) becomes: S(f) = [f(1), f(2), f(3), ... , f(N-1), f(0)]. The sum or the product of two functions is obtained by adding or multiplying each element, namely: f*g = [f(0)*g(0), f(1)*g(1), f(2)*g(2), ... , f(N-1)*g(N-1)]. and to apply a function f on a function g then becomes: f(g) = [f(g(0)), f(g(1)), f(g(2)), ... , f(g(N-1))]. Exercise: Show that the extended Leibniz rule in the discrete mathematics: D(f*g) = f*D(g) + D(f)*g + D(f)*D(g), is correct! -- Torgny Tholerus --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
Mathematical methods for the discrete space-time.
When you are going to do exact mathematical computations for the discrete space-time, then the continuous mathematics is not enough, because then you will only get an approximation of the reality. So there is a need for developing a special calculus for a discrete mathematics. One difference between continuous and discrete mathematics is the rule for how to derívate the product of two functions. In continuous mathematics the rule says: D(f*g) = f*D(g) + D(f)*g. But in the discrete mathematics the corresponding rule says: D(f*g) = f*D(g) + D(f)*g + D(f)*D(g). In discrete mathematics you have difference equations of type: x(n+2) = x(n+1) + x(1), x(0) = 0, x(1) = 1, which then will give the number sequence 0,1,1,2,3,5,8,13,21,34,55,... etc. For a general difference equation you have: Sum(a(i)*x(n+i)) = 0, plus a number of starting conditions. If you then introduce the step operator S with the effect: S(x(n)) = x(n+1), then you can express the difference equation as: Sum((a(i)*S^i)(x(n)) = 0. You will then get a polynom in S. If the roots (the eigenvalues) to this polynom are e(i), you will then get: Sum(a(i)*S^i) = Prod(S - e(i)) = 0. This will give you the equations S - e(i) = 0, or more complete: (S - e(i))(x(n)) = S(x(n)) - e(i)*x(n) = x(n+1) - e(i)*x(n) = 0, which have the solutions x(n) = x(0)*e(i)^n. The general solution to this difference equation will then be a linear combination of these solutions, such as: x(n) = Sum(k(i)*e(i)^n), where k(i) are arbitrary constants. To get the integer solutions you can then build the eigenfunctions: x(j,n) = Sum(k(i,j)*e(i)^n) = delta(j,n), for n the grade of the difference equation. With the S-operator it is then very easy to define the difference- or derivation-operator D as: D = S-1, so D(x(n)) = x(n+1) - x(n). What do you think, is this a good starting point for handling the mathematics of the discrete space-time? -- Torgny Tholerus --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~--~~~~--~~--~--~---
