Johnny5

I do that very well. Let's see... Well I am trying to understand what a ring is, that's the point of this thread. As I already know the field axioms, this shouldn't be a hard thing to do, since a field is more complex than a ring. Yet I am still fuzzy on a few things, so lets see just what. At least this way if I make any errors, you can point them out. Lets start from Dave's definition of a Ring: A ring [math] \mathcal{R} [/math] is a set of elements which is defined axiomatically, as follows: There are two binary operations +,* on any arbitrary ring, which obey the following axioms: Let x,y,z denote arbitrary elments of [math] \mathcal{R} [/math]. [math] \text{Axiom I:} x+y=y+x [/math] [math] \text{Axiom II:} (x+y)+z=x+(y+z) [/math] [math] \text{Axiom III:} \exists 0[0+x=0] [/math] [math] \text{Axiom IV:} \exists (-x) [x+(-x)=0] [/math] (Thus, for those who know what an abelian group is, a ring is an abelian group on at least one of the ring operations, possibly both, but necessarily at least one.) [math] \text{Axiom V:} (x*y)*z=x*(y*z) [/math] [math] \text{Axiom VI:} x*(y+z)=x*y+y*z [/math] The above axioms are all necessary axioms of any ring [math] \mathcal{R} [/math]. Now, the first question is, are the axioms above sufficient to prove that 0 is the unique element of the ring, with the property listed above? If not, we have to add another axiom. The following proof was provided by Dave Proof that 0 is unique (Post 2 above) Assume you have two zero elements, [math]0[/math] and [math] \widetilde{0} [/math]. We know: [math] 0 = 0 + \widetilde{0} [/math] Sending [math] \widetilde{0} [/math] through the ring axioms provides: [math] \text{Axiom III:} \exists 0[0+\widetilde{0}=0] [/math] And we have assumed that [math]\widetilde{0}[/math] satisfies the same property that 0 satisfies in the axioms so also we have: [math] \exists \widetilde{0}[\widetilde{0}+x=\widetilde{0}] [/math] And since 0 is a necessarily a ring element we have: [math] \exists \widetilde{0}[\widetilde{0}+0=\widetilde{0}] [/math] So we have: [math] 0+\widetilde{0}=0 [/math] [math] \widetilde{0}+0=\widetilde{0} [/math] Now ring [math] \mathcal{R} [/math] is an abelian group under +, which means that its a group under +, with the additional axiom that + is commutative. Commutativity of + is axiom 1. So the following is true: [math] 0+\widetilde{0}=\widetilde{0}+0 [/math] Thus, by the transitive property of equality we have: [math] 0=0+\widetilde{0}=\widetilde{0}+0=\widetilde{0} [/math] From which it follows that: [math] 0=\widetilde{0} [/math] Contrary to the assumption that they were different elements. Hence the ring element 0, asserted in axiom 3 is unique.QED (Without commutativity of +, this would not have been provable) Thus, the following is a theorem of any Ring theory: [math] \text{Theorem:} \exists! 0[0+x=0] [/math]

I've been trying to fix an error that propagated through my original attempt, and unfortunately you only have six hours to edit any given post, but the information has to be kept in one place to follow the argument. What I am trying to do, is come up with a method of solving a cubic equation which is better than Cardan's formula. Here is what I have so far: Now, this is where i left off yesterday. So I am going to pick up the argument from here today, and try to finish it off by the end of the day. To quickly recapitulate, suppose you are given the following random cubic equation: [math] 3n^3-9n^2+15n-1 =0 [/math] You want to find three roots, r1,r1,r3, such that: [math] (n-r_1)(n-r_2)(n-r_3)= 0 [/math] Taking the steps outlined above, you discover the following: [math] r_1+r_2+r_3 = 3[/math] and [math] 9r_1^2+r_1(-12+3(r_2-r_3)^2)+1=0 [/math] Now, we started off by assuming that we knew r1, but didn't know r2 or r3. Now, lets reverse this. Suppose we don't know r1, but we do know r2, and r3. We could now figure out r1, from the formula above. Divide both sides of the equation above by 9, in order to obtain: [math] r_1^2+r_1 \frac{(-12+3(r_2-r_3)^2)}{9}+ \frac{1}{9} =0 [/math] In order to lighten up on the reading, define xi such that: [math] \xi \equiv \frac{(-12+3(r_2-r_3)^2)}{9} [/math] Thus, we can rewrite the equation as; [math] r_1^2 + \xi r_1 + \frac{1}{9} = 0 [/math] I think i've finally got a handle on the arithmetical mistake i made in earlier posts. Previously I had: [math] r_1^2 + \xi r_1 + 1 = 0 [/math] But this is wrong. What I should have had was: [math] r_1^2 + \xi r_1 + \frac{1}{9} = 0 [/math] Ok so... using the quadratic formula, the roots of the equation above are given by: [math] \text{root 1} \frac{-\xi + \sqrt{\xi^2 - \frac{4}{9}}}{2} [/math] [math] \text{root 2} = \frac{-\xi - \sqrt{\xi^2 - \frac{4}{9}}}{2} [/math] Now, r1 is a root of the principle equation which was: [math] 3n^3-9n^2+15n-1 =0 [/math] And we cannot have two solutions for r1, as the above roots, labeled 'root 1' and 'root 2' suggest. root 1 = root 2 if and only if [math] \frac{\sqrt{\xi^2 - \frac{4}{9}}}{2} =0 [/math] Which is true if and only if [math] \sqrt{\xi^2 - \frac{4}{9}}} =0 [/math] Which is true if and only if [math] \xi^2 - \frac{4}{9}}} =0 [/math] Which is true if and only if [math] \xi^2 = \frac{4}{9}}} [/math] Now, there are two possible solutions to the equation above, they are: [math] \xi_1 = \frac{2}{3} [/math] [math] \xi_2 = -\frac{2}{3} [/math] Recal the definition of x [math] \xi \equiv \frac{(-12+3(r_2-r_3)^2)}{9} [/math] Hence [math] \frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math] or [math] -\frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math] Lets just deal with x1 for now. So we have: [math] \frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math] which leads to [math] 6 = -12+3(r_2-r_3)^2 [/math] which leads to [math] 18 = 3(r_2-r_3)^2 [/math] which leads to [math] 6 = (r_2-r_3)^2 [/math] Which leads to: [math] r_2-r_3 = \pm \sqrt{6} [/math] That is: [math] r_2-r_3 = \sqrt{6} [/math] or [math] r_2-r_3 = -\sqrt{6} [/math] Now, lets consider x2 for now. We have: [math] -\frac{2}{3} = \frac{(-12+3(r_2-r_3)^2)}{9} [/math] This leads to: [math] -6 = -12+3(r_2-r_3)^2 [/math] Which leads to: [math] 6 = 3(r_2-r_3)^2 [/math] Which leads to: [math] 2 = (r_2-r_3)^2 [/math] Which leads to: [math] r_2-r_3 = \pm \sqrt{2}[/math] That is: [math] r_2-r_3 = \sqrt{2}[/math] or [math] r_2-r_3 = -\sqrt{2}[/math] Now, if the two roots found by using the quadratic formula are necessarily complex, then it is meaningless to speak of one root being greater than the other. On the other hand, if they are both real, than one of them is greater than the other. Assume it is the case that r2 > r3. Thus, r2-r3 > 0 Therefore: [math] r_2-r_3 = \sqrt{6} [/math] and [math] r_2-r_3 = \sqrt{2} [/math] Thus, we have two equations in two unknowns, so we can solve for r2,r3. From the second equation we have: [math] r_3 = r_2-\sqrt{2} [/math] Substituting that into the first equation we have: [math] r_2-(r_2-\sqrt{2}) = \sqrt{6} [/math] From which it follows that: [math] \sqrt{2} = \sqrt{6} [/math] Which is false. Hence, it is not the case that r2>r3. Now, assume that r3>r2, hence r3-r2>0, hence r2-r3 is negative. Therefore: [math] r_2-r_3 = -\sqrt{6} [/math] and [math] r_2-r_3 = -\sqrt{2} [/math] Again, we arrive at an impossibility, hence the two roots r2,r3 are complex. In earlier lines of work we arrived at the following two equations: [math] 2r_2 = (3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math] [math] 2r_3 = (3-r_1)-\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math] We added them together, subtracted them from one another, now lets multiply them together, to see what we get: [math] 4 r_2 r_3 = (3-r_1)^2 - [(-3+r_1)^2- \frac{4}{3r_1}] [/math] Which leads to: [math] 4 r_2 r_3 = (3-r_1)^2 - (3-r_1)^2+ \frac{4}{3r_1} [/math] Which leads to: [math] 4 r_2 r_3 = \frac{4}{3r_1} [/math] Which leads to: [math] r_2 r_3 = \frac{1}{3r_1} [/math] Which leads to: [math] r_1r_2 r_3 = \frac{1}{3} [/math] Let's treat r2,r3 as complex numbers. They would have shown up from using the quadratic formula, so that we would have: [math] r_2 = a + bi [/math] [math] r_3 = a - bi [/math] Now, complex roots would show up in pairs, hence r1 would have to be a real number. Thus, we would have to have this: [math] r_1r_2 r_3 = \frac{1}{3} = r_1(a + bi)(a - bi) [/math] From which it follows that: [math] r_1= \frac{1}{3(a + bi)(a - bi)} [/math] From which it follows that: [math] r_1= \frac{1}{3(a^2 + b^2)} [/math] Now, we also know that [math] r_1+r_2+r_3 = 3 [/math] hence we must have: [math] \frac{1}{3(a^2 + b^2)}+(a+bi)+(a-bi) = 3 [/math] From which it follows that: [math] \frac{1}{3(a^2 + b^2)}+2a = 3 [/math] From which it follows that: [math] \frac{1}{3(a^2 + b^2)} = 3-2a [/math] From which it follows that: [math] 3(a^2 + b^2) = \frac{1}{3-2a} [/math] From which it follows that: [math] 3(a^2 + b^2) (3-2a) = 1 [/math]

Answer: Mathematical theory of probability. Treatment of electron's position as a random variable, in the center of mass reference frame of a single hydrogen atom. Born Interpretation of [math] \psi^* \psi [/math] I can actually run through it for you, its really complicated though. The quickest way to answer your own question, depending on how much you already know, is to find a site that discusses QM, and ask about some of the formulas you see there. You will get different answers from different people, i assure you.

I do, I studied d`Alemberts solution to the wave equation, done Fourier integrals, Fourier series in QM. I've seen a whole lot. I am not sure how much I would suddenly 'recall' but what exactly are you thinking of, when you say, "solution to the wave operator." Also, i don't want the thread to stray too far from rings, so if the answer is long, then move it to another thread. PS: For what it's worth, most of this stems from trying to understand quaternions. which i still dont.

I didn't think it had to be position, i just wanted to choose an example, and that was the first that came to mind. I understand that x is a variable, very well i understand that. The differential operator given by Dave, could act on any number of 'items' one of which is position. I am keenly aware of that, i assure you. You don't have to interpret t as time, i understand that you are a 'maths' person, and you think mathematically, which is good. But it is also good to be able to apply mathematics to physical reality. Much maths was developed in order to solve problems in physics. And i am sure the opposite has happened as well, that math was developed in a pure sense, and then applied to physics. Regards

Dave this looks strangely familiar for some reason. It resembles a power series, which hasn't operated on anything yet. Or a Maclaurin series expansion I should say. I was wondering if there was some physical interpretation that we could give it. Like for example, suppose that it acts on position variable x (you can treat x as position vector r if you are so inclined, but I am just going to use x). Then you have lets see... a1x+ a2x`+ a3x``+ a4x```+ So I am wondering about units here, and how the coefficients manage to keep them the same. For example, x` is interpreted as velocity, x`` as dv/dt which is acceleration, and then x```` would be da/dt, which can be nonzero, as for many many terms which follow. Yet, in physics we generally ignore the higher terms. Can you add anything to this? (specifically I am wondering if i could do physics using all the higher terms) (I thought about this long long ago, but forget what i did. I was thinking along these lines at the time, though this was very long ago... saying that the total motion of a particle was described by an infinite series of the form given by Dave, yet to avoid actual infinity, the formula for the coefficients of the series, eventually went to zero, so that there was a highest term, so that the series wasn't 'really' infinite. The main thing is, you would have more complex things than acceleration in the kinematics.)

I guess my question would be, is it provable within the group axioms, that the identity is unique. I don't think that is true, so that you would have to add that as an axiom. But i will try to answer my own question right here, in a moment.

What is the definition of the quaternion norm?

Ok, this is good. A ring R is necessarily a group under +, but furthermore, it must be an abelian group under+. (Now I have a side question... if ring R was a group under +, but not an abelian group under +, could you still have proved that 0 is unique?) So any ring R is an abelian group under +. Now, * has to be associative. 1) If a*b=b*a then the ring is commutive. 2) If 1*a=a*1 then the ring has an identity. 3) if a*b=b*a and 1*a=a*1 and (for any a in R, other than 0, there is 1/a in R such that a*(1/a) = 1 then R is a division ring??? Can you check my 3 above?

That's what i was thinking YT, but i think its best not to ask.

I want this thread to be about Rings. I think the main issue for the thread should be, what is/are the purpose/purposes of defining a ring. Let me start off with the definition Dave gave, of a ring, in a different thread. A ring is a set R, with two operations defined on it, which are symbolized +, *. The ring R, must be a group under at least one of the two operations. It doesn't have to be a group under both, but it could be. Suppose that the ring is a group under +. That will mean the following four things are true about elements of R: 1. There is a zero element of R, denoted by 0, such that for any element x of the ring, 0+x=x. 2. Associativity: For any x,y,z elements of the ring R, we have: (x+y)+z = x+(y+z) 3. (Inverse elements) For any element X of the ring R, there is an element (-x) of the ring, such that: x+(-x) = 0 4. Commutativity: For any x,y elements of the ring R, we must have: x+y=y+x (Thus R is a group under +). Now, as for the other operation, namely *, we have to have associativity. That is: 1. Associativity of * For any x,y,z elements of R: x*(y*z)=(x*y)*z Lastly, there has to be two distributive laws: a*(b+c) = a*b+a*c (b+c)*a = b*a+c*a A couple more questions. Question: Using the ring axioms above, (assuming i got them right that is), can you prove that the element 0 is unique? Question: Must the second operation *, be commutative too? If so, then the second distributive law above is redundant, and can be omitted from the set of ring axioms.

They are unit vectors, einheitsvektors, versors, vectors with magnitude one. Why not do it? As in here is the following standard orthonormal set of basis vectors in R^3: [math] \mathcal{f} \hat e_1, \hat e_2, \hat e_3 \mathcal{g} [/math]

This is a branch from another thread. Matt, it's been a long time since I've used poles and residues. I remember the method of images, and some other things here and there, but its been a long time. One question.... What is a meromorphic function, and how does it tie into the concept of infinity? Ok, analytic functions. That was the entire first half of complex variables. Analytic functions. As I recall, a function is analytic at a point, if it's differentiable there?

[math] Z^3+AZ^2+BZ+C= 0 [/math] Presume that you already know one root r1. It follows that you can factor the equation like so: (Z-r1)(Z^2+DZ+E)=0 Now, distribute to get: [math] Z^3+DZ^2+EZ-r_1Z^2-r_1DZ-r_1E = 0 [/math] Now, group things in descending powers of Z, to obtain: [math] Z^3+(D-r_1)Z^2+(E-r_1D)Z-r_1E = 0 [/math] And the given expression was: [math] Z^3+AZ^2+BZ+C= 0 [/math] from which it follows that: [math] A = (D-r_1) [/math] [math] B = (E-r_1D) [/math] [math] C = r_1E [/math] Since you already knew one root, it follows that you can now figure out the remaining two roots... r2 and r3 using the quadratic formula. In other words you already know this: (Z-r1)(Z^2+DZ+E)=0 So now focus on the following quadratic equation: Z^2+DZ+E=0 Using the quadratic formula, the remaining two roots are given by: [math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math] [math] r_2 = \frac{-D-\sqrt{D^2-4E}}{2} [/math] At this point, i wanted to solve an example problem. You are given the following cubic equation, and asked to find all three roots: [math] 3n^3-9n^2+15n-1 =0 [/math] The first thing you do, is make the leading coefficient equal to 1, like so: [math] n^3-3n^2+5n-\frac{1}{3} =0 [/math] Let A= -3 B=5 c=-1/3 We can now write the equation as: [math] n^3+An^2+Bn+C =0 [/math] Let us presume that we already know r1, and can factor it out like so: [math] (n-r_1)(n^2+Dn+E) = 0 [/math] Now, the roots of the quadratic are: [math] r_2 = \frac{-D + \sqrt{D^2-4E}}{2} [/math] [math] r_3 = \frac{-D - \sqrt{D^2-4E}}{2} [/math] And the relationships between D,E and A,B,C are: [math] A = (D-r_1)= -3 [/math] [math] B = (E-r_1D)= 5 [/math] [math] C = r_1E = -\frac{1}{3} [/math] Hence: [math] D=r_1-3 [/math] [math] r_1D= E-5 [/math] [math] E= \frac{1}{3r_1} [/math] Hence: [math] r_1D= \frac{1}{3r_1}-5 [/math] From which it follows that: [math] \frac{1}{3r_1} -5 = -3r_1 + r_1^2 [/math] From which it follows that: [math] r_1^3-3r_1^2+5r_1-1 = 0 [/math] Now, we also know that: [math] r_2 = \frac{-D+\sqrt{D^2-4E}}{2} [/math] [math] r_3 = \frac{-D-\sqrt{D^2-4E}}{2} [/math] Hence, we can replace D by -3+r1, to obtain: [math] r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2-4E}}{2} [/math] and [math] r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-4E}}{2} [/math] And we can replace E by [math] \frac{1}{3r_1} [/math] to obtain: [math] r_2 = \frac{-(-3+r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}}{2} [/math] and [math] r_3 = \frac{-(-3+r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}}{2} [/math] So, we now have the following three formulas: [math] r_1^3-3r_1^2+5r_1-1 = 0[/math] [math] 2r_2 = (3-r_1)+\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math] [math] 2r_3 = (3-r_1)-\sqrt{(-3+r_1)^2-\frac{4}{3r_1}}} [/math] The question is, "can we now find the specific numbers r1,r2,r3, from the three equations immediately above?" Suppose that we add two of the three equations together, like so: [math]2r_2 +2r_3 = 2(3-r_1) [/math] From which it follows that: [math] r_1+r_2+r_3 = 3[/math] If we now subtract 2r3 from 2r2 we obtain: [math] 2r_2-2r_3 = 2\sqrt{(-3+r_1)^2- \frac{4}{3r_1}}} [/math] Dividing both sides by two, and then squaring we obtain: [math] (r_2-r_3)^2 = (-3+r_1)^2- \frac{4}{3r_1}} [/math] If we multiply both sides by 3r1 we obtain: [math] 3r_1(r_2-r_3)^2 = 3r_1(-3+r_1)^2- 4 [/math] Now, the RHS is a cubic in r1, whereas the LHS is going to be quadratic in r2,r3. [math] 3r_1(r_2-r_3)^2 = 3r_1[9+r_1^2-6r_1]-4 [/math] [math] 3r_1(r_2-r_3)^2 = 27r_1+3r_1^3-18r_1^2-4 [/math] From an earlier line of work we know that: [math] r_1^3-3r_1^2+5r_1-1 = 0 [/math] Solving for r1 cubed gives: [math] r_1^3 = 3r_1^2-5r_1+1 [/math] Therefore it follows that: [math] 3r_1(r_2-r_3)^2 = 27r_1+3(3r_1^2-5r_1+1 )-18r_1^2-4 [/math] From which it follows that: [math] 3r_1(r_2-r_3)^2 = 27r_1+9r_1^2-15r_1+3 -18r_1^2-4 [/math] Combining like terms on the RHS leads to: [math] 3r_1(r_2-r_3)^2 = -9r_1^2+12r_1-1 [/math] Which leads to: [math] 9r_1^2-12r_1+1 +3r_1(r_2-r_3)^2 =0 [/math] Therefore: [math] 9r_1^2+r_1(-12+3(r_2-r_3)^2)+1=0 [/math] The previous equation is quadratic in r1, so in principle, if we knew r2, and r3, then we could solve it for r1 using the quadratic formula. Divide both sides of the equation above by negative three to obtain: [math] r_1^2+ ( \frac{-4 + (r_2-r_3)^2}{3} ) r_1 +1 =0 [/math] Now, to lighten up on the reading, define xi such that: [math] \xi \equiv( \frac{-4 + (r_2-r_3)^2}{3} ) [/math] And rewrite the equation above as: [math] r_1^2+ \xi r_1 +1 =0 [/math] The roots of the equation above, by the quadratic formula are: [math] \text{root 1} = \frac{-\xi + \sqrt{\xi^2 - 4}}{2} [/math] And [math] \text{root 2} = \frac{-\xi - \sqrt{\xi^2 - 4}}{2} [/math] Now, r1 is a root of the principle equation which was: [math] 3n^3-9n^2+15n-1 =0 [/math] And we cannot have two solutions for r1, as the above roots, labeled 'root 1' and 'root 2' suggest. So one of them will lead to an explicit contradiction. Hence root 1=root 2 if and only if [math] \sqrt{\xi^2 - 4} = 0 [/math] Which is true if and only if [math] \xi^2 - 4 = 0 [/math] Which is true if and only if [math] \xi^2 = 4 [/math] Which is true if and only if [math] \xi = 2 [/math] Which is true if and only if [math] \frac{-4 + (r_2-r_3)^2}{3} = 2 [/math] Which is true if and only if [math] -4 + (r_2-r_3)^2 = 6 [/math] Which is true if and only if [math] (r_2-r_3)^2 = 10 [/math] Which is true if and only if [math] r_2-r_3 = \sqrt{10} [/math] And this result is new. We now have two equations in three unknowns. [math] \text{Equation 1}: r_1+r_2 + r_3 = -3 [/math] [math] \text{Equation 2}: 0r_1+ r_2-r_3 = \sqrt{10}[/math] Now, we just need one more equation in the same three unknowns, and we can use the techniques of linear algebra, to find the unique solution.

Matt, it's been a long time since I've used poles and residues. I remember the method of images, and some other things here and there, but its been a long time. One question.... What is a meromorphic function, and how does it tie into the concept of infinity?

Ok thank you. We can get back to working on GR whenever you'd like. I'd like to learn it, regardless of whether or not i think its correct, because I now find myself wanting to learn tensor calculus for reasons that have nothing to do with GR. Specifically, i am trying to learn about quaternions, and how to use them to derive rotation matrices. I am also trying to understand Foucalt's pendulum, coriolis force, as well as the mathematics of gyroscopes, including the parallel axis theorem. While trying to learn about those things, i discovered that Hamilton was the one who coined the term 'tensor,' and he had nothing to do with Ricci, or the invention of the Ricci calculus, which is now apparently called "tensor analysis" due to Einstein. So there is some confusing stuff going on. Nevertheless, apparently Hamilton broke up quaternions into two parts, one part called a tensor, and the other part was either a vector, or a versor or a scalar. Not sure yet. But my linear algebra is good enough to use it. Anything I don't happen to recall, only takes a moment to look up. Regards PS: I am perfectly content to continue the discussion in the other thread, where revprez was involved, but whenever you're ready.

All right here is an article which explains how to add and multiply two quaternions: Addition and multiplication of quaternions Let Q1= (a+u)=quaternion 1 Let Q2 = (b+v)=quaternion 2 a is the scalar part of quaternion 1, and u) is the vector part. b is the scalar part of quaternion 1, and v) is the vector part. So if the above article is correct, then in the case where a=b=0 the answer should be Q1Q1= -(dot product of u with v) + (cross product of u with v)

You translated = as equal to. This will raise some questions as to what is meant by two elements of Á being equal. There is a simple answer to that, which has to do with simultaneity being an equivalence relation on Á. But I will let that come out naturally. For right now where we are is good. Now, let me analyze (2): [math]\forall Y \exists X [ \neg(X=Y)\Rightarrow \text{X before Y} ] [/math] Let me translate. For any moment in time Y, there is at least one moment in time X, such that if X is different from Y, then X comes before Y. That is not what I meant. Yes you are correct, I am referring to 1. In fact, to say that time did not have a beginning, i would formulate that as follows: [math] \neg \exists X \forall Y[ \neg(\text{X simultaneous to Y})\Rightarrow \text{X before Y} ][/math] Which is equivalent to: [math] \forall X \exists Y[ \neg(\text{X simultaneous to Y}) \text{and} \neg (\text{X before Y}) ][/math] Which is equivalent to: [math] \forall X \exists Y[\text{Y before X}] [/math] Translation: Given any moment in time X, there is at least one moment in time Y, such that Y before X. The temporal logic which I am using, is based upon the fact that the following statement is necessarily a tautology: [math] \forall X \forall Y[\text{X simultaneous Y or X before Y or Y before X}][/math] Undefined binary relation on Á: before Definition: For any x,y elements of Á X simultaneous Y if and only if (not(X before Y) and not(Y before X)) Using nothing but first order logic, and the definition above, you can prove that the following statement is necessarily a tautology: [math] \forall X \forall Y[\text{X simultaneous Y or X before Y or Y before X}][/math]

All i remember is this: [math] G_{\mu}_{\nu} = \frac{8 \pi G}{c^4} T_{\mu}_{\nu} [/math] The LHS is the einstein tensor, and the RHS is a constant (whose units are inverse force) and T is the stress-energy tensor. Also, the Einstein tensor can be written in terms of the Ricci tensor and the Ricci scalar, as follows: [math] G_{\mu}_{\nu} = R_{\mu}_{\nu} - \frac{R}{4}g_{\mu}_{\nu} [/math] R mu nu is the Ricci tensor, R is the Ricci scalar, and g mu nu is the metric tensor. Someone else can be more specific. But the answer lies in the meaning of the stress tensor. And I don't understand it yet. But at least this post might get an answer for you started. Regards

Let Q1 denote quaternion (w1,x1,y1,z1). Let Q2 denote quaternion (w2,x2,y2,z2). Let the product of Q1 and Q2 follow the normal rules of multiplication of one polynomial by another, so that we have: (Q1)(Q2)= (w1,x1,y1,z1)(w2,x2,y2,z2) (Q1)(Q2)= (w1+x1i+y1j+z1k)(w2+x2i+y2j+z2k) Consider the case where w1=w2=0. (Q1)(Q2)= (0+x1i+y1j+z1k)(0+x2i+y2j+z2k) Following the rules of multiplication of one polynomial by another we have: x1ix2i+x1iy2j+x1iz2k+ y1jx2i+y1jy2j+y1jz2k+ z1kx2i+z1ky2j+z1kz2k Now, here is the formula for the vector cross product: V1 X V2 = <a,b,c> X <d,e,f> = i(bf-ce)-j(af-cd)+k(ae-bd) Here is Q1Q2 again: x1ix2i+x1iy2j+x1iz2k+ y1jx2i+y1jy2j+y1jz2k+ z1kx2i+z1ky2j+z1kz2k Which is equivalent to: x1ix2i+y1jy2j+z1kz2k + x1iy2j+x1iz2k+ y1jx2i+y1jz2k+ z1kx2i+z1ky2j

This was something i was working on awhile back, and wanted to see through to the end, but then Latex went down. Since Latex is finally back up, its been about a month, I'm going to resume this. The whole issue had to do with Cardan's formula. It's been so long that I'm going to begin again fresh. It began in post 8, with this: So I am going to pick it up from here.