Johnny5

What is just psi? Before psi star psi? If [math] \psi^* \psi [/math] is a probability, what is just psi? Oh nevermind.

Thanks Basically, all I wanted was to have a look at what it was about, which is what that did for me. Thanks

After talking with Rev, it became apparent, that I need to review linear transformations. Last night, I spent about 4 hours reading through 3 different books on linear algebra. All I want to do right now, is fully and competently understand what a linear transformation is. Here is where I am at, and would appreciate any help. Ok first of all... The first thing that became apparent to me, was that I need to be able to logically express the concept of "exactly one," as opposed to merely "there is at least one." And I am a stickler about my logic. So for right now, I would like know exactly how some of you address the issue of "exactly one" using symbolic logic. Let it be presumed that someone knows first order logic, and wishes to use the symbols from it to define the concept of a linear transformation. Now, in thinking of the graph of a function, you can see that for any x, there is one and only one f(x). But I want to say this using first order logic. So here is how I started off: [math] f: A \to B [/math] f,A,B, are sets. A is the domain of the function, B is called the codomain. [math] \forall x \in A \ \exists y \in B [ (x,y) \in f \ \& \ \neg \exists z \in B [(x,z) \in f \& \neg (z=y) ]][/math] Translation: For any x an element of the domain A, there is at least one element y of the codomain B, such that: The ordered pair (x,y) is an element of set f, and it is not the case that there is at least one z in the codomain B, such that (x,z) is in f, but z is different from y. (the z stuff was to make sure that any point x maps to one and only one f(x). ) Question: Did I express the meaning of function correctly with that? I don't want to know if you like the symbolism, I want to know if I got the meaning right. Thank you

Ok cool. How do you make the logical symbol for 'AND'? It looks like ^ only bigger.

Linear transformation I remember the formula right away If f(ax+by) = af(x)+bf(y) then f is a linear transformation. But to be proper' date=' I need to be clear what my domain of discourse is. And at the wikipedia site, there is reference to fields vector spaces And i know how complicated that stuff is. I want to keep things as simple as i can, while i learn. Rings, fields, modules, submodules, a lot is coming back to me. A field is that for which the algebraic axioms are true. 1. closure under addition. 2. closer under multiplication. 3. commutativity of addition 4. associativity of addition. 5. existence of additive identity 6. existence of additive inverses 7. commutativity of multiplication 8. associativity of multiplication. 9. existence of multiplicative identity 9. existence of multiplicative inverses 10. distributive axiom 11. non triviality not(0=1) and a few other minor ones. For example... Let the domain of discourse be the field know as the real numbers. Let [math'] \mathbb{R} [/math] denote the reals. We postulate two processes on real numbers, addition, and multiplication. The process of addition is denoted by the symbol + and the process of multiplication is denoted by the symbol [math] \cdot [/math] Postulate 1 (Closure under addition): [math] \mathbb{R} [/math] is closed under + What this means is that if we add two real numbers together, the answer is also a real. We can say this using logic: for any symbol x, and any symbol y: [math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R}) \Rightarrow x+y \in \mathbb{R} [/math] Translation: If x is an element of the real number system, and y is an element of the real number system, then x+y is an element of the real number system. Postulate 2 (Closure under multiplication): For any symbol x, and any symbol y: [math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R}) \Rightarrow x \cdot y \in \mathbb{R} [/math] Translation: If x is an element of the real number system, and y is an element of the real number system, then [math] x \cdot y [/math] is an element of the real number system. What this means is that if we multiply two reals together, the answer is also a real. Postulate 3(Commutativity of addition): [math] \forall x \forall y [/math] [math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R}) \Rightarrow x+y=y+x[/math] Postulate 4(Associativity of addition): [math] \forall x \forall y \forall z [/math] [math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R} \& \ z \in \mathbb{R}) \Rightarrow x+(y+z) = (x+y)+z [/math] What this postulate says, is intuitive. Suppose we are only permitted to add two numbers at a time, but we wish to add three numbers together. What this says is that it doesnt matter which two we add together first. In other words it tells us this 5+(7+8) = 5+(15) = 20 (5+7)+8 = (12)+8 = 20 In other words it tells us 5+(7+8) = (5+7)+8, but it tells us that we can do this for any three real numbers, not just 5,7, and 8. Postulate 5 (existence of additive identity): [math] \exists 0 \in \mathbb{R} \ \forall x \in \mathbb{R} [ 0+x=x ] [/math] Translation: There is at least one real number 0, such that for any real number x, 0+x = x. We leave uniqueness of zero as a theorem.

Thanks

How do you write the logical symbol for 'not' using latex. The symbol looks like the letter L, only sideways. Thank you And another question. What is the proper symbolic way to refer to the angle between a and b? I know that the following... [math] \angle \alpha [/math] will be read as "angle alpha" and I know that the following... [math] \angle ABC [/math] Indicates that there are three points in a figure, labeled A,B,C, and that the vertex of the angle is the point B. But what is the best way, or proper way, to discuss the angle between a, and b. In other words, I've been writing this... [math] \vec A \bullet \vec B \equiv |\vec A| |\vec B| cos(\vec A ,\vec B) [/math] And I expect it to be read as "... cosine of the angle between vector A, and vector B." Is there a better way to symbolically refer to the angle between two vectors?

Rev, I would like the notation that I end up using, to be the one the most amount of physicists are already used to, if that tells you anything. Which is more common, alpha/beta or mu/nu?

[math] G^\alpha^\beta [/math] is called the Einstein tensor. [math] T^\alpha^\beta [/math] is called either the stress-tensor, or the energy-momentum tensor. The Einstein general relativistic field equation is: [math] G^\alpha^\beta = \frac{8 \pi G}{c^4} T^\alpha^\beta [/math] The coefficient of the stress-tensor is just a scalar, a number, and its units are inverse force. G is the Newtonian gravitational constant, and c is the speed 299792458 meters per second. The LHS of the field equation has units of meters. The stress-tensor has units of energy. The Einstein tensor can be written in terms of the Ricci scalar, the Ricci tensor, and the metric tensor as follows: [math] G^\alpha^\beta = R_\mu_\nu - \frac{1}{2} g_\mu_\nu R [/math] [math] R_\mu_\nu [/math] is called the Ricci tensor. [math] g_\mu_\nu [/math] is called the metric tensor. R is called the Ricci scalar. [math] R^\mu_\alpha_\mu_\beta [/math] is called the Reimann curvature tensor.

I'm not following you. You are using the letter U to denote what? Speed apparently. Don't use geometrized units yet. I don't know how to write the stress-tensor in terms of energy density rho yet.

pressure = force/area Let p denote pressure. energy density = energy/volume Let r denote energy density. Now [math] T_\mu_\nu [/math] (the stress tensor) has to have units of energy' date=' so that the LHS of the general relativistic field equation has units of meters, units of distance. p= pressure = force/area = (force)(distance)/(volume) = (energy)/(volume) I am reading this right now. And also this.

A vector in Euclidean space can be written, using the coordinates of a frame. Suppose we have frame S. That means we have three axes, which are mutually orthogonal. One is always called the X axis of the frame, and the others are the Y axis, and the Z axis of the frame. And there are three unit vectors, called orthonormal basis vectors. Now, a point in a frame, is represented by a three tuple. Instead of an ordered pair, its an ordered triple. So here is how to specify a unique location in the frame: (3,2,4) But order matters in the above notation of that location, it is a three tuple. The first number you see, 3, is the point which is three units of distance away from the origin of the frame, in the positive i hat direction. Do not forget that, i^ is a unit vector, whose tail is at the origin of the frame, and whose head is one say meter away from the origin an on the positive X axis of the frame, where we have chosen the unit of distance to be the meter. The other numbers 2, 4 are interpreted similarly. Now, the theory of special relativity introduced a fourth orthonormal basis vector t^. Thus, the coordinates of a point in four dimensional Minkowski space need to be represented by a four tuple, instead of the more logical three tuple used for points in Euclidean space. I'm reading this now. A point in Euclidean three space is represented as follows: (x,y,z) A point in Minkowski space is represented as follows: (t,x,y,z) The previous point is a four tuple. Here is a general example of a four tuple: [math] x^\mu = (x^0,x^1,x^2,x^3) [/math] Now, here is a general example of a four-vector: [math] \left[ \begin{array}{c} x^0\\ x^1\\ x^2\\ x^3\\ \end{array} \right] [/math] In the theory of relativity, the first element in the four tuple is the time component of the "position four vector of SR", the second element in the four tuple is the x coordinate of the ...., the third element in the four tuple is the y coordinate, and the fourth element is the z coordinate. At wolfram, the position four-vector of SR theory, is written as follows: [math] \left[ \begin{array}{c} ct\\ x\\ y\\ z\\ \end{array} \right] [/math] It's a vector that can be represented with only one indice. Let's choose mu for the indice, and write the equation at wolfram. [math] x^\mu = \left[ \begin{array}{c} x^0\\ x^1\\ x^2\\ x^3\\ \end{array} \right] [/math] So for SR theory, the first component of the four vector is ct. So when we see ct, we know we are dealing with SR just by looking at the symbolism. So merely by writing the following [math] x^\mu = \left[ \begin{array}{c} x^0\\ x^1\\ x^2\\ x^3\\ \end{array} \right]=\left[ \begin{array}{c} ct\\ x\\ y\\ z\\ \end{array} \right] [/math] We know the symbol on the RHS, is the position-four vector of SR theory. The earlier notation should be connected to this one. Let us have chosen the following symbol for a vector in m dimensional-space. [math] \vec V [/math] If we choose to leave summation explicit, then we can write: [math] \vec V = \sum_{i=1}^{i=m} v^i \hat e_i [/math] If we want to have summation implicit, we can use the Einstein summation convention, and write the arbitrary m dimensional vector as follows: [math] \vec V = v^ie_i [/math] Notice, we have lost information about the dimension of the space, because we lost the symbol m. At any rate, here is how we know there is a summation of the dummy indice i, one of the indices is superscripted, the other subscripted, and the indice symbols are identical. Now, when dealing with SR theory... the sum is started from 0, and goes up to 3, so that there are four coordinates for a point in the space. And specifically, when dealing with SR theory, the zeroth component corresponds to the time coordinate. To be quite honest, here is the notation I like the best for this nightmare: [math] \vec V = \sum_{i=0}^{i=3} v^i \hat e_i [/math] In other words, leave the summation explicit. Then expand the series... [math] \vec V = \sum_{i=0}^{i=3} v^i \hat e_i = v0 \hat e_0 +v1\hat e_1 +v2 \hat e_2 +v3\hat e_3 [/math] Then in order to switch to SR theory, the e0 unit vector must represent the unit time vector, whose dot product with itself is negative one, unlike the orthonormal basis vectors of ordinary Euclidean space, which when dotted with themselves, have their dot product equal to 1. Now, in SR theory we have: [math] \hat e_0 = \hat t [/math] [math] \hat e_1 = \hat i [/math] [math] \hat e_2 = \hat j [/math] [math] \hat e_3 = \hat k [/math] And the space is called Minkowski space, and a point in Minkowski space is represented as follows: (ct,x,y,z) So in SR theory we have: [math] \vec V = \sum_{i=0}^{i=3} v^i \hat e_i = v0 \hat e_0 +v1\hat e_1 +v2 \hat e_2 +v3\hat e_3 = ct \hat t +x\hat i +y \hat j + z\hat k [/math] And we can write that as a column vector, as follows: [math] \vec V = \left[ \begin{array}{c} ct\\ x\\ y\\ z\\ \end{array} \right] [/math] I guess there is a small problem, in that the above notation doesn't lead to t^ dot t^ = -1. A simple way to cure this, is to demand that the unit time vector be imaginary. In other words we have this kind of thing going on here... [math] \hat i \bullet \hat i \equiv |\hat i| |\hat i| cos(\hat i,\hat i) [/math] And since the magnitude of unit vector is equal to one, we must have: [math] \hat i \bullet \hat i = cos(\hat i,\hat i) [/math] And the angle between ihat and itself is zero, and the cosine of zero is equal to one so that we have: [math] \hat i \bullet \hat i = 1 [/math] And the angle between i^ and j^ is 90 degrees, and cos(90)=cos(pi/2) = 0, therefore i^ dot j^ =0, and for the same reason, i^ dot k^ = 0, and for the same reason, j^ dot k^ = 0, so that we must have: [math] \hat i \bullet \hat i = \hat j \bullet \hat j = \hat k \bullet \hat k= 1 [/math] and [math] \hat i \bullet \hat j = \hat i \bullet \hat k = \hat j \bullet \hat k = 0 [/math] Now, consider the unit time vector t^. Using the definition of dot product we must have this: [math] \hat t \bullet \hat t \equiv |\hat t| |\hat t| cos(\hat t,\hat t) [/math] Now, if we stipulate that the magnitude of the unit time vector is equal to the square root of negative one, and that the angle between the unit time vector and itself is 0, then we get this: [math] \hat t \bullet \hat t \equiv |\hat t| |\hat t| cos(\hat t,\hat t) = \sqrt {-1} \sqrt{-1} cos 0 = i^2 = -1[/math] But now if we do that, then the magnitude of a unit vector no longer has length equal to 1. And I believe the magnitude of a unit vector is stipulated to be one. But apparently not so, for the "unit time vector" of SR theory. On the other hand, suppose that we stipulate that the magnitude of the unit time vector must equal one. Then the only way to have the dot product come out to be negative one, is to have cosine (t^,t^) = -1 cos(0)=1 cos(90) = 0 cos(180) = -1 cos(270)=0 cos(360)=1 cos(450)=0 cos(540)=-1 cos(630)=0 cos(720)=1 and so on... the values of cosine which yeild negative 1, occur when the angle is 180,540, 900, 1260, and so on... and also -180, -540, -900, and so on... and we can write this in terms of pi. [math] \pi radians = 180 degrees [/math] So use the following symbol for the integers... [math] \mathbb{Z} [/math] Now, let n denote an arbitrary element of [math] \mathbb{Z} [/math]. In other words: [math] n \in \mathbb{Z} [/math] So, |t^|=1 and cosine (t^,t^) = -1 Provided [math] cos(\pi + 2n\pi) [/math] for some n an element of [math] \mathbb{Z} [/math]. And there are no other ways to make the dot product of the unit time vector with itself come out to equal negative one. Therefore, let us close the scope of the assumption right here. If [math] \hat t \bullet \hat t [/math] = -1 then ( [math] |\hat t | = \sqrt{-1} [/math] or [math] cos(\hat t,\hat t) = cos(\pi + 2n\pi) [/math] Let's clean the previous statement up, using logical symbols. [math] (\hat t \bullet \hat t = -1) \Rightarrow (|\hat t | = i \vee cos(\hat t,\hat t) = cos(\pi + 2n\pi) ) [/math] Where: [math] n \in \mathbb{Z} [/math] Translation: If the dot product of the unit time vector with itself is equal to negative one, then either the magnitude of the unit time vector is equal to the square root of negative one or the cosine of the spatial angle between the unit time vector and itself is equal to the cosine of pi radians plus a multiple of 2pi radians. So, for the sake of argument, suppose that: 1. The dot product of the unit time vector with itself is equal to -1. 2. The magnitude of the unit time vector isn't root -1. Then it must be the case that the angle between the unit time vector and itself is equal to: [math] \pi + 2n\pi [/math] For at least one integer n. In other words, the following is a fact: If the dot product is defined in the conventional way, and the dot product of the unit time vector with itself is equal to -1, and it is not the case that the magnitude of the unit time vector is equal to the square root of negative one, then it must be the case that there is at least one integer n, for which the spatial angle between the unit time vector and itself is equal to [math] \pi (2n+1) [/math] And the symbols above, are just a fancy way of saying, "an odd multiple of pi." Definition: The dot product of two vectors [math] \vec A, \vec B [/math] is defined as follows: [math] \vec A \bullet \vec B \equiv |\vec A| |\vec B| cos(\vec A,\vec B) [/math] Let [math] \theta [/math] denote the angle between the unit time vector and itself. Therefore, theta is an element of the following set: [math] \mathcal{f} ...,-5\pi, -3\pi, -\pi, \pi, 3\pi, 5\pi, ... \mathcal{g} [/math] where I have used the roster method of set denotation. For a discussion on set theory, see here set theory . Now, the natural question to be raised, would be how in the world can a temporal vector, have a nonzero spatial angle between it and itself? The correct answer is that it cannot. I will come up with a proof for this later. So, it is not the case that there is at least one integer n, for which the angle between the unit time vector and itself is equal to: [math] \pi(2n+1) [/math] We can use the existential quantifier to help us say this symbolically... [math] \neg \exists n \in \mathbb{Z} [(\hat t, \hat t) = \pi(2n+1)] [/math] Translation: It is not the case that there is at least one integer n, such that the angle between the "unit time vector" and itself is equal to pi times (2n+1) times. Recap After a bit of work, the following statement can be proven: If the dot product of the unit time vector with itself is equal to -1, and the magnitude of the unit time vector isn't [math] \sqrt{-1} [/math], then it must be the case that the angle between the unit time vector and itself is equal to: [math] \pi + 2n\pi [/math] For at least one integer n. And later, it will be proven that it is impossible for the consequent of the conditional above to be true. So the negation of the antecedant of the conditional above is false. [math] \therefore [/math] Either the dot product of the unit time vector with itself isn't equal to negative one, or the magnitude of the unit time vector is equal to the square root of negative one. Right now, it has been stipulated that the dot product of the unit time vector with itself is equal to negative one. Therefore, we are forced to conclude that the magnitude of the unit time vector is equal to the square root of negative one. The strange thing about this, is that a unit vector is supposed to have a magnitude of 1, not root -1.

Yes inverse force, the gravitational constant G is in the numerator, and that has units of kilograms in the denominator, therefore inverse force. Yes, thank you. LHS units of distance... for that to be true RHS must have units of distance also. Therefore... the stress-tensor must have units of... Force times distance.... i.e. units of "energy" What does "volumetric component" mean?

[math] \vec V = V^ie_i = \sum_i V^ie_i [/math] For non SR: [math] \vec V = V^ie_i = \sum_i V^ie_i = \sum_{i=1}^{i=3} V^ie_i = v1\hat e_1 + v2\hat e_2 + v3\hat e_3 [/math] Where: [math] \hat e_1 = \hat i [/math] [math] \hat e_2 = \hat j [/math] [math] \hat e_3 = \hat k [/math] i hat is a unit vector along the positive x axis, with its tail at the origin, j hat is a unit vector along the positive y axis, with its tail at the origin, and k hat is a unit vector along the positive z axis, with its tail at the origin. For SR: [math] \vec V = V^ie_i = \sum_i V^ie_i = \sum_{i=0}^{i=3} V^ie_i = v0 \hat e_0 + v1\hat e_1 + v2\hat e_2 + v3\hat e_3 [/math] I could subscript v0,v1,v2, etcetera, but I don't want to. Now, in SR, e0 hat is the "unit time vector," that's what I'm calling it for now. Since it's special I am going to represent it by t hat. [math] \hat e_0 = \hat t [/math] Hence: [math] v0 \hat e_0 + v1\hat e_1 + v2\hat e_2 + v3\hat e_3 = v0 \hat t + v1\hat i + v2\hat j + v3\hat k [/math] So now we can go back and forth from one notation to the other. Right now, the coefficient of the unit time vector is meaningless. The vector above, is an example of a four vector. .

I need to catch up to you. I am going learn this as fast as I can. The rate you are going is good. I am going to post things as I learn them, and possibly questions. Oh and I didn't think it was the same as the Maxwell stress tensor. Thank you

Let me think about this ant thing. You say the ant is a two dimensional being.... mmm well ok. And that he is constrained to move along the surface of a three dimensional cylinder. Thats a little better but... He isn't really a two dimensional being, if his body is bent. in other words, any tiny patch of surface area, on the surface of a cylinder, is not flat, even locally. You can take your limit, and it shouldnt matter. Now matter how close you are to a point on the surface of the cylinde, if... You let R denote the vector from that point to wherever you are... then There is a point on the vector, which isn't on the cylinder. I know how to say what I mean better...

Let me ask you this, is simultaneity absolute, if SR is correct?

Strange but ok. Manifold mathematically precise substitution for word 'space'. Here is what I meant by strange... the ant thing You have two types of ways to look at what is only one thing in reality. Extrinsically Intrinsically Extrinsic geometry Intrinsic geometry

I understood that. Good. When the amount of distance away from a point is small, the Pythagorean theorem is true. But apparently, the truth value of statements in GR, vary in space. In other words, if you pick a point (x2,y2,z2) very far away from (x1,y1,z1), then If GR is correct, and the space is curved then The distance between the points is not given by the Pythagorean theorem. Suppose that space is not curved, and instead 'flat', then no matter how far away (x2,y2,z2) is from (x1,y1,z1), the Pythagorean theorem is true. That would mean that the distance D between the points is given by the following formula: [math] D = \sqrt{(x2-x1)^2 +(y2-y1)^2+(z2-z1)^2} [/math] Now I am saying that is true, but relativists entertain an impossibility, so ok whatever. How did you reach the conclusion that in SR, space is globally flat?

yes I know that. But in the end n=3. But I can remember that anyway... Say it this way... If the fundamental postulate of SR is correct, then the Pythagorean theorem isn't globally true. Perhaps locally, but not globally. The Pythagorean theorem is correct, but there seems to be some confusion on this. What you have said is this... The Pythagorean theorem isn't globally valid on a curved manifold. I don't know what a manifold is, so I am having trouble understanding this, but I do know what the term 'curved' means. It means not straight. Suppose you have a three dimensional rectangular coordinate system. Here are your basis vectors... [math] e_1 [/math] [math] e_2 [/math] [math] e_3 [/math]

Ok, I need to have a look at this... Definition: The dot product of two vectors is defined as follows: [math] \vec A \bullet \vec B = |\vec A||\vec B| cos(A,B) [/math] Therefore, by definition: [math] \hat i \bullet \hat i = |\hat i | |\hat i | cos (\hat i, \hat i ) [/math] Now, a basis vector must have unit length. Therefore: [math] |\hat i | = 1 [/math] Therefore: [math] \hat i \bullet \hat i = |\hat i | |\hat i | cos (\hat i, \hat i ) = 1*1*cos (\hat i, \hat i ) = cos (\hat i, \hat i ) [/math] Therefore, the dot product of i hat vector with itself is equal to the cosine of the angle between the i hat vector and itself. The i hat vector, in some frame of reference S, is The vector whose tail is the origin of the reference frame, and whose head is located on the positive X axis of the frame, one unit of distance away from the origin. In other words, the tail of the i hat vector is located at the point (0,0,0) in frame S, and the head of the i hat vector is located at the point (1,0,0) in frame S. Now, consider the dot product of the j hat vector of frame S with the i hat vector of frame S.

Yes ok, Rev's question. By tomorrow morning I will know more. I will continue posting the things I learn/recall, so that you can see a bit of what I know just by reading what I do between now and then. Thank you very much. I was working on expressing a vector, using various different notations. I am going to continue with that for awhile.