Johnny5

What does that mean Matt? "Isn't algebraically closed." Real number system is "closed" under addition, and multiplication.

I tried to prove it once too, you learn something about straight lines, and distance.

Agreed. From memory: Michelson Morely experiment' date=' late 1800's, Albert Michelson wanted to measure the speed of light. He used a device which was a modified version of the device used by Fizeau to measure the speed of light. It involved a spinning mirror. The rate of revolution was known. Through some, perhaps specious line of reasoning, there was to be a difference in the diffraction pattern being observed. I remember this too... That it was thought, by Michelson, that the speed of light would be c in some special reference frame, but that in others it would be different. Using Michelson's train of thought, if the earth were at rest in the special frame, then the speed of a particular electromagnetic wave would be c. But... If instead, the earth was moving relative to that same photon, parallel to its direction of motion relative to earth, then if v denoted the relative speed between the rest frame of the earth, and the rest frame of the photon, then 1. if the earth was moving at the photon, the speed of the photon at impact would be v+c, and 2. if the earth was moving away from the photon before impact, then the photon would strike whatever measured it with speed c-v. Albert Michelson thought, like others, that light was an electromagnetic wave (I suppose this was due in part to Maxwell's theory, in which the solution to Maxwells equations lead to wave equations, where the speed of an EM wave is [math'] 1/\sqrt{\epsilon_0 \mu_0}) [/math]. The photon idea came later, with theoretical physicist Albert Einstein's explanation of the photoelectric effect. So Michelson performed his experiment at a time of year when the earth was moving in one direction relative to our sun, which emits "light". And, six months later, he repeated the measurement. Now the earth travels a roughly circular path around the sun. So therefore, six months later, the earth was moving in the opposite direction it was moving six months earlier (in the rest frame of the sun). So... Let v denote the tangential speed of the center of inertia of earth in the sun's rest frame. Look at the solar system from above. Let the center of inertia of the earth, and the center of inertia of the sun be located in the XY plane of a three dimensional rectangular coordinate system. Let the origin be the center of inertia of the sun. Thus, the earth is moving circularly, about the origin of the frame, and the motion is constrained to be in the XY plane. At one moment in time, the velocity of the earth is given by: [math] \vec v = -v \hat i [/math] And six months later, its velocity is [math] \vec v = v \hat i [/math] So its tangential speed has been presumed to be constant, roughly so, since the gravitational force is approximately constant, and the earth is travelling through a vacuum, which we assume does not impede the foward motion of the earth in any way. Now, I don't perfectly remember the device, because it isn't ever explained clearly, and furthermore there was some kind of error in the derivation. I remember there was, something strange about the S and S` thing in the derivation. The book was Serway, Moses, Moyer, modern physics. At any rate, I will re-analyze the Michelson experiment, and show you where the error is. Moving on... Oh yeah... Suppose Michelson's basic idea was right, and that he should be able to detect the earth's motion through the luminiferous ether... We get something like this... In order for the Michelson experiment to work, the experiment must be carried out during the daytime. Now the earth is spinning on its axis... we all know it's spinning, because we don't all burn up. 12 hours of day, followed by 12 hours of night. And the sun regularly traces out the pretty much the same path in the sky, day after day... rises in the east, and sets in the west. So lets see what part of the earth the Michelson experiment took place in, I want to say New York... No, I was incorrect, it was Annapolis, so Maryland. The experiment took place in 1879. So, lets think about the earth's motion in the XY frame. Lets stipulate that the earth is moving counterclockwise in the frame.

By setting the constants of nature to 1, you are losing information. Cant you see that? Regards

You say it is based on that principle, I have not seen that yet. So far, all that I have seen is notation, which is a necessary thing that must be learned, so this is perfect for a beginner, which is what I am. As for the theory of General relativity, I have always understood that its fundamental postulate is the principle of equivalence (not the constancy of the speed of light in vaccum), and there are strong, and weak versions. And I understand that as either gravitational mass = inertial mass, OR a reference frame at which the center of mass of some gravitating body is at rest, and some object is at rest relative to the body, that the laws of physics where that observer is, in the gravitational frame, are the same as if the observer is uniformly accelerating in deep space. Now, I have never seen anyone go from the assumption of GR (which is the principle of equivalence) directly to the field equations, and I certainly would like to see this done, because... I have an argument which leads to the conclusion that the two kinds of frames are not equivalent, and that argument is based upon formulas from electrodynamics. I have heard that the general theory of relativity leads to the special theory of relativity, as a special case, but this has never been proven to me. Regards

I'd prefer to leave them in the formulas for now, if that's ok with you.

Alright, take your time, I am learning this painfully slowly I'm afraid. But I am doing my level headed best. And you are a good teacher, thank you. Kind regards

Ok, one at a time. What is the Riemann tensor?

What if that matters? I don't understand this "geometrization" thing which keeps getting referred to. If it doesn't alter the truth value of any statements, then I will learn how to switch to it. Actually, I can learn how to switch to it no matter what. But what do most physicists do?

It's ok to be a bit sloppy with notation, as long as you explain it as you go, I can follow. I understand that e is reserved for basis vectors. Ok so... for me... let me see where I am at here: (this will all be from memory) Start off with Einstein's field equation of GR, which is this: Thus, we can also write: (N M) to denote the tensor. [math] G_\alpha_\beta = \frac{8 \pi G}{c^4} T_\alpha_\beta [/math] In the equation above, the LHS is called the Einstein tensor, and the RHS is the product of a scalar (with units of inverse Newtons (in SI) ) times the stress-energy tensor. G is the Newtonian gravitational constant, and c is the speed of light relative to an emitter. Tensors are linear operators, that means this... For any mathematical operator [math] \text{L} [/math]: [math] \text{L} [/math] is a linear operator if and only if: [math] \text{L}[a f(x) + b g(x)] = a\text{L}[f(x)] + b \text{L}[g(x)] [/math] In the formula above, a,b denote constants, whereas x denotes a variable quantity. In the years from 1884-1894, Gregorio Ricci Curbastro worked out absolute diffferential geometry, later called the Ricci Calculus, and now referred to as Tensor analysis. This was the mathematics used by Albert Einstein, in the formulation of the general theory of relativity. The Einstein tensor can be written in terms of the Ricci scalar, and the Ricci tensor, and the metric tensor as follows: [math] G_\alpha_\beta = R_\alpha_\beta - \frac{1}{2} R g_\alpha_\beta [/math] In the expression above, R sub alpha sub beta is referred to as the Ricci curvature tensor, and R denotes the Ricci curvatur scalar, and g sub alpha sub beta denotes the 'metric tensor'. Now, here is one notation for a tensor with N "one forms" and M vectors for its arguments: [math] \left( \begin{array}{c} N\\ M\\ \end{array} \right) [/math] In the notation above, N denotes the number of "one-forms" and M denotes the number of vectors that are the arguments of the tensor. And there is also a similiar notation, in which we write out the elements of the tensor horizontally. This will lessen the usage of latex as well: (N M) Ok lets see what else do I remember... Ok, there are Christoffel symbols of the first and second kind, here is what they look like: [math] \Gamma_i,_j,_k [/math] The symbol above is a Christoffel symbol of the first kind. And there is also... [math] \Gamma^k_i,_j [/math] And the symbol above, is a Christoffel symbol of the second kind. Lets see, what else is there... Ok, I need to know about Einstein summation convention... Suppose we have some arbitrary vector in three dimensional Euclidean space... the most common notation for an arbitrary vector is: [math] \vec V = V1 \hat i + V2 \hat j + V3 \hat k [/math] Now, the LHS makes sense in the affine plane, that is a plane without coordinates, or an affine space I should say, which would be a frame without coordinates, but the RHS requires that your frame have coordinates. let me explain this: Let us define an affine frame as follows: Definition: An affine frame is three mutually orthogonal infinite straight lines. Definition: A non-affine frame, or simply frame for short, most commonly called a rectangular coordinate system, is a frame in which the lines have coordinates on them. In other words, the axes of a frame are are number lines, instead of just lines. And because we have mapped the set of points on an axis to real numbers, we can pick a unit of distance, like the meter for example, to be our unit of distance measurement. Oh, in post 141, you said... Raised indices denote covariant... Lowered indicied denote contravariant... "metric inverse metric" respectively... In the general theory of relativity, the following symbol denotes the metric tensor... [math] g_\alpha_\beta [/math] Alpha, beta are called contravariant components. And the following symbol denotes the inverse metric tensor... [math] g^\alpha^\beta [/math] In the previous symbol, alpha,beta are covariant components of the tensor. When a comma is used in the notation, that means that subsequent indices are to be differentiated, rather than previous ones.

As I am learning something old Matt' date=' I am [i']simultaneously [/i] working on something new. You know what that word means dont you Matt Kind regards What is CHAR 3? PS: Oh and I don't doubt that I am confusing things, but that's only temporary. How in the world, can vector (1,1,1) have length zero? Tw=3w I am slowly following you. How do you reach the conclusion that there are an infinite number of maps?

Thats a bit better. Thank you very much. Kind regards

I just looked at it, but didn't see a way to make a smooth arc.

Yes, something like that. Is there something with a smoother curve?

Can latex be used to make an 'arc' symbol. I want to refer to arc AB, and put a curved line above AB. Can you do that with latex? Thank you

Maybe this will help. Right now' date=' I am doing a study of classical geometry, and restricting myself to only those things allowed by Euclid's postulates. What I have been doing, is busily constructing an extremely long proof sequence. Recently, I posted a thread on similarity of triangles. The reason I did that, is because if I have AA, then I must also have AAA, because the sum of the interior angles of a triangle are 180 in Euclidean Geometry. Now, once I have AA, and likewise AAA, then I can quite easily prove that the ratio of the circumference to the diameter of [i']any [/i] circle, is a dimensionless numerical constant, using a construction that I have found. Its quite a simple construction, here let me describe it to you: [math] \text{Theorem:} [/math] Let A, B denote circles chosen at random. Without loss of generality, let them lie in a common plane, and let them be concentric (having the same point as center). Ok so... Using a postulate of Euclid, we can pick a point in space at random, and describe two circles with that point as center, but different radii. Let C1 denote the circumference of the smaller circle, and let r1 denote the radis of the smaller circle. Let C2 denote the circumference of the larger circle, and let r2 denote the radius of the larger circle. We now want to set out and prove the following statement is true: [math] \frac{C_1}{D_1} = \frac{C_2}{D_2} [/math] Where D1 denotes the diameter of the smaller circle, and D2 denotes the diameter of the larger circle. Recall [math] D_1 = 2r_1 [/math] [math] D_2 = 2r_2 [/math] So let two circles have been described in a common plane, with a common center. Now, from the center of the circle, produce two rays in the plane in any direction, long enough to cut both circles. (Euclids first postulate allows us to construct a finite straigth line between any two points, and his second postulate allows us to make the straight line segment longer, from either extremity, but still in a straight line. So an angle has been chosen at random. Thus, once the theorem is proven, it will hold for any arbitrarily chosen angle. Now, a person could look at the figure and see two angles, one larger than the other, but in what follows, I am referring to the smaller one. So we have a sector of a circle we can focus upon. Focus now, on the ratio of the smaller arc length, to the larger, in the sector. Right now, these lines are curved, instead of straight, but to use AAA, we need triangles with straight lines as sides. Without a diagram, this is going to be somewhat hard to communicate, but I can see the picture without even drawing it, so I will convert it into words just the same... Let the center of the circles be denoted by A. Let Ray1 cut the smaller circle at the point B, and the larger circle at point C. Let Ray2 cut the smaller circle at point D, and the larger circle at point E. Thus, we have two arcs: [math] \overset{\frown}{BD} [/math] [math] \overset{\frown}{CE} [/math] The radius of the smaller circle is such that: [math] r_1 = \overline{AB} = \overline{AD} [/math] And the radius of the larger circle is such that: [math] r_2 = \overline{AC} = \overline{AE} [/math] Now comes the interesting part. We are going to construct two straight lines, whose lengths are equal to the lengths of the curved arcs. These two lines are going to be sides of different, but similiar triangles. In other words, we are going to measure that which is curved, using that which is straight. The whole point is this... we are going to use knowledge of triangles, to learn about circles. Now, I just have to explain what is actually an extremely simple construction to carry out. Incidentally, once this construction has been carried out, then it can be connected to another thread, which I am still working on, which has to do with the formula for the area of a circle. In that thread, I start off with the definition of the area of a square, then move on to the area of an arbitrary circle, the pythagorean theorem gets proved, and next, I am going to cover the formula for the area of an arbitrary regular n-gon. And then in the limit as n approaches infinity, I am going to obtain the following formula for the area of a circle: [math] A_{circle} = \pi r^2 [/math] But, in order to properly succeed in that argument, it is necessary that I have previously proven the theorem I am proving here in this thread, using the construction I am about to give. At the point B, construct the tangent line to the smaller circle; the smaller circle is the circle with arc BD upon it. At the point C, construct the tangent line to the larger circle; the larger circle is the circle with arc CE upon it. (If you don't know how to construct the tangent, that is quite easily explainable. Knowledge of how to construct the tangent line is considered as part of the a-priori knowledge of this reasoning event going on here.) So let those two lines have been drawn. Now, focus upon [math] \overset{\frown}{BD} [/math] Suppose that there was a finite straight line BF, whose length was exactly equal to the length of the arc [math] \overset{\frown}{BD} [/math]. Then you could describe a circle with radius [math] \overline{BF} [/math] which would cut the tangent line lying upon the smaller circle at the point F. Suppose the point F was successfully found. Similarly, if there was a finite straight line [math] \overline{CG} [/math] one could describe a fourth circle, such that the point G is on the tangent line currently lying upon the larger circle. Suppose the point G was found. Thus, we have formed two right angles. One right angle has side AB, and side BF. The other right angle has side AC, and side CG. We know the angles are right angles, because the tangents to a circle must be perpendicular to the radius drawn to the point of tangency. If this were not so, then the tangent would cut the circle at more than one point, and therefore not be a tangent to the circle, it would be a secant. Now, since the sum of the interior angles of a triangle must equal two right angles, it is necessarily true that: Angle BFA is equal to angle CGA. But this could use a bit more explanation: In order to use this construction to conclude that the ratio of circumference to diameter is a numerical constant, we need to be certain that: The straight line AF and the straight line FG lie in one common infinite straight line. Suppose we already are certain that AFG line in one infinite straight line. We will then be certain that: [math] \angle BAF = \angle CAG [/math] We are already certain that the three points A,B,C lie on the same infinitely long straight line, because that was stipulated during the construction. And we know that we constructed two right angles, so we know this too: [math] \angle ABF = \angle ACG = \mathcal{R} [/math] Where I have used the symbol [math] \mathcal{R} [/math], to denote a right angle. Now, if we hadn't have chosen an arbitrary angle, if instead we choose the perfect angle for segment BF to equal segment AB, then triangle ABF would be isosceles. And if we hadn't have chosen an arbitrary angle, if instead we choose the perfect angle for segment CG to equal segment AC then triangle ACG would be isosceles. And base angles of isosceles triangles are congruent. The angle at which this occurs, is called one radian. But, if we choose our angle ahead of time, then the angle chosen wasn't at random... it wasn't arbitrary. Since the sum of the interior angles of a triangle is [math] 2\mathcal{R} [/math], it would follow that: [math] \bigtriangleup ABF [/math] is similiar to [math] \bigtriangleup ACG [/math] And here is why... Both triangles have a right angle inside of them. Focus now on isosceles triangle ACG. The right angle cannot be a base angle, because then the other base angle would also be a right angle, and the sides of the triangle wouldn't meet at the apex, which they clearly do. Thus, the following two angles are the base angles of triangle ACG, and they are equal as Pappus ingeniously proved several hundred years after it appeared as proposition 5 of Euclid. [math] \angle GAC [/math] [math] \angle AGC [/math] In other words, each of the angles above is a base angle of triangle ACG. Therefore, they are equal in measure. Thus, the following statements would be true had we chosen a specific angle of one radian: [math] \angle GAC = \angle AGC [/math] [math] \angle FAB = \angle AFB [/math] You know what, this argument still isn't clear enough, because we have assumed the conclusion. In other words, by saying that we chose the perfect angle, we cannot conclude we chose the perfect angle. And this construction is supposed to be for the purposes of reasoning correctly. Perhaps I should start over, and be clearer. ---------------------------------------------------------------------- Goal: To prove, as a theorem of some unknown set of axioms that, the ratio of circumference of any circle, to its radius, is a numerical constant. Suppose that we have the ability to rotate circles and lines out of a common plane, into the third dimension, as well as translate their centers through space. Then, given any two circles in the universe, we could move them from where they initially are, into a common plane, and then cause them to have the same center. Now, circles and straight lines don't really exist in reality, but in geometry, we pretend they do, so that we can reason about space. Now, in the theorem we are attempting to prove here, the theorem starts thusly: For any circle .... Now, in the construction being carried out, it is stipulated that we start off with two concentric circles lying in a common plane. So someone can come along, and argue that we haven't proven the theorem is unconditionally true, that what was proved was this: If two circles are coplanar, and have the same center, then the ratios of their circumferences to their respective diameters are both equal to the same numerical constant. But that isn't what we want to prove. Let us proceed as though we have as axioms, that we can translate the center of a circular figure through space. Let us also proceed as though we have the ability to rotate a geometric figure out of the plane of construction, into the third dimension. Thus, as long as there is a sequence of steps that can be carried out, which translates the center of a circle far away from a given circle, all the way to where its center coincides with the given circle, and then we can rotate it so that it lies in the same plane as the given circle, then we will be justified in saying that we proved that "for any circle" And that what we proved is unconditionally true. So we have found two unknown axioms, which will permit the particular construction I have in mind, to actually succeed in proving the theorem. So we are going to be moving circles and straight lines around in our mind, so the question arises as to whether or not this takes place in time. In other words, are the transformation formulas functions of time or not. Historically, this is of some significance. The reason why has to do with the special theory of relativity, in which moving objects shrink if they are moving in a reference frame, as compared to a measurement made in their rest frame. If it can be shown that simultaneity is absolute, then SR is wrong, and we don't have to worry about this. But some reasoning agents will not know a priori to this reasoning event, that SR is wrong. For that reason, the motion here is considered to take place at the same moment in time. Thus, the geometric figures which we move around in our minds, cannot change size. This means that the transformation formulas we are looking for are abstract. They do not represent real transformations about the center of mass of real objects, with real centers of inertia. And there should be no objection to this, inasmuch as real circles don't exist anyway. So what I have in mind is something very very simple. There is a theorem here, which is to end up being true for any circle in three dimensional Euclidean space. First lets state the theorem to be proven clearly: [math] \text{Theorem:} [/math] The ratio of circumference to diameter, of any circle, is a numerical constant. Since the constant is unique among all numbers, let us reserve a special symbol for it, which is the universally accepted symbol [math] \pi [/math]. So let C denote the circumference of an arbitrary circle, and let D denote its diameter. Symbolically, here is the theorem we are setting out to prove: [math] \forall O[ \frac{C_O}{D_O} = \pi ] [/math] There is also another way to write the theorem we are trying to prove. Suppose we are given two arbitrary circles, circle A, and circle B. Each circle has its own diameter, and each circle has its own circumference, and these are commonly thought of as properties of circles. We can avoid being abstract, if we choose a unit of distance, so I will choose the meter. So the diameter of circle A, is some amount of meters, and the circumference of circle A is also some number of meters. And the same goes for circle B. The diameter of circle B is some number of meters, and the circumference of circle B is some number of meters. Now we shouldn't think that the lengths in question must be an integral number of meters. For example, circle A could have a diameter of .2 meters, and circle B could have a diameter of 1.52954345 meters. We are being entirely general, and thorough in this analysis. So let us denote the diameter of circle A, as follows: [math] D_a [/math] And let us denote the circumference of circle A, as follows: [math] C_a [/math] And let us denote the diameter of circle B, as follows: [math] D_b [/math] And let us denote the circumference of circle B, as follows: [math] C_b [/math] So we have two circles, chosen at random from the set of circles. So, using the symbols above, the theorem we are trying to prove is: [math] \frac{C_a}{D_a} = \frac{C_b}{D_b} [/math] Since the circles were chosen at random, it would thus be true that for any circle A, and any circle B: [math] \frac{C_a}{D_a} = \frac{C_b}{D_b} [/math] But we need to have adopted translation and rotation axioms.

Yes I know the formula is in radians, it's the one I committed to memory 20 years ago.

Let's see... [math] S = r \theta [/math] where theta is in radians. S is arc length' date=' and r is the length of the radius of the circle. So le me have a look at your formula here... [math'] \frac{A}{360} C = \frac{A}{360} 2 \pi r = L [/math] A is "degree of arc" L is arc length. So here is a relationship between degrees, and radians: [math] \frac{degrees}{180} = \frac{radians}{\pi} [/math] So using the formula above, you can switch back and forth, i know that. So of course it follows that: [math] \frac{degrees}{360} = \frac{radians}{2\pi} [/math] Therefore: [math] \frac{2 \pi}{360} = \frac{radians}{degrees} [/math] So you have L= arc length, I have S=arc length, therefore S=L Therefore: [math] \frac{A}{360} C = \frac{A}{360} 2 \pi r = S [/math] Therefore: [math] \frac{A}{360} C = A \frac{radians}{degrees} r = S [/math] So with A in units of degrees, r in units of meters, S will have units of radians times meters?

Oh, it is exactly one radian. There are a little more than six radians, in any full circle. mmmm 6.28.. something or other, the number being transcendental. Which means that it isn't the root of an algebraic equation with rational coefficients. The first human being to demonstrate that pi, and hence 2pi, is a transcendental number, was German mathematician 1852-Carl Louis Ferdinand von Lindemann-1939. This was in the late 1800's, i forget the exact year. So yes, the answer is exactly one radian, but I want the answer in degrees. Regards I don't want a trivial answer. I was thinking of something that can be pictured. The steps involved in figuring out the answer.