Johnny5

Jaques, would you mind deriving the formula for 'doppler effect' for me? I seem to have trouble remembering it. Thank you PS: I know it resembles the effect that happens when a car is coming towards you with its horn on.

Thanks for the site. I clicked on the hypercube. Highly interesting animation. Yet there is nothing in reality which rotates in that manner. Yet still... it is interesting. Regards PS: What is most interesting, is that anyone at all had a formula for it, which could then be worked into a program. One of the ways i can see that it cannot exist, is that while it rotates, certain things vanish. And also pass through each other, as if it isn't solid.

A student comes to you, and asks you where did they get the formula y=mx+b from? How do you answer the student? Thank you

Is this right? Let V denote a vector in SR. We write: [math] \vec V = V^\alpha \hat e_\alpha [/math] ?

Here is exactly what I want... A logical proof' date=' but you can draw whatever you want to assist in the explanation. I started out with one right triangle. Then i picked a point at random on the hypotenuse. Then, i constructed a perpendicular to one of the legs. So then I had a right triangle inside a right triangle. I then labled the legs a,b,A,B. Now, I am just lookin for a proof that a/b=A/B. If you prove that aB=Ab, then this will suffice for me. aB is the area of a particular rectangle in the figure i started out with, and Ab is the area of a different rectangle in the figure i started out with. But this drawing just led me in circles. Then i started off with the assumption that [math'] \exists r \in \mathbb{R} [ ar=A] [/math] So r is some real number. So if the smaller triangle has a base of 10 feet, and the larger triangle has base of 100 feet, then r=10. I am having trouble finding a proof for it. The shorter and simpler the proof is the better. I intend to use this as a lemma in a more complicated theorem.

In this part here I'm gonna run through the vector space axioms. Actually, I think I'm gonna turn them into theorems. Let [math] \vec u [/math] denote an arbitrary vector. Let [math] \vec v [/math] denote an arbitrary vector. Definition: Vector addition is defined geometrically. Theorem: [math] \vec u + \vec v = \vec v + \vec u [/math] Let S denote an arbitrary three dimensional Euclidean frame. Therefore, the vectors in the theorem can be expressed using their coordinates in the frame as follows: [math] \vec u = u1 \hat e_1 + u2 \hat e_2 + u3 \hat e_3 [/math] [math] \vec v = v1 \hat e_1 + v2 \hat e_2 + v3 \hat e_3 [/math] Or in a more compactified notation: [math] \vec u = u1 \hat e_1 + u2 \hat e_2 + u3 \hat e_3 = \sum_{n=1}^{n=3} u_n \hat e_n [/math] [math] \vec v = v1 \hat e_1 + v2 \hat e_2 + v3 \hat e_3 = \sum_{n=1}^{n=3} v_n \hat e_n [/math] Or, using Einstein summation convention, we can write: [math] \vec u = \sum_{n=1}^{n=3} u_n \hat e_n = u^n \hat e_n[/math] [math] \vec v = \sum_{n=1}^{n=3} v_n \hat e_n = v^n \hat e_n[/math]

You are given two right triangles, one with a larger area than the other. Each of the triangles has the angle [math] \theta [/math] in it. The legs of the smaller right triangle are a, b; with a>b. The legs of the larger right triangle are A,B; with A>B. I am trying to prove that: [math] \frac{a}{b} = \frac{A}{B} [/math] I just can't find a simple way to prove it. Anyone know how? Thanks

Swansont, what is a perihelion shift exactly? Regards

You have to do things using both words, and math/logic symbols. Words are for communication to those who speak your language, but the symbols allow you to communicate to others who do not. (also, and most importantly, logical symbols allow you to reason correctly) Right now I'm addressing an issue which bothered me since the very first time I saw it, which was well over ten years ago. Why was the distance from a point to itself, defined at all? There are an infinite number of paths from a point to itself, and no one of them is shorter than all the rest. It is only in the case of two distinct points, that "shortest path" has any meaning. It's a minor point perhaps, but once I solve it to my satisfaction, I can move onto something else, with better approach. I guess, I am just trying to decide whether I am going to adopt axiom 1 or not, above. This one: [math] \forall A \in \mathbb{R}^3 [d(A,A) = 0] [/math] The others are fine. I have already accepted the truth of the triangle inequality, and distance between two points has nothing to do with direction, its positive, and hence I have no problem with d(A,B)=d(B,A)>0 for any points A,B such that not(A=B). If there is no reason not to adopt it, then i will, but I don't want it to screw anything else up, and it might, since frames can move relative to one another. I just want to check so that I can be sure. Perhaps it will help you understand more, if I explain to you what I am trying to do. Take a given Euclidean frame of reference. It has a euclidean metric, given by the Pythagorean theorem. Let me explain this. All I mean by Euclidean frame of reference, is really two things... First that there are three mutually perpendicular coordinate axes. That constitutes a frame, and the distance between any two points in a single frame S, is given by the generalized Pythagorean theorem. So, I approach the problem, as always, using first order binary logic. I ask myself, what my domains of discourse are. In the case that there is only one domain of discourse, things are rather easy. In the case of multiple domains of discourse, I know the symbolism is going to increase in complexity, but necessarily so. Now, there are already firmly established symbols in place for me to use. The most commonly used symbols are: [math] \forall \equiv \text{For any}[/math] [math] \exists \equiv \text{There is at least one} [/math] [math] \neg \equiv \text{not} [/math] [math] \wedge \equiv \text{and} [/math] [math] \vee \equiv \text{or} [/math] [math] \Rightarrow \equiv \text{if-then} [/math] [math] \Leftrightarrow \equiv \text{if and only if} [/math] Now, there is some redundancy in the use of all of them, but that is irrelevent. They are all studied, so all can be known. Now, the logical operators above, have precise logical definitions, in terms of the two truth values true,false, and their usage must be in accordance with those definitions. But those symbols are defined on one and only one set, which I will denote as [math] \mathbb{S} [/math] So when you teach logic, one of your domains of discourse is [math] \mathbb{S} [/math] An element of S is usually called a statement, but also sometimes called a proposition. I pretty much stick with the term statement. The criterion for X to be an element of [math] \mathbb{S} [/math], is that it be either true or false, at any moment in time. And this idea can be formulated, using set-builder notation, as follows: [math] \mathbb{S} \equiv \mathcal{f} X: |X|=0 \ \text{XOR} \ |X|=1 \mathcal{g} [/math] Translation: [math] \mathbb{S} [/math] is the set of all X, and only X, such that, the truth value of X is true (XOR) the truth value of X is false. The meaning of XOR comes from its truth functional definition: AB A XOR B 00 0 01 1 10 1 11 0 Thus, the statement denoted by (A XOR B) is true when A is true, or B is true, but not both. Now I am accustomed to using temporal logic, which means that the truth value of some statements can vary in time. This actually matters in what I am doing, because there are frames which can go from being inertial frames, to non-inertial frames, in an ultimate reality sense. I dont really know how else to say it. In other words some frame S, can go from being an inertial reference frame, to a non-inertial reference frame, in what I'm working on. So that I have to deal with statements whose truth value can vary in time. It's harder than it sounds. It's not hard at really. In fact it makes more sense, then thinking only about statements whose truth value is constant in time. At any rate, astronomers believe that space is expanding. So whether they know it or not, they need to analyze that idea using temporal logic, since the distance between two points in one frame is a function of time. Contrast that with the Euclidean space with which everyone is familiar. In Euclidean space, the distance between two points in one frame, cannot vary in time. So if I say the distance between point X, and point Z, in frame S is 3 meters, that statement is true forever. Contrast that with the idea that space can stretch. Using that idea, the statement that the distance between X, and Z is not constant in time. So that the statement can go from being true in frame S, to being false in frame S. The only way to succeed at logical analysis, using this kind of complexity, is to know binary logic thoroughly. So while it may seem like I am stuck on one stupidly simple statement, which happens to be: [math] \text{for any reference frame S, and any point A in S} [d(A,A) = 0] [/math] I can't just casually mark it off as true forever, without first checking out how it impacts the massive logical system which I've been steadily working on. If you didn't understand, it doesn't matter, eventually i will figure out what I want to know, and then that will be the end of it. Regards PS: I can be more precise. Suppose that some ruler is in its own rest frame S, permanently, no matter whether it is at rest in some other frame S`, or moving at a constant speed in S`, or accelerating in S`. Then if the theory of special relativity is correct, its length is a frame dependent quantity. Here is the formula for its length: [math] L = L_0 \sqrt{1-v^2/c^2} [/math] In its own rest frame, its speed is 0. So using that formula, its length in its own rest frame is: [math] L = L_0 [/math] Now, a ruler is that which can measure distance. Suppose that the center of inertia of this ruler is moving through the coordinates of inertial reference frame S` at speed c. Then using the formula above, the length of the ruler in this frame is given by: [math] L = L_0 \sqrt{1-c^2/c^2} = 0[/math] And this would certainly cause some kind of contradiction, which could be made explicit if I wanted. Mathematicians think nothing of setting the distance from a point to itself equal to zero. But if distance is operationally defined, then they cannot do this. The only way I would easily agree to it, is if they proved that it reduced proof complexity. What's bugging me is strings. Strings are things with length. Now, no matter how much you twist a string, its length is pretty much constant. But now focus on the ends of the string. If I ask what the distance is between them, you can take a ruler and measure it, no thought necessary. You don't need the Pythagorean theorem you just need a ruler. But the key to straight line, is that of all the strings that could go from one point to another, the straight line is the shortest. The other lines have a 'distance' just not a 'straight' distance. BUt the Euclidean metric is the one which represents a minimization of "all possible string lengths from one point to another different point." So if distance is associated with length, then allowing distances of zero, might lead to a contradiction to something else.

I was just thinking of it as isotropic over all solid angle. I used to know all the formulas, for antenna gain, etc. They can come back if i need them to solve this. Here is a link to Stefan's law Here is Stephan's constant: [math] \sigma = 5.76 x 10^{-8} \frac{J}{s m^2 K^4} [/math] The units are joules per second, per square meter, per degree Kelvin to the fourth. So if we multiply stephan's constant by the thermodynamic temperature to the fourth, then the final units will be Joules/second per square meter. Joule per second is unit of power called watts. so.. Units will be Watts per square meter, which is a unit of intensity. And in the formulas I used to use for EIRP, we use the letter psi in order to denote the flux density of an emitter, which has units of Watts/square meter. So using psi to denote the flux density, we have: [math] \psi (T) = \sigma T^4 [/math] What I now need, is a formula for psi, in terms of the distance from the radiation source. The surface area of a sphere is: [math] 4 \pi R^2 [/math] The radius of earth is: 6,378 kilometers 6,378,000 meters Perform the analysis in the rest frame of the sun. Imagine a sphere of photons radiated outwards. Denote the number of photons emitted per second by N. The speed of a photon in the rest frame of the emitter is: 299792458 meters per second Now, the total number of photons on an expanding sphere is an S frame constant, since the speed of the photons in the frame are all the same, namely c=299792458 m/s. Now, eventually, the photons will encounter the earth. And when they do, only one side of the earth will be facing the sun. From the photons point of view, there is a circular area in their way, of area pi R^2, where R is the radius of the earth. Now, we have measured the intensity of the sun at the surface of the earth, as approximately 1400 watts per square meter.

I want to compute the number of photons emitted by the sun per second. Assume that the sun is an isotropic radiator. Denote the energy per photon E as: [math] E = \hbar \omega = hf [/math] Now, we know the approximate earth-sun distance. Using google we find that: [math] R_{earth-sun} = 1.5 \times 10^{11} \text{meters} [/math] Which is equivalent to 150 million kilometers. Now, as the photons move outwards, the intensity falls off. The photon density decreases, over any spherical patch. Just think of a pin cushion. When you are near the object with the pins in it, there are say ten needles passing through an area of say one square millimeter. Now, move your tiny one millimeter square box away from the pincusion, and soon less needles will be passing through that same square. The number of needles per unit area, decreases as you move away from the pin cushion. You experience this basic effect, every time you move your face towards a light bulb. The number of photons hitting your face is larger when your face is nearer to the light bulb then when it is further away. I am trying to remember the formula for equivalent isotropic radiated power. Here are some formulas about it Antenna pattern measurment

Yes, let me see if I remember, not using google on this If two events E,F are independent then: [math] P(E \cap F) = P(E)P(F) [/math] Now let me see if this is right... yes, that is right, its equation 11 at Wolfram on probablity theory Ok so thank you for that, now let me read the rest of your response. Let E,F denote events. Two events are mutually exclusive if and only if [math] E \cap F [/math] Where we have used set theory to define the events. I need to make sure I have the logic right... Definition: Let E,F denote two events. Let S denote the sample space. [math] E \cap F \equiv \mathcal{f} s \in S| s \in E \wedge s \in F \mathcal{g} [/math] [math] P(E) \equiv \text{probability E occurs} [/math] [math] P(F) \equiv \text{probability F occurs} [/math] [math] P(E|F) \equiv \frac{P(E \cap F)}{P(F)} \equiv \text{probability of E given F} [/math] Let me ask you a question, what is the sample space in your particular problem? Lookin back to your first post, I see that you admit you couldn't define the events properly. That should give you a clue as to where the problem lies. You have formulas which are defined using a sample space, and you have a question whose sample space is impossible to define. Regards PS: But thank you anyways, at least I got to look at the formulas again. I used to use a tree diagram, to do Bayes formula, and I never bothered to remember the formula for total probability, but your post has made me want to go back and recall it again, so thanks.

The answer to the question, must be centered around the concept of inertia then. The amount of gravity due solely to the sun, in the local region of space to the earth, is related chiefly upon the inertia of the sun. This fact will not be ignored in the final solution of the problem. With that in mind, there has to be a connection between the amount of inertial mass of the sun, call it gravitational mass if you want, and the photons being emitted by it, which are the carriers of the sun's gravitational effect upon the earth, which is as a force, which accelerates the earth, and therefore, alters the path of the earth, making it deviate from a straight line, it might otherwise take. The shining is from the photons, but they in turn are emitted in enormous numbers, because there is so much matter inside the sun, and that amount of that matter is related to the total inertia of the sun. I see your point... A ball of matter which isn't shining, such as a black hole, would still have a gravitational effect upon material that enters its nonzero gravitational potential somewhere in space. But photons are that which travel at c, not necessarily those in the visible spectrum. There are two variables that I can think of, that might influence the total amount of gravity, in the field, and they are: Intensity of radiation Energy per photon I have to keep thinking about this. [math] \psi = 1395\frac{watts}{m^2} [/math] This is the intensity of solar radiation at the surface of the earth. Assume the sun is an isotropic radiator. So there will be EIRP formulas, that is, equivalent isotropic radiated power. From memory, they involve natural logs, I forget the exact formulas. Intensity of radiation falls off as 1/r^2. Any help would be appreciated.

Before I attempt to answer this, what is the flux of the sun at the surface of the earth, do you know its value?

I only just started thinking about this recently. No model yet.

Yesterday, I started thinking about the distance axioms, that's where I will pick things up today, nice and fresh. Here are the distance Axioms: [math] \text{Axiom I: }\forall A \in \mathbb{R}^3[ d(A,A) = 0] [/math] [math] \text{Axiom II: }\forall A \in \mathbb{R}^3 \ \forall B \in \mathbb{R}^3 [\neg(A=B) \Rightarrow d(A,B) > 0] [/math] [math] \text{Axiom III: }\forall A \in \mathbb{R}^3 \ \forall B \in \mathbb{R}^3 [d(A,B) = d(B,A)] [/math] [math] \text{Axiom IV: }\forall A \in \mathbb{R}^3 \ \forall B \in \mathbb{R}^3 [d(A,B) + d(B,C) \geq d(AC)] [/math] Axiom four is called the triangle inequality. It's logical formulation comes to you by merely looking at a triangle in an affine plane. An affine plane, is a fancy way to say, the triangle isn't in a coordinate plane, there are no coordinates assigned to axes. The symbol [math] \mathbb{R}^3 [/math] stands for Euclidean three dimensional space. An element of R^3 is called a point. So, for example, to translate axiom 1 into English, we have to say, "For any point A in Euclidean space, the distance from the point to itself is equal to zero." The notation, is actually a compactified form for the following more complicated expression: [math] \text{Axiom I: }\forall X[X \in \mathbb{R}^3 \Rightarrow d(X,X) = 0] [/math] Translation: For any symbol X, if X is an element of Euclidean space, then the distance from X to X is equal to zero. If you additionally know that elements of Euclidean space are called points, then you can translate it this way: Translation: For any symbol X, if X denotes a point, then the distance from X to X is equal to zero. Now, somewhere in the universe is the center of mass of the universe, and this location is at rest in at least one an inertial reference frame. Let us choose one out of the many, and call it S. Now, let S` denote a frame which is in motion through S. What I want to do, is make sure that axiom one is ok. Pick two points X,Y, at random in frame S. It is impossible for these two points to be moving relative to one another. This distance between them is denoted by: [math] d(X,Y) [/math] Now here is the thought which is bothering me. Suppose we stipulate that distance is something which must be measured by a ruler. No ruler can have a length of zero, and so a distance of zero makes no sense. Or, to put this another way, if a line can shrink to a point, then there would be a contradiction of your domains of discourse. In other words, you would have something like this going on... No element of the set of lines, is an element of the set of points, and no element of the set of points, is an element of the set of lines. And if you entertain as a possibility that the following statement is true: [math] L = L_0 \sqrt{1-v^2/c^2} \wedge \exists S [ v=c] [/math] You can reach an explicit contradiction. The first conjuct of the compound statement above, is just the Lorentz contraction formula of special relativity theory. Suppose that you have that as true, and that you entertain as possible, that there is at least one inertial reference frame S, in which the speed of the center of inertia of something is v in frame S, and also v=c in frame S, then what happens is this: Using the formula for length as a function of time, you will see that in frame S, the object which is moving is a point, and not a straight line. And no point is a line, yet the object is a line in other frames, in which it has its proper length. So that the following contradiction is explicit: [math] A \in \mathbb{R}^3 \wedge \neg (A \in \mathbb{R}^3 ) [/math] From which you conclude the following: [math] L = L_0 \sqrt{1-v^2/c^2} \Rightarrow \neg \exists S [ v=c] [/math] But of course if SR is false, you don't have to worry about the previous argument. Moving on... The key thing which defines a straight line, is that it is the shortest of all possible paths from one point X, to another point Y, in a single frame. So now, to any two points in a single frame, there is exactly one real number chosen which corresponds to the distance between the two points. The concept of shortest path is indispensable in physics, and I want to make sure there is no contradiction introduced into the logic of the distance axioms, when they interact with "shortest path" axioms which we may choose to introduce alongside the distance axioms. Distance is applied to "path length" Suppose something is moving in an inertial reference frame S, with constant speed v, and path: straight line. By definition, the speed of the object in this frame is defined to be distance travelled, divided by time of travel. [math] v_s = \frac{D}{\Delta t} [/math]

Actually, I would prefer to answer this question mathematically. Not necessarily. It's getting complicated to answer correctly, very fast. You need a mathematical model of gravity at this point in the discussion. PS: Gravity is an electromagnetic wave, better to say, superposition of them. And then you have to link this to the path traced out in the right frame.

Why would you pick lead? To know the answer to this, to prove you know it, would be very complicated. BUt here.... Let it be stipulated that something emitted by our sun, is what by and large, causes the earth to take its elliptical path around the sun. If you could build something to block that which the sun emits, and shield the earth from that which causes it to move in its curved path, then indeed the earth would fly off, as though you reduced the local gravitational potential to zero. So... If you then took this material whatever it was, and encased yourself in a sphere, and entered outer space, it would block that which would otherwise keep you in orbit about the sun. But in space you are already weightless. But your idea is consistent with the notion that photons mediate the gravitational force. A shield would do just that, shield the interior from external effects.

I'm pretty sure that photons mediate the gravitational force now. Does anyone disagree? He was wondering about how it is that the universe can expand, and take the matter along with it... inquiring into the method through which space couples to matter. But a true vacuum can not ever couple to matter. So, he made me think about what really is responsible for the gravitational force. I know it's not action at a distance. Or I should say,"It isn't instantaneous action at a distance." If the sun suddenly stopped existing, we here simultaneous to that, would still follow an elliptical path around the place in the solar system, which is the center of inertia we currrently orbit. Then, in 8 or so minutes, it would be as if the umbilical cord was cut, and we would shoot off out of the solar system, in a straight line at a constant speed, by Galileo's/Newton's Law of inertia. Why do I say 8 minutes, and not simultaneously? The answer lies in Gravitational field theory, and the "retarded gravitational potential." We know for a fact, that our sun emits something, which we call light. At least that's what it was first called. It gives us heat, and warmth, and makes plant's grow, and allows us to see the surfaces of things. Then with photon theory, we now say particles. The sun emits particles, which get called photons. But whatever is really going on, is going on in the local region of space surrounding the earth, where these 'photons' and the material of the earth finally get a chance to interact somehow. Fourier series can describe it. Some kind of wave superposition going on, which leads to the elliptical orbit of the center of inertia of the earth, in the rest frame of the center of mass of the solar system. Like i said, spyman gave me an idea. Regards