uncool

What? Congratulations; you've drawn a picture. Please, next time write the text out so it can be quoted. First, "If we move the whole set-up in the second diagram to the left at 1 m/s, we get the third diagram" is false. The third diagram has the lone electron show no movement towards the wire (as you've said that it's the same as the first diagram with the nuclei moving to the left instead of the electrons moving to the right), but you labeled the second diagram with the electron moving towards the wire. Second, "The only difference between the third diagram and the first is that the protons are moving to the left instead of the electrons moving to the right" is false, for two reasons. It neglects the linear density of charges. Unless you want to assume Galilean relativity, which would negate the point of the exercise entirely. Third, the final statement doesn't follow. Please try to explain your logic again. =Uncool-

So "obviously" meaning that the person who came up with the explanation would know the relation between the current and the magnetic field? This sentence makes no sense. Please try again. =Uncool-

Wow, another bunch of words that seems to have nothing to do with the post it's responding to. Are you ever going to respond to the papers cited? You are asking me to explain using a theory that I don't use. Not my job. You can feel free to ask me to explain relativity; that is where my experience lies. I like how you decided not to respond to all of the other parts of my post. No explanation of "obviously" or "illogical". Are you ever going to explain? =Uncool-

It would generate a magnetic field. This is not "the accepted explanation for magnetism", but a demonstration of how "a current generates a magnetic field" falls directly out of "an electric field relative to one frame gains a magnetic component relative to another frame". =Uncool- Again, bullshit. You don't ever focus on the physics. You still have yet to address the papers and the data - and as far as I can tell, you still have yet to even acknowledge them. How is this at all relevant to anything Bignose posted? You're welcome, but this is one of the most common explanations in every book on basic electromagnetism in special relativity - the fact that you don't know it shows that you don't know even the basics of special relativity. So again, you are trying to dispute something that you don't know. In order to dispute a theory, you have to know what the theory says in the first place. "Obviously"? Please explain. "Complex"? Fine, but if you're going to refuse to understand it, you have no ground on which to dispute it. "Illogical"? Please explain. And finally, it's easy to test - if you want, I can write the actual numbers to show the magnetic field generated by a wire according to this explanation, and you can check that it's right. Do you want me to do the calculation? That sure is a bunch of statements and a conclusion, but none of the statements supports the conclusion in any way. =Uncool-

The fact that you've added this requirement to the usual probability stuff is unclear. Is there any reason for us to assume it, other than for your formula to work out? =Uncool-

You haven't changed anything in what I used. I didn't use anything that was primed. Further, P(B|A') is not necessarily equal to P(B'|A). So no, my example still shows that your reasoning is incorrect. =Uncool-

For one thing, it's false. Let's say we have that P(A and B) = 0.3, P(A and not B) = 0.2, P(not A and B) = 0.3, P(not A and not B) = 0.2. Then P(A) = 0.5, so v1 = 0c; P(B|A) = P(B and A)/P(A) = 0.3/0.5 = 0.6, so v2 = 0.1c. v3 is then 0.1c, so according to your formula, P(A|B) = 0.6. But P(A|B) = P(A and B)/P(B) = 0.3/(P(A and B) + P(not A and B)) = 0.3/(0.3 + 0.3) = 0.5. You forgot about P(B). =Uncool-

Well, for one thing, you seem to be forgetting about the positively charged nuclei in the wire - those also have an effect, and when there is no current, it's an equal and opposite effect to that of the electrons. That's why when there's no current, there's also no force. Now, a wire here is assumed to lack charge relative to our original frame; this means that the linear density of the positive and negative charges (protons and electrons) is equal relative to this frame. In other words, relative to this frame, the wire looks like a line of stationary positive charges and another line of moving negative charges, such that the spacing between charges is the same for both of these. Relative to this frame, the electrical field from the entire wire is then 0. One of the ideas of relativity is that distance and time are relative, although charge isn't. So let's take a look at what's happening from another frame - the frame where instead of the wire (which basically means the positively charged nuclei) being at rest, the electrons are at rest and the nuclei are moving. Relative to this frame, what is the linear charge density of both the nuclei and the electrons in the other wire? Here, we're going to look at the current going in the same direction. Well, the electrons in the other wire were moving at the same speed as the electrons in this wire. If we do the relativistic calculations, we get that the distance between charges would therefore increase - so the linear charge density for the negative charges would decrease, and so the repulsive force between the negatively charged electrons would decrease. What about the force between the electrons and the nuclei? Well, the nuclei weren't moving, and now they are - so the distance between them has decreased - and so the force between the electrons and the nuclei in the other wire has increased. So if in total, there was no force originally, there now is an attractive force between the two wires. And we call that attractive force magnetism. I could show you the math explicitly, if you want, but it would take me a bit longer to write up. =Uncool-

Bullshit. Both Bignose and I cited original papers and experiments, and offered to explain several things to you. You seem to have forgotten the positively charged nuclei, which are a large part of the explanation here. It does tell us "why" an electric current creates a magnetic field: that the magnetic field is just our name for the force that arises out of the movement of electrically charges objects. What do you mean by "what they really are"? Now who is mysticizing? So you agree that mainstream physicists have "mastered magnetic fields mathematically"? In other words, that they have managed to come up with models that are amazingly accurate? =Uncool-

Source, please. Something other than your site. Why? Why shouldn't the aim be to find accurate models, and only once that's done to find the simplest model? I just gave the simple explanation. It appears that you are able to read, so you should have seen the explanation. I'm curious; since your other trolling seems to be successful, why add in sexism? From this, I'm beginning to doubt that you even understand what epicycles are. Yay, another baseless attack on mainstream physicists. Your "explanation" is what is mundane here. You have a bunch of words that say little and show even less knowledge. =Uncool-

Yay, another baseless attack on mainstream physics. Please, save it until you have something new - like actual evidence for your idea. See, the funny thing is, it's not that hard to explain quantum field theory - just very, very time-intensive and math-intensive. The problem is that people like you are unwilling to learn even the first step towards it, either on the physics side or on the math side. You really shortchange supporters of epicycle theory. A is going around B while B is going around C is really easy to explain. =Uncool-

Not to say that I support newts and his theory, but he's throwing physicists in with creationists and astrologists, and pretending that he's the skeptic of both creationism and astrology. =Uncool-

I'll assume that we are talking about classical physics so far. Newton's 2nd law says that there is a constant proportion between the force on an object and its acceleration. The acceleration of an object can be defined easily; force isn't quite so simple. The third law gives us a better idea of force; it says that any force on one object from another is balanced by an opposite force on another object. Put that with the first law, and we get that the total force on all objects is 0. Put all together, and what we get out is that Newton's three laws are equivalent to the following statement: that there is a constant of proportionality for each object (which we call mass) such that the sum of the velocities of each object multiplied by that constant is conserved over time. That gives us a definition of mass, based on a position function over time. =Uncool-

The definition that's being used is nonsense; however, mumbo jumbo has a different connotation than nonsense. When "mumbo jumbo" is being used to mean nonsense, it means nonsense that's dressed up in pseudo-technical terms and made to appear overly complicated. But in short, the definition being used is the one about nonsense. =Uncool-

We know that (b - c)/2 is nonzero by the paragraph before; if (b - c)/2 = 0, then b = c, so f(b) = f© <= u, and we assumed that f(b) > u. =Uncool-

In more mathematical words: Where the theorem says "Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε", we can rewrite it as: "Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε and δ < (b - c)/2". We can do that by using the following: Since f is continuous, there is a δ1 such that |f(x) - f©| < ε for any x with |x - c| < δ1. Then let δ = min(δ1, (b - c)/2). Then for any x with |x - c| < δ, |x - c| < δ1, so |f(x) - f©| < ε; we then also have that δ < (b - c)/2. So there is such a δ - so we can choose such a δ. =Uncool-

No, it isn't. The value of δ is chosen by us to satisfy a property, and as we can see from the property, if we have a δ1 that satisfies the property, then for any δ2 < δ1, δ2 satisfies the property as well; therefore we can choose δ as small as we want (but still positive). =Uncool-

We already have that c is an upper bound for S. In other words, for any x in S, x < c. Therefore, if we can prove that for every x in S, x is not in (c - δ, c], then every x in S is less than c - δ, and c - δ is an upper bound for S. When it says that there is such a δ, that means that you can choose the δ. So choose one such that that x is less than b. If c = a, then f© = f(a), so u = f(a), which we know is false. Similar proof for b. =Uncool-

They are the result of math - math which includes Newton's law of gravity. Ellipses are not a solution to the differential equations if you use a simple inverse law of gravity, nor if you use an inverse cube. I've already shown you the math that uses the inverse square law to get the ellipse. Initial position and initial momentum. Removing the angle, as I did, leaves us with total energy (kinetic and potential) and angular momentum (I'm not including initial radius and initial derivative of radius because you're looking at orbits, not at points on orbits). It's clear that you haven't seen the actual calculations. The two-body problem under any radially symmetric law of gravity can be reduced to a one-body problem. The conservation laws for the two-body problem reduce to the conservation laws for the one-body problem, so considering the one-body problem is sufficient. No, observations don't depend on theories. Further, when there have been departures from the elliptical orbits, that has given us the ability to predict the existence of more planets - this is how Neptune was discovered. Please explain what you mean by "no one has been able to interpret those solutions". If you mean that people haven't been able to use Newton's laws to predict where planets will be, then you are completely and utterly wrong. You've just asked the same question in 3 not-very-different ways; you're only asking about why a certain angle. And the answer to that is the same as it is for any differential equation: initial conditions. It would be a better use of time and "other resources" to actually learn the theory than to declare it wrong without knowing what it says. If you look at any calculation of the two-body problem, you will see one of two ways it is reduced to the 1-body problem: 1) One body is so massive that it can be assumed to be still, or 2) The problem is looked at with respect to the barycenter. The first is actually a special case of the second, since as the sun is so massive, the barycenter is effectively where the sun is, and barely moves. So your "new" theory is what has already been known - for hundreds of years. Further, it doesn't address ellipses at all. As an addendum: your claim that Newton's law doesn't predict ellipses has been rebutted. If you're going to continue claiming that Newton's law doesn't predict ellipses, then please try answering the math. Otherwise, please accept that your initial claim was false. =Uncool-

One simple reason for you, lidal: Since the set of paths is rotationally symmetric around the barycenter (that is, given a path, if we rotate the entire path at once, we get another path), we can model them using differential equations based entirely off of the radius and the time. Circles then correspond to constant functions. By Newton's 3 laws, this differential equation is a second-order differential equation - so the set of functions that satisfy the differential equation has 2 degrees of freedom. But the set of circles doesn't have 2 degrees of freedom - it has 1. So no, Newton's law of gravity - and in fact, any law of gravity that doesn't reduce Newton's laws to a first-order differential equation - predicts non-circular paths. Now, if you want to see the math that shows that Newtonian gravity leads to elliptical orbits, you could have looked it up in a lot of places. Here is one: http://www.math.wisc.edu/~robbin/234dir/kepler.pdf =Uncool-

How does that satisfy the second property? Are negative numbers allowed? =Uncool-

Devil's advocate answer: Because drugs are much more than a 1-off deal; as they are, they are exploitative and harmful. Therefore, for the same reasons that we have office safety requirements, even though some might be willing to work in a more dangerous environment: that those offering such a job are doing so to exploit the desperation of those who need a job so badly that they are willing to endanger their own lives and well-being. Drugs are much more than just "what we decide to put in our body"; they also change what we fundamentally want - in ways that can easily be undesirable. In short, a prohibition of drugs can be seen as an attempt to prevent exploitation of people's need for a temporary fix. The war on drugs as it is clearly isn't executed in such a way, and I fundamentally disagree with it, but this part of the idea behind it isn't entirely off the mark. =Uncool-

Finding equilibrium in chemical reactions with large numbers of reactants. =Uncool-

Let's unwind this. e^(2*(((ln(pi))^2+2*ln(pi)+1)/(2*ln(pi)+2))-1) e^(2*((ln(pi) + 1)^2)/(2 * ln(pi) + 2) - 1) e^(2 ((ln(pi) + 1)^2)/(2 * (ln(pi) + 1)) - 1) e^(((ln(pi) + 1)^2/(ln(pi) + 1)) - 1) e^((ln(pi) + 1) - 1) e^(ln(pi)) pi You've written a tautology. In other words, if you used any number instead of pi (in every position), the equation would remain true. e.g. 3 = e^(2*(((ln(3))^2+2*ln(3)+1)/(2*ln(3)+2))-1) =Uncool-

The Dirac delta already makes sense on the complex plane - however, the idea of residues would need to be extended to some subset of the set of distributions, rather than only dealing with (ratios of) functions. One property that it should have: the residue of any derivative should be 0, as can be seen by examining the Laurent series or using the residue theorem. =Uncool-