Bignose

There is at least one big huge thing missing -- definitions of what the terms in your equations mean! i.e. what is supposed to be the difference between inertia and total inertia? I'd suggest you define every term -- with the units -- at the top of every page. At the very least, you can't just assume that every reader knows what you mean by all the symbols you use.

Without math, how can you make such a statement?!? Taking the average is necessarily a mathematical operation. And exactly what are you taking the average of? And what experiment or theory gives you these values you are taking the average of? You still haven't answered any of my questions -- like what experiment that has been performed is impossible or poorly described by the standard model that your model does a better job of?

So? The vast majority of modern physics is done via computational programs/simulations. There is still a great deal of math behind the programs, and that math is verified and examined in great detail. The computer program is what is used to solve the math equations that validates the theory or rejects the theory. This is science. What you have is a story. And without some math, how could you get numbers? Anyone can make up numbers. It's the process that gets those numbers that is the important part. Translating your story into math makes it scientific. Just putting out numbers without any given method is essentially meaningless. ------------- Also, you didn't really answer my first question. What evidence is there that the current model -- again which is very well supported by the evidence (both mathematical and experimental) -- is inferior to your idea? Can you cite an experiment where your idea better explains the results? Can you show where the mathematics behind your model predict the results of experiments better than the current mathematical models. I.e. what are the different terms in your model, and why should they be included? What terms should be taken out of the current model and why?

The power tower is different from an exponent -- that's why there is a different notation. Your question, to me at least, isn't terribly clear, because you use a prime, but there are two variables given f(a,b). Is the prime meant to represent differentiation with respect to a or b?

So... any evidence that your theory is superior to the current model for which there is much supporting evidence?

Doesn't this look familiar? http://www.scienceforums.net/forum/showthread.php?t=36912 I responded to your question there. Guess what, I'm still not going to do your homework for you. And I hope that no one else on the forum will either. We can discuss either here or there, but repeating the same question isn't good forum etiquette. Also, since you are re-posting the exact same thing again today, it obviously wasn't "due tomorrow", was it?

flyguy, for the second choice, look up "Archimedes' Principle" in your favorite Internet search Engine

OK, well, thanks for stopping by, don't let the door hit you on the rump on the way out... But, seriously, dude. You aren't the first to play the conspiracy card, and you won't be the last. You really think that there is a cabal of scientists that are huddled together in dark alleys and smokey back rooms trying to make sure that the "truth" that black holes aren't singularities gets out? Why would anyone care?!? Scientists included! It's not like anyone owns the black holes or using them right now. Or will be for hundreds or thousands of years. Who would this conspiracy benefit?!? If you truly believe this, then you have a very poor idea of how the vast majority of scientist work. Most would be very eager to work on a completely new idea -- if and only if that new idea had some evidence that it was the right path. Every scientist I know would relish the opportunity to work on something completely new -- that's the primary reason a lot of people chose science as a career in the first place! Sure, there might be a few people bitter or angry or upset about a paradigm shift -- that's human nature. But, the overwhelming majority would love to jump in elbows deep into a new problem. But, there has to evidence that it is right. People aren't just going to chance their mind with a few paragraphs of text, and a few calculations with questionable assumptions. Those assumptions have to be shown correct, for example. There would have to be experiments to show that the new idea is right. The mathematics behind the idea would have to be checked and rechecked over and over again. And, the idea would have to stand up to questions from everywhere. People from all the disciplines. If you can't even stand up to 23 posts on an Internet forum, then science may not be the game for you. If your idea cannot even stand up for 23 posts on an Internet forum, then it may not be as good as you think. The questions asked here have been good, and if want or need time to research and answer them, then please take that time and come back with good answers. There will then probably be more good questions. This is how science works. What doesn't work is copping an attitude, and getting all huffy when people don't immediately call you the greatest visionary since Einstein. What doesn't work is accusing all of us to be part of a conspiracy to protect black holes. So, if you want to come back and act like a scientist, this forum will welcome you with open arms. Answer questions that are asked of you as best you can -- "I don't know" is an acceptable answer. Name calling and claiming conspiracies aren't. I sincerely hope you will continue to participate.

Why are you taking the ratio of the two temperatures here? This would be a unitless number now. The Stefan-Boltzmann equation has a constant in front of it that converts the units of K^4 to Joules per meter squared per second. The Stefan Boltzmann equation also doesn't have the ratio of two temperatures in it -- only the temperature of the radiating body. Also, heat can move around an object in different ways that just radiation. Especially in a body that is a fluid/plasma like the sun. I imagine convection and conduction are also important. Also also, it isn't any different than a chemical reaction occurring in a catalyst particle or in a droplet. The reaction can occur everywhere and the combination of radiation/conduction/convection determines the heat distribution along with the rate of reaction. I imagine the sun is the same way, in that the distribution of heat is due to the rates of fusion across the entire sun, not just due to the 2 millionth of a kg of material at the center of the sun. I don't even know what this has to do with anything. All in all, granpa, this is highly speculative. At the very least, your numbers need units on them. I can't follow what any of this means because almost all the numbers don't have units on them. Nor are a lot of your numbers cited properly. I.e. who says that "the energy in the gravitational field of a proton is 10^36 times less than the energy in the electric field of a proton." provide a cite for this please. Who also says "the energy required to heat the corona is 1/40,000th the total output of the sun." provide a cite for this too. Why? You can't just say something "should" equal something else unless you have a reason why. And finally, the obligatory question, do you have any evidence to back up your theory? I.e. an experiment to prove the existence of the gravity waves as you think they exist?

This forum doesn't do homework for people, but we will help you. Which one do you think it is, and why? If you don't have any idea, which one(s) can you definitely eliminate and why? Are there any properties of the quantities being graphed that helps you determine which graphs can be eliminated?

Google yields many good results. The first result Google presents is the wikipedia article where the first section is called "definition and calculation" http://en.wikipedia.org/wiki/Standard_deviation

Dennis, What Klaynos is trying to say here is that while you may indeed have the kernel of a good idea, it is impossible to know without the ability to make predictions and test them against the evidence. And, so, since it is impossible to know, there really is nothing to discuss. You may be right, or any of an infinite number of other ideas may be right, too. We don't know. Just because something sounds good, doesn't mean that it is so. So, without anything to discuss, an Internet forum -- 100% based on the idea of discussion -- this may not have been the best place to post this. Because this forum is based on discussing things in a scientific manner, and without evidence there is nothing to discuss here. Maybe this would be better posted on a philosophy or literature forum, but on a science forum, evidence is going to be asked for every time.

At the barest minimum, they are a very compact notation. Sure, turning [math]y_1 = a_{11}x_1 + a_{12}x_2 +b_1[/math] [math]y_2 = a_{21}x_1 + a_{22}x_2 +b_2[/math] into [math]\mathbf{y}=\mathbf{A}\cdot\mathbf{x}+\mathbf{b}[/math] doesn't look like much savings, but what if there were 1000 variables? I'd much rather write out [math]\mathbf{y}=\mathbf{A}\cdot\mathbf{x}+\mathbf{b}[/math] than 1000 equations with 1000 variables. And yes, people do work with 1000 variable systems. Computational fluid dynamics is one area where many thousands of equations are solved simultaneously. In short, matrices are used quite a lot in mathematics. Studying them is well worth it.

Mathworld is usually a good place to start: http://mathworld.wolfram.com/ConicSection.html

This isn't quite true, though. Since [math]x^2-4 = (x-2)(x+2)[/math] the [math](x-2)[/math] terms cancel and the expression is simply [math]x+2[/math], which of course is pretty easy to find a limit for.

This is what happens when I drive my computer without my glasses. I can see the big things, but sometimes p's becomes w's...

I hate myself for asking, but this cannot be left alone: African or European swallow?

Two little things: 1) you can't just "divide by [math]\cos\theta[/math]" You have ensure that whatever operations you do result in no change or that the same change is performed to both sides of the equation. For example, if you multiply a term by 1, that results in no change. And, multiplying by a clever choice of the form of 1 is what you are really doing here: [math]1=\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta}}[/math]. It is a small thing, but, technically you aren't "dividing by [math]\cos\theta[/math]." and 2) you might want to write it in LaTeX next time. It is pretty easy to learn and makes reading the math many, many times easier. [math] \frac{\sin\theta + \mu\cos\theta}{\cos\theta-\mu\sin\theta}[/math]

der..... ya, me did. Me dumb. Me only gets that title 'cuz me likez the color green.... OK, enough silliness What exactly do you want to want say here, Fortissimo? 1) The forum's policy is to not to homework for other people. Even more than that, it is my personal policy not to do someone else's homework for them. So, I asked waht work CrazCo had done to try to solve the problem on his own. When I was tutoring and when I was a teaching assistant, I never helped a student who had done nothing at all. They always had to show me that they put some effort into the problem. I knew how to do those problems -- I was the tutor or TA after all -- I didn't need to practice them anymore. The students need to practice them, and they don't get any practice doing them if someone else does it for them. I think that the forum -- espeically the homework help section -- is similar. I'm not doing some one else's homework for them, Maths Expert or not. 2) I then asked a question that hopefully would lead CrazCo in the right direction. If he could answer the question I asked, then he can do his homework problem. It really is as simple as that. If he came back and didn't see the connection, I'd give another hint (or in this case the tree gave another hint). I'd help get them going, but again, I'm not doing someone else's work for them.

If the "zero dimension" has no measurable quantities, why does it matter if it exists or not? Furthermore, if it has no measurable quantities, how could you ever either prove or disprove its existence?

I think that after 100+ posts, you probably know by now, we aren't just going to do your work for you. So, what work towards an answer have you done so far? What does is mean for a quadratic equation to have two equal roots?

The problem is that if left unchecked and unchallenged, it is an implicit acceptance by the forum members. And, that's not the kind of forum we want to run here. In addition, I personally delve in there to try to help when I can. I try to take an approach where I ask questions so that the poster can see the flaws in their arguments. I fully admit that it doesn't always work. Ok, it rarely works. But, there have been at least 2 times I remember where we got someone turned around. There is always the hope that the person on the other end does really want to learn and take criticism and develop their idea in a correct manner. Sure, a lot of the times any attempt to reach out and help is rebuffed pretty quickly, but I figure that's what the forum is here for -- to extend the offers of help.

These kind of problems are made for separation of variables: Let [math]f(x,t) = X(x)T(t)[/math] Note that we are letting the unknown function, f, be a product of two unknown functions, X & T, and that X(x) is only a function of x, and T(t) is only a function of t. Now, put this into the eqn: (I'll do the RHS, you can do the LHS) [math]D\frac{\partial f(x,t)}{\partial t} = D\frac{\partial X(x)T(t)}{\partial t} = D X(x) \frac{\partial T(t)}{\partial t}[/math] The X(x) part comes out, just like the T(t) part will come out of the left hand side. Now, divide the equation by X(x) and T(t) and you should get LHS that has only derivatives and functions only of x = [math]\frac{D}{T(t)} \frac{\partial T(t)}{\partial t}[/math] Now, since the LHS is only a function of x and the RHS is only a function of t, the only way that is possible is if both sides are equal to a constant, call it C. LHS = C = RHS This actually is now two ODEs (LHS=C & C=RHS) which hopefully are easier to solve. The constant is determined by boundary conditions and initial conditions.

Just to second farmboy here, but there is a lot of research, some good, some not-so-good, that can be done without math. This is even prevalent in some traditionally very math-centric fields, like engineering and physics. There have been many seminars and talks at national engineering and physics conferences where the slides have only 1 or 2 equations in the entire talk. Many presentations would even have no equations. Coincidentally or not, these non-mathy talks are almost always bio-technology or nano-technology related. Now, are these people doing physics and engineering with math? That's probably a different question. Nevertheless, it is possible today to be a researcher in physics and engineering without being necessarily strong in math. You have never had to be very strong in math to do research in chemistry or biology. There are the subfields of mathematical biology and the like, but most biologists only grudgingly took calculus and a stat class or two that's about it. No differential equations, no theory and derivations.

Atheist makes a good point. You can find a 104th degree polynomial that will fit all of your data exactly, but you probably don't really want a 104th degree polynomial fit because in between the points is probably going to be very poor. Does the physics of situation give you any insight into what kind of curve it may be? Exponential? Logarithmic? Linear? Quadratic? Once you have a decent candidate function or two, then the easiest and most often used method is called least squares: http://en.wikipedia.org/wiki/Least_squares Basically, you draw your curve, find the distance between each point from the data and your curve, square that distance (so that you don't have negatives), add up the squares of all those distances between the data and the curve, and minimize the total of all those squares. There are more advanced versions, especially if you have good measures on the confidence of each of your data points, but least squares is a good starting point.