Bignose

I hope that there is a lot more that just one pretty picture as "proof". Any quantitative predictions at all? Where's the mathematics that makes those predictions? Come back and make some predictions somewhere in the neighborhood of the current well-supported theory and then get back to us. edited to add: Saw the additional comments and just needed to add a few more things. RE: "This is why light always travels rectilinearly (sic) " Light does NOT always travel in a straight line. It is affected by gravity, and can be bent. Gravitational lensing a a very powerful technique http://en.wikipedia.org/wiki/Gravitational_lens RE: "This also explains universal gravitation and why it appears to act "instantaneously"." Gravity doesn't act instantaneously -- As near as we can tell today, the speed of light is the speed of gravity. http://en.wikipedia.org/wiki/Speed_of_gravity Definitely not instantaneous at the very least. RE: "Electricity occurs when an atom spins its magnetic thread faster" please cite some proof that the spin rate of an atom causes electricity. Take a cylinder of oxygen gas. The molecules (and hence the atoms) will collide with one another, and those collisions, rarely being square, cause the molecules to spin. And yet the cylinder doesn't spontaneously create electricity. Spin and yet no electricity. I find this idea (it has not earned the word theory yet) very, very lacking compared to reality as we know it today.

mahela, you may be interested in post#10 of this thread: http://www.scienceforums.net/forum/showthread.php?t=35615&highlight=restitution I wrote out all the equations for a collision with a coefficient of restitution -- and the energy term involving that coefficient is in there. The rest of the thread is a lot of the OP making a bunch of mistakes and spouting nonsense, but it may be helpful for you to read the replies to the garbage, too.

I strongly second this. But, there are some hints that may get you to look in the right direction: The sum or difference between two different normal distributions is again normal -- you just have to take care in defining what the new mean and variance is.

Pot, meet Kettle. Com'on, you've been at least as guilty as anyone else in this thread in jumping topics, hoping to find one that isn't thoroughly debunked. And, asking people to discuss a complicated topic with "small" posts is pretty silly, too. Keeping things short to just "sound bites" may be a large part of the reason you are so misinformed about the current body of knowledge about evolution today. This isn't a simple topic, and limiting to just short responses doesn't do it anywhere near the justice it deserves. To correct beliefs that are wrong, you have to point out the errors and then explain the situation as we best understand it today. That it most likely going to take more than just a "small" post. This is going to sound blunt I think it needs to be written: If you really cared, you'd take the time to read all the posts in as much depth as necessary until you understood them, but so far I think that it is obvious that you are here just to repeat many of the same tired old arguments that are refuted by a great deal of evidence if you'd just actually look at the evidence ("small" or not).

But, J.C., doing the complete momentum conservation isn't the way the principle is applied very often at all. When water in a pipe goes around an elbow, no one takes into account the momentum of the fluid that pushes into the pipe wall, which then pushes against the brackets and hardware that hold the pipe, which then push against the building, which then pushes against the earth. Etc. etc. This is not how anyone solves this problem. They solve the problem with the Navier-Stokes equations, which are an expression for the conservation of momentum in a Newtonian fluid, and include the gravity and pressure forces as part of the equation. The real applications of the conservation of momentum include forces as sources and sinks.

Here's hoping the OP wasn't waiting 4 1/2 years until he got this last post (or in other words, why bring up this ancient thread?)

I think you have this backward, it is very easy to define randomness. Let x be distributed uniformly across the set (H,T). This would represent a truly fair coin. Or let x be distributed across the set (1,2,3,4,5,6) which would represent a truly fair 6 sided die. Or the previous example of the normal distribution -- a random variable doesn't have to be uniform to be well defined -- it just needs a definition that obeys some regulations (like the integral over all values has to have a total of probability of 1). From these definitions you can work out all the properties you want about these events. No, again, I think that it is fair to make comments how there is no such thing as a truly fair coin, or a device that would be able to act as a truly fair coin flipper, or a fair die or a fair die roller. Or a be able to write a computer program to truly simulate random events. etc. etc. But, that doesn't preclude us from being able to define a variable that is truly random. The definition exists, and it is a solid definition. The definition isn't the problem -- it's the applicability of the definition to the real world

J.C.MacSwell, sure, but in the case of a typical ball dropped on the Earth -- how much does the Earth move? A trillionth of a meter? The amount of velocity and movement and force on the earth is so small it is for all purposes zero. Technically non-zero, but effectively zero.

In a broader sense, the sources and sinks are part of the "conservation". The full conservation equation looks like: [math]\frac{\partial \rho\mathbf{v}}{\partial t} +\mathbf{v}\cdot \rho \nabla\mathbf{v} = \nabla^2 \mathbf{T} + \mathbf{f}[/math] where v is the velocity of the object [math]\rho[/math] is the density T is the stress tensor and f is the body forces. This equation simply says that the rate of change of momentum = rate of momentum advecting in - rate of momentum advecting out + rate of momentum diffusing in - rate of momentum diffusing out + forces acting. Those forces are included in the conservation equation, and simple show how much the momentum changes per unit time. ------ Consider a simple example: You hold a ball perfectly still in your palm with your arm outstretched while standing. The ball is still and hence has no velocity. A time t=0, you drop the ball. At t>0, the ball is moving -- it has gained momentum. The force of gravity acted on the ball and the ball gained momentum. So, external forces are part of the momentum conservation as well. A opposite example is true, too. You putt a golf ball, with the instant of impact being t=0. As the ball rolls over the ground, it slows down and at some t>0, the ball will come to a standstill. In this case at t=0, the ball definitely had momentum, and then some time later it is still and of course has no momentum. The force of friction acts on the ball to lower its momentum down to zero. Forces are part of the momentum conservation as well.

Maybe, maybe not. Again, we just don't know. Turbulent fluid motion is random down to very small scales -- small enough that the random fluctuations of the atoms themselves could be having an effect on macroscopic turbulence quantities. And, as near as we can tell, those fluctuations as describes by Brownian motion are indeed completely random -- perhaps as a quantum mechanical effect. If someday we figure out that QM is deterministic, then maybe, but at the moment, the best information we have is that QM has a strongly random component, and that the exact result for a single event is unknowable -- all we can talk about is what happens over a large number of samples.

And I never said it did, thief. The equations can predict the most probable location of where the dandelion seed would fall, however, or the region where 95% of the seed will fall. And that's pretty powerful, really. Part of it is in your use of the terminology -- because when you write "exact" that has different meanings. That p(x) equation I wrote above is exact in that p(x) is exactly the probability density function of a random variable that distributed normally with mean 0 and variance 1. It doesn't --- and doesn't claim to be able to -- predict what any single value of that random variable will be. It predicts what a large number of samples from that random variable will be. And that's it.

This is dead wrong. 0.99999... is less than 0.1? In what universe? where did I post the use of the function 1-1/x? This has nothing to do with what I wrote.... This isn't any proof at all! You can't just throw in a y and call it "indeterminate" and call that a proof. Besides that, there is a pretty obvious error here, too. 0.1 x 0.099........ would be equal to 0.00999.... Finally, I didn't lecture you. I tried to make an appeal for you to use precise language so that everyone understood exactly what you mean. I am trying to understand your posts so that I can respond intelligently. You've never answered some of my questions, and you've made some pretty basic errors in those you did answer (0.99999 less than 0.1 !!!), so I guess it wasn't worth my effort. You don't have the be a mathematician to use the words correctly. I'm not a mathematician, but I know what words I use I use correctly. If you aren't aware of their specific meanings, there is no shame in that -- just look up the precise meanings or ask and this forum will help. I'll help, I promise. Look, it's no different than when you first learn a game, you learn the terminology used in that game. Take baseball for example. If you started using your own words for balls and strikes for example, no one would know what you are talking about. "The count is 1 banana and 2 envelopes." Or if you used different terms for safe and out. "That runner was clearly polka dotted!" Or, take chess for example. You can't make a move like "Duke to Cardinal's 23rd, check". It doesn't mean anything in the accepted terminology of the game. Math is the same way. There are terms that have been accepted and are used in one way and only one way. Again, if you don't know them currently, then it only takes a tiny bit to learn how to use them right and then use correct language -- everyone benefits because then there won't be several posts asking you to clarify what you write because you are using terms that aren't defined. I don't mean this to sound like a lecture -- It is just an appeal to use terms properly so that everyone can understand what you are trying to say. Because, I still don't know what most of your first posts in this thread were trying to say.

http://mathworld.wolfram.com/DeltaFunction.html The delta function is there to mathematically represent a point-source. It is mathematical abstraction, because there is no such thing as a point-source, but very often the mathematical results computed using a point-source are exceptionally close to reality so they are used quite a bit. It's biggest property is that at the value of x in the delta function that the input of the function is equal to zero, the function has a value of [math]+ \infty[/math]. Everywhere else it has a value of 0. The really useful part of this come from the fact that when you integrate over the point where its input is zero, the value of that integral becomes equal to 1. So. [math]\delta(x-y)=0[/math] everywhere except when x=y, then [math]\delta(0)=+\infty[/math]. And the integral property: [math]\int_\Omega \delta(x) dx = 1 [/math] if x is in the domain [math]\Omega[/math]. If x isn't, then the integral evaluates to zero. This integral value makes the Dirac delta function a selector of sorts because the integrand only has a value at one point. e.g. [math]\int^{+\pi}_{-\pi} \sin x \delta(x-0.2) dx = \sin 0.2[/math] Even though you are integrating from [math]-\pi[/math] to [math]+\pi[/math], the only point where the integrand is non-zero is at 0.2 as selected by the delta function, so the integrand less the delta function itself is the value of the integral so long as the selected value is in the domain of the integration. So, in EM, if there is a point source of charge, this is represented by a delta function, as the delta function is the mathematical representation of the charge. So, when you integrate over the entire domain, there is no sources except that 1 point, and that's what comes out.

Reference frame is the point of view you take. Consider a universe with only 3 balls in it (each with mass of 1 kg). All three are moving. Now, consider a point of view (or observer or reference frame) wherein you travel with ball A. From ball A's point of view, ball B moves at 1 m/s, and ball C moves at 10 m/s. Because we are travelling with ball A, it appears that ball A is not moving at all (0 m/s). So, from the reference frame of moving with A, it appears that the universe has 101(=10^2 + 1^2) J of kinetic energy. Now, change to a reference frame where B is stationary -- A is now moving at -1m/s and C is now moving at 9 m/s. In this frame, it appears that the universe has 82(=1^2 + 9^2) J of kinetic energy. Finally, in the reference frame were C is stationary, A is moving at -10 m/s and B is moving at -9 m/s. In this frame, the universe has 181(=10^2 + 9^2) J of kinetic energy. So, which one is right? They all are. Each is right from it's own frame. This is why kinetic energy is very, very observer dependent, very, very reference frame dependent. So, it is important to point out what reference frame you are working in when doing these kinds of problems. It is also a very important result to understand that there is no such thing as a "preferred" reference frame. That any and all reference frames are equally valid and the laws of physics are necessarily reference frame independent. Sources and sinks are part of the conservation equations. If I had a box (assume it is perfectly rigid and non-deformable), conservation of mass shows that the amount of stuff that goes into the box must be equal to the amount of stuff that comes out. Conservation of momentum and energy is similar -- if you draw a box around fluxes in and out, what comes in must come out. However, unlike mass, it is easy to imagine sources or sinks for momentum and energy. A source for momentum may be something like a pressure gradient in a fluid that acts like a source for momentum in the fluid mechanics equations. In this case, more momentum may "come out" than went in because the pressure force accelerated the fluid. A sink for momentum may be that same fluid being pumped vertically and the gravity force will act on the fluid decelerating it. A sink for energy is something like a chemical reaction that may be either exothermic (source for energy) or endothermic (sink for energy). Similarly, what happens in inelastic collisions is that the inelasticity acts as a sink for momentum and energy. The two inelastic objects that collide deform and fail to rebound and that momentum and energy is effectively lost. It isn't "gone", it gets turned into heat and noise and whatnot, but it isn't affecting the movement of the objects anymore.

Firstly, welcome to the forum Secondly, your statement need a slight amendment: "The kinetic energy that an object possesses in a certain reference frame is dependant (sic) on it's speed according..." The "in a certain reference frame" is a very important qualifier because kinetic energy is very highly observer dependent. Thirdly, you have to remember that a momentum conservation does have sources and sinks -- specifically forces. When an inelastic collision occurs, there are forces that have to be accounted to deform the two particles that are colliding. This deformation is both a momentum sink and an energy sink, and if the amount if momentum lost is a constant, you can use a coefficient of restitution formulation to get exact numbers for the momentum and energy losses.

I doesn't appear to converge in that it is tending toward a single number, but it is a converging oscillation between 0 and 1. This isn't chaos in a proper mathematical term. Chaos mathematically is when a small change in inputs makes dramatic changes in outputs. E.g. consider some function f(x) If f(1.0)= 10 most functions behave reasonably well in that f(1.0001) = something like 10.04 or 9.95 or even just 11 or 9. and f(1.0002) would then be 10.1 or 9.9 etc. A chaotic function would be something like f(1.0) = 10, f(1.0001) = -765.9, f(1.0002) = 387.9, f(1.0003) = 22.8, f(1.0004) = -98.11, etc. Where a very small change in input leads to a very large change in output. This is almost exclusively due to nonlinearity of the function (I don't know of a single linear chaotic example). This is a decent article: http://www.imho.com/grae/chaos/chaos.html

Just a quick note here that be sure to go to that Mathworld link and look at the specific definition of L. (Not that I am blaming you herman, because you got it from the webpage, but what I thought was L from "length of the hole" was NOT what L actually was!) The diagram on the Mathworld site cleared it up quite a bit for me.

Thief, there is a whole field of mathematics -- stochastic calculus -- that is the calculus of random variables. They are "exact" equations written for variables that involve randomness. For that matter, even saying x is distirubted normally with mean 0 and variance 1 has an exact definition: [math]p(x)=\frac{1}{2 \pi } \exp(-0.5x^2)[/math] This equation is the an exact mathematical formula for the distribution of samples of x. So, you need to be a bit more careful in explaining exactly what you mean, because the mathematics of probability, statistics, and stochastic calculus is very rich and very well defined.

Firstly, welcome to the forum Secondly, The guy the theorem is named after is Stokes, not Stoke, so it would be properly written Stokes' Thirdly, the link is broken (extra http://) in it Fourthly, Stokes' Theorem isn't the same as the fundamental theorem. Stokes' Theorem is how to turn surface integrals into line integrals. The Gauss-Green-Ostrogradsky divergence theorem (has many names depending on the reference you are looking at) is how to turn volume integrals into surface integrals. Divergence is how to turn 3-D volume into 2-D surface and Stokes' is the 2-D to 1-D equivalent. Neither one covers the same ground as the fundamental theorem which proves the truly fundamental property that differentiation and integration are intricately related and are inverse operations (to an arbitrary constant).

This one gets my vote by and far. Not a whole lot gets done without this being true. I guess that's why it's been named "fundamental" eh?

I had a fear that this was the real gist of these posts. 0.9999... infinitely repeating 9's = 1. They are the same number. There are several proofs. Let x = 0.999999999999... (where the ... after indicates infinitely repeating). Therefore 10*x = 9.9999999999... 10x-x = 9x = 9.99999... - 0.999.... = 9.000000000000... Divide by 9 and x = 1.00000.... QED Another one: If 0.99999... and 1 are not the same number, then by definition if [math]a /ne b[/math] then there is some number c that is between a and b. That is there exists c such that [math]a > c > b[/math] or [math]b > c > a[/math] depending on whether a>b or b>a. So, if a=0.99999... and b = 1, and [math]a /ne b[/math], then what is c? Conversely, if no such c exists such that [math]a > c > b[/math] or [math]b > c > a[/math], then there is not other conclusion than a=b. So, 1) what is the hole in the first proof, and 2) what is a number c such that 0.999... < c < 1? If 0.999... and 1 really aren't the same number, then these should be easy questions to answer. Merged post follows: Consecutive posts merged Again, I am sorry, but I don't know what this means. How do you calculate "how many 0.1's would be needed to make up the deficit of .0001"? What deficit? What do you mean by deficit? How many makes up I read as division: 0.001/0.1 = 0.0001 --> which is how many 0.1's "make up" 0.001. But, I don't that this is what you are looking for. And rounding is an entirely different mathematical operation. So, once again, you are using words I am familiar with in very unfamiliar ways. When you write about science and math, you need to be very exact in the word choices you make, because each word has a very specific definition associated with it. This is different than regular English where words have different meanings -- in math, words only have one meaning so that everyone knows exactly what is meant when that word is said. Mathematically, when I write "multiply" everyone knows that I mean the multiplication operator and not the exponentiation operator, even though they are related. Whereas when writing a novel or story, you can be loose with terms like multiplication or rounding or deficit -- when you write math, you need to be very precise in what you mean. So, I ask, please be more precise, more exact in what you mean when you write these things so that I and everyone else familiar with the mathematics will understand you and your points better.

No, I'm going to say I still don't understand because you aren't making yourself clear. Anyone else is free to jump in here and try to make it clearer, but I still don't understand your point. Maybe if you came right out and said exactly what your point is instead of trying to dance around the issue, it would be clearer. Sure, 0.1*10 = 1 But, 3*0.3333.... = 1 as well. And you can do this in an infinite number of ways. So, again... what is the point? You never did clarify by what you meant when you wrote "But divide 4 by 3 and you get 1.3333333 remaining not 1" You are using words I am familiar with, but not in a familiar way. I am simply asking you to clarify exactly what you mean when you use words in a confusing way. Similarly, I don't know what you mean by "The 0.1 starts at ZERO." There is no start or end to a single number. It only is a number. If you want a start and end, you have to define a range like [math]x \in (0,0.1)[/math], then I think you can say x starts at 0. But a single number, like 0.1, doesn't start or end anywhere (unless you want to say it starts and ends at 0.1, because that is the only value)!

ok But what does this even mean? Your question doesn't make sense to me... "But where do these .1's begin at O or 1?" Where does a 0.1 begin at anywhere? 0.333333... does tell about the distance between 0 and 1 -- it is 1/3 of the distance between 0 and 1. Ok, so what? It is a number just like any other number. What exactly is wrong with it?

What do you mean by "remaining not"? I don't get at all what you are saying. 4/3 = 1.333333 repeating because 0.3333 repeating is the decimal representation of 1/3. That's all.

sterologist, you are mistaken about what Excel refers to it's [math]R^2[/math] value as. Excel's [math]R^2[/math] is simply a "goodness of fit" measure (and an awfully poor one at that). Do your example up in Excel -- put in y values that don't change with changing x values, and ask Excel for the best-fit line and the [math]R^2[/math] value will equal exactly one. [math]R^2[/math] in Excel is simply a measure of the sums of the squares between the "best-fit" line and the actual data. http://en.wikipedia.org/wiki/Coefficient_of_determination has a more thorough discussion (as well as it's statistical roots which you are referring too) But, you should not use it to determine which model is best (i.e. between linear and logarithmic and quadratic, etc.) because you can always manipulate the results to whatever [math]R^2[/math] value you want. I.e. if you have 20 data points, a 19th order polynomial going through every point will have an [math]R^2[/math] value of 1, but is in all likelihood a terrible representation of what is actually happening. Too many people place way too much value on [math]R^2[/math] measures. There are many more meaningful measure of goodness of fit out there. A chi-squared ([math]\chi ^2[/math]) is one of the simpler and much more meaningful measures of goodness of fit.