Bignose

A dimensional analysis doesn't have plugged-in numbers in it. All it tells you is what units the final results are in. i.e. [math]\mathbf{F} = m\mathbf{a}[/math] let the symbol [=] indicate "has units of" m [=] m, mass (no big surprise there) a [=] [math]\frac{l}{t^2}[/math] length per time squared That means that F [=] [math]\frac{ml}{t^2}[/math] F has units of mass*length per time squared. That is all. The equation above boils to to having units also of mass*length per time squared. It says absolutely nothing about the correctness of the specific equation. Think of it this way: A correct equation MUST have the correct units, but an equation with the correct units doesn't mean that the equation itself is correct. Here is another example. Dimensionally, work and torque both have units of Newton*meter (or mass squared * length per time squared if you prefer). However, they are definitely not the same thing, even though dimensional analysis of both equations would show that they are dimensionally equivalent. Dimensional analysis is an excellent start, but it doesn't actually solve anything in the end.

The wikipedia article on it is pretty good, really: http://en.wikipedia.org/wiki/Newton%27s_method It is a computational method wherein you make an initial guess. Then, the derivative of the function at that guessed value provides you with the next guess. And then you repeat until you get the desired accuracy. It isn't a perfect method. For functions that oscillate a lot, the next guessed value may overshoot the target value by too much and the subsequent guesses never converge. There are ways of testing and minimizing these issues, however. Furthermore, the theta versus arc length function will be nicely behaved. It won't dip or oscillate or anything like that. It is monotonically increasing, so I think that Newton's method is guaranteed to be convergent for a monotonic function.

Right, as I said and Amr said, it is incredibly unlikely that it will be in a "nice" form. I won't say impossible, but nothing pops out at me. However, if you want to use it for computations, a computer program can do the trick for you. A straightforward "brute force" method to find what the angle is for a given arc length would be to guess a theta, and then add or subtract from the guess based on whether the guessed theta resulted in an arc length above or below the value you are aiming for. Then iterate following that same plan until you achieve the accuracy you desire.

You'll have to try to find the inverse. Given the form of the function, I highly doubt that it will be a "nice" form. However, since the formula is 1 to 1 valued (a given [math]\theta[/math] results in 1 and only 1 value of s), an inverse should exist.

Amr, in the future, please note that this forum tries not to just give answers to posters, especially when the problem seems very home-work like. We try to help the questioner help themselves, by giving them ideas to look at, different points of view, asking them what knowns and unknowns they have and what equations relate them all, etc. We also will look over work that they post, and point out mistakes if we see them. But, we usually try to not spoon feed answers. In short, no one gets much of anything out of that. Look around the Homework Help section to get an idea of what I am writing about here. The "not doing homework" rule applies to all sections, not just the homework help section. Thanks

If you are going to differentiate with respect to n, you cannot let n! be a constant. n! is the term with the greatest rate of change for large n in that expression. To do it right, you'd use the Gamma function as the extension of the factorial to the reals, and use the derivative of the Gamma function.

I want to take at least one crack at this. The calculations allow objective determinations of the validity and accuracy of an answer. Words can be very descriptive, and very interesting. There are many tremendously well pieces of written literature in existence today that while reading it is very easy to imagine being transported to the world the author created. But in the end, they are only words. Words have different meanings to different people. Words are inexact. Words are not objective. In short, words are imperfect. Whereas when mathematics gives you a definite answer, it is an objective measure. If my theory says that the ball will land 146m from the catapult, that 146m is exactly 146m whether you measure at your house, at my house, on the moon, or anywhere else, because everyone is using the same unit of measure. Using only words, you cannot describe where the ball lands as exactly as saying "146m". If your theory says 146m and the ball lands at 298m, then clearly your theory needs work. But, if you just use words and say "a moderate distance away", your moderate may be a lot different that what I consider moderate. And the other calculations of physics are the same. You create a model of the internal processes of a star, and you predict the color temperature of the star. And, then you observe a star and see how closely your prediction fits with the observation. You create a model of the universe after the big bang, and run the clock forward -- your model better predict the current intensity of the cosmic background radiation, or else why even bother to look at what else the model predicts? You create a model of fluid flow around a bend, and compare the pressure drop predicted by the model. If the pressure drop isn't right, you don't even both to look at the predicted velocity distribution, because the model has an error somewhere. And that is the theme, over and over and over. You make a prediction, you compare to observation, and the agreement or disagreement between the two is the objective measure of how well the model makes predictions. Or, to take it further, if one model makes better predictions than another, the assumptions or theories behind the better model gain a little more credibility as being closer to reality. Mathematics are the tool that allow us to make these predictions and comparisons. But, as elaborated on by other responses, it isn't the only part. It is an important part, but definitely not the only one.

The equations themselves are incomplete without the appropriate boundary and initial conditions that allow them to be solved (i.e. give the values of E and B). Please don't take this to be rude, but in your first post you say "I do have education in classical physics" and you have forgotten about about initial and boundary conditions? Look, here's a simple example. Velocity = dx/dt. Sure, you can integrate this, given a function for velocity. Let's take a really simple example, V=1. You can integrate this, x(t) = 1 + x(0). The value x(t) is completely unknowable without knowing what its value at time 0 was, x(0). It is exactly the same for Maxwell's equations. You have to have the appropriate boundary and initial conditions to compute the values. The initial and boundary conditions are just as important and necessary as the equations to solve. In a slightly different vein here, I am confused about where this thread is going or was meant to go. It started off innocently enough asking about how to solve some equations, but it seems to have turned into some sort of anti-Maxwell's equations rant -- using words like "farce" when describing the equations. Which I don't get at all, because Maxwell's equations are among the most verified equations known to mankind. The computer you are using wouldn't work unless Maxwell's equations didn't work, as just one of many, many examples. The problem as I see it, is a lack of understanding of the vector operations, and a lack of how to apply them, specifically applying them via Maxwell's equations. A good advanced book on the topic may go a long way to clearing this up. From the table of contents, Maxwell's Equations and the Principles of Electromagnetism by Fitzpatrick seems like an excellent place to start. If you aren't familiar with vector equations, Shey's Div, Grad, Curl and All That is extremely excellent. But, if I might ask a favor, if an Anti-Maxwell rant is your agenda, could you just lay it all out on the table now instead of trying to be coy about it?

No, it is the other way around. Coloumb's Law and Biot-Savart Law are special cases of Maxwell's equations. Maxwell's equations are based on Ampere's and Faraday's laws. Maxwell's laws at their basis are just conservation laws that apply to anything that can be conserved. Mass, energy, momentum, angular momentum, charge, magnetic field, etc. etc. Then, using Ampere's Law and Faraday's laws as constitutive equations (like Newton's law of viscosity goes into deriving the Navier-Stokes equations of fluid mechanics), you get Maxwell's equations. This is a decent write up: http://www.engr.uconn.edu/~lanbo/DeriveMaxwell.pdf I completely agree that Maxwell's equations are not just in wires, however. They are full three dimensional field equations. They predict resultant magnetic field and electric fields in any geometry at all. That is why they are so powerful, they are not constrained to just simple shapes.

No, E and B are all coupled together and you solve them all together. You will have to use some sophisticated computational techniques, but it can be done. The simplest idea is some sort of sequential solver. That is, you start with an initial guess for B0 and E0. Then, treating E0 as a constant, you solve for B, and call that solution B1. Then, treating B1 as a constant, you solve for E1. And repeat, probably a great number of times. There are all sorts of constraints on how large of a jump between iterations you can let your solution take for the simulation to remain constant, what kind of method you use to solve, etc. etc., that are beyond the scope of a forum post. There is a huge body of work on the computational solution of nonlinear equations, however. I am most familiar with the methods to solve fluid mechanics equations, but the high-level view is the same. Both fluid mechanics and Maxwell's equations are conservation equations, coupled together and nonlinear. If the situation does not allow simplifying assumptions, then you may have to solve the entire system computationally.

check this again. And you can write out pi on this forum using the math tags: [_math_]\pi[_/math_] = [math]\pi[/math]. Just remove the _'s (hover over the pi symbol to see the LaTeX code needed to generate it.)

Introduction to Chemical Engineering Thermodynamics by Smith, Van Ness, and Abbott is a good start. Chemical Engineering Thermodynamics by S. Sandler is excellent, and a good second book. These will give a good practical introduction to thermodynamics. Any book with something like "Heat Transfer" in the title will be good about how energy and heat naturally move around. Then, once you've mastered the above, you can get into the deep theories as espoused in books like Truesdell's Rational Thermodynamics and finally Thermodynamics and Its Applications by Tester and Modell. If you can master Tester and Modell, you will have an expert's understanding of thermodynamics.

It depends entirely on the process. If a large reason why the temperature fluctuates is random (i.e. a large variation in the ambient atmosphere causing large variations in the convective heat loss), then it may be independent of previous time. If the random component is small, then it may not be. You'll have to determine for your own process whether the assumptions regarding a Markov process as applicable or not.

I really only have 1 question: once you go out and get a real job, who are you going to pay to do your work for you then?

It is something related to the time derivative of the data, which certainly can have a lot of meaning. It may also have some meaning as to whether the sample of time t is independent or dependent on samples of time t-1 and earlier or not -- that is, whether it is a Markov process or not. However, generally, if you are sampling a population, the order in which you sample should not matter, so the difference between specific samples usually carries no meaning.

Yes. An "Euler" fluid is an extrapolation to a perfectly inviscid fluid, and can serve as a useful approximation in some circumstances. Nevertheless, an invisicid fluid still responds to pressure differences, so imposing a pressure difference can still cause a fluid to flow.

These things aren't all that uncommon when doing an expansion via a perturbation method. Basically, at some point you have to truncate or approximate the expansion. I.e. decide that the 6th derivative and further are constants or negligible or some other assumption. Knowing the pathology of the solution can be very useful to know what and when such approximations are needed/appropriate. Nayfeh's Perturbation Methods is a absolute classic on perturbation methods and is the standard by which all other books on the subject are judged. You may want to poke through that book and see if it can help.

[math]\sqrt{2}[/math] is an irrational number. It will have an infinitely long non-repeating non-pattern forming decimal representation. All irrational numbers have this property. Like [math]\sqrt{3}[/math] or [math]\pi[/math] or [math]e[/math]. http://en.wikipedia.org/wiki/Irrational_number

And exactly how do you propose to do this? There aren't cameras out there above the orbital plane videotaping the Earth going around the Sun. Simulation is going to be the best you are going to get. The fact that simulations match exactly what is observed has to mean something. Those models are very accurate, and the nice thing about the model I gave you a link to is that you can play with the things like position (height), mass (weight) and velocity. You can stop the simulation and change the mass, position, or velocity of any object in the simulation.

You don't need youtube... again look at this: http://www.arachnoid.com/gravitation/ And yes, once the space shuttle is up high enough with the correct velocity, then it stays up there without assistance. The moon stays up there without assistance.

Nevertheless, you used the language COMPLETELY wrong. You HAVE to use the language correctly if you want to describe physics. Just like you have to speak the same language of the person who is listening, otherwise you won't be able to describe anything. And please don't (thinly veil) insult to me. I never insulted you at all. I have been very cordial.

These three (crude) drawings should explain it all.

gafferuk, this is a meaningless statement here. mph is NOT a unit of force. So, saying "the gravitational force acting on it is 100 mph" is as meaningless as saying "the gravitational force acting on it is 100 bananas" or "the gravitational force acting on it is 100 lengths from your pinky toe to your nose". Units of force are Newtons (SI) or pounds-force (English). You HAVE to get the units right when discussing physics, or else it breaks down incredibly quickly into nonsense.

I think that there was some confusion. I agree that things falling to the Earth are going to fall at the same acceleration. But, you can't just say that "gravity causes all things to fall at the same rate" because it matters if you are talking about falling towards the Earth or falling towards the moon or falling towards the Sun. I.e. what happens when you change m_1. THEN mass definitely is important. And, I agree that there was a rounding error in the Open Office calculations, that the answers are equal, it was just machine precision errors that accumulated to make the last digit off.

You were told something that is true for all intents an purposes for everyday objects here on Earth, but not true in general. Here's Netwon's Law of Universal Gravitation (in a scalar form): [math]F= G\frac{m_1 m_2}{r^2}[/math] F = force of gravity between the two objects, labeled 1 and 2 G = gravitational constant = 6.67428 * 10^-11 m^3/(kg*s^2) m_1,_2 = mass of objects 1 and 2 r = distance between the objects Now, let's look at an orange and a brick dropped here on earth. Mass of the Earth is 5.9742 * 10^24 kg Mass of a brick is: 2.7 kg (from http://wiki.answers.com/Q/What_is_the_weight_of_a_red_clay_brick_in_Kilograms first on a Google search of "weight of a brick") Mass of an orange is: 9-11 oz. (from http://wiki.answers.com/Q/What_is_the_average_weight_of_an_orange again, first on a Google search of "weight of an orange) Let's call that 10 oz. that is equal to 0.28 kg Let's do the math. Let's say that the brick and the orange are both dropped from a height of 10 m above the Earth. In this case, r = 10 + 6378100 (average radius of the Earth in m). Force on the orange = 2.35 N Force on the brick = 26.5 N But, the speed at which the orange or brick actually moves is the acceleration. Force = mass * acceleretion. So, we divide the above forces by their masses and we get Acceleration orange = 9.799792 m^2/s Acceleration brick = 9.799792 m^2/s identical down to 6 decimal places In fact, Open Office calculates the numbers to be identical for the 1st 14 decimal places given the numbers I put in (about 6 sig figures), and it isn't until that 15th decimal place that 1 digit is 1 greater than the other. Why is this.... quite simply, the mass of the Earth is so much greater than the mass of the brick or the orange, that the gravitational effect of the Earth is for all purposes a constant. Hence, your being told that a brick and an orange do fall at the same speed on Earth. However, if you start putting things in like the mass of the moon, 7.36*10^22 kg or the mass of the sun 1.99*10^30 kg, then the masses do become important and can be calculated from Newton's law given above.