D H

First, a correction to post #29. The β+ decay that steevey described in post #28 is 11C to 11B. It doesn't matter whether you used wikipedia directly, some site that rehosts wikipedia content, or some other site entirely. The important thing is that you don't know the difference between β+ decay and pair production. Given that, who do you think comes off as more credible, you or technical paper after technical paper after technical paper on pair instability supernova published in peer-reviewed journals? I cited two such articles in post #25 above, one published in Nature ('nuff said) and the other in ApJ (no slouch of a journal either). Think about it for a second. At this point I call troll.

Yes. First off, you fudged the numbers (i.e., you cheated) to make the ratio look closer to an integer than it is. The sidereal year is (or was, epoch J2000.0) 365.256363004 days long. The sidereal month is, on average, 27.321661 days long. That is a ratio of 13.3687466. If you want to pretend that that is "almost exactly 13", have at it. Second, there is no reason for such a coupling. The number of months per year is a function of the distance between the Earth and the Moon. The Moon is retreating from the Earth and has been since it first formed due to angular momentum transfer from the Earth's rotation to the Moon's orbit. The mechanism that drives this retreat is the tides. The Earth tides are responsible for some of the transfer, but the bulk of the transfer is due to the ocean tides. The transfer rate due to the ocean tides is not constant. It depends a lot on the configuration of the continents. You are playing numerology, and doing so badly. The ancients did a much better job than you. They saw that there are almost exactly exactly equal to 235 synodic months in 19 tropical years (1 tropical year is 365.24219 days). This is the Metonic cycle. They also saw that 223 synodic months, 242 draconic months, and 239 anomalistic months are almost all the same length of time, to within a couple of hours. This is the Saros cycle. Numerology still, but very useful numerology (at least for mystics).

Nice. This, however, gives me pause: O RLY? I suspected a problem in the parser/interpreter re precedence rules. There is no doubt as to what 3x^2+3x should calculate. Do they give exponentiation an incredibly low precedence that they feel compelled to have this caveat? Nope. They must assume that the typical user doesn't know the precedence rules.

Nonsense. You don't need relativity to explain the tides. All that relativity does in this case is to add unneeded complexity. Good old Newtonian mechanics does an excellent job of explaining the tides. George Darwin (Charles' son) and Arthur Doodson pretty much wrapped up the theory of the tides in a nice package, complete with ribbon and bow, and did so solely within the construct of Newtonian mechanics.

Taking this one step further yields [math]x=2^{x/5}[/math], and this makes for a good fixed point iterator for finding that first root. The original equation, [math]x=5\log_2x[/math], makes for a good fixed point iterator for finding the second root.

This reference: Arthur Sinton Otis, Ph.D, "Light velocity and relativity: the problem of light velocity, disproof of the Einstein postulate" ? Two words: Crack Pot. He wasn't even a physicist, and had no advanced training in physics. Otis was a psychiatrist or psychologist who apparently designed a school aptitude test. In his own field he may have been very good, but not in physics. Sorry' date=' this isn't an argument, only hand-waving. [/quote'] No, claiming that experiment after experiment after experiment after experiment has shown that light does travel at c in vacuum regardless of the relative velocity between the source and the receiver and regardless of the frequency of the light is not hand-waving. Word to the wise: Stop reading crackpots and start reading about the physics of relativity.

The only logic that isn't working here is yours. You just described the beta decay of one particular isotope, [math]^{11}_6C[/math], into [math]^{11}_5B[/math]. I take it that you cut that right out of the wikipedia article on positron emission. Other than the fact that both pair production and beta decay are both described by quantum physics, the one has little or nothing to do with the other. Some advice: Stop embarrassing yourself. I am starting to wonder whether you are a troll.

The wikipedia article on negative temperature, http://en.wikipedia.org/wiki/Negative_temperature, quotes Kittel and Kroemer in defining the temperature scale from cold to hot as "+0 K, . . . , +300 K, . . . , +∞ K, −∞ K, . . . , −300 K, . . . , −0 K" The reason these negative temperatures arise is from the thermodynamic definition of temperature, [math]\frac 1 T = \frac{dS}{dE}[/math] In most systems, increasing the energy of the system increases the number of states in which the system can be. The number of possible states and entropy are closely aligned concepts. Thus in most systems, increasing the energy increases the entropy. However, for a system that have a maximum possible energy, increasing the system energy will at some point result in a decrease in the possible number of states. [math]dS/dE[/math], the multiplicative inverse of temperature, is zero at this point and is negative at higher energy levels. Thus the thermodynamic temperature is infinite at the point where the number of possible states reaches its maximum and is negative for even higher energy levels. A negative temperature does not mean that the system has suddenly become very cold. It instead means that it has become very, very hot.

That won't happen. While various ceramics are rather important in the field high temperature superconductivity, only a select few such as this Joe Eck character have specialized in the study of cracked ceramics. There's just something about cracked ceramics that prevents reports such as this from making it into peer-reviewed journals.

No.

To some extent, yes. Let's take this to the extreme of expelling photons, e.g., laser propulsion. This is, simply put, a silly notion. The energy and momentum of a photon are related by E=pc, or p=E/c. Multiply this by the number of photons emitted per second and you get F=P/c, where P is the power of the photon stream. Suppose we had a laser propulsion spacecraft whose power production was equal to the entire world's electric energy production, 6.25×1019 joules per year (about 2 terawatts) for 2005. The amount of thrust produced by that laser propulsion spacecraft? 6,600 newtons. Compare that to the 66,700 newtons produced by a single RL10 rocket (which is rather old technology). So, back to ion propulsion: First, a link to the article that Mr Skeptic quoted from: http://en.wikipedia....toplasma_Rocket. This is potentially a very worthy contender. One of the biggest challenges with ion propulsion has been getting any substantial amount of thrust out of the engine. Ion propulsion engines prior to VASIMR have had a very high specific impulse but also an incredibly low thrust. Even though SMART-1 was incredibly light (367 kg launch mass), it took the vehicle over a year to get from its initial 7,035 × 42,223 km geostationary transfer orbit to past the Earth-Moon L1 point. It took another seven months to get from this point to a low lunar orbit. The reason is the incredibly low thrust produced by the ion propulsion engine used on SMART-1. The ion propulsion engines used on Deep Space 1 and Dawn (the ongoing mission to Vesta and Ceres) are also quite wimpy, producing less than 0.1 newtons of thrust max. VASIMR might just get past that problem.

Oh please. In post #21 you said that antimatter production is completely irrelevant. In post #24, after being asked to google the term "pair instability supernova" you apparently softened your stance. That is "moving the goalposts". So let's look at post #21. There is so much that is wrong with this post that it needs to be dissected point by point. So what? It doesn't to survive for long to result in extreme havoc. Wrong. You apparently are talking about virtual pairs. The pair production in supermassive stars instead involves a real gamma ray colliding with a real atomic nucleus, resulting in a real electron and a real positron. That positron will soon annihilate with some other electron, but that annihilation does not occur immediately. The result is that pair production slows down the transport of energy from the star's core to its surface. The star begins to lose hydrostatic equilibrium. In supermassive stars the onset of pair production triggers a runaway reaction. This happens well before the star runs out of fuel. Observed by whom? Wrong. While this is descriptive of the prototypical core collapse supernova, it is not the only reason a supernova happens. A pair instability supernova happen well before fusion ceases. It is the onset of carbon-carbon fusion that triggers the pair instability supernova. Another example where your statement is blatantly false is a type IA supernova, which happens when fusion restarts (explosively) due to accumulation of mass stolen from a binary pair.

You are moving the goal posts here. You previously disputed that antimatter could have anything to do with a star going supernova. Now you are disputing the concept of a true pair instability supernova, which is lesser claim. That said, here is what you asked for: Gal-Yam, A. et al, "Supernova 2007bi as a pair-instability explosion," Nature 462, 624-627 (3 December 2009) Full-text at nature.com: http://www.nature.com/nature/journal/v462/n7273/full/nature08579.html Pre-print at arxiv.org: http://arxiv.org/abs/1001.1156 Ofek, E.O. et al, "SN 2006gy: An Extremely Luminous Supernova in the Galaxy NGC 1260," ApJ 659:1 L13 (2007 April 10) Abstract at iop.org: http://iopscience.iop.org/1538-4357/659/1/L13/ Pre-print at arxiv.org: http://arxiv.org/abs/astro-ph/0612408

Try again, but only do so after you have googled the phrase "pair instability supernova".

Neither is nonsense. The tides can indeed be attributed to the Moon and to the Sun, and solar tides are smaller than lunar tides. And the point of this is?

Not a very representative poll, is it? 61% of the respondents were 45 or older, and 77% were white. Given the demographics, the low numbers for Jon Stewart are not all that surprising, but the huge negative impact numbers for Limbaugh are "wow". He is indeed a train wreck.

Utter nonsense. See post #6 for the correct explanation. (Topic was split. Post #6 is http://www.scienceforums.net/topic/52968-tides/page__p__574204#entry574204)

Utter nonsense. In a sense, the Moon does indeed orbit the Sun. One obvious sense is that the specific energy of the Moon in a Sun-centered frame, [math]v^2/2 - GM_{\mbox{sun}}/r[/math], is negative. Another sense is that suppose we could magically make the Earth vanish. The Moon's orbit about the Sun would remain more or less the same (more or less meaning within about 10%). On the other hand, that the Moon's acceleration toward the Earth is about 1/2 that of the Moon toward the Sun means that the nearby presence of the Earth is a bit more than a perturbative effect. Effects should be less than an order of magnitude smaller than the primary effect to be considered 'perturbative'. The presence of the Earth is considerably more than just a perturbation of the Moon's orbit about the Sun. So, does the Moon orbit the Earth? Certainly. First off, the specific energy of the Moon in an Earth-centered frame is once again negative. The Moon is gravitationally bound to the Earth. Just as before, if we could magically make the Sun vanish, the Moon's orbit about the Earth would remain more or less unchanged. From the perspective of an Earth-centered frame, the effect of the Sun's gravitational influence on the Moon's is the inertial frame acceleration of the Moon toward the Sun less the inertial frame acceleration of the Earth toward the Sun. This is greatest when the Earth is a perihelion, the Moon is at apogee, and the Moon is new. Even then, it is only about 1.4% of the Moon's gravitational acceleration toward the Earth. This is a small perturbative effect. Yet another way to look at it is in terms of spheres of influence. There are two competing definitions of a gravitational sphere of influence of a planet on some object, Laplace's sphere of influence and the Hill sphere. With either definition, the Moon's orbit about the Earth is well inside Earth's gravitational sphere of influence. The Moon orbits the Earth (and the Sun, and the Milky Way, and the local group, and ...).

You can solve Kepler's problem to any desired degree of accuracy. A very high degree of accuracy might require some sophisticated techniques, but that is just an implementation detail. Your "sixth order symplectic solver" is only an approximation. If you are dealing with a two-body problem you should be viewing the orbital elements solution as the analytic solution against which you can judge the accuracy of your numerical integration. If you are dealing with something beyond the two-body problem, such as our solar system with multiple planets that interact gravitationally with one another as well as the Sun, a satellite in orbit about the non-spherical Earth, a satellite making a hyperbolic gravity slingshot fly-by of a planet, then you are stuck in a situation in which no analytic solution does exist. You need to use some numerical integration technique. Even then, using an orbital element approach can be "better" than your presumably cartesian approach. One approach is to use an Encke-style integrator. Another is to integrate Lagrange's planetary equations or some variant. There are small number of metrics in the "my integrator is better than yours" game: What is the best that a given integrator can possibly do? Every numerical integrator has to trade off between two extremes. Errors inherent in the integration scheme dominate when the time step is too large while numerical errors dominate at small time steps. The optimal step is a balance between these extremes. The error at this optimal step size varies from integrator to integrator. This optimal time step is one of two main factors in determining the the amount of CPU time needed to achieve some desired degree of accuracy. The other is the number of derivative calculations per step. Putting these together yields the number derivative calls to achieve some desired degree of accuracy for a given period of time. This varies dramatically from integrator to integrator. Every numerical integrator starts producing garbage eventually. The time span over which a numerical integrator produces meaningful results varies from integrator to integrator.

You can go the other way around. I just showed you how to do so.

Tough. Learning to write clearly is not a specialty. It is an essential skill. Write clearly if you want a response to something you wrote. If you continue to post gibberish you won't last long here. If you want to make any progress in real life you will need to learn to spell correctly, write complete sentences, and form these sentences into cohesive paragraphs. This is a science board, and clear communications is an essential element of any science.

Your opening post is wrong in the sense that it does not agree with experimentally observed results. Garbage in, garbage out.

You are being silly. Scientists are neither pure nor "generally biased". Scientists are human and don't quite know what they are doing. "If we knew what we were doing, it wouldn't be called research, would it?"

A couple of caveats first. One, there are many different sets of orbital elements. I will discuss a few of the variants. Secondly, Keplerian orbits are an approximation. They are only valid only in the Newtonian two body problem. The existence of multiple bodies (e.g., the solar system), a central body with a non-spherical shape (e.g., the Earth) and general relativity (e.g., relativistic precession) mean that Keplerian orbits are not quite exact. The canonical set of Keplerian orbital elements is [math]a[/math], the semi-major axis of the orbit, [math]e[/math], the eccentricity of the orbit, [math]i[/math], the inclination of the orbit relative to some reference plane, [math]\Omega[/math], the right ascension of the ascending node referenced to that reference plane, [math]\omega[/math], the argument of periapsis, and [math]T_{\mbox{peri}}[/math], the time of periapsis passage. You also need to know the gravitational coupling parameter of the central body. Naively this is the product of the universal gravitational constant G and the central body's mass M. For the Sun and most of the planets, the product [math]\mu=GM[/math] is known to a much higher precision than are G and M. You can find a list of standard gravitational parameter at http://en.wikipedia.org/wiki/Standard_gravitational_parameter. The mean motion [math]\dot m[/math] is a function of the the coupling constant [math]\mu[/math] and the semi-major axis length [math]a[/math]. For an orbital body of inconsequential mass, this is given by [math]n = \sqrt{\frac{\mu}{a^3}}[/math] If the mass of the orbiting body is a significant fraction of the mass of the central body (e.g., Jupiter), it is better to use [math]n = \sqrt{\frac{\mu_c+mu_p}{a^3}}[/math] Some variants of the above list of orbital elements include Instead of [math]T_{\mbox{peri}}[/math], some specify an epoch time [math]T_0[/math] and the mean anomaly [math]m_0[/math] (or the true anomaly [math]\nu_0[/math]) at that epoch time. You can calculate [math]T_{\mbox{peri}}[/math] via [math]T_{\mbox{peri}} = T_0 - m_0/n[/math]. If you are given [math]\nu_0[/math] instead you will have to compute the corresponding eccentric anomaly and then compute [math]m_0[/math] using Kepler's equation. Instead of [math]a[/math] and [math]\epsilon[/math], some specify the perifocal and apofocal distance or the perifocal and apofocal altitude. Going from these to semi-major axis and eccentricity is again simple math via [math]r_p = a(1-e)[/math] and [math]r_a = a(1+e)[/math]. You will need to incorporate the planet's radius if you are given altitudes instead of distances. Instead of [math]\omega[/math], some specify the argument of argument of latitude [math]u_0=\nu_0+\omega[/math]. Assuming you have that canonical set, to calculate the Cartesian position at some point in time [math]t[/math] you will need to Calculate the mean anomaly at that time: [math]m(t) = n(t-T_{\mbox{peri}})[/math] Calculate the corresponding eccentric anomaly. The relation between mean and eccentric anomaly is called Kepler's equation, [math]m = E - e sin E[/math] Note that this specifies [math]m[/math] as a function of [math]E[/math] and [math]e[/math]. You want [math]E[/math] as a function of [math]m[/math] and [math]e[/math], the inverse of Kepler's equation. Since Kepler's equation is transcendental, the inverse cannot be expressed algebraically. Solving for [math]E[/math] given [math]m[/math] and [math]e[/math] is called "Kepler's problem". Using a Newton-Raphson iterator works very well or small eccentricities. For largish eccentricities you will need to do something more advanced such as using Bessel's solution. Calculate the corresponding true anomaly via [math]\sqrt{1-e} \tan \frac{\nu} 2 = sqrt{1+e} \tan \frac E 2 Calculate the position in the canonical plane via [math]r = \frac{a(1-e^2)}{1+e\cos\nu}[/math] [math]x_c = r\cos\nu[/math] [math]y_c = r\sin\nu[/math] The canonical plane is a related to your reference axis as a rotation about the z axis by the the right ascension angle [math]\Omega[/math] followed by a rotation about the rotated x axis by the inclination i. Reverse this to transform the canonical position [x_c,y_c,0] to the desired [x,y,z] position. I hate, loathe, ..., expletive deleted this new system. The last two elements of the last list in the above post should be Calculate the position in the canonical plane via [math]r = \frac{a(1-e^2)}{1+e\cos\nu}[/math] [math]x_c = r\cos\nu[/math] [math]y_c = r\sin\nu[/math] The canonical plane is related to your reference axis as a rotation about the z axis by the the right ascension angle [math]\Omega[/math] followed by a rotation about the rotated x axis by the inclination i followed by a rotation about the rotated z axis by the argument of periapsis angle [math]\omega[/math]. Reverse this to transform the canonical position [x_c,y_c,0] to the desired [x,y,z] position.

You must admit that Lisi's "exceptionally simple theory" has garnered a lot of interest. Whether that idea has any legs: It's too early to tell. Whether he was an amateur: That's fuzzy. He definitely had the education. However, teaching at a junior college whose physics curriculum is limited to freshman physics is, in my opinion, further removed from being a working physicist than was Einstein's stint as a patent clerk.