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You can ask the same question of a flexible rope or chain whose ends are attached to two fixed points. Regardless of the location of the two end points and regardless of how much slack is in the rope, the rope will form a section of a catenary. This curve is the minimal energy configuration of the rope. How does the hanging rope "know" to follow the minimal energy configuration? Does the rope do calculus to form itself into a catenary? What you are looking for is called the principle of least action. The development of this principle has a long history, starting with Newton and Leibniz. Physicists now use Hamilton's principle: A dynamic system follows the path that represents an extremum of the time integral of the difference between kinetic and potential energy. This extremum is almost always a minimum. Hamilton's principle is widely used in classical and quantum mechanics. In classical mechanics, Hamilton's principle bypasses the need to represent all of the forces acting on a system. Suppose you have some classical problem readily soluble by means of Hamilton's principle. If you can represent the forces in the system and if you can solve the problem using the Newtonian force formulation, the resulting solution will be the same as that arrived by applying Hamilton's principle. Hamiltonian and Newtonian mechanics are mathematically equivalent. You can look at Hamilton's principle as magical: Some magical agent does the calculus. The early developers of the principle essentially viewed it as magical: God does the calculus. There is no need for God here; you can also look at at Hamilton's principle as an abstraction of some deeper but much more difficult to manipulate principles. For example, consider the case of your light beam. In quantum electrodynamics, a photon simultaneously takes all possible paths from point A to point B. Sum all of these possible paths and one path will arise as the "winner" by virtue of interference. This winning path is the time minimizing path.

You can ask the exact same question of other fundamental quantities such as distance. Time and distance are axiomatic in physics.

Stop trolling. Moved to pseudoscience.

There never has been a right wing dictatorship. If there had been we wouldn't be having an election in less than a month.

Good lord, why is this silly thread still here? If I had a appropriate fishing net, I would cast it into this thread and have a tasty meal of red herring for dinner. To answer one of the multiple questions in this thread, yes, the Earth is moving away from the Sun as the Sun loses mass. Easily calculable: The Sun is losing 9*10-14 solar masses per year due to radiation and solar wind. This mass loss won't affect kinetic energy, but it will affect potential energy. To conserve energy the Earth has to move away from the Sun at the same rate: 9*10-14 astronomical units per year, or 1.3 centimeters per year, or 1.3 meters per century. There are too many other confounding factors to make the change over a course of a year measurable. The 1+ meter change over the course of a century is measurable -- perhaps.

You're answer is correct, but I don't think that what question BA asked. To elaborate, for objects near the surface of the Earth, the potential energy of some object with mass m due to gravity is approximately [math]mgh[/math], where h is the height above the surface of the Earth. More generally, the potential energy of some object with mass m due to the gravity resulting from a point mass with mass M (or any spherical distribution of mass with mass M) is [math]U = U_0 -\,GMm/r[/math], where r is the distance from the point mass and U0 is an arbitrary constant. The same expression applies in the case of a spherical mass rather than point mass (in which case r is the distance to the center of the sphere), but it is only valid for points outside the mass. About that arbitrary constant U0: The potential energy corresponding to some conservative force field is some scalar function whose gradient yields the force field. Suppose you have some function [math]\phi(\boldsymbol {x})[/math] such that [math]\nabla \phi(\boldsymbol {x}) = \boldsymbol F[/math]. Add an arbitrary constant to the function phi and you get another function whose gradient is also the same force field. Potential energy is indeterminate with respect to an arbitrary constant. There is no problem here, as it is only the difference between two values of the potential that is of interest. Physicists are free to choose any value for that arbitrary constant. For a point not necessarily close to a spherical mass distribution, one obvious choice is U0=0. So, how does one go from [math]U_0-\,GMm/r[/math] to [math]mgh[/math]? They certainly don't look at all like one another. Suppose that [math]r=r_0 + h[/math], where r0 is the radius of a spherical mass and h is the height of some object above the sphere. Then [math]U=U_0-\,GMm/(r_0+h)[/math]. Rewriting the denominator [math]r_0(1+h/r_0)[/math] and approximating [math]1/(1+h/r_0)[/math] as [math]1-h/r_0[/math], valid for [math]h \lll r_0[/math], the potential energy for an object of mass m near the surface of a spherical mass is approximately [math]U \approx U_0 -\frac{GMm}{r_0} + \frac{GM}{r_0^2}mh[/math] Choosing [math]U_0 = \frac{GMm}{r_0}[/math] (it's an arbitrary constant; I can choose whatever value I want so long as I stick with it), [math]U \approx \frac{GM}{r_0^2}mh[/math] That first factor, [math]GM/r_0^2[/math], is the acceleration due to gravity at the surface of the object, which in the case of the Earth is g. Thus [math]U \approx mgh[/math] for objects near the surface of the Earth.

A similar phenomenon occurs when one is developing an ad hoc model of some process. Should one use all of the data at hand to generate the model, or reserve some of the data as a test against the generated model? The answer is, it depends. If one can use all of the data and still construct some kind of meaningful statistical measure of "goodness of fit", use all of the data. If, on the other hand, the model generation process is so complex (e.g., machine learning techniques such as neural nets and ant colony optimization) that a goodness of fit constructed from the data is suspect, it is much better to reserve some of the data. Doing so most likely will reduce the accuracy of the model when applied to the full data set, but it will certainly give an idea of how well the model performs when applied to data drawn from outside the data set used to construct the model.

From the thread on "Best powerpoint colors": This little interchange reflects a huge difference between Gen Y and previous generations. Assuming you work in a multigenerational team, walk through the offices of your employer. You will see some desks strewed with paper and others that are immaculately clean. Discounting the clean freaks (a disease that can strike any generation), those immaculate desks inevitably belong to someone from Gen Y. Their desks are clean not because they are afflicted with some obsession to keep their desks unsullied but rather because "Pr... int" is a foreign word to them. They have finally achieved the goal of a paperless office. All prior attempts at making a paperless office have failed because we older farts would feel lost without paper. To our Gen Y (and younger) members: Do you feel more comfortable looking at something on the screen rather than on paper? How in the world do you scribble math without paper? I personally would be completely lost without my pad of green engineering paper. To our more aged members: Does your work environment support IM? Do you use it? Do you even know what IM means? If your boss threatened to take all of your paper away would you print your resume immediately?

You have a very common misperception of Newton's third law here. Newton's third law talks about forces exerted on a pair of bodies. If body A exerts some force on body B, Newton's third law says that body B exerts a force on body A that is equal but opposite to the force A exerts on B. For a person standing on the Earth, the Earth as a whole is pulling on the person gravitationally and the little patch of Earth under the person's feet is pushing the person up. This is not a third law interaction. Suppose you skydive from a stationary helicopter. There is no wind resistance when you first jump out. The only force acting on you is gravity. So what is the third law interaction here? Simple: The Earth is falling toward you gravitationally! Let's say for the sake of argument that your mass is 75 kilograms. The Earth is exerting a force of 735.75 newtons on you. You are exerting a force of 735.75 newtons on the Earth. While the forces are equal but opposite, the resultant accelerations are anything but. You fall toward the Earth at 9.81 m/s^2, and the Earth falls toward you 1.23×10-22 m/s^2. While the Earth's acceleration toward you is immeasurably small, the effects are observable for a much larger mass -- the Moon, for instance. The exact same thing is happening when stand on the Earth. The Earth is still exerting a gravitational force of 735.75 newtons on you, and you exert a gravitational force of 735.75 newtons on the Earth. Yet another third law interaction occurs when you are stand on the Earth. It is called the normal force. You aren't really touching the surface of the Earth. Your feet are hovering a tiny, tiny distance above the surface. The electrons in your feet repel the electrons in the little patch of Earth just below your feet. Pick up a heavy book and that hover distance will shrink a tiny bit. The slightly closer distance means the Earth is exerting a slightly larger normal force on you. By Newton's third law, you are exerting a counter force on the Earth. If you instead stand on a scale, things get even more complex. The third law interactions now are You and the Earth attract each other gravitationally You push down on the top of the scale and the top of the scale pushes up on you (normal force) The top of the scale pushes down on the top of the spring and the top of the spring pushes the top of the scale up (rigid forces) The top of the spring moves downward, compressing the spring (elastic force). The spring elastic force makes the top of the spring push upward and the bottom of the spring push downward. The bottom of the spring pushes down on the bottom of the scale and the bottom of the scale pushes up on the bottom of the spring (rigid forces) The bottom of the scale pushes down on the surface of the Earth, which pushes up on the bottom of the scale (normal force). Whew! If you poke a bit deeper, those things I called normal force, rigid force, and elastic force all arise from the electromagnetic force.

General relativity says singularities exist. General relativity is the best model of spacetime we have. Some even more bizarre things than singularities (e.g., wormholes) might arise, but saying they are fact is just wrong. They are purely conjectural. You appear to be young and you appear to have read some scientific popularizations. One problem with these for-the-masses books is that they present things as known fact that are in fact far removed from fact. Hawkings, like many other physicists, doesn't like the idea of singularities and has been conjecturing ways around them. It is, however, just that: conjecture. (Then again, so is the SciAm article that is the subject of this thread.) If Hawkings is currently the best in cosmological physics at the moment, cosmological physics is in a big heap of trouble. Hawkings has a huge handicap, and no I am not talking about ALS. His huge handicap is a very common ailment, and is one that no scientist has ever been able to overcome: He is old. Old physicists (and by old, I mean over 45) do not make meaningful contributions to the body of physics. We cannot apply quantum mechanics all the way back to the beginning of time. To do that we would need a theory that weds quantum mechanics and gravity. This grand unification theory currently does not exist. Different groups of people are taking different routes to get to that theory, but none of them have achieved it (yet).

Later in your post you have no problem with viewing multiple forces acting on a body that together yield a net force of zero. Force and acceleration are both vectors, related by [math]\mathbf F = m\mathbf a[/math]. What then is the problem with viewing multiple accelerations acting on a body that together yield a net acceleration of zero? [math]\aligned \mathbf F_{\text{net}} &= \sum_i \mathbf F_i \\ \mathbf a_{\text{net}} &= \frac{F_{\text{net}}} m = \sum_i \frac{\mathbf F_i} m \endaligned [/math] There is no reason mathematically that one must compute the net force first and only then compute the acceleration. There is no difference between computing the net force and then the acceleration versus first dividing each force by the mass to yield the acceleration due to that force and summing these acceleration terms to yield the net acceleration. The two approaches are mathematically identical, so both views are equally valid. In the case of somebody standing on a scale, the gravitational force results in a downward acceleration, the normal force results in an upward acceleration. The two cancel, yielding a net acceleration of zero. Obviously not, because the counter force (and hence the acceleration) would go away. The normal force is a constraint force. If gravity were not pulling the person down, there would be no normal force.

Whoa! There are a lot of problems here. To name a few, You are conflating the big bang singularity, which is what the SciAm article addresses, with black hole singularities, which is the subject of the post you referenced. There is too much aggrandizement of Hawking in your post. Stephen Hawking is not the be-all and end-all in cosmology. Moreover, he has not reconciled quantum mechanics and general relativity. Nobody has, yet. That wormhole statement is, well, bizarre. The Penrose-Hawkings singularity theorems do indicate that gravity is singular, assuming general relativity is true. You have Hawking's work exactly backwards. Also note that Hawking's name is second in these theorems.

In most other industries that involve significant risk, a separate group (and sometimes multiple separate groups) independently double-check the work of the designers and developers. This independent assessment is a core concept in the aerospace, electrical power and transportation industries, just to name a few. Problems happen even with safeguards in place (e.g., NASA's Challenger and Columbia disasters). For one thing, the safety experts have to be treated as key partners rather than as interlopers. The fault in the Challenger and Columbia disasters was deemed to lie primarily with NASA management - even though NASA had safety checks, independent assessments, a hoard of safety experts, ... The financial industry: Safety checks: The financial industry required the developers of the financial products to do their own safety checks. That should cover the safety check issue, right? Independent assessments: That would have required the financial institutions to hire another group of "Nobel-track" physicists whose sole job purpose would have been to get in the way of their existing group of "Nobel-track" physicists who were making money hand-over-fist. In the end, I predict the financial crisis will be viewed as not only a failure of management, but as a criminal failure of management. I also predict that much of the money those executives paid themselves will be seized by authority of the RICO act.

Nobody here has claimed that the feather and hammer experiment had any scientific merit. This demonstration was targeted at a lay audience, many (or even most) of whom have a pre-Archimedean view of the world. A Cavendish-style experiment would not have the visual impact that the feather and hammer did. That is not correct. If Jupiter was a lot smaller (for example, the size of one of its moons), the mini-Jupiter would take longer to orbit the Sun than does our big honkin' Jupiter. The orbital period of two bodies about each other is, ignoring all other influences (i.e., other gravitational bodies), [math]T = 2\,\pi\,\sqrt{\frac{a^3}{G(m_1+m_2)}}[/math] Ignoring the mass of Jupiter in the above equation (or making it very small) yields a period of 4,335.4 days at Jupiter's distance from the Sun. Accounting for the mass of Jupiter but still using the above simple two-body problem equation yields a period of 4,333.3 days. The actual value: 4,332.589 days. Accounting for Jupiter's mass removes 75% of the error.

No. They are called economists only if they have a degree in economy as well as a degree in engineering, math, or science. By the time someone has gone through grad school and received a non-economic PhD they are not too likely to go back to school to study economics, particularly when financial institutions will hire top-notch PhDs with little or no economic training. The generic term for this specialty is "quantitative analyst", or "quant" for short. Therein lies part of the problem. Quants don't necessarily know much about the economics of the systems they are modeling, and the management in those companies don't necessarily know much about the models developed by the quants.

From the early 90's on, I have known quite a few engineers, mathematicians, and physicists who have abandoned the real field of "rocket science" to go into IP law or to become "rocket scientists" -- that is, a bright boy in finance. A six figure salary (and a high six figure salary too boot) was a rather compelling reason, plus it was all legal. They certainly helped contribute to the financial crisis. Was it their fault? No. Their management chose to ignore that the mathematics was over their heads when the quants were making their employers billions (trillions?) of dollars in profits. Management could have asked about the underlying modeling assumptions, could have had a different group of "Nobel-track" physicists assess the models against risk, and could have had yet another group of "Nobel-track" physicists independently verify and validate the models. Did they do any of these things? Not that I can see. For those of you who don't like videos (e.g., me), The transcript of the 60 Minutes report A 1999 Physics World feature article on the exodus of physicists to finance A 1999 Physics World editorial on the same A Sept 16 Physics World feature article on the collapse of Lehman Bros. Did Physics PhDs cause current financial crisis, a Sept 16 blog by a physicist

Larry Niven, "What Can You Say About Chocolate Covered Manhole Covers?" Phillip Pullman, "His Dark Materials" Arthur Clarke, "The Nine Billion Names of God" The first in particular is a dead ringer for the OP.

In other words, it is ad hoc. It also disagrees with some very important observations: We live, as far as we can tell, in a three dimensional universe. Newton's law of gravitation has gravitational force between two objects inversely proportional to the square of the distance between the objects. This squares very well with the three dimensional nature of space. The surface of a sphere is proportional to the square of the radius of the sphere. Naively, the gravitational gravitation force that emanates from some object should decrease as the surface area of spheres centered on the object increases. The planets have nearly elliptic orbits about the Sun. With the exception of Mercury, Newton's law of gravitation accounts for the deviation from true elliptical orbits by virtue of inter-planetary gravitational interactions. Suppose gravitation somehow had something other than an inverse square relationship with respect to distance: [math]F=kr^{\alpha}[/math]. Here, [math]k[/math] is a proportionality constant and [math]\alpha[/math] is the power that relates distance to force. There are only two powers that yield closed elliptical orbits: [math]\alpha=1[/math] (e.g., a spring) and [math]\alpha=-2[/math] (e.g., gravitation). Your 1/r relationship will not yield closed elliptical orbits. Newton's third law. Newton's law of gravitation, in which the relative acceleration between two objects is proportional to the sum of the masses of the objects, is consistent with Newton's third law. Yours is not.

Your calculator does have scientific notation, doesn't it?

Units, please. One of the signs of a crackpot is someone who doesn't know a m/s^2 from a kg/m. The reason you lose is that science has hundreds of years of observation that confirms their models of gravitation. You, on the other hand, have nothing to offer. You claim to have performed some experiments. Please explain your experimental setup, how you made measurements, how your experimental setup and procedures addressed bias and error, how you analyzed your data, and the results of your experiment. Am I being mean? No. The scientists who have performed real experiments have had to endure much harder questioning. You claim to have a model of gravitation, A=(L-S)/R2. This equation doesn't even have the right units. As it stands it is invalid. The units problem can of course be solved by adding a proportionality constant. Suppose you do just that: A=k*(L-S)/R. Some questions need to be addressed at this point: What is the basis for this model? Is there anything behind it, or did you just make it up out of thin air? What tests can be be made that distinguishes your "theory" from others? With this new, improved equation you should be able to estimate the masses of Jupiter and Saturn from the orbits of their relatively tiny natural satellites; the masses of Mercury, Venus, Earth, Mars from the incredibly tiny artificial satellites in orbit around them; and the mass of the Sun from the planets. Your new, improved theory should agree with things outside the solar system as well, things such as 61 Cygnii. Finally, your estimated values must be consistent with one another. The problem with your "theory" is that that won't happen.

Red herring. Red herring. Red herring. Red herring. Particle physicists used to flaunt the measured accuracy of the fine structure constant over lesser branches of physics such as dynamics and astronomy. No more. The verification of the equivalence principle currently stands as the most accurate result in all of physics. See this Physics World article for details. We know the equivalence principle is true to within a few parts in 1013. If this principle is not exactly true, the contradictions lie in those few parts per 1013. Your "theory" contradicts the equivalence principle, and not just to within a few parts per 1013. Any "theory" that assumes the relative gravitational acceleration of two objects toward one another is proportional to the difference of the masses will fail miserably in predicting the behavior of 61 Cygni.

If that's the case, your equation A=(L-S)/R2 doesn't even have the right units. Acceleration is not mass/distance. The stereotypical scientist is an asexual nerd who prefers blondes on the rare occasion that he does feel some stirrings in his loins. Getting serious now, scientists are overwhelmed with crackpots' outrageous claims. Why should they waste their time on a claim that is patently *wrong* (e.g., yours)? Cavendish' experiment did not measure the acceleration of a moving object. It measured the displacement of a non-moving object. Measuring acceleration is rather difficult to do, isn't particularly precise, and is subject to confounding errors. Measuring displacement is comparatively much easier to accomplish, is much more accurate, and introduces fewer error sources. It is still a tough experiment to set up correctly. Red herring. I looked them up. The people who did the actual work used several things to determine the gravitational parameters1 for Jupiter and the Sun. First off, both Jupiter and the Sun have many small bodies orbiting them. Secondly, they did indeed assume that our best model of gravity2 is correct, at least as far as we know. If the models were not correct, contradictory results would arise in modeling the orbits of various satellites around their planets and the planets around the Sun. The results, instead, are self-consistent. Footnotes: 1 Note that I said "gravitational parameter" rather than mass. The gravitational parameter for some body is the product of the universal gravitational parameter G and the mass of the body. The gravitational parameter is highly observable; we know the gravitational parameter for the Sun, the Earth, the Moon, and Jupiter to many decimal places of accuracy. The universal gravitational constant, on the other hand, is notoriously hard to measure, and thus so are the masses. Astronomers work with G*M directly rather than G and M separately. 2 I mentioned that those at JPL used the best model of gravity to assess the mass of the planets. This is not Newton's law of gravitation. I also mentioned that contradictory results would have arisen if the model was incorrect. An example of a contradictory result: Newton's law of gravity predicts a different value for the precession of Mercury than is observed. Our best model of gravity to date is general relativity, not Newton's law. GR explains the precession of Mercury; Newton's law does not.

Something doesn't jibe here. In short, I don't believe you have performed a Cavendish experiment.

Nice one. He could also stay a little closer to home and look at the solar system. In particular, explain Jupiter's orbit. If one just used Kepler's third law (which ignores the mass of the planet), one arrives at a sidereal period of 4,335.4 days for Jupiter. Accounting for the mass of Jupiter but still using the simple two-body problem equation [math]T_{\text{jupiter}}=2\,\pi\,\sqrt{\frac{a_{\text{jupiter}}^3}{G(M_{\text{sun}}+M_{\text{jupiter}})}}[/math] one arrives at a period of 4,333.3 days. Using the difference between rather than the sum of the masses yields a sidereal period of 4,337.4 days. The actual value: 4,332.589 days. Relevant data: G*Msun = 132,712,440,018 km3/sec2 G*Mjupiter = 126,712,768 km3/sec2 Jupiter mean semimajor axis = 5.20336301 AU Sources: The gravitational parameters are from Table III in http://iau-comm4.jpl.nasa.gov/de405iom/de405iom.pdf. The mean semimajor axis is from http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html.

No, it is not. First off, you need to define your terms. Secondly, you have a sign error. Yes, there is a sign error. The acceleration due to gravity of two massive objects toward one another is [math]\frac{G(m_1+m_2)}r^2[/math] Units, please. Your "theory" is wrong. There is a simple test, however. The Moon has a mass of about 0.0123 Earth masses. If your "theory" were correct, the sidereal period of the Moon's orbit around the Earth would be about [math]2\,\pi\sqrt{\frac{(384400\,\text{km})^3}{G\,M_e (1-0.0123)}}\approx 27.62\,\text{days}[/math]. With the sign correction this becomes 27.28 days. The Moon's sidereal period is actually about 27.32 days. Third body effects (e.g., solar gravity) account for the difference between the simplistic 27.28 day calculation of the length of the sidereal month and the observed 27.32 day-long sidereal month. The difference between 27.62 days and 27.32 days is not explainable by any theory. Your "theory" is wrong.