DrRocket

It should be obvious that those operations are invertible, with an inverse of a type contained in your list of allowable operations, and that any solution of the original set of equations is a solution after any of the named operations are performed.

And it is gravity that caused the agglomeration of materials that we call the sun and that resulted in the conditions of temperature and pressure that are conducive to fusion, and that continues to be a major factor in the physics of that star.

Two resistors are in series if the current that flows through one must also flow through tne other. Two resistors are in parallel in voltage impressed across one is also impressed across the other.

To keep the calculation simple let's consider the case of a constant force applied to a mass starting from rest. Then [math]F=ma[/math], [math] d=\frac {1}{2} at^2[/math], [math] v= \frac {d}{t}[/math] and [math] a = \frac {v}{t}[/math] So [math] F \times d = ma \times \frac {1}{2} at^2 = \frac {1}{2} ma^2t^2=\frac {1}{2} m \frac {v^2}{t^2} t^2 = \frac {1}{2} m v^2[/math] Note that no other mechanism and no notion of "internal energy" are involved. The kinetic energy arises quite simply from the application of the force[math] F[/math] to the mass [math]m[/math]. You can gussy this up with line integrals and time-varying force, velocity and acceleration, but the end result is the same.

aha

[math] \frac {\sqrt {2x}}{\sqrt x} = \frac {\sqrt 2 \sqrt x}{\sqrt x} = \sqrt 2 [/math] [math] \overline {a + bi} = a-bi[/math] and [math] \overline {a+b} = a+b[/math] where [math]a,b \in \mathbb R[/math] and [math] i = \sqrt {-1}[/math]

First it is not force acting at a distance that imparts energy. It is force acting over a distance. Second it is a logical consequence of Newton's laws of motion that a force [math]F[/math] acting on a mass [math]m[/math] over a distance [math]d[/math] results in a velocity [math]v[/math] that satisfies [math]F \times d = \frac {1}{2} mv^2[/math] and [math] \frac {1}{2} mv^2[/math] is defined to be the kinetic energy of the body. There is no notion of "internal energy" or indeed any internal structure in the particle picture of Newtonian mechanics. So to accept your theory one must reject established mechanics or adopt a completely new definition of energy. You cannot "speculate" regarding a definition, and if you reject mechanics which is rather well supported by a mountain of empirical evidence, then whatever you are talking about is not physics. Passing to special relativity does not change the picture materially.

If you accept Newton's laws of motion then it is a logical consequnce that the application of a force [math]F[/math] over a distance [math]d[/math] to a body of mass[math] m[/math] results in [math]\frac{1}{2} mv^2 = Fd [/math] and [math]\frac{1}{2} mv^2 [/math] is defined to be the kinetic energy of the body. So to reject force over distance as imparting energy you must logically either reject Newton's mechanics or you must impose a decidedly non-standard definition of energy. Going to special relativity does not materially change this picture. In any case to reject this implies that you are no longer talking about physics. There is no room for speculation when it comes to definitions.

Yep. Mg will burn under water once ignited. Aluminum is nearly as reactive. In powdered form it is the primary fuel in large solid rocket motors. The Mg/Al wheels on your car are not a hazard. But the Al superstructure on British ships in the Falklands war burned nicely when hit with an exocet missile. Fires involving large quantities of powdered magnesium are pretty spectacular. Fighting such fires is not recommended. You not only don't want to put water on the fire, you don't want to put powdered magnesium (or aluminum) in water -- the evolved hydrogen is quite hazardous. I have seen a magnesium fire with water sprayed on nearby oxidizers and don't recommend that either. I have also seen a small buildings loose its roof from hydrogen evolved from a water-filled scrap bucket. Things that one can do with very small quantities of explosives and incendiaries in a laboratory can result in tragedy when larger quantities are involved -- and thos larger quantities need not be very large. A quarter pound of magnesium/teflon is quite enough to enable a fatal accident.

Bold added. Do you disagree that application of force over distance imparts energy ? This is basic Newtonian mechanics.

Application of a force does not necessarily imply energy transfer. It is application of force over a distance, work, that transfers energy. There is this branch of science called physics and within in there is a subject called thermodynamics. The term "heat" has a well-established meaning within thermodynamics and that definition is the answer to your question. If you choose to ignore that definition, which you may certainly do, then what you are discussing is something other than physics. But there is no point in usurping a widely recognized term, like heat, and applying it to something else. That is poor communication, and just plain silly. Call it Oscar, and don't confuse the discussion of Oscar with a discussion of physics. You really do need to read a book on thermodynamics.

"It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so. "– Mark Twain

[math]2x/(1+4x) = (2x/x)/(1/x+4x/x) = 2/(1/x+4) \ne 2(x/1)+2(1/4) = 2x+1/2 [/math]

Absolutely correct. Magnesium is still used in incendiary devices. It is extremely dangerous in the hands of amateurs, and even professionals have been badly injured with relatively small quantities. Injuries are comparable to war wounds -- third degree burns over large areas, inner ears burned out, digits on the hands burned off, lungs badly damaged, ... If you are lucky in an accident you are killed . Safe handling of ground or powdered magnesium, particularly if oxidizers are in proximity, takes expensive equipment and expertise not available to amateurs. Been there. Done that.

Right. For those who think that all you need to do is focus the sun's light to a point, the problem is that the sun's rays are not perfectly parallel and won't focus to a point. The best that you can do, in the ideal case, is a disc that replicates the surface temperature of the sun. The second law wins again. This issue is not too different from the situation in which a resistor is heated by current from a (lower temperature) battery. Call it work and all is well.

Your question is rather poorly phrased and makes little sense. But perhaps this will help. [math]PV=nRt[/math] for an ideal gas. [math]PV[/math] is proportional to the total internal energy of an ideal gas. T is proportional to internal energy per mole, which by the equipartition theorem is proportional to the kinetic energy of translation per mole.

That depends on how the problem is formulated. It is rather typical in gas dynamics to speak of both static temperature and stagnation temperature. The former is the local gas temperature in a frame of reference at rest with respect to the local flow. Stagnation temperature is the temperature in rest frame if the gas were to brought to zero velocity isentropically. The difference is the "kinetic energy of motion of the entire system". While static conditions would be of interest to a chemist, both static and stagnation conditions are of great interest when designing rockets and rocket components, where kinetic energy of the entire system can be of great interest.. Chemists, physicists and engineers tend to look at thermodynamics in slightly different (but not inconsistent) ways. Chemists and to a lesser extent physicists tend to consider closed systems (i.e. systems without mass transfer). Engineers tend to consider open systems (systems allowing for mass transfer). No confusion arises so long as the assumptions are clearly stated, but communication does require clear agreement on assumptions and terms. It is worth remembering that the formulation of classical thermodynamics by J. Willard Gibbs predates both quantum theory and relativity, and indeed the acceptance of the atomic theory. Classical thermodynamics is rather abstract and independent of the atomic hypothesis, the link being the modern theory of statistical mechanics. Thus it is rather easy to throw atomic phenomena into the mix that at first blush are difficult to classify in classical terms. What is amazing is how well the theory has held up despite the revelations as to how nature works since the time of its invention.

String Theory and M-Theory: A Modern Introduction by Becker, Becker and Schwarz Superstring Theory by Green, Schwarz and Witten (2 vols)

-1/3x3 + 3/2x2 = -((2x3 - 9x2)/6)

Right [math]dU=dQ-dW[/math] (Q is heat flow into the system and W is work done by the system) and the difference between Q and W is somewhat in the eye of the beholder, so long as one is consistent. When Gibbs formulated classical theremodynamics lasers were in short supply. Whether you call incoming laser light heat or work is unimportant so long as you account for it and are consistent. To be consistent with text books and the literature, per Obert's book, "Heat is energy transferred, without mass transfer across the boundary of a thermodynamic system because of a temperature difference between the system and its surroundings." However, as you observed the system does not really care whether the incoming photon originated in an incandescent lamp or a laser, and the effect on internal energy is the same whether it is heat flow into the system or work done on the system. But it must be one of the two and it cannot be both. Convention results in it being called work.

That is not at all what swansont is suggesting. What he is stating very clearly is that thermodynamics does not allow what you are suggesting. There are many good texts on thermodynamics. Two good ones are Thermal Physics by Morse and Elements of Thermodynamics and Heat Transfer by Obert and Young. You are in desperate need of reading a book on the subject.

Whatever he is drinking is powerful stuff.

Sort of puts the old kibash on refrigeration technology doesn't it ?

This is correct. It is also a matter of definition, and arguing about definitions is futile. There are, of course, meanings that lie outside of physics, but this is a physics forum. http://www.nba.com/heat/

It is pretty easy to see that if the limit exists then the limit must be 1.