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No need for that. Severian replied the way he did because, based on the post both he and I quoted, you gave every appearance of being a crackpot (of which we have more than our fair share at this site and at this time of year). Specifically, You specified a range for the photon mass which excludes zero. That made it appear that you think photons have a non-zero mass. That photons having a non-zero rest mass is one of the more common crackpot conjectures. How you use words is important. You specified a mass without specifying units. Crackpots are notoriously sloppy with units. You missed a negative sign on the exponent, assuming you meant grams. Crackpots are notoriously bad with scientific notation. Some advice: Stop being so flip, take the chip off your shoulder, and please a bit more careful in what you say.

That doesn't make sense since you didn't specify units. It makes even less sense with units. It makes even less sense as you specified a range for the photon mass. A rest mass of zero is within experimental error in every experiment to date to assess the mass of a photon. The experimentally ascertained upper limit on the photon mass is dropping as experimental techniques improve. You are probably referring to experiments by Jun Luo and his associates, described in http://www.aip.org/pnu/2003/split/625-2.html A new limit on photon mass, less than 10-51 grams or 7 x 10-19 electron volts, has been established by an experiment in which light is aimed at a sensitive torsion balance; if light had mass, the rotating balance would suffer an additional tiny torque. Note the difference. The report specifies a number along with units, and specifies that the result is an upper limit. The lower limit is zero. Careful, now. There are three distinct meanings of the word "weight". Legally and colloquially, weight is a synonym for mass and has units of mass. In this sense, yes, if a photon has mass it has weight. In pre-university and some lower level university classes, and to airplane designers, weight is gravitational acceleration times mass (units of force). This is not a very useful definition as there is no way to measure this definition of weight. In this sense, if a photon has mass, it has weight, but only if it is under the influence of an (observable) gravitational field. In some undergraduate physics classes and almost all general relativity classes, weight is what an ideal spring scale weighs. In classical physics, this is the net non-gravitational force acting on an object; in general relativity, this is the net force acting on an object (gravity is a pseudo-force and doesn't count). Even if a photon does have mass, it has no weight per this definition. As soon as something acts on a photon it ceases to exist. Bottom line: Scrape the idea of weight. If a photon has non-zero mass there are a whole lot of things in physics that need to be patched up.

1 joule is, among other things, The work done on some object by a force of one newton as the object moves a distance of one meter, from [math]W=\oint \mathbf F \cdot d\mathbf l[/math] The kinetic energy of a 2 kilogram mass moving at the non-relativistic velocity of 1 meter per second, from [math]KE=\frac 1 2 m v^2[/math] These expressions have the same fundamental units and the same numerical value: [math]1\,\text{joule} = 1\,\text{newton}\,*\,1\,\text{meter} = 1\,\text{kg}\,\text{meter}/\text{second}^2\,*\,1\,\text{meter} = \frac 1 2\,*\,2\,\text{kg}\,*\,(1\,\text{meter}/\text{second})^2[/math] Energy has units of mass*velocity squared. The expression [math]mc^2[/math] has units mass*velocity squared -- energy.

Roughly speaking, yes. However -- What definition of the calorie? There are several definitions, ranging 4.182 to 4.190 joules. One big reason for the different definitions is that the heat capacity of water is not constant. It drops slightly from 0oC to 34oC and then rises slightly from 35oC to 1000C. The variation is small but measurable.

There are two incompleteness theorems. The first says that any sufficiently powerful mathematical system cannot be both consistent and complete. The second says that any sufficiently powerful that contains a proof of its own consistency and completeness is inconsistent. To understand these theorems you need to understand what is meant by "sufficiently powerful", "consistent", and "complete". "Sufficiently powerful". A simple mathematical system can be consistent and complete. Gödel's incompleteness theorems pertain only to systems that can reproduce the natural numbers and their properties: addition, multiplication, and induction. Remove induction from the Peano axioms and you still get an incomplete theory because the numbers are defined recursively. You have to get rid of induction and at least one of the concepts of zero, one, addition or multiplication to get a consistent, complete (and completely boring) theory. Consistent. While "a foolish consistency is the hobgoblin of small minds" (RWE), mathematicians are sticklers that a mathematical body of knowledge be consistent (or minimally, not provably inconsistent). A mathematical system is inconsistent if it contains a contradiction. Suppose some mathematical system allows a proof of some logical statement P made in that system and another proof of the logical negation of P. With these two contradictory statements at hand, one can prove any statement in that system is true. This is not good. Mathematical systems must be consistent or they are absolutely worthless. Complete. A mathematical system is complete if every statement that can be made in that system is either provably true or provably false. Completeness is a very "nice to have" feature. Unfortunately, consistency and completeness together are possible only for mathematical systems of rather limited capability (and hence limited use). One of the first practical applications of the incompleteness theorems was that they put the kibosh on one of the key goals of early twentieth century mathematics. In 1900, the mathematician David Hilbert presented a list of 23 problems in a keynote speech at the International Congress of Mathematicians to kick off the new century. The second problem was to prove that the axioms of mathematics are consistent. A mathematical system can be proven to be consistent, but not within itself. The proof can only lie in a larger system. Another practical application is Hilbert's first problem, to prove the continuum hypothesis. Gödel himself proved that the continuum hypothesis is consistent with the Zermelo-Fraenkel axioms (specifically, he showed that the negation of the continuum hypothesis is not provably true). Paul Cohen later proved that the negation of the continuum hypothesis is consistent with the Zermelo-Fraenkel axioms plus the axiom of choice. Both Gödel and Cohen assumed that the axioms are consistent.

Suppose we discover that not only matter is discrete, but so is space-time. Said discovery will make calculus and the reals physically non-realizable. We will still be able to use calculus and the reals to describe the universe, but only in an approximate sense. Full disclosure: I am of the opinion that mathematics is invented, not discovered. However, because we humans have rather limited imaginations, most of our mathematical inventions are motivated by physical reality. That said, Cantor had a dang near unbounded imagination. What, pray tell, would the physical realization of infinitary logic look like?

From the link, It works quite fine. Your job as a proponent of an alternate explanation is to find some observed electromagnetic phenomenon that is not explained by the standard model of physics. Just because you don't understand it doesn't mean its not true. Theoretical physics uses mathematics to describe observed phenomena. You have no mathematics, therefore you have no theory. Once again, just because you don't understand some theory does not mean that the theory is not valid. All it means is that you don't understand it. Thread moved to pseudoscience. It can be moved back to a non pseudoscience forum when (if) this is made a bit more concrete and preferably backed up with experimental observations.

No, it's not. There's that factor of 10-36, which is a quantum mechanic result. There are at least two things wrong here. First off, the assumption that thermal gravitons will carry away 1/1036 of the thermal radiation does not follow from the fact that the gravitational attraction between a pair of protons is 10-36 times weaker than electrostatic repulsion between a pair of protons. Second, mixing classical and quantum mechanics approaches in describing thermal emissions of photons leads to big problems (google ultraviolet catastrophe). There is no justification for your very first step, and your entire concept depends on this unjustified assumption.

A somewhat simplistic view of heat is that heat is the transfer of energy between an object and its surroundings due to temperature differences. To understand heat you first need to understand the concept of temperature. The zeroth law of thermodynamics says that two objects in thermal equilibrium with each other have the same temperature. Objects that aren't at the same temperature will come into thermal equilibrium by transfer of heat. One way to look at temperature is as a part of the total internal energy of an object. A lot of failed explanations of heat didn't work out because while objects can indeed have a temperature, objects do not have "heat". What does work out is describing heat as a energy transfer process between a system and its surroundings. Suppose some system does some work on its surroundings, absorbs heat from its surroundings, and changes its internal energy as a result. The change in the system's internal energy ([math]\Delta U[/math]) is related to the heat transferred into the system ([math]\Delta Q[/math]) and the work done by the system ([math]\Delta W[/math]) via [math]\Delta U = \Delta Q - \Delta W[/math] This is the first law of thermodynamics. The [math]\Delta Q[/math] is heat transfer. It is a process variable. Temperature and mass are state variables. You can take some object's temperature with a thermometer and measure its mass with a scale. A state variable depends only on an object's current state (hence the name). Suppose you have a pile of rocks with a mass of 10 kilograms. If you add 5 kilograms and then 10 kilograms of rocks to the pile the pile will now have a mass of 25 kilograms. The same mass results if you had reversed the order, or added 5 kilograms to the pile three separate times, or plopped all 15 kilograms at once. The order (process) in which you added by which you added mass doesn't matter; all that matters is the total mass added. The same does not apply to the first law of thermodynamics. The heat transfered into or out of a system depends on the path taken between states. That is why I said earlier that heat is a process variable. Your air conditioner and car engine use that different paths between states result in different heat transfers to their advantage.

What Sayonara was hinting at is that we already know a good deal about why the planets orbit about the Sun and rotate about their axes. At a minimum you need to understand Newtonian mechanics. If you want to add something new to the body of knowledge you need to understand relativity.

OTOH, Obama couldn't have bought better exhoneration with all the non-deleted expletives Blago that hurled at Obama regarding Senate Candidate #1.

The person standing on the surface of the Earth undergoes (nearly) uniform circular motion due to the rotation of the Earth. There is necessarily a net force directed toward the axis of rotation. For a person with a mass of 160 pounds (mass) located at 45 degrees latitude, this force is [math]F=mr\omega^2 = 72.57\,\textrm{kg}\,\times\,\cos(45^\circ)\,\times\,6367\,\textrm{km}\,\times\, (2\pi/\textrm{sidereal day})^2 = 1.74\,\textrm{newtons} = 0.39\,\textrm{lbf}[/math]

Insane_alien and Sisyphus have it right: third law reactions always take place between different objects. Some primary/secondary teachers mistakenly teach that the earth pulling down on you (gravity) and the earth pushing up on you (normal force) as an example of Newton's third law. This is wrong for several reasons. Third law reactions involve the same force acting on two different bodies in an equal-but-opposite manner. The Earth does pulls you down gravitationally; suppose for argument that this force has a magnitude of 160 pounds-force. The third law reaction: Your body is exerting a 160 pound gravitational force on the Earth. The Earth also does push up on you with a normal force. The third law reaction: Your feet exert an equal-but-opposite normal force on the surface of the Earth. Another reason that this high school example is wrong is that the normal force is not equal but opposite to the gravitational force. If you are standing at 45 degrees latitude, the normal force is neither equal to (159.7 pounds force, not 160) nor opposite to (179.9 degrees apart, not 180). There are some third law reactions going on here, but the earth pulling on you and pushing on you is not a third law reaction pair. No. Look at the above example. Suppose a person located at 45 degrees latitude measures his weight with on an accurate spring scale to be 159.7 pounds. The gravitational force on that person is 160 pounds (force). The vector sum of the normal and gravitational forces is a 0.39 pound force directed about 45 degrees north of down (i.e., toward the Earth's axis of rotation). The resulting motion is circular rotation about the Earth's axis with a period of one revolution per sidereal day. Note well: Neither the gravitational force nor normal force has a magnitude of 0.39 pounds (force).

Chase, To be brutally honest you have set your expectations far too high. Amateur groups have sent rockets into space (100 kilometers altitude), but at the cost of tens of thousands of dollars and *a lot* of experience in amateur rocketry. You should also beware that there are a lot of rules and regulations that pertain to high powered amateur rockets. Do not venture into high powered rocketry without learning about these rules and regulations. These aren't small regulations, pay a fine and you are done. Some of the penalties are felonies. Low powered model rocketry is more-or-less exempt from FAA and ATF regulations. I suggest you find a local model rocketry club.

A bit belated; welcome back!

Obviously, it

Repeating the same thing over and over does not make it correct, nor does it communicate. You are not talking about directions, you are talking about regular simplexes (simplices?). A 0-simplex is a point, a 1-simplex is a line segment, a 2-simplex is a triangle, and so on. These simplexes are the simplest geometrical shapes in N-space (hence the name).

I'll use the standard definition of the natural numbers, starting with [math]0 \in \mathbb N[/math]. The natural numbers are endowed with the operations '=','succ', '+', '<=', and '<' (there are others, I'll ignore them here). Equality is defined to be reflexive ([math]a=a[/math]), symmetric ([math]a=b\,\Rightarrow\,b=a[/math]), transitive ([math]a=b\,\textrm{and}\,b=c\,\Rightarrow\,a=c[/math]), and closed ([math]a \in \mathbb N\,\textrm{and}\, a=b\,\Rightarrow\,b \in \mathbb N[/math]). [*]The successor primitive recursively defines the natural numbers: [math]a \in \mathbb N \Rightarrow \textrm{succ}(a) \in \mathbb N[/math]. [*]A natural number and its successor are not equal to one another. [*]Addition is defined as the successive application of the successor function: [math]a+b \equiv \underbrace{\textrm{succ}(\cdots(\textrm{succ}(a)\cdots)}_{b\;\textrm{successors}}[/math]. [*]Less than or equal to is defined by given [math] a,b \in \mathbb N, a\le b \Rightarrow \exists c\in \mathbb N[/math] such that [math]a+c=b[/math]. [*]Less than is defined as [math]a<b \Rightarrow a\le b\,\textrm{and}\,a\ne b[/math]. Define [math]1=\textrm{succ}(0)[/math]. [math]b=\textrm{succ}(a)\,\Rightarrow\,a<b[/math]. Proof: [math]b=\textrm{succ}(a)\,\Rightarrow\,b=a+1\,\Rightarrow\,a\le b[/math]. Since a number and its successor are not equal to one another, [math]a<b[/math] by the definition of '<'. Since [math]1=\textrm{succ}(0), 0<1[/math].

Mathematical theorems start from a set of axioms (i.e., assumptions) and proceed to prove the theorem via a set of rules of deduction/induction. You have been introduced to one set of axioms if you have taken secondary school geometry: Euclid's postulates. There is not one set of axioms that constitute the basis of all mathematics. Geometry is a good example. Euclidean geometry postulates that given a plane containing a line and a point not on the line, exactly one line can be drawn through the point that does not intersect the line anywhere (Playfair's postulate). If you replace this axiom with something different you get a different geometry. Hyperbolic geometry says an infinite number of lines pass through the given point that do not intersect the line in question. Elliptical geometry says that all lines intersect somewhere.

A quick word of warning: Wikipedia is not a scientific source -- not to put Wikipedia down; neither is the Encyclopedia Britannica. There is a definition involved, which is to denote the cardinality of the power set of some set [math]\mathcal S[/math] as [math]2^{|\mathcal S|}[/math]. Thus the cardinality of the power set of the natural numbers is by definition [math]2^{\aleph_0}[/math]. The real numbers can be put into a one-to-one correspondence with the power set of the natural numbers. Thus [math]|\Re|=2^{\aleph_0}[/math] is a derived identity, not a definition. There are a countably infinite number of primes. In short, they can be indexed by the natural numbers.

The error in the original post ultimately lies with the implicit use of [math]\sqrt{(-1)^2}=-1[/math]. There is no need for imaginary numbers here. For example, using [math](-1)^4 = 1 = ((-1)^2)^2[/math], [math]\aligned 1 &= \sqrt{1} \\ &= \sqrt{(-1)^4} \\ &= \sqrt{(-1)^2}\sqrt{(-1)^2} \\ &= \sqrt{(-1)^2}\sqrt{1} \\ &= (-1)\times 1 \\ &= -1 \endaligned[/math] The square root function takes a branch cut at x=0. Proofs that -x=x involve an abuse of this branch cut. Don't do that!

You are implicitly assuming that multiplication of transfinite cardinals works exactly like multiplication of finite numbers. It doesn't.

[math]\left.\frac{x^2-4}{x-2}\right|_{x=2}[/math] and [math]\lim_{x\to 2}\frac{x^2-4}{x-2}[/math] are different expressions. The former expression is not defined while the latter is defined -- and is equal to 4. The value of a function at a point is equal to the limit of a function as the argument approaches the point in question are equal to one another if and only if the function is continuous at the point in question.

There are several ways to look at this problem: 1. As a combination of rotational and translational motion. In general, the motion of any object can be expressed as a combination of rotation about the center of mass plus translation of the center of mass. You haven't done that in this case. 2. As a purely rotational problem. If you choose the right representation you won't have any translational energy to worry about. 3. As a pure translation problem. The expression [math]1/2 I \omega^2[/math] is derived by looking at the contribution to kinetic energy by each infinitesimal mass element in a solid object. ======================= There is no reason to separate the velocity of some point on the rod into x and y components. All you care about is the square of the magnitude of the velocity vector. What is the speed at which some point on the rod is moving given that the rod has angular velocity [math]/omega[/math]?

You are attempting to apply the relation [math]a^ma^n=a^{m+n}[/math] to a situation where this relation does not apply: complex numbers. You are also abusing the square root symbol, which strictly applies only to non-negative reals.