Tom Mattson

1. Yes, zero is a real number. 2. No, infinity is not a real number. 3. No, infinity is not the reciprocal of zero. Zero has no multiplicative inverse.

Start by looking at a vector v=<x' date='y> in the xy-plane. Then rotate the axes by an angle A and see what the new components are. You should get: [b']v[/b]'=<xcos(A)+ysin(A),-xsin(A)+ycos(A)> You should be able to write the RHS of the above equation as a matrix times the original vector v. That matrix is: [cos(A) sin(A)] [-sin(A) cos(A)] Now consider that what you have really done here is rotate the coordinate system about the z-axis. Recognizing that if our vector v had a z-component so that v=<x,y,z>, the z-component would clearly be unchanged. So we have: [cos(A) sin(A) 0] R[sub]z[/sub](A)=[-sin(A) cos(A) 0] [0 0 1] A similar analysis can be used to obtain Rx(A) and Ry(A).

Not likely, since rotations don't include translations, which also figure prominently in classical mechanics.

Mathematically they are special only in the sense that they are a special case of O(3). One could just as well regard as "special" the subgroup of O(3) whose determinants are -1.

Here's a handy fact if you are interested in generating Pythagorean triples. Consider the Fibonacci sequence: 1,1,2,3,5,8,... (generate the next term by adding the two previous terms) You can generate a Pythagorean triple by using the following algorithm: 1. Choose any 4 successive terms in the sequence. 2. Multiply the 1st and 4th. 3. Double the product of the 2nd and 3rd. 4. Add the squares of the 2nd and 3rd. The results of steps 2,3, and 4 always form a Pythagorean triple.

Yes. Orthogonal matricies have the property that RTR=RRT=I, where T stands for transpose and I stands for the identiy matrix.

It's a subgroup of O(3), the set of 3x3 orthogonal matrices under matrix multiplication. Elements of O(3) can have a determinant of either +1 or -1. The "S" in SO(3) stands for "Special", and it is indicative of the fact that elements of SO(3) are those elements of O(3) which have a determinant of +1 only.

You would write down the Hamiltonian for all the atoms in the body. Now for the real question: Why on Earth would you want to?

Why don't you look it up yourself? It's not that hard to find. Is your objective here to learn something or is it to test everyone's patience? What you are being asked to do is to learn the statement of the theorem so that you stop with these ridiculous strawman arguments (and if what you're doing isn't childish, then I don't know what is). The theorem does not say that every nth degree polynomial will have n points of intersection with the x-axis. That is the result of you forcing the word "distinct" in between "n" and "roots" in the following statement: Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers. Simply put: You are arguing against a statement that is not equivalent to the statement of the Fundamental Theorem of Algebra. Since the thing you are arguing against is a non-theorem, it's no surprise that you are able to topple it so easily. No one is impressed.

No' date=' but only because I didn't think that you were quite this daft. I stand corrected. You don't say.

You don't even need a proof, all you need to do is come up with a single counterexample. Can't wait to see it.

On Take One you got 1b and 1f right. You're looking good now. But more importantly, do you understand the reasoning process?

Their utility, aside from their natural, obvious application to the harmonic oscillator in quantum mechanics, is that they are used (in part) to describe quantized fields.

Not so hot. Let's take a look... As long as that region is more than a single point, then this is true. In fact this would be true if V were any constant, because d(constant)/dx=0. So you got this one right. Think about this. If the potential V(x) is zero at some point then that just means it has an x-intercept at that point. Is it necessarily the case that the function is neither increasing nor decreasing there? No, of course not. Look at any diagram of electric field lines. The lines always go from positive charges towards negative charges. What does that tell you? No matter what V(x) is, you can always modify it by adding a constant c. This is because V(x) and V(x)+c give the same electric field. Since it is the field that is measureable, that is all we care about. So the two potentials above are physically equivalent. And c can always be chosen to be -V(x0) for an arbitrary x=x0. What does that tell you? You got this one right. Capacitance depends on geometry only. Not on q or V.

Your questions are difficult to answer what with LaTeX being broken. I am going to refer you to the following page of a free online relativity textbook. It has the Lorentz transformation for accelerated motion. Chewing on this should keep you busy for a while! http://www.geocities.com/zcphysicsms/chap5.htm#BM5_4 No problemo!

There's really no reason to complicate things by going into GR. SR is perfectly capable of handling accelerated motion.

Right. And to do that he had to apply a force to do the work necessary to change his velocity. In other words, he expended energy, just as one would expect. So this should answer your orignial question: Movement does not require energy. Changing one's state of motion does.

The first 2 pages of the following document explain it succinctly. http://www.cithep.caltech.edu/~fcp/physics/quantumMechanics/secondQuantization/SecondQuantization.pdf

Commuting operators are called 'bosonic' because the creation and annihilation operators for bosons must commute. The reason for that is that in experiments it is observed that any number of bosons can occupy the same quantum state, and so the eigenvalues of the bosonic number operator must be allowed to take on any whole number value. This is achieved by using commutating operators to represent the creation and destruction of bosonic particles. Anticommuting operators are called 'fermionic' because their creation and annihilation operators must anticommute. Experimentally it is known that at most one fermion can occupy a quantum state (Pauli Exclusion Principle). So the number operator for fermions can only have eigenvalues of 0 or 1, which is achieved with anticommuting operators.

I'll try to get back to the thread with that. You can see my username, and you can safely assume that that is what I would like to be known as.