Tom Mattson

Good luck!

I don't understand why people have such an urge to identify mathematical objects with physical objects, or to try to make mathematical arguments by analogy with physical arguments. It simply does not work. Why is it so difficult to accept that mathematical objects are their definitions?

Sometimes I do.

Can you divide by duck? Can you divide by table? Can you divide by styrofoam? Can you divide by blue? Can you dividy by excited? Of course not. And do you know why? Because none of those things are real numbers. And neither is "undefined".

In addition to that, you might want to keep looking at David Bachman's website. He is the author of the book on differential forms (downloadable) that I am basing my thread on forms at Physics Forums (I promise, I really will get back to that thread this week!). It's a very readable book and manifolds are covered at the end. Bachman has promised that symplectic forms will be treated in the next edition. Keep watching!

Assuming your arithmetic is correct: Yes. In an expeirment of this type you must correct for the flight time of the light from the clock face. Of course if you wanted to know at what time the flash occured, you would have to make this correction even if Galilean Relativity held. But Special Relativity predicts that even after this correction is made, there will still be a discrepancy, while Galilean Relativity predicts that there will be none.

No, it isn't. Time dilation has been observed with atomic clocks and with subatomic particle decays. It is so well-verified and well-understood that particle physicists actually use it on a practical level. If a charged particle is very short-lived then expeirmentalists can accelerate a beam of them to very high speeds so that they live longer in the lab frame, and hence they can be better studied.

I said a lot of things in that thread' date=' and I can't possibly hope to recall all of it without re-reading it, but the gist of what I was saying is that there is no definition of vector division [i']as a multiplicative inverse[/i], with reference to the already-defined multiplicative operations with vectors. Sure you can give meaning to the symbols S=A/B, and you can even call it a quotient if you want to. But if we want to interpret division by B as multiplication by B-1, then what is B-1? And what is the multiplicative operation? Is it? I didn't look at the whole thing, but where exactly does he do anything more than simply write S=A/B? You did that much when talking about Newton's 2nd law (m=F/a).

You forgot to put -A into the sine function. So the second factor on the RHS should be: ( cos A - isinA)

About this whole "wave-particle duality": It doesn't exist. That expression is just a way to attempt to make quantum theory understandable to laymen, in terms of familiar concepts. If you look at QM or QFT you do not see a dual treatment of either particles or photons. You see a single, unified treatment of them.

I would not start with any of the things you mentioned. I would start from simple kinematics. It would begin with the general expression for the derivative of a vector. If A is some (time-dependent) vector in some reference frame that has angular velocity w then dA/dt is given by: dA/dt=(dA/dt)rot+wxA. Where: (dA/dt)rot is the derivative of the vector as observed in the rotating frame.

Every presentation of group theory I have seen defines a group <G,*> as a set G with some binary operation * defined on it. And every presentation contains the following 3 group axioms: 1. Associativity must hold. 2. There must exist an identity element. 3. There must exist an inverse under * for each element in G. The only possible source of conflict I can imagine is that some references list a fourth axiom: 4. G must be closed under *, whereas other references include that as part of the definition of a binary operation on a set, rather than as part of the definition of a group itself.

Are you sure about that? An isomorphism is a homomorphism that is both one-to-one and onto. But there is a two-to-one map of SU(2) onto SO(3). For that reason Sakurai (Modern Quantum Mechanics) says that SU(2) and SO(3) are only locally isomorphic.

That's true but v is not squared in the other factor in the Lorentz transformation.

I know that the atom can deexcite one level at a time because transitions from state n to state n-1 are allowed transitions. So there is the possibility of it occuring that way. Photons that have energies that are equal to that of any allowed transitions from a lower level to a higher level can be absorbed. If a photon does not have an energy corresponding to an allowed transition then it will not be absorbed. The energies quoted in my post are unique to the hydrogen atom. The energy spectrum of each atomic species is different, and in principle it can be determined from quantum mechanics. I say "in principle" because only for hydrogenlike atoms does QM admit an exact solution. For more complex atoms one must invoke approximations. Energy spectra for the most complicated atoms are determined empirically.

We aren't talking about a spinning frame, we're talking about a rotated frame. The angular velocity of the rotated frame is zero.

But the state of the universe is not what is meant by "the physics" of the universe. That term refers to "the laws of physics", which determine the way in which physical states evolve in time. Obviously a rotation of a reference frame results in a different description of a physical state from the point of view of that frame. The point is that, as long as there is rotational invariance, the physics is unchanged after rotating the physical system by an angle A, and that this is no different than rotating the axes by an angle -A. But none of this is about SO(3), which is what we were talking about. To see that rotating the axes by an angle A is the same as rotating the vector by an angle -A all you have to do is show that the rotation matrices are identical (and they are).

What physics? We're talking math here. Anyway your statement that the physics will be different is not true in general. Any closed system (that is any system that is not influenced by work, forces, or moments from external sources) is rotationally invariant, and hence rotating the system has no effect on the physics of that system. For these systems there is no difference between rotating the physical apparatus and rotating the axes we refer it to. This symmetry of rotational invariance is in fact what gives rise to conservation of angular momentum. If your statement were true in general, and the physics of a system really did always depend on the angular orientation of that system, angular momentum would never be conserved.

No' date=' it isn't. First, that does not answer the question. Obnoxious asked if it is possible to divide by infinity, not if it is possible to divide infinity by some number. Second, an infinite quantity divided by another infinite quantity is indeterminate. Consider the following function: f(x;n)=(xn+1)/(x2+1) Then take the limit as x approaches infinity for n=1,2,3. You will get 3 different answers.

Do we have an ETA on when LaTeX is coming back? There are some things I've been holding off on posting until it returns.

Whoops, I sure did change the problem from one post to the next, didn't I? The matrices I cited are in fact for rotating the axes counterclockwise by an angle A. If the angle is negative it means that you are rotating clockwise. Note that a counterclockwise rotation of the axes will give you the same answer as a clockwise rotation of the vector. Gotta go.