Tom Mattson

I'm not exactly sure of what it is you're having a problem with. Are you trying to justify why we use those two solutions for a repeated root? If so then the simple answer is that we do it because it works. It's just an ansatz that happens to get the job done.

Are you talking about breaking up the time dependence of operators and state vectors according to Dirac's so-called "interaction pitcure"?

You can put both the k and the x in the exponent by typing e^{kx} in math tags. That should be [math]\sum_{i=0}^{n}A_ik^i[/math]. This doesn't make sense, because k shouldn't depend on x. Don't you mean [math]y=e^{kx}[/math] and [math]y=xe^{kx}[/math]?

Not unless Dirac has some other version of QM of which I am not aware. Dirac's quantum mechanics is relativistic and specifically it describes spin-1/2 particles. And actually, you've left out Bohm's quantum mechanics. From what I've read it describes nonrelativistic quantum phenomena just as well as Schrodinger's and Heisenberg's. For nonrelativistic applications I've only ever used Schrodinger's QM, so Heisenberg and Bohm are off the list for me. For relativistic applications we don't get to choose between Dirac and the various nonrelativistic applications. Rather, we choose between quantum mechanics and quantum field theory. I've only ever used relativistic QM when required to by a professor in a homework problem. When doing particle physics, I apply QFT instead. I haven't studied the other nonrelativistic versions of QM in any detail, so dunno.

Nope. [math]v^{\prime}[/math] does not appear in the right hand side of the formula for integration by parts.

Say what? The [math]I[/math] term does not cancel out. And you don't get [math]I_{n-1}[/math] after integrating by parts once. That's because you get a cosine (not a sine) in the integrand after doing it once. You have to integrate by parts twice to get back to the sine function.

You generate systems of linear equations in resistive networks by doing mesh current analysis (as you have indicated above) and/or node voltage analysis. Here's another nifty application. When dealing with sinusoidal inputs, dealing with the trigonometric functions can be a pain, especially when sines and cosines start cropping up in the same expression. To make life easier, electrical engineers often switch from the sine/cosine functions to complex exponentials. In linear algebraic terms, the set [math]\{\sin(kx),\cos(kx)\}[/math] spans the same vector space as the set [math]\{\exp(ikx),\exp(-ikx)\}[/math]. The switch from one set to the other is a change of basis.

You don't even need the full blown Sylow theorems. You just need Lagrange's theorem.

The statement is true whenever the magnitude of a vector changes, but not its direction.

The statement is false. I can give you a simple counterexample from physics. Consider a mass on a rope that is swung in a circle at a constant speed. The velocity vector [math]\vec{v}[/math] constantly changes, so [math]\frac{d\vec{v}}{dt}\neq 0[/math], but the speed is constant so [math]\frac{d|\vec{v}|}{dt}=0[/math]

That can't be the complete problem statement. Where do A and a come from? Do they both belong to the commutative ring? And are you sure that aA=A for all a not equal to zero? Seems to me it would make a lot more sense if it were for all A not equal to zero. The way you've worded it we get the following. aA=A aAA-1=AA-1 a1=1 a=1 Mind you, that's a=1 for all a not equal to zero. That doesn't make much sense to me.

I used to teach for TPR, and I've been thoroughly indoctrinated in the "Joe Bloggs Method".

A former Princeton Review student I take it?

Pi is less than 4, so it's definitely not infinitely large. The problem is not its magnitude, but rather the fact that it's irrational.

Why does the fact that it is multi-valued make it not well-defined? The relation f is basically a set of points, and for any point P it is certainly the case that P is either in the set or not in the set. Also, you could turn it into a single-valued relation by a change of coordinates that rotates the graph by 90 degrees (so you have y=x^2). Does it make sense to say that a coordinate transformation can turn an ill-defined relation into a well-defined one?

My comment is that the whole thing is a sloppily tossed word salad. SR is certainly compatible with an infinitely long past, but it doesn't suggest it. And GR seems to suggest that the universe is not infinitely old. You could keep winding the time coordinate back, but eventually you would run into a singularity. And even if the universe were infinitely old, that would not mean that present, future, and past exist simultaneously. If that were the case then all events would happen at once, which is absurd. What relativity suggests is that one's notion of time is the result of changes in the physical state of the universe, as he observes them. If it weren't for bodies in motion, there wouldn't be any such thing as time. Time exists only as a relation between moving objects. There's nothing "logical" about any of this, as there are no valid deductions from premises to conclusion. The author of this piece doesn't seem to know the meaning of the words "logic", "force", "energy", or "metaphysical". Say what? Dualism is the metaphysical school of thought that there exist two fundamental substances: the mental and the physical. Nowhere in parts 1 through 3 did the author ever show that he understood the physical, and me didn't even mention the mental.

90% accuracy? Err...As compared to what?

Well if you want to get into it, then may I ask: Where did you leave off with mathematics? I am sure I have some free online resources that can get you up to speed.

Since you mention field theory' date=' let me offer a suggestion. Let's pretend the photon [b']does[/b] have a mass [imath]\mu[/imath]. In that case we can write down the Klein-Gordon equation to describe the dynamics of this massive spin-1 field as follows. [math](\nabla^2-m^2)\phi=-\frac{\partial^2\phi}{\partial\phi^2}[/math] If we look at the static limit then the time derivative goes to zero, and we have the following. [math](\nabla^2-m^2)\phi=0[/math] For a spherically symmetric point charge we have the following solution. [math]\phi=C\frac{e^{-kmr}}{r}[/math] I don't remember the exact combination of constants in the exponent, so I put the "k" in there to absorb any missing factors. It may well be that k=1. Anyway that's not the important part. What is important is that the above is the classical potential for an interaction mediated by a massive spin-1 field. If m=0 then we recover the old familiar Coulomb potential. So you could take the negative gradient of this potential and derive what the electrostatic force would be if the photon were massive. Then you could in principle perform sensitive tests on violations of the inverse square force law to try and find the photon mass.

I like Halzen and Martin, but I think that the book by Griffiths entitled Introduction to Elementary Particles sits even one step closer to undergraduate level QM than Halzen and Martin does.

OK, now I'm confused. I was taught that the definition of a group states the existence of the identity element, and that a theorem establishes the uniqueness of the identity element. Is that not the mainstream view? If not, where am I going wrong?

The motivation for applying Laplace transforms to differential equations is to turn them into algebraic equations, because the latter are easier to solve. But if you try to take the Laplace transform of the two sides of this differential equation you are going to create more work for yourself, not less. That's because the equation is nonlinear. Fortunately it is separable, so you can straightforwardly solve it by integration.