revprez

The square of the timelike component of a massive particle's motion in any physically known field is always larger that the sum of the squares of the spatial components. Rev Prez

Correct me if I'm wrong, but are you and Tom proposing that measurement itself is a map from the complex numbers to the reals?

Yeah, my bad. Should be |X + iY| Isn't the distance preserved in a transformation between [x y] and [x iy]? Right, got that. And the total energy has to be real anyway. I'll get back to you on that. Rev Prez

But it does equal <x,y>ei(theta). That's just Euler's formula. Well, it doesn't have to, but of X and Y are parameterized by some t then it should, right? Can't the Hamiltonian decompose into complex kinetic and potential energies, and in fact isn't that what we'd expect in a quantum harmonic oscillator? Rev Prez

Yeah, for a Hermitian operator A+ is just AT. So X + iY or (<x,y>)eiwt is some rotation operator, correct? Rev Prez

Just to add on Tom's explanation, a Hermitian operator is its own complex conjugate.

All right. So what's an example of a complex-valued observable. The Hamiltonian obviously isn't. Yeah, sucks doesn't it? What are they using as their converter? Texvc? Rev Prez

Can you just briefly explain the notation. I'm assuming you're not saying AA+ = A+A = 0. BTW, is LaTeX working for you?

Wait, I'm confused. I thought the definition of a normal operator is one that commutes with its Hermitian adjoint. Rev Prez

Just curious. I'm working through mostly lecture notes and some homework problems. BTW, Latex isn't working for me. It seems to be doing fine for everybody else, but I'm getting "[LaTeX Error: couldn't open URL]" on all machines. Anybody got a clue as to what's going on? Rev Prez

Wow, you're not H=16, are you? Something tells me you were wrong a lot more times than that. Rev Prez

I'm curious, what math and physics materials are you working through?

I have to agree with you. I think that's the way to go about it.

And just for clarity's sake, the [math]\left(\frac{2M}{r}\right)[/math] term is chosen with units such that G and c = 1.

Wait, let me try this out. I'll get back to you.

Photons mediate the electromagnetic force.

The one problem I have is I can't prove that all positive mass-energy distributions can be represented as a diagonal matrix of the form [math]diag(\rho, p...)[/math]. When I come back later I think I'll kick this over to the appropriate math forum as well. Maybe we can get some feedback from formally trained folk. Rev Prez

Not generally. The range is usually just a subset of the codomain. We say that f:A->B where B is the codomain (i.e., reals, complex, some ordered set), and the range are the values in B allowable under the transformation. For something like f = x for x in A, then the codomain and the range are clearly the same thing (in fact, its clear that the codomain and the domain are the same as well). Rev Prez

It looked good to me. Only thing I can add is a clarification. The Euclidean 3-space is described by the 3x3 spatial subset of the 4x4 metric (just take the lower right subset of the 4x4). But I think its clear to anyone who's familiar with the Minkowskian from special relativity ([math]diag(\eta_{\mu\nu}) = (-1,1,1,1)[/math]). Beyond that, I think you're more comfortable with the math and phyiscs--I'll defer to someone more knowledgable. Rev Prez

I don't think so, not without writing some lecture notes on linear algebra. You might want to check out Wikipedia. The best I can say in a short amount of time is that if you consider a linear function that takes a column vector and spits out a scalar, it should key in your mind that this function is a row vector matrix multiplied on the argument. Visa versa, too. You can think of contravariant and covariant tensors as generalized linear functions on each other. Does that help? Look at your above reply to h=16, I'm guessing you've got some sense of what a linear map/function/operator is. Rev Prez

I don't know what the break down is, but let's just say that MIT, Wikipedia and Mathworld like lower indices, and Hofstra likes upper indices. You can use either convention, so long as you're consistent. After all, a covariant tensor is just a linear operator on a contravariant one and visa versa; so long as you distinguish between the two in notation it doesn't matter which gets the upper or lower indice. Either way, just look at what you wrote down. Does it seem right to you that the linear combination of two covariant tensors should equal some scalar multiple of a contravariant one? Rev Prez

Careful here. If you're going to use the upper indice, alpha beta convention, then carry it through to the other side as well. In this case, [math] G^{\alpha\beta} = R^{\alpha\beta} - \frac{1}{2} g^{\alpha\beta} R [/math], and the Riemann tensor should be [math]R_{\lambda}{}^{\alpha\lambda\beta}[/math].

And energy density is what? Force * distance / volume. The components [math]U_{\mu}U_{\nu}[/math] (or [math]U^{\alpha}U^\beta}[/math] if you like that convention) also give you units of [math]\left(\frac{m}{s}\right)^2[/math].