Tom Mattson

Raider: "It doesn't have rest mass. It has energy, so it does have mass." No, real photons are massless. If they were not, then gauge invariance could not hold, as the vector potential A would appear as a physical field in Maxwell's equations. Radical Edward: "you're saying photons emit gravitons?" I presented the (non-quantum) general relativity, and did not say anything about gravitions. GR says that all nonzero energy/momentum densities affect the curvature of spacetime, and thus give rise to the illusion we call "gravity" (in GR, gravity is not a force, but a geometrical effect). I don't know much about quantum gravity, but I would imagine that gravitons would have to couple to photons for the theory to be accepted. This should be clear from the way that gravity affects light (eg: light cannot escape a black hole). Tom

Yes, the basic equations of GR are the following system of coupled, nonlinear differential equations: Ruv-(1/2)Rguv=8:pi: GTuv These are a system of equations in the metric tensor, guv, which contains the geometry of spacetime. Tuv is the energy-momentum tensor, and acts as a source term. Any nonzero energy-momentum density is implied, and this includes EM radiation. Explicitly, the EM energy-momentum tensor is: Tuv=(1/4:pi: )(FurFvr-(1/4)guvFrsFrs) The important thing to note is that it is nonzero. edit: Changed from Greek to Roman indices to make equation more readable. They still stand for Lorentz indices (0,1,2,3). Tom

He doesn't--he is just mistaken. See my previous post. All matter and radiation contributes to the curvature of spacetime. Tom

No, energy does indeed gravitate. If it did not, then mass and energy would not be equivalent! In GR, all mass and energy are entered into the Einstein tensor, which then determines the curvature of space time. Tom

As promised... http://xxx.lanl.gov/PS_cache/gr-qc/pdf/0212/0212076.pdf That should keep you busy for a while! Tom

double post deleted

It doesn't. The quandry is resolved in special relativity, in which the equivalence of mass and energy is deduced. The energy associated with the mass is entirely accounted for. For e+e- annihilation, the conservation law would be written: 2mec2+KEi=hv1+hv2 me=electron mass KEi=total initial kinetic energy v1=frequency of photon 1 v 2=frequency of photon 2 c=speed of light h=Planck's constant edit: subscript bracket Tom

I do not agree with the opening post, as every point seems to address the theorist, and not the theory. There is no need to examine the theorist's actions or career to determine if his work is pseudoscience. We recently had a discussion about pseudoscience at physicsforums, and I would like to share my posts from that thread. And then I gave an example: Hope that helps, Tom PS: Metaphysics is not pseudoscience, but a legitimate branch of philosophy. I am not sure of why it is included in the forum title. Also, I think you meant to name the Physics/Astrological Sciences department Physics/Astronomical Sciences. Astrology really is pseudoscience. My $0.02.

:slaphead: Clicking on "get more" sure helps. Thanks, Tom

Could someone please show me how to do stuff like this: Is there a website that contains all this information? Thanks, Tom

Hi, it's me Tom from Physics Forums. (I get around a lot ) The curvature of spacetime around a black hole demands a relativistic treatment. Since the Schrodinger equation is nonrelativistic, it cannot be used. You can, however, use relativistic quantum mechanics (RQM), which is the quantized Hamiltonian: H2=(pc)2+(mc2)2 Amazingly, it is quantized according to the same rules as the nonrelativistic case, namely: H-->i(hbar)d/dt p-->-i(hbar)grad RQM in curved spacetime has been worked out. I will come up with some references when I get home from work. Tom

Logic, by David Baum Fields, by Warren Siegel Quantum Electrodynamics, by Walter Greiner

No, it doesn't suggest a photon mass, it suggests that there is something about gravity which would cause no sign change in that particular internal quantum number. For instance, the mass of the anti-u quark is not the negative of the mass of the u-quark, even though all the other "charges" (electric, flavor, color") are conjugated under the action of 'C' operator. Of course, we know that there is something special about gravity, namely that it is a spin-2 field, as opposed to the other 3 forces, which are all spin-1. I'm just wondering if that is the cause of the difference. Unfortunately, it is not that I forgot the answer, it is that I never knew it. I never did learn anything about quantum gravity, although hopefully I will be able to fix that in the future. Tom

It is, but that's an external quantum number. The internal numbers are the ones that have to do with gauge symmetries, and thus interactions. Now that I think more about it, another problem pops up: Why isn't mass included in that list? That is, after all, the "gravitational charge". Perhaps it has to do with the fact that the gravitational field is not a rank-1 tensor, but rank-2...

The photon is its own antiparticle. Let |e,i> represent the state vector of some particle with 'e' standing for a collective index of the external quantum numbers (parity, energy, momentum, polarization, etc.), and 'i' standing for a collective index of the internal quantum numbers (charge, flavor, color, etc). Let C represent the charge conjugation operator that transforms particle states into antiparticle states and vice versa. An empirical fact is that particles differ from their antiparticles only by a difference in sign in the internal quantum numbers, so the operation: C|e,i>=|e,-i> transforms any particle into its antiparticle. However, for the photon, all quantum numbers represented by the collective index 'i' are zero. Tom

There were only 2 QM courses offered to undergrads at my school, too. Even so, beyond that, I took 2 grad courses in QM (Quantum Mechanics I from Physics and Quantum Chemistry from Chem). Aside from that, I took all the 'applied' courses I could get my hands on: 'Introductory Nucleonics', 'Applied Atomic and Nuclear Physics', 'Particles and Nuclei', etc. Even so, QM II and III kicked my butt (in a good way). Bare minimum to survive grad courses (beyond what you listed): 1 semester Linear Algebra (proving theorems, not matrix calculations) 1 semester Advanced Calculus (with Vector Analysis) 1 semester Complex Variables 1 semester Partial Differential Equations If I were sitting on an admissions committee, I would have serious doubts about passing anyone who did not have these basic prerequisites. Then why don't you plan on taking General Relativity as an undergrad? Forget that "Physics of Stellar Systems" stuff. I took that, and it's boring as hell. It won't tell you anything interesting about cosmology or relativity. What you need is a good course in GR, from Ohanian and Ruffini or some book at a similar level. Also, you need a math course in Tensor Analysis.

My $0.02? You should rethink this. You're thinking of majoring in science, which means you're going to have to go to grad school. I don't know anything about neuroscience, but I am certain that the plan you propose will not prepare you for graduate study in physics. For starters, it does not have nearly enough math. And only two semesters of quantum? Fuhgeddaboutit. I took 4 semesters of QM as an undergrad, and I was still breathing heavy in grad school. I think you should major in one science, and if you're really hot for both of them, minor in the other one. Tom

Nah, I'm not challenging you. It's just that all the threads here seem to go something like this: Person 1: Here's an interesting fact.... Person 2: Thanks, I didn't know that. Person 3: That's cool! Person 4: I think so too. Person 5: I'm confused. Talk about :zzz: I was just trying to squeeze something more out of this one. Yeah, the antiparticles are conspicuously absent from the chart. As for the other particles, I would not have included them anyway. I would have dispensed with all the hadrons altogether, cuz the title of the chart refers to 'fundamental particles'. This one I think we can let slide. The delta is just an excited state of the nucleons, which are included. It really isn't a seperate particle, any more than a hydrogen atom in, say, an n=2 state is a different atom from the ground state. Tom

Such as...?

Whatever gave you that idea? The momenta of photons and of particles are indeed of the same nature. This should come as no surprise, because momentum is a defined concept, and the photon momentum is defined in such a way as to make it usable in conservation of momentum problems. Look at the derivation of the Compton shift and you'll see what I'm talking about.

Actually, momentum is very much a property of light. If it weren't, you wouldn't have the Compton or photoelectric effects, for example. The relativistic energy-momentum relation is: E2=(pc)2+(mc2)2 So, even for m=0, we still have E=pc. Quantum mechanics tells us that the momentum of a photon (or of any wave) is p=h/(lambda)

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It happens because of the Bernoulli effect. The air moving past the curtain is faster inside the shower, thanks to the falling water. Bernoulli says that if the airspeed over one surface is greater, then the air pressure must decrease. Thus, the pressure on the outside of the curtain is greater than on the inside, and the curtain is pushed in. It's the exact same reason that airplanes fly. Tom