Severian

I am still not following. So where do the divergences actually go then? The lattice should just give a regularisation of the divergences (just like putting in a momentum cut-off) but unless you cancel them with something they will just come back again when you take the continuum limit. Since the big problem with quantum gravity is that it is non-remormalizable (ie that there is nothing to cancel the terms once they are regularized) I don't see how this works. Presumably there must be some other feature of the theory itself which restores renormalizability, but then why not regularize in the usual way?

Sometimes I just want to curl up and die....

YT: I am like that when Christmas shopping. Sometimes the shops are just so busy that I have to run out of the store just to get away from the people.

You guys have been playing too many video games and watching too many rambo movies. Keep out of their way and call the police. My TV's insured but my life isn't...

Callipygous: that is a lot more work than doing it by hand. Since you know 11*d+7 <=750 (with d an integer) you must have d<=67, so you start with 67 and see if (11*d+7-5) is a multiple of 7. The second number you try (d=66) gives the right answer....

I don't understand what you mean by a 'traingulation' in this context. What is the traingulation supposed to represent? I understand Feynman's path integrals (I think I may have to teach this next year) but your triangulation have me confused. Are the triangles somehow supposed to be a discretization of a 2D space time? i.e. building a 2D surface out of little triangles? If so, wouldn't this be 'cheating' in that one would be regulating the ultraviolet divergence by introducting a finite triangulation size? I suppose approaching the continuum limit would bring back the divergence in the usual way (since the whole problem with Quantum GR is that it is non-renormalizable) - are they advocating that the finite triangulation is real, thus regulating the divergence automatically?

This is included in QFT (but not QM) and has nothing to do with gravity (per se). In fact' date=' this is one of the major differences between QM and QFT. A (virtual) photon for example can turn into an electron positron pair [math']\gamma^* \to e^+e^-[/math], which is a manifestation of QED. Although the formulation and evolution of wave equations in QM/QFT is local (ie. to work out what it looks like at time t+dt, you only need to know what it looked like at time t within (c dt) of the point of interest) the collapse of the wavefunction into an eigenstate isn't. To be honest, I don't think this collapse is satisfactoraly described in QM/QFT: it is simply axiomatic in the theory, which I don't like. Anyway, this is intrinsically non-local because a measurement at one point affects the wavefunction instantaneously everywhere, but does not violate relativity since no information is passed at speeds > c.

Lol - only the second sentence is partially true. Sentence 1: QM decribes only a one-particle state. The extension to many particles is Quantum Field Theory. Sentence 2: QM or QFT describes the physics at very small distances, while relativity describes physics at very high speeds. Since very small objects can move very fast, you are correct that they are connected. This is done in relativistic quantum field theory. The problem is how to combine general relativity (ie. gravity) with QFT. Sentence 3. Sub-atomic particles do not transmit information instantaneously. As you point out this would break relativity. Sentences 4 and 5 fall over with Sentence 3...

The light never goes faster than c - only the position of the intersection of the light with the clouds. If the light is at an angle [math]\theta[/math] to the ground and the cloud cover is say a distance d above' date=' then the horizontal distance from the point directly above the source is [math']s=d \cot \theta[/math] and the speed with which this moves is [math]\dot s = -\dot \theta d/\sin^2 \theta \approx -\dot \theta d/\theta^2[/math] where [math] \dot \theta[/math] is the rate at which the angle changes, and the approximation is valid when [math]\theta[/math] becomes small. So if we reduce the angle at a constant rate, when we get below an (approximate) angle [math]\sqrt{-\dot \theta d/c}[/math] the end of the beam will be moving faster than c. (Remember [math]\dot \theta<0[/math].)

Has anyone played Rome:Total War? The campaign map is strategy, while the battle map is tactical.

The first 2 should be answered by the LHC in a few years. The last one is a bit more tricky but there are steps to at least rule out certain GUT models at the LHC.

In chronological order: 1. The discovery of the Higgs boson (or other method of Electroweak symmetry breaking) 2. The discovery of Supersymmetry (or other solution to the hierarchy problem) 3. The formulation and test of a consistant GUT model (E6?)

The classic example is a spotlight moving over low clouds. Lowering the spotlight beam, as the angle of the beam to the plane of the clouds becomes small, the beam moving over them becomes arbitrarily fast. Faster than c! Bit since the end of the spotlight is not an object as such, there is no information flow and no causility problem.

I don't know about other subjects but to do any physics degree you will need at least AS maths. However, admissions may be slightly lower in some Universities. At Glasgow you could do it only with a maths AS: From http://www.gla.ac.uk:443/studying/ug/prospectus2005/entryfaculty.cfm?faculty=7

Yes - you are right, not your friend. The impulse which your friend gives the rod will travel down the length of the rod at a speed <c for pretty much the reasons you say. This shows by the way that there is no such thing as a 'rigid body'.

I know Mike Peskin very well; he is an excellent physicist and I respect his opinions. But there are thousands of particle physics papers published each year, so the 'top 50' is not a good respresentation either. Quantity is not always a good thing.

You have to be very careful with the Spires citation rankings since they count self citations (so someone who writes a lot of papers on one area will have more than someone who writes on lots of different areas) and there is no distinction between 'This paper relies heavily on the excellent work of Ref.[4]' and 'This paper demonstrates that the assumptions made in Ref.[5] are not valid.'.

You might find this thread useful.

In particle physics, matter is usually defined as the fundamental representation of symmetry groups. So an electron is matter (since it is a fundamental representation of the U(1) symmetry) but the photon is not (since it is an adjoint representation). In practice this means everything is matter except for the particles which transmit the forces (photon, gluon, etc). The definition of mass given my MolecularMan14 is 'gravitational mass' (eg in Newtonian gravity it is the m's in F=Gm1m2/r2). 'Inertial mass' is a quantity which governs how the body accelerates when a force is applied (so in Newtonian mechanics it is the m in F=ma). In 1900 or so Eotvos showed that they were the same (to the accuracy he measured) and Einstein used this idea to construct general relativity.

I am sure that is what he meant. A bosonic particle can never have negative mass; in the Lagrangian the mass term is proportional to mass squared, so if it were negative the Lagrangian would be identical to if it were positive and you would just redefine it to be positive.

Sorry, I made a brain fart there. I didn't really mean open and closed strings when refering to heterotic. I meant whether the theory treats left and right handed states the same.

Your article does not mention gravity. I repeat, the Higgs boson has nothing to do with gravity. The Higgs mecahnism is simply a mechanism for breaking the electroweak symmetry. As the article points out, breaking this electroweak symmetry with the Higgs mechanism, causes the fundamental particles to aquire mass (proportional to their coupling with the Higgs boson). But this is not relevant to gravity for two reasons: 1. Gravity does not care about mass. Mass is only in Newton's theory because the slow moving particle he observed have most of their energy as mass. Massless particles are still affected by gravity. 2. The masses provided by the Higgs boson are only for the fundamental particles. A large proportion of the masses of particle bound states (such as the proton) comes from QCD. So even without the Higgs mechanism, there would still be mass; just not at a fundamental level.

Instead of particles being points, they art extended objects like strings (hence the name). It was originally hoped that the different vibrational modes of the strings would correspond to different particles, so that, for example, the muon and an electron might be described by the same string vibrating at different frequencies. This didn't quite work, since the excited modes tend to give very very big masses to the particles. There were also problems with describing fermions as strings, until it was realised that one could incorporate them by including "supersymmetry". The result was supersting theory. One of the most interesting aspects is the difference between open and closed strings. Open strings are when the string just stops (like a small piece of string). Closed strings are when the string forms a closed loop. Whether they are open or closed leads to different properties. Also some string theories treat left handed polarisations differently from right handed ones, and are called "heterotic". Hence the fashionable modern string theory is "heterotic string theory".

The Higgs boson has nothing to do with gravity.

That's a tachyon.