JonathanApps

Cheers. That makes some sense Second question: In the NG action we vary the coordinates of the string in the embedding space X^{\mu} to find the extremum of the action. That's fine for spatial components but I don't get how we can vary X^{0} (the time component). Explicitly this is \delta t(\tau, \sigma) where \tau and \sigma are timelike and spacelike coorinates in the 2D string repectively. In the static gauge, at least, \tau = t, everywhere on the string so how on earth can we VARY the function t(\tau)? It doesn't make sense. Even if t and \tau aren't the same (some other gauge), I can't see that \delta X^{0} is physical or necessary. At most it would indicate a reparameterisation of \tau and \sigma....? No Latex here - I've temporarily forgotten how to get it. Apologies, and promise to get it working in the future. Cheers.

I was wondering if there are any string theory experts on here. I'm studying the subject from a textbook (Barton Zweibach) and am likely to have a lot of questions. If there is anyone, I would be eternally grateful for any help they can give from time to time. First up: The action for a relativistic string (Nambu-Goto action) is the (2D) area of the string. This contrasts with the space-time action in general relativity, which is the volume integral of the Ricci scalar. Intuitively, I would have thought that the two actions should be the same. I know they're different objects, but why the curvature integral for one and just the area / volume for the other? Cheers, Jonathan

What's k? You have K1 and K2 in the original question? Do we include the effect of gravity?

or indeed anything beyond simple undergraduate physics. Was this his only gaffe? I suspect there are others.

There's something about it in The Gunslinger by Stephen King as well.

I'm not sure what it has to do with Hooke's Law though. That is about a mass on the end of a spring: T = l x constant. This is a more complicated situation, although in the example elfmotat gave above, I guess you're applying a version of Hooke's law to the infinitesimal string elements.

It's a bit fiddly to derive: the wave equation on the string is [math] \frac{\partial^{2}\psi}{\partial t^{2}} = \frac{T}{\mu}\frac{\partial^{2}\psi}{\partial x^{2}} [/math] where [math]\mu[/math] is the mass density and [math]\psi(x,t)[/math] is the displacement, either along the string or perpendicular to it, depending on what kind of wave it is (longitudinal or transverse). You might be able to work out the speed from that

Yes; it's just that |i> and |f> seem unrealistic when the QFT courses decide they're going to be definite momentum.

Ah, thanks. You had me very worried for half an hour. I'll look at the LSZ stuff. It seems basic QFT courses don't go anywhere near justifying their claims in this area. Not very impressive. Thanks for all your help.

OK, sure, but not in the Interaction Picture, surely? The field equations are the same; it's the quantum state that's time-dependent.

Those 2 lines look the same - k.x is a 4D scalar product, right? I thought in the IP the time-dependence of the field operators was only due to the non-interacting Hamiltonian, as for no interaction. So the time-dependence of the "a"s should be unchanged....? Are you in the Interaction Picture here?

Cheers. But so then the commutation relations [math] [a^{\dagger}_{p},a_{q}]=\delta(p-q) [/math] won't be true for all time, and then we mess up the field theory, don't we? I've heard of switching H_{I} on gradually, and then off again, by multiplying it with something f(t) that starts at 0, goes to 1 for a bit, then ends at 0, but again I don't see how you can do this without messing things up. We would screw up the formula for the S matrix in terms of H_{I} because it would now be f(t)H_{I}...... PS perhaps the answer to the above is in the LSZ link. It looks good - I'll have a look at it. Cheers.

Thanks; that makes some sense. But the interaction should typically take place over a time of about Lm/(p-q), where L is the length of the box and m is electron mass (there might be a gamma factor in there!) We do the integration from t = -infinity to infinity. That seems to be giving the particles a lot more opportunity to interact than they have.....?

I don't think anything is _causing_ it, right at the moment. If you throw a ball up in the air, gravity isn't causing the upward movement. It is slowing the ball down over time of course, then eventually reversing it. What we have here is space-time expanding. Think of it as a rubber sheet that all the galaxies / stars etc. are sitting on. They aren't moving over the sheet, but the sheet is increasing in the amount it's stretched. The initial momentum for the expansion I guess comes from the Big Bang but I'm not sure. The gravitational force between galaxies should slow down the expansion over time, much as gravity slows down the ball that got thrown upward. I think current observaions show that the expansion is ACCELERATING, i.e. increasing, which is confusing. This could be due to an extra term in the equations for the way space-time behaves, called the Cosmological Constant, which can drive either acceleration or deceleration depending on whether it's positive or negative. I'm not an expert.

[math] a^{\dagger}_{p}b^{\dagger}_{q}|0> [/math]

In Quantum Field Theory we calculate the matrix elements which give the probability that an initial state |i> will turn into a final state |f>, e.g. |i> could be an electron and a positron, and |f> could be two photons. I don't really understand why this works the way it's done. The interaction process is meant to be a collision: the initial particles start off far away from each other so they don't interact, then come together briefly and react to produce the final result; the resultant particles move apart again. It's a scattering process. So why do we look at the matrix elements for states which have definite momentum? These are spread all over space so there's no "collision" as it were: they always have the potential to interact, for all t. A typical inital state that gets considered is something like a^{\dagger}_p b^{\dagger}_q |0>, i.e. 2 particles with definite momenta p and q. What we surely need is packet states that represent particles with a reasonably well-defined position, moving towards each other. These would have a range of momenta. I can't understand why the definite momentum states work, and correctly predict experimental results in real particle accelerators. Sorry I'll try latex: [math] a^{\dagger}_{p}b^{\dagger}_{q}|0> [/math] What I tried was "[math] a^{\dagger}_{p}b^{\dagger}_{q}|0> [\math]" Didn't work.

Would we expect (approximately) the same radiation from say a charge in free fall at the Earth's surface and a charge accelerating at g in empty space? To me they seem totally different phenomena. For the first, the charge isn't really accelerating and an observer sitting on the particle will see a Minkowski metric with zero first derivatives (but not second). In other words, it's a tidal gravity effect, surely..... For the latter, an observer on the particle will see a Rindler metric or whatever it's called. The first derivatives won't be zero. So it should be a completely different order of effect.....?

"So there we have it: a charge sitting on the ground WILL NOT radiate." Except that it's in free-fall with respect to the Sun. So we might expect some strange movement of any charge stationary with respect to Earth, i.e. it won't quite follow the Earth's orbit. No idea if that would be measurable.

So they do exist. Thanks! DAMTP just down the road from me....

The Lagrangian for a scalar field (eg charged pions) interacting with an EM field is roughly (\partial_{\mu}\phi - e A_{\mu}\phi)(\partial^{\mu}\phi - e A^{\mu}\phi) + m^2 \phi^2 or something like that, for charge e. So my question is: one term in this is proportional to \phi\phi\ A^2. So do we get Feynman graphs with 4 lines (2 pion, 2 photon) leading to one vertex? I've never seen a graph like that and it seems odd. I've also never heard anyone mention them. Do we get them or am I missing something (i.e. these 4-point vertices somehow disappear from the matrix elements?) Cheers, Jonathan

If I'm right she's saying Hawking radiation will decrease the mass enough to compensate for the decrease in "radius" so that a black hole never forms? But HR is a quantum effect proportional to Planck's constant and collapse is not. Is that all that's needed to debunk it?

Buych778: It's probably more helpful to think of fundamental, sizeless particles as waves or clouds rather than billiard-ball-like objects, which is what I think you're doing. An electron or quark in a usual state will be spread over a certain region of space. With an electron, it's often spread over an atom, a region of size about 10^(-10)m. With a quark, we've only seen them spread over the interior of a proton / neutron / meson, a region of size about 10^(-15)m. In fact (and this is Quantum Mechanics here - you'll have to accept it's weird), at least an electron can be distributed over a much larger volume in some circumstances. You could put it in a box of size 1 mile x 1 mile x 1 mile and the electron "cloud" would be spread over this 1 mile^3 volume. But this is not what we mean by "size" in this discussion. Here, "size" means the MINIMUM SIZE WE CAN GET IT DOWN TO. With an electron or quark THIS IS IN THEORY LIMITLESS. What this means is that by measuring its position accurately enough, we could in theory reduce that cloud / wave / whatever you call it down to a size of, say, 10^(-30)m or even smaller. (We can't actually do this because we don't have accurate enough instruments). So, in fact, what you'll get from the above if you read it carefully is the following: ***Measurement of a particle to within an accuracy of distance d will reduce the size of its cloud to that distance.*** If you think about that, it's VERY weird. But that's Quantum Mechanics for you. In general, measuring some characteristic of a particle will cause it to have a definite value of that characteristic where before the value was not definite. Another example is energy. A particle could be in a state where it has 2 energies at once, E1 and E2. (Yes that's even weirder - it's Quantum Mechanics). If we then measure the energy, it will change to having a definite energy, either E1 or E2. (Which one we get is as far as we know totally random). This is getting long so I'll stop there but I hope that gives some idea of what "size" means in Quantum Mechanics. As many have said, if QM doesn't shock you, you haven't understood it. Jonathan

An electron has no size at all, at least according to the Standard Model. It's got ~1/1800 the mass of a proton though, as you said. No-one knows why electrons and quarks have the mass they do as far as I know. The next lepton up, the muon, has a mass about 212 times the electron, and AFAIK no-one knows why. Size is no guarantor of charge. Quarks, like leptons, are sizeless. So you've got a group of sizeless particles all with closely related charges (electron -1, up quark +2/3 etc.) and I don't think we've established the reason. I Googled it and there seems to be some discussion about quantum gravity / complicated supersymmetry theories causing the charge correspondence but I'm ignorant of it. And I didn't call you a noob ;-)

Electrons aren't made of quarks at all. They are their own fundamental particle. (In fact, look at the title over your photo box (and mine) "Lepton", which I guess means you haven't made many posts and the forum is comparing you to the most basic particle of the lot ;-) ) A quark is also fundamental as far as we know, so that's the two types: quarks and leptons. I honestly don't know why electrons have exactly the same size charge as protons. I don't think it even comes out of the standard model of particle physics but I could be wrong. It's obviously not a coincidence.

"You're kinda turning everything upside down." Ah well, that's (ex) theoretical physicists for you. Never overestimate an ex theoretical physicist. :-/ Certainly it ultimately comes from experiment - for electrons we can measure the spin and yes, I guess that lead Dirac to hypothesise a spinor field. And after googling it you can measure photon spin by EM interaction of atoms. However for gravity I don't think we've measured anything and all we can do is theorise a spin of 2 based on the 2nd rank field. Could be wrong I suppose. I don't know if I even believe gravity has quanta; I think it's totally different from the other "forces". But most people would disagree I suspect.