Jump to content

JonathanApps

Senior Members
  • Posts

    30
  • Joined

  • Last visited

Profile Information

  • Location
    Cambridge, UK
  • College Major/Degree
    Manchester University, Physics B.Sc, Ph.D
  • Favorite Area of Science
    Quantum Field Theory, General Relativity
  • Occupation
    Programmer

JonathanApps's Achievements

Quark

Quark (2/13)

4

Reputation

  1. 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.
  2. 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
  3. What's k? You have K1 and K2 in the original question? Do we include the effect of gravity?
  4. or indeed anything beyond simple undergraduate physics. Was this his only gaffe? I suspect there are others.
  5. There's something about it in The Gunslinger by Stephen King as well.
  6. 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.
  7. 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
  8. Yes; it's just that |i> and |f> seem unrealistic when the QFT courses decide they're going to be definite momentum.
  9. 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.
  10. OK, sure, but not in the Interaction Picture, surely? The field equations are the same; it's the quantum state that's time-dependent.
  11. 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?
  12. 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.
  13. 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.....?
  14. 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.
  15. [math] a^{\dagger}_{p}b^{\dagger}_{q}|0> [/math]
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.