# Tag Info

18

In QED there are 4 kinds of divergences: Ultraviolet divergences. Naive calculations depend on the cut-off in such a way that they go to infinity as the cut-off do. However, QED is a perturbatively renormalizable theory so that non-naive, well-done computations (see regularization and renormalization) give sensible results. Landau pole. The coupling ...

14

Let us take standard fluid dynamics as a model of continuum media physics. It is obvious what is its "short-distance" physics isn't it? These are atoms and molecules. Let's apply your questions to this case -- suppose that we know nothing about atoms and molecules. in that case we can still use hydrodynamics, don't we? Do we replace the unknown/wrong ...

10

A "plane wave" generally refers to an infinite and perfectly flat wavefront, which cannot exist in reality, of course. However, there is nothing at all impossible about a plane wave of finite extent. Such a wave will experience diffraction at its edges, of course, but can still propagate over long distances before losing its planar nature. The problem with ...

10

Yes, the effect you're looking for is called Schwinger pair production. It requires immensely strong electric fields (of the order of $10^{18}$ V/m) for a constant field. One of the methods for computing the rate is the worldline method, described briefly here. To follow it, some knowledge of effective action methods are required.

8

The other answers explain that there's no paradox but they don't explain why the particular particle called photon is massless. It's massless because it is the messenger particle responsible for electromagnetism which is a long-range force. Its range is infinite so the mass has to be zero. One may view the Coulomb potential as the zero-mass limit ($m\to 0$) ...

8

Hwlau is correct about the book but the answer actually isn't that long so I think I can try to mention some basic points. Path integral One approach to quantum theory called path integral tells you that you have to sum probability amplitudes (I'll assume that you have at least some idea of what probability amplitude is; QED can't really be explained ...

8

Just because $F^{\mu\nu}$ has two indices does not mean that it represents a spin-2 particle. Note that the metric $g^{\mu\nu}$ is a symmetric two indexed object while the EM field strength $F^{\mu\nu}$ is antisymmetric. In fact, the metric $g^{\mu\nu}$ is analogous to potential $A^\mu$ in EM and the field strength of gravity is the four indexed Riemann ...

8

The possibility of spontaneous Lorentz symmetry violation due to the infrared problem of the Dirac-Maxwell equation was conjectured a long time ago by Frohlich, Morchio and Strocchi, in references [1,2] mentioned in the given Balachandran and Vaidya article. In perturbative QED, we usually assume that the scattering states are free eigenstates of the number ...

7

Dear Claude, you are extrapolating electromagnetism way too high. You're going from low energies to the Planck scale, assuming that nothing qualitatively changes, but this assumption is wrong. The fine-structure constant is essentially constant below the mass of the electron - the lightest charged particle - which is 511,000 eV or so. You are extrapolating ...

7

Worrying about the walls can be misleading. See A blackbody is not a blackbox for an illuminating account of the derivation of the Planck spectrum without enclosing the field in a box. If you cant get the published version, see the arxiv version. EDIT (25 March 2012) Planck's Radiation Law: A Many Body Theory Perspective discusses blackbody radiation ...

7

P&S almost seem to argue that they need to be included for the simple reason that, if they didn't include them, they would get a nonsensical (i.e. infinite) result. Well, I am confident that Peskin and Schroeder not only "seem" to argue in this way but they explicitly and comprehensibly enough write this fact because it is both true and important. ...

7

Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states  |\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ...

7

(1) The completeness relationship for a basis of vectors orthonormal with respect to $\eta_{\mu\nu}$ is $$\eta_{ij}\epsilon^{(i)}_\mu \epsilon^{(j)}_\nu = \eta_{\mu\nu}$$ This normalization convention is picked for Lorentz invariance... I know you said you didn't want that answer but the point is that the normalization of these ...

7

Comments to the question: First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like. As Frederic Brünner mentions a ...

6

This is a typical trade-off between the position properties of the purple electron and its wave properties. If the orange electron is sufficiently far so that it doesn't influence the purple electron much, everything will continue as before: the purple electron will produce an interference pattern while the orange one will draw one point on the photographic ...

6

Dear Peter, there is no freedom to "make" the (anti)commutator of your fields whatever you want, so there is no freedom to "make" it gauge-invariant, either. The canonical momentum is obtained as the derivative of the Lagrangian density with respect to the time-derivative of the canonical coordinate, and the (anti)commutator of these two is just the ...

6

Let's start with nuclear beta decay as a similar example. 12C, with 6 protons and 6 neutrons, is stable. As Marek has pointed out, it's not enough just to say that neutrons are unstable. If all neutrons were unstable with respect to beta decay, then 12C could decay into 12N. The reason this can't happen is that the mass of a 12N nucleus is higher than the ...

6

Here is a rough estimate why the muonium atom is unstable against the muon beta decay (in contrast to a neutron in a deutron). The binding energy of the electron in a muonium atom neglecting the reduced mass effect is approximately equal to its binding energy in a Hydrogen atom according to the Bohr's model, i.e., 13.6 ev. The muon performs a beta decay to a ...

6

The velocity gets into the spinor via the boost operator. At rest $\psi_L$ and $\psi_R$ are equal. After a boost they are multiplied by. $\psi_L ~\rightarrow~ \Lambda\psi_L ~~=~~ \exp\big\{-\eta\cdot\frac{\sigma}{2}\big\}$ $\psi_R ~\rightarrow~ \Lambda\psi_R ~~=~~ \exp\big\{+\eta\cdot\frac{\sigma}{2}\big\}$ So the momentum is indeed doubly "encoded" in ...

6

Michael Dine's response, quoted with permission: I now have to think back, but the argument in QED is based on the spectral representation ("Kallen-Lehman representation"). The argument purports to show that the wave function renormalization for the photon is less than one (this you can find, for example, in the old textbook of Bjorken and Drell, ...

6

The first I believe was the Pound-Rebka experiment in 1959 in a tower at Harvard. The best experimental proof we currently have available is the daily successful operation of the Global Positioning System (GPS).

6

$\chi$ is a real-valued function. This is part of the definition of the gauge transformation, since $U(1)$ is a one (real) dimensional group. In general, when talking about gauge transformations in particle physics, group parameters are restricted to be real by convention. In principle, I suppose you could perform a transformation on the wavefunction that ...

6

Is there a reason that you are interested in QED in particular? For processes within current experimental ranges, the standard perturbative treatment is incredibly accurate, and simulations are not really necessary. A sector of the standard model where simulations are incredibly important is in strong interactions (QCD) for which a perturbative treatment ...

6

Drawing from Feynman's and Wheeler's memoirs: Feynman was originally motivated to produce a theory of EM without the infinities of self-interaction, but he then needed a mechanism to reproduce radiation reaction, the loss of energy of an accelerating electron. He thought that a nearby electron could back-react to achieve the effect, but his advisor ...

6

One of the candidate explanations of the QCD color confinement involves the distinction between the Yang-Mills field electric and magnetic components. This model of confinement was qualitatively proposed in the 70s, and according to which, the quark confinement is explained by assuming the QCD vacuum to be composed of a magnetic monopole condensate in a ...

6

Starting with the Lagrangian for a massive $U(1)$ vector boson $A_\mu$ which like you said has 3 DOF: $\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 A^\mu A_\mu$ now if we change variables to $A_\mu\rightarrow A_\mu - \partial_\mu \theta$ and we have (Note that $F^{\mu \nu}$ and hence $F_{\mu \nu}F^{\mu \nu}$ is invariant under this ...

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