Hot answers tagged quantum-electrodynamics
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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
The best way to explain renormalization is to consider what at first looks like a complete detour: Mandelbrot's fractal geometry. Developed in the 1960s and 1970s, Mandelbrot's geometry is the key idea behind major advances in statistical physics in the early 1970s, pioneered by Leo Kadanoff (mostly independently), but also associated with Alexander ...
10
The main important idea of Feynman Wheeler theory is to use propagators which are non-causal, that can go forward and backward in time. This makes no sense in the Hamiltonian framework, since the backward in time business requires a formalism that is not rigidly stepping from timestep to timestep. Once you give up on a Hamiltonian, you can also ask that the ...
9
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
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 ...
7
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 ...
7
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
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 ...
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
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 ...
6
This is a very good question, but it is really two completely different questions in one.
Feynman's propagator
The probability amplitude for a photon to go from x to y can be written in many ways, depending on the choice of gauge for the electromagnetic field. They all give the same answer for scattering questions, or for invariant questions involving ...
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
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
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 ...
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
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. ...
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 ...
6
You're asking what would happen if we could view things with an unlimited high resolution. You view the emissions of the synchotron photons as discrete events and you ask is the path linear between these emissions. The problem is - quantum particles do not have trajectories so it's not meaningful to ask about the actual path followed by the particle. All ...
5
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$) ...
5
This is true. The simple explanation is this: For calculating the decay rate of an excited state, you use Fermi's Golden Rule, which involves the matrix element
$$|\langle f | V | i \rangle|^2$$
where $f$ and $i$ denote the final and initial state, respectively.
Since the final state contains the electron in its groundstate together with a photon created ...
5
Expanding on what Vladimir said: the thing that is changing with energy is $e$ (the others are not constants so much as conversion factors between length and time, time and energy, etc.). The reason the charge can vary is that the vacuum is not entirely empty. Sloppily speaking, near a charge, the electric field interacts with virtual (electron/positron) ...
5
Virtual photon clouds are responsible for potentials, not electric and magnetic fields, and this is what makes the explanation of forces in terms of photon exchange somewhat difficult for a newcomer. The photon propagation is not gauge invariant, and the Feynman gauge is the usual one for getting the forces to come out from particle exchange. In another ...
5
Experimentally it has been found that neutrinos travel in a straight line through matter, unless they interact directly with nuclei on the way into a spate of particles. Their path is not affected by the magnetic field of a bubble chamber for example. (The OPERA neutrinos travel kilometers in a straight line unaffected by the magnetic field of the earth).
...
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