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First, note that you don't have a trace yet. You get a trace when you do the spin sum for the external fermion lines for the square of the amplitude; c.f. Peskin and Schroeder p. 132. Second, you need two photon propagators; one for each line in the diagram. One will couple to your $\gamma^\mu$ and to a new vertex with some $\gamma^\rho$ on the loop, and ...


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How did he get it? How did he get what? Which part of the calculation is confusing you. You have listed multiple equalities, which one is at issue? Eq 3.55 of the pdf online version of Peskin and Schroder shows that: $$ u^\dagger u=2E_p\xi^\dagger \xi $$ and since $$ \gamma_0^2=1 $$ we know that $$ \bar u\gamma_0u=u^\dagger u $$ and, in the ...


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$$ \frac{\eta^{\mu\nu}}{\Box} $$


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If we write $A_\mu(x)=\varepsilon_\mu(p)e^{ipx}$, the polarization vector should satisfy $\varepsilon_\mu p^\mu=0$, which is a Lorentiz-invariant relation, and is necessary to make sure that we have an irreducible representation of the Lorentz group (actually, the little group that leaves the momentum invariant). This knocks down the number of D.O.F to 3. ...


3

The longitudinal mode decouples from all physical processes as a consequence of gauge invariance, which in turn forces the Ward identity $$ k^\mu \mathcal{M}_\mu = 0$$ where the S-matrix element decomposition $\mathcal{M}^\mu$ is obtained from the polarization vector $\epsilon^\mu(k)$ by $\mathcal{M} = \epsilon^\mu(k) \mathcal{M}_\mu$. This decoupling (and, ...


1

Now, what I'm wondering is this- do we have a single model/theory whose equations accurately describe the behavior of light in all scenarios? Yes, for all intents and purposes: QED. Clearly, we can not say that QED works at all scenarios (e.g., very tiny length scales and very high energy scales), since we can not test it in all scenarios. But, as ...


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There used to be two models: wave and particle. QM mechanics does not pick particle over wave. QM actually picks neither and says either (all 3 interpretations are equally valid) it is both at the same time or it is something we can not yet understand.


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Instead of the word light it would be better to use the word electromagnetism. Newton and Young were fascinated about the decomposition of white light into it colors and about fringes behind edges. Since Maxwell it was obvious that light was only a small part of the electromagnetic spectrum. Later were discovered the weak and the strong nuclear power. But in ...


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Light exists in nature therefore physicists were bound to be interested in it at some point. But light is special in a sense. All we see, we see with light. Our eyes are sensitive to electromagnetic fields which make up this light. Furthermore, the main force which keeps the stars together is electromagnetism, which is the interaction of matter and light ...


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Light is a rare phenomenon as it has no mass. Most particles have mass and therefore cannot act in the way that light is, which is to move at the speed of light. At this speed, due to relativity light moves at the same speed regardless of your perspective; no matter how fast you are able to move you can never catch up to light, and even as you try harder to ...


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It's funny that all the answers so far forgot the simple and elegant following criterion : quantum mechanics appears when $$\dfrac{\hbar\omega}{k_{B}T} > 1$$ with $\hbar\approx6,63.10^{-34}\text{J}\cdot\text{s}$ the Planck constant and $k_{B}\approx1,38.10^{-23}\text{J}\cdot\text{K}^{-1}$ the Boltzmann constant, $\omega$ and $T$ being the (angular) ...


1

since this expansion follows directly from maxwell's equations in Lorentz gauge, it should still hold identically for the cavity because the former also holds. However what changes are the boundary conditions so that the sum over momentum goes from being over $\mathbb{R}$ to being over $\mathbb{N}$ for $p = 2 \pi n\hbar/L$


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Two things are actually mixed in the question. Simple answer, one photon kicks out one electron in photoeffect, that (together with the hole) you can detect as a current. 1/ Gamma (photon) interaction with matter: you can have three scenarios - pair production (electron+positron), Compton effect and photoeffect. 2/ In photoeffect - electron kicks out an ...


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A link is given in the comments the question where there is a publication that calculates the photoelectric effect using Feynman diagrams. Here are two screen captures as it is hard to copy from pdf. this one reduced so as to capture the Feynman diagram. Anybody interested should go to the link above.


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From the Maxwell-Dirac Lagrangian $$ \mathcal L = -\frac{1}{2}F^2 + \overline{\psi}(i\gamma^\mu D_\mu +m) \psi $$ where $D_\mu$ is the gauge covariant derivative it is clear that the 4-current that acts as the source term in Maxwell's equations is $$ j^\mu_D = q\overline{\psi} \gamma^\mu \psi. $$ Using the Dirac equation it can be shown that (see, e.g., ...


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I think you bring up an interesting question, I'm not sure why there were some derogatory comments to this... First off, the fermionic current doesn't couple to the gauge field due to its dimension. The Dirac field is dimension $ 3/2 $ and the current is dimension $ 3 $. Therefore, the coupling of the fermion to the gauge fields is of higher order. On the ...


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Maxwellian electrodynamics fails when quantum mechanical phenomena are involved, in the same way that Newtonian mechanics needs to be replaced in that regime by quantum mechanics. Maxwell's equations don't really "fail", as there is still an equivalent version in QM, it's just the mechanics itself that changes. Let me elaborate on that one for a bit. In ...


2

Anna V is wrong when she says Maxwell's equations are inconsistent with black body radiation. In quantum mechanics, even if you ignore radiation there is a charge density which you can calculate (in principle) from Schroedinger's Equation. In a warm body, this charge density is fluctuating by random thermal motion. If you track the time evolution of the ...


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Classical electromagnetic waves emerge from the underlying quantum electrodynamic description in a smooth and consistent manner. The quantum framework means that the classical waves are built out of photons, and the only time one has to worry about more detail than what Maxwell's equations provide is at the level of particle physics and wherever quantization ...


3

The amplitude for emission of n photons with polarizations $\epsilon_{\mu_j}$ written as $\epsilon_{\mu_1}\ldots \epsilon_{\mu_n}\mathcal{M}^{\mu_1 \ldots \mu_n}$ satisfies $k_{\mu_1} \mathcal{M}^{\mu_1 \ldots \mu_n} = k_{\mu_2} \mathcal{M}^{\mu_1 \ldots \mu_n} = \ldots =0$ due to gauge invariance (do you know this fact? This is a consequence of ward ...


0

What oscillates in the e.m. field is not the amplitude of some chord, but the electric field vector and the magnetic field vector. For visualizing your formula, see first of all the animated graph in this site. You see there an e.m. field propagating, and the electric and magnetic field oscillating in time between a maximally positive amplitude to a ...


3

If there were a $t$-channel or $u$-channel diagram for this process, it would have to involve a vertex where an electron changes into a muon and some other particle. There is no such vertex in the standard model.


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In QED there are vertex only with the same two lepton line and one photon line. I mean that there is not vertex when $\mu\to e\gamma$ in QED.


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Yes, this is actually often used in a spectroscopic technique called REMPI -- see the image on this wikipedia page https://en.wikipedia.org/wiki/Resonance-enhanced_multiphoton_ionization There are some important physics techniques that rely on interaction with two photons -- two photon spectroscopy (http://cua.mit.edu/8.421_S06/Chapter9.pdf). Some other ...


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The key to the effect is the idea - really just a conjecture, and as the paper states, "these ideas have not been met without controversy" - that there are materials that are opaque to gravity waves in the same way that conductors are opaque to EM waves. If that's true, then the cavity between parallel plates will contain a reduced set of modes of the ...


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charge is an intrinsic property of any particle. we in principle cannot change the intrinsic property of any particle. photons are the carriers of electromagnetic interaction(action at a distance).


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I have read the paper and the gravitational explanation of the Casimir effect is not the main point, it's the experiment that provides evidence of the graviton's existence. The paper starts by drawing a macroscopic analogy between Maxwell's equations and Einstein's linearized field equations (known as gravitoelectromagnetism). The author uses GEM because of ...


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The point is that a Dirac field isn't really a solution for an electronand positron but instead an electron and the conjugate of the positron. This means that the Dirac electron field does indeed obey, \begin{equation} e ^{ i \alpha (x) Q } \psi = e ^{ - i \alpha (x) } \psi \end{equation} Such misconceptions are dissolved if one works in the Weyl ...


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I cannot answer to all the questions but would like to stress something regarding what the Casimir effect tells us and what it doesn't. If you look at how it is derived for the usual EM interaction, an experimental verification of the standard Casimir effect tells us that: the EM field can have standing waves between two plates and outside them There ...



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