# Tag Info

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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. ...

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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, ...

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To justify giving mass to a would-be massless particle, scientists were forced to do something out of the ordinary. They assumed that vacuums (empty space) actually had energy, and that way, if a particle that we think of as massless were to enter it, the energy from the vacuum would be transferred into that particle, giving it mass. A mathematician named ...

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The Higgs field is actually two complex scalar (spin 0) fields so there are two particle and two anti-particle excitations (quanta). The pair of fields transform as an electroweak doublet which essentially means that the Higgs field quanta interact with the electroweak gauge field quanta (W and B bosons). In addition, the Higgs field has a peculiar ...

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Yes, an electron is just some wave, as you say, in the electron field, as it is for any particle. You can also interpret in a broad sense that a field needs to be perturbed at a particular point in spacetime for you to have a non-zero odd of measuring it a that point, although this simple picture is complicated by quantum phenomenas. The energy of a ...

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This is the Euclidean classical action $S_{cl}[\phi]=\int d^{4}x (\frac{1}{2}(\partial_{\mu}\phi)^{2}+U(\phi))$. It would be nice if somebody could explain the structure of the potential. I don't understand why $\phi$ is used instead of a position vector $\textbf{r}$. Also, how can $(\frac{1}{2}(\partial_{\mu}\phi)^{2}$ be interpreted as ...

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If you have a result stating that the non-linear completion of a free massless spin-2 field must be general covariant, then that means the action must only depend on curvature invariants like $R$, $R_{ab}R^{ab}$, $R_{abcd}R^{abcd}$, ... Because we started with the action for a free massless spin-2 field, we can narrow down the allowed dependence on the ...

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You are free to define the vector $\bf{F_{EM}}$, but I don't believe this vector would have any value. It wouldn't obey any simple laws, and it would not be found to have any practical use in the lab.

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I will find the Maxwell word in the Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics by Hestenes, on pages 25/26 That formidable text presents a better math formalism for physics, imo. Starting with $F(x,t) = E(x,t) + i B(x,t)$ ... The 4 equations of Maxwell (64..67) that describe two viewpoints ( E and B ) of a single ...

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Let me try this more clearly than the other answers, which aren't wrong. You ask: So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition? There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable ...

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If the field is not stationary, curl of $\vec{E}$ does not vanish. So generally you cannot identify electromagnetic field with the curl-free part of the decomposition. However, you can indeed introduce a complex vector combination of electric and magnetic field, in a certain system of units it is $\vec{E}+i\vec{H}$. This is the so-called Riemann-Silberstein ...

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If fits very well so we can write that electromagnetic field is equal $\bf{F_{EM}}=\bf{E+B}=-\nabla\phi+\nabla\times\bf{A}$ or can we? No! For the love of god, no! Do not just add those fields together... it's not a useful quantity. In the SI system $E$ and $B$ have different units. Another good indicator that you don't want to just add ...

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Actually the electromagnetic field can be seen as a tensor. The combination Wikipedia talks about is this, $E$ and $B$ are organised in an antisymmetric matrix $F_{\mu\nu}$ with $\mu,\nu = 0,\ldots, 4$ so the number of independent components is $6$.

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Take 2D SPT as an example. When the gauge fields are not dynamical, physically it really means that we are just modifying the Hamiltonian to a certain gauge field configuration (i.e. by changing the coupling of some of the terms, for example). These "gauge fields" are just extrinsic parameters in the Hamiltonian. To detect the SPT, we need to introduce ...

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The mathematical concept that I was searching for in this question is the following: http://en.wikipedia.org/wiki/Hodge_dual I will not elaborate more, but except to say that, the Hodge dual allows you to define a conserved current corresponding to any choice of the "time axis". (Sorry for the vagueness).

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This is frequently good enough, but in your specific case its actually much easier to show this holds exactly in the finite case. I won't do this for you, but note that this is a global symmetry, i.e. alpha has no dependence on x. At most, you might also need to use the Baker-Campbell-Hausdorff formula.

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Doesn't the result follow from a simple diagram? If you change the X,Y position by a small amount dx, dy, the change in length of the hypotenuse is simply \begin{align} &= \sqrt{(a+dx)^2 + (a+dy)^2} - \sqrt{a^2 + a^2}\\ &= \sqrt{2a^2 + 2a dx + 2a dy + ...} - \sqrt{2a^2}\\ &= a\sqrt{2}\left(1 + \frac{dx}{a} + \frac{dy}{a} - 1\right)\\ &= ... 2 I'm going to go a bit overboard here and give you the sketch of how vectors are geometrically constructed, since I think it's helpful to know. While writing this I found I was phrasing things very carefully, which means: you may need to reread parts of this in a quiet corner if it doesn't all make sense at first. Suppose that you have a set of scalar fields ... 2 Rule of thumb is that things with indices are tensors, the order being given by the number of free indices you have. By free index I mean any index for an object which is not repeated, hence not involved in a sum (or contraction). For exampleg_{\mu\nu}$$indicates a covariant tensor of order (or rank) 2. If this represents a metric, its inverse is the ... 3 The derivative \mathrm{d}\phi is a proper (co)vector:$$ \left(\begin{matrix}\frac{\partial\phi}{\partial x^1} \\ \frac{\partial\phi}{\partial x^2} \\ \frac{\partial\phi}{\partial x^3} \\ \frac{\partial\phi}{\partial x^4}\end{matrix}\right)$$with components \partial_\mu \phi and \partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi. This works also in ... 0 Not all conserved charges are obtained by integrating the time component of some conserved current. For example, momentum and angular momentum are conserved charges and are obtained by integrating a spatial component of a conserved current. So the equations and interpretation for conserved charges in a Euclidean theory are the same as in the ... 2 Assuming no quantum gravity, \eta^{\mu\nu} is a constant and can be pulled out of the derivative and what remains looks like a \delta^k_{\mu} or \delta^k_{\nu}-type expression (in the sense of a Kronecker \delta), pulling the k into the \partial^\mu or \partial^\nu respectively. If you are confused about where the minus sign comes from, I ... 1 Once you fix a coordinate system X, then for any two points A and B the distance d(A,B) between the two can be defined and calculated from the metric in that particular coordinate system, allowing you to define limits of this type: For a sequence of points P_n and a point Q, P_n\rightarrow Q if and only if d(P_n,Q)\rightarrow0. It's true that ... 1 I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the p_3 direction so that p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$$ \delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3} = ...

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Any Lorentz transformation will leave $p^2$ unchanged, hence the mass of the Dirac $\delta$ will remain at $m^2$. Observe that, for physical reasons, one usually only considers the component of the (full) Lorentz group that is connected to the identity. Any transformation in this proper subgroup leaves the sign of $p_0$ (which by the spectral condition is ...

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I think I have came up with the answer, which I hope is correct. Although, neither internal nor the gauge symmetry operations affect the space-time coordinates, there is a big difference: Internal symmetry is an actual symmetry of the system (field): two physically distinct field configurations (or in QM, two physically distinct states in Hilbert space) ...

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This is indeed the same problem, from different viewpoints. The deep source of the problem is that the number of $A_\mu$ fields is higher than the actual physical degrees of freedom. Here are a few remarks: The operator you are trying to invert does not have an inverse because it is just a projection operator(you can check this by direct computation, the ...

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The answer to this question is no assuming dimensions higher than 1+1. This can be seen observing that the equation of motion for the fermionic field is just the limit of mass going to infinity of a scalar field coupled to a fermionic field. This can be seen in the following way. Consider the Lagrangian  L = ...

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As I expected, this simple question calls for no more than a little sleight of hand, as pointed out by the sole comment above. It looks as if @Meng Cheng won't make it an answer. Thanks to him. And here I confirm that it works well.

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