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OP asks (v1): How one can know the gauge field emerging from the local gauge invariance is actually the EM field? Assuming that OP is pondering about gauging theoretical models (rather than concerned with our actual world and phenomenological inputs) then the answer is: One cannot know. For starters, the gauge group $G$ could be different than $U(1)$. ...


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Actually, evolution equations are even more than just second order in time : they don't depend naively on first order derivative, that is, on "velocity". This can be easily understood as the fact that there exists no privileged inertial frames. The change (that is, what is absolute) is given by acceleration and not velocity. If it depended naively on some ...


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Your way of thinking is, essentially, correct. When it comes to this $\tilde\lambda\Lambda^2$, there is this famous quote (citing from memory, don't remember which book it's from, but it's famous), "Even though it is infinitely large, we will assume that it is finite, and that is furthermore infinitely small." In most QFTs the perturbative series diverges ...


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Let us consider a real, scalar field theory for simplicity (and metric signature $+ - - -$). In the free theory, one can use the mode expansions of the field $\phi(x)$ and its canonical conjugate momentum $\pi(x)$ to derive the following expressions for the creation and annihilation operators: $$ a(p)=i\int d^3x\ ...


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First, the obvious explanation for the sign is that if $J$ has a minus sign in (1), then there should be a minus sign in (4). For some reason your $G$ turned into $\phi_i$. Assuming that they are the same thing, then I'm not sure I understand your problem. We didn't use the homogeneous KG equation to get the delta function; we used the inhomogeneous one, ...


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The neutron is not a fundamental particle. It carries no electric charge, yet it can interact with photons as its components - the quarks - carry electric charge and thus couple to photons. Macroscopically/classical, these interactions cancel out since its net charge is zero, but quantumly, there is a very big difference between objects with charged ...


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As has already been said: the orientation of the field drawn in your book is arbitrary. Once you have decided that the wave is say going into the page, then all that is required is that the E-field be drawn on the page (i.e. at right angles to the direction of wave propagation). You can put it up, down, left, right, diagonally, it doesn't matter, it would be ...


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There is a problem with diagrams like this Yes if that was a physical object and you looked at it end-on it would look like a plus-symbol. However it isn't intended to be interpreted physically like that. These diagrams shouldn't be interpreted as showing a vertical or horizontal displacement What is being shown is field strength and direction at a ...


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The direction of the electric field is called polarization. The direction of the electrical field in the free space lies in the plan perpendicular to the direction of propagation, and if this direction is unique for all the beam, it is said to be linear polarization. So, for linear polarization, the electric field can point in whatever direction that lies in ...


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Vertical with respect to whatever arbitrary set of coordinates you devise. If most of the pictures you see have the E-field pointing "up", that's just some kind of cultural bias. The E-field can point in any direction at all.


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Stack all the $\phi^i$s into a column "vector" $\vec\phi$. The mass term $m^2\vec\phi\cdot\vec\phi$ is obviously invariant by $R^{-1}=R^T$. The same with the kinetic term $(\partial_\mu\vec\phi)\cdot(\partial^\mu\vec\phi)$ because $\partial_\mu R=0$. It is $SO(n)$ invariant because I take it $i$ runs over $n$ values. Thus your $r^i_j$ generates $SO(n)$. ...


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Take for example the $n = 13$ line on the graph, which intercepts the $y$ axis at $650\text{cm}^{-1}$. Converting $650\text{cm}^{-1}$ to a wavelength by taking the reciprocal and dividing by $100$ gives the corresponding wavelength as $\lambda = 15.38\mu\text{m}$, and converting to an energy using $E = hc/\lambda$ gives $E = 1.292 \times 10^{-20}\text{J}$ or ...


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Consider the partition function $$Z = \int D\phi ~ e^{-S_0 - S_I},$$ where $S_0$ is the Gaussian/free part and $S_I$ is the interaction part of the action. Within a perturbative framework we may aim to systematically include the contributions of fast modes to the (effective) action for slow modes. For this we expand in the interaction strength as $$Z = \int ...


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Maxwell's thermodynamics and E&M were both atomistic. For example, in his Treatise on Electricity and Magnetism §255, he found it extremely natural to suppose that the currents of the ions are convection currents of electricity, and, in particular, that every molecule of the cation is charged with a certain fixed quantity of positive electricity, ...


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I believe that this may be what you’re looking for. I don’t claim to be an expert in renormalization, so all members of the Physics Stack Exchange Community are welcome to correct my answer. To make the relationship between the two more precise, let’s call the on-shell renormalization scheme the ‘on-shell subtraction scheme’ and the BPHZ renormalization ...


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This looks an incredibly difficult way of solving what is quite an easy problem. If you use Faraday's law in integral form, constructing a small, rectangular loop that goes into and out of the interface, it is easy to show that the component of the E-field that is perpendicular to the normal surface vector (i.e. the E-field parallel to the interface plane) ...


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You can find an explanation of scalar fields and associated quantum effects in the Schwarzschild background in chapter four of these lecture notes. The article also contains references which might be of use to you.



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