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The usual use of an image like the one you posted is to identify nonuniform regions of the material. The amount of variation in polarization at any location on the protractor is related to the amount of strain, or nonuniformity, in the material itself. As explained in this post , there are a couple underlying optical principles which cause the color ...


0

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

we know that E(electic field magnitude) =B(magnetic field magnitude)*c(c is the speed of light in vacuum)... from that, B=E/c here c is very large ie, approx= 3*10^8 so magnetic field magnitude is one by 3*10^8 times the electric field intensity... so compared to electric field magnitude,magnetic field magnitude is very low, hence negligible


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You question is a bit unclear. The red region of the P-E diagram that you show comes from the equation for Gibbs free energy in a ferroelectric, with the electrostatic term: $G= \frac{1}{2} \alpha_{1} D^{2} +\frac{1}{4} \alpha_{2} D^{4} +\frac{1}{6} \alpha_{3} D^{6}-E*D$ Taking the derivative with respect to D: $\frac{ \partial G}{ \partial ...



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