Tag Info

New answers tagged

0

In solid state physics: Any transition to ferro- or antiferro-magnetically ordered state breaks the time inversion symmetry (1'), because the spontaneous magnetic moment on each atom changes the sign at 1'-operation.


1

I don't know if this is standard, but consider a pendulum that can swing a full circle in a plane. Vibrate the point of suspension up and down at the appropriate frequency. The pendulum will gain energy and spin either clockwise or counterclockwise.


2

the parity operation is $P: (t,x) \rightarrow (t',x') = (t,-x)$ lets see how this effects this term... $\epsilon^{\mu \nu \lambda} a_\mu \partial_\nu a_\lambda = \epsilon^{\mu 0 \lambda} a_\mu \partial_0 a_\lambda + \epsilon^{\mu j \lambda} a_\mu \partial_j a_\lambda$ $=\epsilon^{\mu 0 \lambda} a_\mu \partial'_0 a_\lambda - \epsilon^{\mu j \lambda} a_\mu ...


1

You are right that that the symmetry breaking breaks all three symmetries of $SU(2)$. Thus the $SU(2)$ generators give you three goldstone bosons in the theory with broken symmetry. However, we have not yet considered all of the symmetries of the original theory. We know that the full symmetry group has six generators and that five of them must be broken. ...


3

In the spirit of the original post, let $k,x$ be 4-vectors and $\mathbf{k}$, $\mathbf{x}$ the spatial components. Then a quantity of the form $$\phi(x) \propto \int dk[ a(k)e^{ikx} + a^*(k)e^{-ikx}]$$ is manifestly Lorentz invariant because it does not explicit contain any free Lorentz indices. What Srednicki does is that he performs the $k^0$ integration, ...


1

JakobH's comment as an answer: The electroweak gauge group $\mathrm{SU}(2)_L \times \mathrm{U}(1)_Y$ is broken into the electromagnetic $\mathrm{U}(1)_\text{em}$ by the Higgs field acquiring a non-zero vacuum expectation value, granting masses to the quark and $W^\pm,Z$ bosons. Thus, at the scale where up- and down-type quarks have very different masses, we ...



Top 50 recent answers are included