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0

No that is not what you must prove. It is not true that if $u^a$ is hypersurface orthogonal then $\nabla_a u_b = \nabla_{(a}u_{b)}$. In fact this is only true if $u^a$ is geodesic. If $u^a$ is hypersurface orthogonal then, by definition, $u_{[c}\nabla_b u_{a]} = 0$. Writing this out we have $$ u_c \nabla_b u_a - u_b \nabla_c u_a + u_a \nabla_c u_b -u_c ...


3

Yes, you are right. Indeed, since $z$ is a rotational symmetry axis, it defines an eigenspace of the inertial operator $I$. Since $I$ is a symmetric linear operator, it admits an orthonormal basis of eigenvectors, one such vector is ${\bf e}_z$. This unit vector can be completed into a basis of eigenvectors just adding some pair of unit orthogonal ...


2

If physics isn't an issue, you can add arbitrarily many terms. Once the physics comes in though, you will encounter a few restrictions : As said by Gennaro, it is assumed that the Poincaré symmetry applies. Higher derivative terms (second derivatives and above) are generally bad news. They can cause vacuum instability (energies can be arbitrarily ...


4

When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe. It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are ...


0

Yes, parity is really violated, even if neutrinos are massive. You seem to be confusing the relationship between parity, helicity, and chirality in the modern standard model with the physical symmetry operation of spatial inversion. Wu's experiment did not measure neutrino helicity. Wu and collaborators prepared a thin layer of a beta-emitting nucleus ...


5

In physics there is no general criterion on how to write down suitable Lagrangians, rather than a posteriori check on the equations of motions: all the Lagrangians generating the same dynamics are equally correct. For example, as an exercise, you may try to write down all the possible Lagrangians giving you back $F_j = m \ddot{x}_j$. This said, to directly ...


0

While the accepted answer is very clear, I'll write an operator proof. The $\hat{p^2}$ in $\hat{H}$ commutes with $\hat{\mathbb{P}}$ (the parity operator). So, to show that $\hat{H}$ and $\hat{\mathbb{P}}$ commute, we have to show this: $[\hat{V},\hat{\mathbb{P}}]=0$ Note that since $V(x)$ is a symmetric function i.e. even function, it is an eigenfunction ...


0

To study Higgs mechanism, we can use a Lagrangian of the form: \begin{align} \mathcal{L}=(D_{\mu}\phi)^2-\frac{1}{4} F_{\mu\nu}F^{\mu\nu}-V(|\phi|) \end{align} Where: \begin{align} V(|\phi|)&=-2v^2|\phi|^2+|\phi|^4 \\ &=(|\phi|^2-v^2)^2-v^4 \\ D_\mu \phi&=\partial_{\mu}\phi+\mathrm{i}e A_{\mu} \phi \\ F_{\mu\nu}&=\partial_{\mu} ...


1

You could use a little outside knowledge and observe that for any Lie group with lie algebra $\mathfrak{g}$ with $X\,Y\in\mathfrak{g}$: $$\exp(Y)\,\exp(X)\,\exp(-Y) = \exp(Z);\\\\ Z= X + [Y,\,X] + \frac{1}{2!} [Y,\,[Y,\,X]]+\frac{1}{3!}[Y,\,[Y,\,[Y,\,X]]]+\cdots$$ This is another form of the so called braiding formula $\mathrm{Ad}(e^Y) = ...


0

I don't think it's proper to say that the spontaneous breaking of anything is attributed to the Higgs mechanism. It's a postulate of the theory that the potential takes a certain shape, and that shape leads to a symmetry, and that the symmetry is spontaneously broken. The Higgs mechanism is a consequence of that. The symmetry that is broken is the local ...


2

Noether's theorem states that there exists a conservation law for every continuous (in fact, differentiable) symmetry. Reflection is a discrete symmetry, so the theorem is not applicable here. But, in quantum mechanics, you have the parity operator $P$, that reflects the coordinates $$P\psi(\vec{r}) = \psi(-\vec{r})$$ Since $P^2 = I$, the operator $P$ has ...


2

Rotation is a relative quantity. If you rotate by an angle $\theta$ relative to me that means I rotate by an angle $-\theta$ relative to you. So rotating a system of charges by an angle $\theta$ is exactly the same as leaving the charges stationary and rotating yourself by an angle $-\theta$. If you leave the charges stationary then obviously the electric ...


1

Look at the connection between the electric field and the charge distribution: $$\nabla \vec{E}(\vec{r}) = 4\pi\rho(\vec{r}).$$ If you rotate the charge distribution with some $R\in SO(3)$, i.e. $\rho(\vec{r})\longrightarrow \rho(R^{-1} \vec{r})$ and the charge distribution stays the same, i.e. $\rho(R^{-1} \vec{r})=\rho(\vec{r})$, then for the electric ...


1

The "unique" here is, IMO, not a good, evocative word. A better one would be preferred direction or privileged direction. Another way of looking at this is all directions are equivalent. A further confusing subtlety is that there is also something else that the author is assuming without telling you. There are no privileged directions for a problems with ...


3

The object for which you need to find the electric field is a uniformly charged sphere. Uniformly charged means that at every point on the sphere the charge density is same. Suppose someone blind-folded you and then he rotated the sphere in some arbitrary fashion about the origin(assuming your sphere has origin as the center). Then he takes off the blind ...


0

Here it explains what I think you are asking: We work with a formulation of Noether-symmetry analysis which uses the properties of infinitesimal point transformations in the space-time variables to establish the association between symmetries and conservation laws of a dynamical system. Here symmetries are expressed in the form of generators. We ...


1

I think I can remember the derivation for a conservative force field in classical mechanics, wich is a somewhat stronger assumption than pure time-translation invariance. Let $\vec{F}$ be a conservative force field, that is $$ \nabla \times \vec{F} = 0 $$ or alternatively $$ \phi := -\int_\gamma \vec{F} \cdot d\vec{a} $$ does not depend on the path $\gamma$ ...


0

You will get a better answer than this but, assuming all you want is a rough analogy compared to your chair idea, here goes. For a local gauge symmetry analogy, forget the chair. Imagine you are building a house on rough surface, with the ground sloping all over the place. Obviously you want the floor to be level, no matter how uneven the ground is. So ...


1

You're correct. To find the equations of motion, we have: \begin{align*}c_i&=\frac{\partial L(v^2)}{\partial v_i}\\ &=L'(v^2) 2 v_i \end{align*} so that $L'(v^2) v_i$ is constant for all of time. Firstly, you could imagine a world in which all paths ${\bf x}(t)$ are valid mechanical paths. Then the Galilean transform of a valid mechanical path ...


0

It follows from $L$ being a function $\propto\dot{x}^2$. With this at hand, you are left with two choices: $\left(\nabla_{\dot{x}}L\right)'\sim\left(\dot{x}\right)'=0$ implies $\dot{x}=\rm const.$ $L=0$ implies $\dot{x}=0=\rm const.$ Either way, you get that the velocity is constant in time (for this particular, free-particle case).


1

This "criteria" is neither sufficient nor necessary. There are many symmetry-breaking transitions which are not continuous. For example, it is well-known that $n$-state Potts model has a thermal symmetry-breaking phase transition, and when $n>4$ it is first order. For a more realistic one, I think melting transition is first order. Basically there is no ...


0

I believe this is discussed in Nonlinear Dynamics and Chaos by Strogatz. I don't remember the details, but it's worth looking at his discussion of an energy function.


1

(After possibly introducing more variables) then OP is essentially considering an autonomous system of $n$ coupled 1st order ODEs $$\tag{1} \frac{d\vec{z}(t)}{dt}~=\vec{f}(\vec{z}(t)), \qquad f: U \to \mathbb{R}^n , \qquad U\subseteq \mathbb{R}^n, $$ i.e. without explicit time dependence, so that the system (1) possesses time translation symmetry. OP is ...


0

As the above only deals with the gauge boson side of things This is a wrong assumption. The format represents the total knowledge from innumerable data of particle physics that have been fitted with SU(3)xSU(2)xU(1) . The particles are slotted into representations of the groups and there are rules of how the interactions happen within the structure of ...


0

Given that the title of the paper mentions valley contrasting physics, in the two cited paragraphs the authors try to motivate such a notion from basic principles, before delving into details. First they say that if a valley contrasting magnetic moment is to exist, it must be expressible in the form ${\frak m}=\chi\tau$ (where $\chi$ is an irrelevant ...


0

This is not true that closed orbits exist only for harmonic and Keplerian potential. This is indeed true if we stay in the case of central potentials (see Bertrand's theorem). Otherwise, there are many other examples. See, e.g. "On higher symmetries in Quantum mechanics", Fris, Mandrosov, Smorodinsky, Uhlir and Winternitz. $V(x,y)=a (x^2+y^2) ...


2

Comments to the question (v2): Let there be given a Lagrangian $$\tag{1} L(q,v,a,t), \qquad v^i~:=~\dot{q}^i,\qquad a^i~:=~\dot{v}^i,\qquad \jmath^i~:=~\dot{a}^i, $$ that depends on up to second time derivative. Let $$\tag{2} \delta q^i~=~\varepsilon Y^i(q,v,a,t) ,$$ be a (global, vertical) quasi-symmetry of the Lagrangian, i.e. there exists a ...


1

At least when the change of frame is given by some canonical flow the situation appears to be quite nice. If $\phi_t$ is the canonical flow generated by some $J$ and we make the change of frame $$ x' = \phi_t(x) \,, $$ the Hamiltonian in the new frame is given by $$ H' = H \circ \phi_t + J \,. $$ Given I didn't make any mistakes deriving this.



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