New answers tagged

2

You already got your answer, all right, several times over, but I will emphasize the central puzzle of your question which you only got indirect answers for, connected to the peculiar special structure of SO(4). Any self-respecting text introducing the standard model more or less has it. I'll skip all superfluous issues like lagrangian terms, the U(1)s, etc....


1

Answer of this question is quite subtle. First let us consider the most general Higgs potential which is renormalizable and invariant under $SU(2)_{L}\otimes U(1)_{Y}$ gauge transformations, which has the form \begin{equation} V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2} \end{equation} Where \begin{equation} \phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+...


0

After a bit of discussion I believe there is actually a $SU(2)\times SU(2)$ symmetry in a sense. So in principle there is a $U(2)$ symmetry if $\phi=(\phi_1,\phi_2)^T$, $\phi^\dagger=(\phi_1^*,\phi_2^*)$ and the lagrangian $$\mathscr{L}=\partial_\mu \phi^\dagger\partial^\mu \phi-m\phi^\dagger\phi-\lambda(\phi^\dagger\phi)^2,$$ simply sent $\phi\to U\phi$, ...


-2

If the field is a simple complex scalar field, than the symmetry is just $U(1)$. For a higher symmetry, $\phi$ need to be higher dimensional too, for instance you can add a vector index $\phi_i$ with $i=1,2$ for simplicity, which means that you add an additional complex field. If these two fields interact, you can have two cases now: Each field has a $U(1)$ ...


2

This is just supplementing Qmechanic's answer. I think the notations here need to be addressed. OP might be confusing Lagrangian (normal $L$) with Lagrangian density ($\mathcal{L}$). Formally, we have three fundamental relations: $$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3x$$ $$S = \displaystyle\int dt \space L = \...


3

Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives $$ \delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$ where the Lagrangian $$L[q(\cdot,t),v(\...


5

There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as $$R = \sum_{i=1}^n p_i\dot{q}_i - L$$ You can learn more about it by clicking this link for wikipedia's description of it.


3

It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ...


2

That the Hamiltonian is zero is completely correct. The system is time-reparametrization invariant - changing $\tau$ to $\xi(\tau)$ transforms $$ n(q(\tau))\mapsto n(q(\xi)),\quad \dot{q}\mapsto \frac{\mathrm{d}\xi}{\mathrm{d}\tau}q', \quad \mathrm{d}\tau\mapsto \frac{\mathrm{d}\tau}{\mathrm{d}\xi}\mathrm{d}\xi$$ and the action is invariant under this ...


0

The Lagrangians are not identical, but they only differ by a total derivative. In other words, you get from the one to the other using partial integration. For example, for the first term: $$ -\frac 1 2 (\partial_\mu A_\nu) (\partial^\mu A^\nu) = -\frac 1 2 \partial_\mu \left( A_\nu \partial^\mu A^\nu \right) + \frac 1 2 A_\nu\, \partial_\mu \partial^\mu A^\...


3

It seems relevant to mention the importance of distinguishing between explicit, implicit, and total time-dependence. The Lagrangian $L=L(q,v,t)$ depends implicitly on time via the position $q$ and the velocity $v$. The total time derivative of the Lagrangian $L=L(q,v,t)$ is $$\underbrace{\frac{dL}{dt}}_{\text{total $t$-derivative}}~=~\underbrace{\frac{\...


1

The Lagrangian only depends on the potential energy and the kinetic energy. What the statement you quoted means is that if both the potential and kinetic energies are constant w.r.t. time, then so is the Lagrangian. This makes a lot of sense. Usually, we have: $$\mathscr{L}=K(x,t)-P(x,t)$$ Where $K$ and $P$ are the kinetic and potential energies. But if ...


0

Classical Mechanics: Systems of Particles and Hamiltonian Dynamics by Walter Greiner. This is a very good book for the same reasons that all the books belonging to the series of books written by Greiner are good. They are clear, they do not shy away from mathematics (they are written for people who want to pursue theoretical physics) and they have many ...


0

Grassmann-odd variables provide a classical description of Grassmann-odd quantum operators in the same way that Grassmann-even variables provide a classical description of Grassmann-even quantum operators. The classical super-Poisson bracket $$\{\psi^i,\psi^j\}_{PB} ~=~ -i (T^{-1})^{ij} \tag{A} $$ is related to the super-commutator$^1$ $$\hat{\psi}^i\hat{\...


1

The gauge symmetry group associated to the SM is $SU\left(3\right)_{c}\times SU\left(2\right)_{L}\times U_{Y}\left(1\right)$. Then we can not build the lagrangian of the SM with terms of the form $m\bar{\psi}\psi$ because they are not gauge invariant. A term of this kind mix the right and left handed parts, which transforms differently. In order to give mass ...


0

I have talked with people more knowledgeable than I on SUSY, and they agree that something isn't right here. I'm posting my correction to the above in case someone has this problem in the future. This is missing from the errata. We start with (2.60) $$\mathcal{L}=-\frac{1}{2}W^{jk}\psi^{\alpha}_{j}\psi_{\alpha k} + W^{j}F_{j} +h.c.$$ We have the following ...


1

First of all, recall that one may vary the velocity $v$ independently of the position $q$ in the Lagrangian $L(q,v,t)$. In fact, the (Lagrangian) canonical momentum is defined as $$\tag{A} p(q,v,t)~:=~\frac{\partial L(q,v,t)}{\partial v}. $$ This is explained further in e.g. this, this, and this Phys.SE posts. Let us define for later convenience $$\tag{B} F(...


4

The first form is true whenever the Forces are derivable from a scalar, i.e when $Q_i=-\frac{\partial V}{\partial q_i}$ The second equation however is true even when none of the forces can be derived from a scalar The third is true when some of the forces are derivable from a scalar and some are not, i.e. $L$ contains potential of the conservative forces ...


2

As pointed out in the comments, you only need to introduce one conjugate momentum density: $\pi = \frac{\partial \mathcal{L}}{\partial \dot{y}} = \rho \dot{y}$ The Hamiltonian density becomes \begin{equation}\mathcal{H} = \frac{1}{2\rho}\pi^2 + \frac{T}{2}y_x^2\end{equation} Next, the variational equations are: \begin{eqnarray}\dot{y} = \frac{\delta\mathcal{...


2

Okay, let's give it a try. $SU(2)$ sector of Standard Model Lagrangian is rather involved, so we will take a look at something simpler. Neutron-proton interaction comes to my mind. In low energy limit it is mediated by a massive scalar particle — a pion. We will be very qualitative about this, in reality there are a lot of details. Lagrangian will look ...


4

The underlying reason for OP's flawed argument is that a premature use of EOMs in the stationary action principle $$ S~=~\int\!dt ~L(r,\dot{r};\theta,\dot{\theta}), \qquad L(r,\dot{r};\theta,\dot{\theta})~=~\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r),\tag{A}$$ destroys the variational principle. Concretely OP is implicitly assuming that (3) is a ...


2

Effective potential is defined by the formula $E=T_{radial}+V_{eff}(r)$. Your calculation shows that once you make this identification it is not true that $\mathcal L = T_{radial}-V_{eff}$, but that is fine. This happens because centrifugal term (i.e. the one with angular momentum) is really kinetic term and not a true potential. Hence it must enter the ...


0

Because in an isotropic space x is equivalent to -x, this is just Noether theorem. If you want to see how parity violation is explained in the standard model look here


2

Well, if approaching the issue from general principles, there is a reason why the equation of evolution of mechanical dynamical systems is of second (or higher) order: all inertial reference frame are equivalent to formulate the laws of physics (not only the laws regarding mechanics). Since these reference frames have arbitrary relative constant velocity, ...


4

The latter functional would not be very useful in physics, because without the $\dot{q}$ dependence there would be no way to capture the particle's kinetic energy in your Lagrangian. There would be no dynamics, as the functional would be extremized simply by setting $\partial{L}/\partial{q} = 0$. For a typical Lagrangian of the form $L = T - U$, this would ...


1

Here is the short version: Physics is often about calculating the time evolution of a dynamical system. There the kinetic term $T(\dot{q})$ plays an important role. In contrast, in static problems, the kinetic term $T(\dot{q})$ is absent or can be neglected, and we minimize just the potential energy $V(q)$.


4

I am going to write a short answer, because no one has answered yet. The left hand side should be $$\eta^{\mu \nu} \partial_\mu \phi \partial_\nu \phi.$$ It is important you only sum raised indicies with lowered indices and you never sum two lowered indices together. Summing two lowered indices gives you a result that is not lorentz invariant. The right ...


3

Comments to the question (v3): OP is essentially asking about the Lagrangian field-theoretic formulation of a relativistic fluid in an external electromagnetic background $A_{\mu}$. Fluid dynamics have both a Lagrangian and an Eulerian picture. (Note that the word Lagrangian is used in two different meanings.) In the relativistic context, there is also ...


4

You cannot just add a term to the Lagrangian to give the usual electromagnetic gauge theory magnetic charge. The reason is rather simple: The equation of motion for a magnetic four-current $j_m$ is $\mathrm{d}F = j_m$. But $\mathrm{d} F = \mathrm{d}\mathrm{d} A = 0$ independently of the equations of motion. So simply adding a term doesn't work. The first ...


0

Yes there is. Notice that the expression $$ \frac{d}{dt}\frac{\partial{T}}{\partial\dot{q}_\alpha}-\frac{\partial{T}}{\partial{q}_\alpha}+\frac{\partial{V}}{\partial{q}_\alpha}=0 $$ is very close to the Euler-Lagrange equations (ELeq) $$ \frac{d}{dt}\frac{\partial{L}}{\partial\dot{q}_\alpha}-\frac{\partial{L}}{\partial{q}_\alpha}=0 $$ except that the ...


0

Explanation of doubt 1: Consider a double pendulum made up of two massive rods. Here there are two degrees of freedom and hence two generalized coordinates. Assume that 5 separate non-constraint (applied) forces act on this system of double pendulum: ${\vec {F_1}}$ and ${\vec {F_2}}$ on the upper rod (rod 1) and ${\vec {F_3}}, {\vec {F_4}}$ and ${\vec {...


0

The system is supposed to consist of $n$ points at which the forces act. If no force acts at the $i$-th point, just set $F_i = 0$. The assertion that "this can only be true if external forces are the impressed forces" is false (of course all constraint forces are ultimately external). It is precisely the content of d'Alembert's principle that one may omit ...


6

There is no intuitive way to grasp this because it is not true! The Lagrangian is NOT the energy of the system. The energy of the system is $(KE+PE)$, of course. I can define lots of quantities with the units of energy: $KE$, $PE$, $(KE+PE)$, $(KE-4PE)$, $({KE}^2/PE)$, $(KEk_bT/PE)$. Of course, only one of them is the total energy. Some of them are still ...


0

In a few words, as Luke Pritchett writes, the Higgs mechanism provides us description of particles mass without breaking of the unitarity, i.e., breaking gauge symmetry explicitly. It is interesting fact that even if You start from electroweak theory in the broken phase and don't know about Higgs boson, $W/Z$-boson and existense of hidden gauge invariance (i....


2

Because "spontaneous symmetry breaking" does not actually break any symmetries. This is a pretty important principle that is not always adequately taught. In spontaneous symmetry breaking the symmetry in question is always a full symmetry of the theory. The difference between a spontaneously broken symmetry and an unbroken symmetry is just in how the ...


2

Comments to the question (v2): The Minkowski spacetime can be generalizes to a Lorentzian manifold $(M,g)$. We choose the Minkowski signature $(-,+,+,+)$ and put the speed of light $c=1$ equal to one. OP evidently knows that the action $$S = -E_0~ \Delta \tau\tag{1}$$ for a massive point particle is minus the rest energy $E_0=m_0$ times the change $\Delta \...


0

Your derivation is close. The Polyakov action is $$ S[X,\gamma] = T\int d\tau\int d\sigma (-\gamma)^{1/2}\gamma^{ab}\partial_aX^\mu \partial_bX_\mu, $$ for $T = -(4\pi\alpha')^{-1}$. The variation with respect to the string $X_\mu$ then gives $$ \frac{\delta S}{\delta X^\mu} = T\int d\tau\int d\sigma\left[\frac{\delta}{\delta X^\mu}\left((-\gamma)^{1/2}\...


0

The gravitational force is (by definition) vertical. If it were not, a horizontally unsupported body would accelerate laterally, and this is not seen. You may be confusing the purely gravitational forces from other, mechanically coupled, forces.


0

Yes but with lots of subtleties to be described in (2). The main questions to be settled are: whether we count just the fields (configuration space) or fields and their derivatives (phase space) whether we count the fermions along with bosons, or separately (I will count them separately), and whether we double the number for them because the Standard Model ...


3

The corresponding symmetry group is the Lorentz group and yes we can use Noether to derive conserved quantities: Invariance under translations $\rightarrow$ momentum conservation Invariance under rotations $\rightarrow$ spin and angular momentum conservation Invariance under boost $\rightarrow$ some strange, not really useful, conserved quantity


1

$A_\mu$ is introduced simply as a tool to assert gauge invariance of the fermion's (in this case) kinetic term. Once this field is added to our lagrangian, we recognize that we must add a kinetic term for the gauge field itself. This is the point where we will make contact with the vector potential of electromagnetism. We introduce a field strength tensor,...


1

The wave equation needs to stay invariant under local changes of phase. The gauge field $A_{\mu}$ that is introduced to enforce local gauge invariance is NOT an arbitrary function, it needs to represent something and it represents the possibility that the particle either emits or absorbs a photon, a quantum of the EM field. The probability that it does so,...



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