# Tag Info

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 ...

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 ...

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 ...

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 ...

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 ...

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

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 $$\underbrace{\frac{dL}{dt}}_{\text{total t-derivative}}~=~\underbrace{\frac{\... 3 Comments to the question (v3): OP is essentially asking about the Lagrangian field-theoretic formulation of a relativistic fluid in an external electromagnetic background A_{\mu}. Fluid dynamics have both a Lagrangian and an Eulerian picture. (Note that the word Lagrangian is used in two different meanings.) In the relativistic context, there is also ... 3 The corresponding symmetry group is the Lorentz group and yes we can use Noether to derive conserved quantities: Invariance under translations \rightarrow momentum conservation Invariance under rotations \rightarrow spin and angular momentum conservation Invariance under boost \rightarrow some strange, not really useful, conserved quantity 2 Because "spontaneous symmetry breaking" does not actually break any symmetries. This is a pretty important principle that is not always adequately taught. In spontaneous symmetry breaking the symmetry in question is always a full symmetry of the theory. The difference between a spontaneously broken symmetry and an unbroken symmetry is just in how the ... 2 Comments to the question (v2): The Minkowski spacetime can be generalizes to a Lorentzian manifold (M,g). We choose the Minkowski signature (-,+,+,+) and put the speed of light c=1 equal to one. OP evidently knows that the action$$S = -E_0~ \Delta \tau\tag{1}$$for a massive point particle is minus the rest energy E_0=m_0 times the change \Delta \... 2 As pointed out in the comments, you only need to introduce one conjugate momentum density: \pi = \frac{\partial \mathcal{L}}{\partial \dot{y}} = \rho \dot{y} The Hamiltonian density becomes $$\mathcal{H} = \frac{1}{2\rho}\pi^2 + \frac{T}{2}y_x^2$$ Next, the variational equations are: \begin{eqnarray}\dot{y} = \frac{\delta\mathcal{... 2 Effective potential is defined by the formula E=T_{radial}+V_{eff}(r). Your calculation shows that once you make this identification it is not true that \mathcal L = T_{radial}-V_{eff}, but that is fine. This happens because centrifugal term (i.e. the one with angular momentum) is really kinetic term and not a true potential. Hence it must enter the ... 2 Well, if approaching the issue from general principles, there is a reason why the equation of evolution of mechanical dynamical systems is of second (or higher) order: all inertial reference frame are equivalent to formulate the laws of physics (not only the laws regarding mechanics). Since these reference frames have arbitrary relative constant velocity, ... 2 Okay, let's give it a try. SU(2) sector of Standard Model Lagrangian is rather involved, so we will take a look at something simpler. Neutron-proton interaction comes to my mind. In low energy limit it is mediated by a massive scalar particle — a pion. We will be very qualitative about this, in reality there are a lot of details. Lagrangian will look ... 2 There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as$$R = \sum_{i=1}^n p_i\dot{q}_i - L$$You can learn more about it by clicking this link for wikipedia's description of it. 2 It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ... 2 That the Hamiltonian is zero is completely correct. The system is time-reparametrization invariant - changing \tau to \xi(\tau) transforms$$ 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$$and the action is invariant under this ... 1 The Lagrangian only depends on the potential energy and the kinetic energy. What the statement you quoted means is that if both the potential and kinetic energies are constant w.r.t. time, then so is the Lagrangian. This makes a lot of sense. Usually, we have:$$\mathscr{L}=K(x,t)-P(x,t)$$Where K and P are the kinetic and potential energies. But if ... 1 First of all, recall that one may vary the velocity v independently of the position q in the Lagrangian L(q,v,t). In fact, the (Lagrangian) canonical momentum is defined as$$\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(...

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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 $V(q)$.

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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 ...

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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