5
To identify the Noether 4-current with the electric 4-current, one would in principle have to show that the Noether 4-current indeed appears as the source term in Maxwell's equations.
The Maxwell equations with sources (Gauss's + Ampere's laws) are derived by adding the Maxwell Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to a minimally coupled, gauge-...
4
Your mistake is in saying that $\overline{\psi_{L,R}} = \overline{\psi}P_{L,R}$.
In fact,
$$ \overline{\psi_{L,R}} = (P_{L,R}\psi)^† \gamma^0 = \overline{\psi}\gamma^0 P_{L,R}^† \gamma^0$$
$$ = \overline{\psi} P_{R,L}$$
To see why that last line is true we have to expand $P_{L,R} = (I\mp \gamma^5)/2$ and note that $(\gamma^5)^† = \gamma^5$, $(\gamma^0)^2 = ...
3
The last term is a total divergence. The second one is not.
3
The standard convention is to divide each term in the Lagrangian with its symmetry factor. Therefore the kinetic term for a real (complex) scalar field is
with (without) a symmetry factor $\frac{1}{2}$, respectively. A complex scalar field $\phi= \frac{\phi^1+i\phi^2}{\sqrt{2}}$ can be viewed as 2 real scalar fields.
3
It is up to a trivial constant the only gauge invariant Lagrangian that reproduces the Maxwell equations. Since the Maxwell equations are linear in $F^{\mu\nu}$ only quadratic functions of $F^{\mu\nu}$ are possible. There are two possibilities, namely $F_{\mu\nu}F^{\mu\nu}$ and $\epsilon_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma}$. The Maxwell equations ...
2
You need check invariance only on linear level, because you consider linear action. Third term is second order.
Integration by parts is incorrect, because ζ is quadratic differential operator.
I recommend you to start with most general quadratic action and find coefficients from diffeomorphism invariance, like in Zee book on gravity:
After that, you need ...
2
There are at least 2 lessons to be learned from OP's set-up:
Noether's theorem is not necessarily about strict symmetry of the action. It is enough if the action has an (infinitesimal) quasi-symmetry, i.e. symmetry up to boundary terms.
There's no free lunch. To prove energy conservation, one must use a non-trivial assumption: In this case, that the ...
2
I think you might be missing something due to the notation used in that article. In Lorentzian signature the quantity $\psi^\dagger \gamma_0 \psi$ is a scalar. Going to Euclidean signature with $\gamma_0 = \gamma_4$, we have that $\psi^\dagger \gamma_4 \psi$ is again a scalar. However, note that the paper defines a new quantity see Equation (44) to ...
2
There is an eazier way to derive the equations of motion(without calculating this nasty derivative), using calculus of variations(varying with respect to $A_{ν}$) and discarding surface terms:
$\delta S=0 \Rightarrow \delta \int d^{4}x\sqrt{-g}F^{μν}F_{μν}=
\int d^{4}x\sqrt{-g}\delta F^{μν}F_{μν} + \int d^{4}x\sqrt{-g}F^{μν}\delta F_{μν}= 2\int d^{4}x\...
2
Just “brute force” the derivative using product rule. No plugging in Klein-Gordon equation.
What you really need is $\partial_{\mu}T^{\mu \nu}=0$ and symmetry of your stress energy tensor (trivial to check).
2
Not sure if this is what you're looking for, but here is a exhaustive search for all gauge invariant objects in electromagnetism:
Say $f[A_{\mu},\partial_{\mu} A_{\nu},\partial_{\mu}\partial_{\nu} A_{\rho},...]$ is a gauge invariant object. Here $f$ could itself have indices which have been suppressed. Under a gauge transform:
$$A'_{\mu}(x)=A_{\mu}(x)+\...
2
Adapting Cosmas Zachos' comment into an answer.
We say that a field is "free" if the Lagrangian is quadratic in the field and its derivative, eg. for a field $\phi^a$ with some kind of index $a$, the Lagrangian looks like $$ \mathcal L=Q_{ab}^{\mu\nu}\partial_\mu\phi^a\partial_\nu\phi^b+R_{ab}\phi^a\phi^b, $$ where $Q$ and $R$ are field-independent ...
1
OP is essentially asking how the Nambu-Goto (NG) action principle works. We should stress that there is no variation wrt. to the target space metric $g_{\mu\nu}(X)$ per se: It is treated as a background metric. The variation is purely wrt. the target space coordinates fields $X^{\mu}(\xi)$. It is straightforward exercise to derive the Euler-Lagrange (EL) ...
1
This is a good question, and I think it highlights a point where textbooks are very confusing.
At heart, a function transform is just a formal map from functions to other functions, usually invertible. For example, the Fourier transform changes a function of time $f(t)$ to a function of frequency $\tilde{f}(\omega)$. These aren't literally "the same" ...
1
If the structure group of the theory is $G$, with Lie algebra $\mathfrak g$, then the (local) Yang-Mills fields, are locally defined differential $1$-forms that take value in the Lie algebra $\mathfrak g$.
If the local gauge neighborhood is $U$, and we suppose that it is also a coordinate neighborhood, then we may write $$ A(x)=A_\mu^a(x)\mathrm dx^\mu\...
1
You can generally couple a continuum action $S_{\rm field} = \int_V \mathcal{L}(\psi(x^\mu)) d^4 x $ to a worldline action $S_{\rm particle} = \int_{\Delta \lambda} L(x^\mu(\lambda),\psi(x^\mu(\lambda))...)d\lambda$ through a delta function on the world-line
$$S_{\rm tot} = \int_V \left(\mathcal{L}(\psi(x^\mu)) + \int_{\Delta \lambda} L(x^\mu(\lambda),\psi,.....
1
So the other answers have given a solid response but it is a little high-level, I wanted to give a more verbose explanation of what has happened.
In your procedure you have an action integral $S = \int_T\mathrm dt~L(q(t),\dot q(t), t)$ over some time domain $T$. You now want to vary the time coordinate. This means that the domain $T$ is changing, so that ...
1
Let's first clarify which formulation of Noether's theorem we'll be using:
The Lagrangian will be a function
$$
L = L(x,v,t)
$$
and the action a functional
$$
S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt
$$
Proposition. If the transformation
$$ t\to t'(t) = t + \epsilon T(t) $$
$$ x\to x'(x,t) = x + \epsilon X(t)$$
$$
q'(t') = q(t(t')) + \epsilon X(t(t'...
1
Edit:
The Noether Procedure instructs you to take $\varepsilon = \varepsilon(t)$ to be time dependent.
$$
\delta q = \varepsilon(t) \dot q \hspace{1 cm} \delta \dot q = \dot \varepsilon(t) \dot q + \varepsilon(t) \ddot q
$$
\begin{align*}
\delta L &= (\varepsilon \dot q) \frac{\partial L}{\partial q} + (\dot \varepsilon \dot q + \varepsilon \ddot q) \...
1
Supersymmetry only plays a minor role here, so we will suppress it in this answer. The Faddeev-Popov gauge-fixing procedure and the background field method
$$ A_{\mu}^a~=~\underbrace{\overline{A}_{\mu}^a}_{\text{background}} +\underbrace{{\cal A}_{\mu}^a}_{\text{quant. fluct.}} \tag{A}$$
are e.g. explained for Yang-Mills theory in Ref. 1, chapter 71 & ...
1
D'Alembert principle is nothing but a prescription of a type of constraints -- also known as ideal constraints -- such that the equation of motion can be written in the form of Euler-Lagrange equations, by using an arbitrary system of coordinates obtained by "solving the constraint equations".
The definition requires that, at fixed time, the total (...
1
Whether the tension is a constraint, or not, depends how you model the problem.
Method 1: consider "the two masses plus the rope" as one body, and use just one coordinate to measure its position. Obviously the "single body" changes shape as it moves, and one mass moves up and the other moves down, but that doesn't affect the general principle of calculating ...
1
What is the constraint enforce by the tension? How does it show up in your generalized coordinates?
As I noted in the comments it is usual to chose a set of generalized coordiantes with a single position for each rope (and the location of the other end found by calulating from there); this form has the constraints built-in, so that there is no way to ...
answered Jan 10 at 17:32
1
The Euler-Lagrange equation following from this lagrangian is the Klein-Gordon equation, which is the field theoretical equivalent of the special relativistic energy-mass-momentum equation.
1
As you say, when building a Lagrangian for a theory, we almost always wish it to be a Lorentz scalar. Obviously, we can take any power of the scalar field, $\phi$, that is, we can have terms of the form,
$$\mathcal L = \dots + \sum_{n\geq 0} c_n \phi^n + \dots$$
with coefficients of our choosing. This means we could consider a term like $\sin \phi$ since ...
1
Comments & hints:
The invariance $\delta L=0$ of the Lagrangian (1) under the infinitesimal transformation $$\delta q^a~=~\epsilon v^a(q), \qquad \delta t~=~0,$$ follows directly from the Killing equation (2).
According to Noether's theorem, the corresponding conserved Noether charge $Q=p_av^a$ is the momentum $$p_a~=~\frac{\partial L}{\partial \dot{q}^...
1
Your radial coordinate should be changing, so
$$\vec r = (r_0-v_r t)\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}$$
Derive it with respect to $t$ to get the velocity $\vec v=\dot{\vec r}$ and plug it into
$L=\frac{1}{2}m\vec v^2$.
1
It seems that OP is not questioning the standard convention to divide each term in the Lagrangian with its symmetry factor, e.g., $${\cal L}~=~-\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{3!}\phi^3.$$
Rather OP is assuming the above standard convention, and asks if the vertex factor$^1$ is $-\frac{i}{\hbar}\lambda$ ...
1
It is conventional to write interactions normalized by the number of permutations of identical fields. So, there will be a $\frac{1}{n!}$ factor for each interaction with $n$ identical fields. This factor is then canceled by the $n!$ ways of permuting the $n$ identical lines coming out of the same internal vertex.
The diagram is therefore associated with ...
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