New answers tagged lagrangian-formalism
9
votes
How does the absence of quadratic terms in the Lagrangian imply massless quanta?
I will expand a bit on one of the answers to be clearer why a quadratic term usually leads to the 'mass' term. Because the coefficient in front of the $A_\mu A^\mu$ term has dimension $2$ doesn't by ...
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3
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How does the absence of quadratic terms in the Lagrangian imply massless quanta?
If by mass we mean the coefficient in front of the quadratic terms, then the answer is straightforward... (although one might ask whether such a situation could be also treated as a case of infinite ...
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7
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How does the absence of quadratic terms in the Lagrangian imply massless quanta?
This is because if you have a term in the Lagrangian that is proportional to $A_{\mu}A^{\mu}$, with some coefficient in front of it, then its coefficient is going to automaticlly be considered the ...
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2
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Tensionless string in Nambu-Goto action
No, it does not make sense to talk about a theory of fundamental relativistic tensionless strings.
Consider the Nambu-Goto action for the relativistic string
$$ S_{NG} = -T\int d\tau d\sigma_{proper} $...
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2
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Nambu-Goto action and the World-Sheet Area
I can't comment on your first question, but for the second one you can just imagine the unit sphere in $\mathbb{R}^3$. Let's parametrize the sphere via
\begin{align*}
X^\mu(\theta, \varphi) = \begin{...
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2
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Calculations with co- and contravariant formalism in QFT
Here is a Hint: start with a term of the Lagrangian density (I will not do the whole thing, I will just give you a feeling of it), say $\mathcal{L}_0=(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})=\...
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9
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What is the full QED Lagrangian with physics units written out?
The Wikipedia article you linked to gives the action, not the Lagrangian or Lagrangian density, so I’ll do the same.
Note that since $x^0 \equiv ct$, we have $d^4x = c \, dt \, d^3x$ and therefore
$$S ...
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3
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Peskin and Schroeder, Linear sigma model, renormalized perturbation theory
That's a good question.
The linear sigma model (11.14) has 3 terms, and hence 3 counterterms, and hence needs 3 renormalization conditions (11.17a+b+c).
Yes, since the $N$th component
$$\phi^N(x)~=~ ...
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0
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Derive Linearized Einstein's equation from Lagrangian approach
Below is a derivation of the linearized Einstein's equation from the given Lagrangian.
Since $h^{\mu\nu}$ does not appear in the Lagrangian, we have $\frac{\delta L}{\delta h^{\alpha\beta}} = 0 $. So ...
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2
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Why do we put factors of zero in a Lagrangian that is to be extremized?
The problem which motivates the method of Lagrange multipliers is constrained optimization. We have a function $f$ of several variables that we would like to optimize subject to a holonomic ...
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2
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Why do we put factors of zero in a Lagrangian that is to be extremized?
The discussion in the comments suggests that the following might help clarify the confusion.
Original problem The original maximization problem is this: Maximize
$$\ln\Omega_n = - \sum_j n_i \ln \frac{...
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5
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Why do we put factors of zero in a Lagrangian that is to be extremized?
I will expand a bit on why this Lagrangian is defined this way. If I want to maximize a function $f(p_1,\dots,p_n)$ over all its parameters I can set the gradient to zero.
$$\nabla f(\vec p)=0$$
This ...
1
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Can you rewrite the QCD lagrangian in terms of hadron?
By "exactly" you seem to imply "directly", or ab initio. People have built bridges between fundamental QCD on the lattice and σ-models summarizing chiral symmetry breaking for ...
- 54.1k
1
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Why do we put factors of zero in a Lagrangian that is to be extremized?
The question is:
So, this logic implies that equation $\mathrm(7)$ should be written as
$$\frac{\partial}{\partial n_j}\left[\ln \Omega_n-\alpha\color{red}{\left(\sum_jn_j-N\right)}-\beta\color{red}{\...
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1
vote
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Two constraints of $\bar\psi$ from equations of motion for Free Dirac Field Lagrangian
(I haven’t checked the equations of motion you have written, I am assuming they are right)
In quantum field theory fields $(\phi)$are operators but positions $x$ are not. So $\partial_\mu$ is also not ...
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4
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Geometrical intuition for Noether's Theorem
The most basic intuitive explanation of Noether's theorem is that it is the extension to generalised coordinates of the principle that since force is defined as the rate of change of momentum, if ...
1
vote
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Why rescale the kinetic term in Wilsonian renormalization?
OP is discussing the Wilsonian effective action (WEA), cf. e.g. my Phys.SE answer here. A path integral contains a freedom to perform field redefinitions and scale fields. Normalization of a non-zero ...
- 184k
0
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Local Lorentz invariance of General Relativity
For a different flavour of answer, you may want to look into the idea of Lie-algebra valued differential forms. This is because the parameter $\lambda^{a}_{~~b}$ that you use for an infinitesimal ...
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How to find Belinfante-Rosenfeld SEM tensor?
Comparing eqs. (i) and (ii) suggests that
$$ B_{\lambda \mu \nu} -(\mu\leftrightarrow \nu)~=~ -f_{\lambda \mu \nu}.\tag{iv}$$
Now eqs. (iii) is an automatic consequence of eqs. (i) and (iv).
How to ...
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0
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Lorentz/rotational invariance parameter doesn't vanish on boundaries
OP and their teacher are using the trick of $x$-dependent parameters in Noether's first theorem.
The variation $\delta x^{\mu}~=~\epsilon^{\mu}(x)=\omega^{\mu}{}_{\nu}(x)~x^{\nu}$ is never ...
- 184k
1
vote
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Virtual displacement in semi-holonomic constraints
Notice that eq. (2.68) follows immediately from eq. (2.66). However OP wants to derive eq. (2.68) from eq. (1). Unfortunately, OP's project is doomed.
On one hand, a virtual displacement $\delta$ (in ...
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1
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Why rescale the kinetic term in Wilsonian renormalization?
As far as I know, the idea is just that generally, you can just redefine the action with a rescaled field without changing the physics. As such this scaling is just a redundancy in your description ...
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2
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Is the divergence of the energy tensor related to the equations of motion?
The conservation of the energy-momentum tensor is indeed related to the equation of motion in the sense that it is only conserved if the equations of motion are satisfied. One way to see this is the ...
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0
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Photon propagator and the Fermi Lagrangian density
Sorry, i think i messed up with terms in the equations, i can try once more to ask my question:
My professor sayd that the Fermi lagrangian density
$$\mathcal{L_{fermi}}-\frac{1}{2}(\partial_\mu A^\nu ...
0
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How Feynman's path integral lead to least action principle? Math proof needed
You can look up "Laplace approximation" for integrals. In the path integral formalism the word "instanton" is also used.
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4
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Does there exist a square root of Euler-Lagrange equations of a field? (Factorization)
The "square root" in the sense of $-(a^2 +b^2)=(ia-b)(ia+b)\,\,$
of the Klein-Gordon's differential operator is the Dirac differential operator. That is the main message:
$$(i\gamma^\mu\...
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3
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Does there exist a square root of Euler-Lagrange equations of a field? (Factorization)
The Euler Lagrange equations are the equations used to minimise the action. So, they do not exist in versions (i.e. square root of EL eqs etc). What you should be after if you would want (for some ...
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0
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Interpretation of $\phi^n$ terms in Lagrangian density
Let us introduce the notation that $S_n$ denotes action terms that depend on the $n$th power of the fields.
A quadratic action $S_2$ (and hence linear EL equations) corresponds to a free (=non-...
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0
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Photon propagator and the Fermi Lagrangian density
Note that the Lagrangian density (5.10) needs the Lorenz gauge fixing condition
$$ \partial_{\mu}A^{\mu}~=~0\tag{5.13}$$
in order for its EL equation to correctly reproduce the Maxwell equations.
...
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3
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Understanding this abstract Lagrangian of effective field theory
The two definitions are the same since in $d=4$ the (power-counting) renormalizable operators are those with dimension $d \leq 4$.
"Local" operator means the operator can be written as a ...
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0
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What is the physical significance of the vector field term $X_{\nu}$ in the improved Noether current $T^{\mu\nu}X_{\nu}$?
Whatever I understand most of the cases, we do not really know whether the $T^{\mu\nu}$ is actually symmetric or not, but, it should be conserved covariantly because improvement terms are in general, ...
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3
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Second-order Lagrangian of Einstein-Hilbert action
You're going in the right direction, but it's not quite correct because you also need to find the first order perturbation of the square root of the determinant:
$$(\sqrt{-g} R)^{(2)} = \sqrt{-g}^{(0)...
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3
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Proof that the axial current is conserved in classical QED
The transformations are not what you have. The field $\bar \psi$ is defined by $\bar \psi = \psi^\dagger \gamma_0$, so they should be
$$
\psi\to e^{i\gamma^5 \alpha}\psi, \quad \bar\psi \to \bar\psi ...
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2
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Vertex factors for Feynman rules in QCD
The interaction vertex with 3-gluons is:
\begin{align*}
S_{I(3G)}=&- ig_s \eta_{\rho \nu} f^{a b c} \int d^4 x \int d^4 p_1 \int d^4 p_2 \int d^4 p_3\,p_{1 \mu} \widetilde{g}^{a \rho}(p_1) \...
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0
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How to expand Maxwell Lagrangian?
The calculation in the question is correct, but incomplete. We can continue as follows
$$
\begin{align}
L&=-\frac{1}{4}[(\partial_\mu A_\nu)^2+(\partial_\nu A_\mu)^2-2(\partial_\mu A_\nu)(\...
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3
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How do I understand the Hodge $⋆$ operator in Yang-Mills Lagrangian?
Intuitively, you can think of the Hodge dual and outer product geometrically as you already do, and this combination as a sort of generalised dot product.
$A\wedge B$ has a magnitude that extracts the ...
8
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How do I understand the Hodge $⋆$ operator in Yang-Mills Lagrangian?
The Hodge dual of a $p$-form (antisymmetric $(0,p)$ tensor) in $d$ dimensions is a $(d-p)$-form whose components are defined by
$$
(\star A)_{\mu_1 \cdots \mu_{d-p}} \equiv \frac{1}{p!} \epsilon_{\...
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