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3 votes
1 answer
89 views

Equation for real/complex $\phi^4$ theory

On wikipedia (see this link), the Lagrangians of the $\phi^4$ equation for real AND complex scalar fields are given. One may derive the Klein-Gordon equation by inserting into the Euler-Lagrange-...
Octavius's user avatar
  • 743
2 votes
3 answers
110 views

Sign conventions for the Lagrangian from the EM Lagrangian density

In Chapter 13.6 of the 3rd edition of Goldstein's Classical Mechanics, Goldstein proposes the Lagrangian density of the electromagnetic field as: $$\mathcal{L} = -\frac{F_{\lambda \rho} F^{\lambda \...
tugboat2's user avatar
-1 votes
2 answers
126 views

Why do we multiply the Euler-Lagrange equations by negative one?

As I've learned classical mechanics from different sources, I've seen both $$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right) - \frac{\partial L}{\partial q_k} = 0,$$ and $$\frac{\...
wlancer's user avatar
  • 183
0 votes
1 answer
90 views

The Klein-Gordon equation and the sign of the mass term

A derivation of the Klein-Gordon equation starts with the following lagrangian for a scalar field ϕ: $$ L=\frac{1}{2}g^{ab}(∇_a\phi)(∇_b\phi)-V(\phi) $$ If we plug this lagrangian in the Euler-...
DrD's user avatar
  • 141
1 vote
0 answers
82 views

Diagrammatic derivation of the Dyson–Schwinger equation

I am following chapter 9 of the Rammer's book on Field Theory (which you can find here: https://www-thphys.physics.ox.ac.uk/talks/CMTjournalclub/sources/Rammer.pdf). I am referring to section 9.2.2, ...
Ruth Murphy's user avatar
1 vote
0 answers
23 views

Definition of energy for a Lagrangian with constraints [duplicate]

Suppose I have the Lagrangian in spherical coordinates: $$\mathcal{L} = T=\frac{1}{2}m (\dot{r}^2 + r^2 \sin^2 \theta \dot{\phi}^2 + r^2 \dot{\theta}^2)$$ Where $T$ is the kinetic energy. Now suppose ...
Jbag1212's user avatar
  • 2,599
1 vote
1 answer
69 views

Sign convention for the Lagrangian of a free massive point particle in general relativity

As far as I understand, the Lagrangian of a massive free particle in the context of general relativity is the following: $$L=-mc\sqrt{g_{\mu\nu}\dfrac{dx^\mu}{dt}\dfrac{dx^\nu}{dt}}.$$ But is this the ...
Wild Feather's user avatar
0 votes
0 answers
61 views

Combinatorics in $\lambda \phi^4$ theory and critical exponents

I am trying to understand critical phenomena from the perspective of Statistical Mechanics. The interacting term in the $\lambda \phi^4$ scalar theory is usually (but not always) multiplied by $\frac{...
Ruth Murphy's user avatar
2 votes
2 answers
273 views

Does the Lagrangian have to have units of energy?

I've read that the standard formulation of the lagrangian is $L = T-V$ which has energy units (Joules). But when I read a bit about electromagnetism and relativity I've seen that the lagrangian is ...
user1127202's user avatar
1 vote
0 answers
46 views

On the metric signature and the energy-momentum tensor [duplicate]

Given a Lagrangian $$\mathscr{L} = -\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - V(\phi),\tag{1}$$ is the metric always with signature $(-, +, +, +)$? It seems to me that This post Sign Convention ...
Geigercounter's user avatar
0 votes
0 answers
31 views

Sign of Lagrangian constraint function in rotations

How can I foolproof decide the correct sign of the constraint function in Lagrangian mechanics, particularly in rotations? In a simple example, lets consider a cylinder rotating around its main axis ...
Daniel's user avatar
  • 51
1 vote
1 answer
50 views

Convention when considering a mathematical pendulum attached to an object with a spring [closed]

Sorry for the horrible picture, it was the best I could do. I am trying to find the lagrangian for this set up, but I have problem with the convention of the potential energy, whether it should be ...
imbAF's user avatar
  • 1,398
1 vote
3 answers
791 views

How did we find the Noether current $j^\mu(x) = \bar\psi(x)\gamma^\mu\psi(x)$ for Dirac equation?

The Dirac Lagrangian reads: \begin{equation*} \mathcal{L} = \bar\psi(i\not\partial-m)\psi.\tag{1} \end{equation*} It's invariant under the transformation $\psi(x)\rightarrow e^{i\alpha}\psi(x)$. Now ...
IGY's user avatar
  • 1,783
1 vote
1 answer
89 views

Why do we multiply $(1+\delta)$, but just add $\delta$ to construct counterterms for a Lagrangian?

Consider the Lagrangian $$ L = \frac{1}{2}(\partial^\mu\phi\partial_\mu\phi-m^2\phi^2)+\bar\psi(i\not\partial-m)\psi-g\phi\bar\psi\psi. $$ I was told when we include the counterterms, it becomes $$ L =...
IGY's user avatar
  • 1,783
0 votes
1 answer
279 views

Lagrangian formalism for boson mass

Why for charged boson, it's mass term is $m^2 W^+_{\mu}W^{-\mu}$? While for neutral boson, it's mass term is $\frac{1}{2}m^2 Z_{\mu}Z^{\mu}$. (Is their a mathematical reason that charged boson mass ...
Daren's user avatar
  • 1,421

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