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1 vote

Worldsheet action in the presence of background fields in complex coordinates

The integration measure is always a top-form. This means that $d^2\sigma$ is really (proportional to) a 2-form $\mathrm{d}\sigma\wedge\mathrm{d}\tau$. $d^2z$ is really (proportional to) a 2-form $\...
Qmechanic's user avatar
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0 votes

Gauge boson interacting with derivative of the Higgs field

Look at a simpler model, with a charged scalar and a $U(1)$ field $A_\mu$. Then we have $$ \mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} + \left(D_\mu \Phi\right)^\dagger \left(D^\mu \Phi\right) $$ ...
Andrew's user avatar
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0 votes

Gauge boson interacting with derivative of the Higgs field

The pages you are referring to are on Spontaneous Symmetry Breaking, and the Higgs particle, h, is "debris" in the process and the Higgs mechanism, mere dross, so your text wisely postpones ...
Cosmas Zachos's user avatar
2 votes
Accepted

Question about functional derivative computation in Quantum Field Theory for the Gifted Amateur

Working at $\mathcal{O}(\epsilon)$ and the binomial expansion you have $$[f(y)+\epsilon\delta(x-y)]^p\approx[f(y)]^p+\epsilon{p\choose 1}[f(y)]^{p-1}\delta(x-y)+\mathcal{O}(\epsilon^2).\tag{1}\label{...
Mr. Feynman's user avatar
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5 votes

Does a massless boson imply an infinite interaction distance?

Using the basic formulae $$\Delta \frac{1}{r}=-4\pi \delta^{(3)}(\vec{x}) \quad \text{with} \quad r=|\vec{x}| \tag{1}$$and $$\Delta \frac{f(r)}{r} =\frac{f^{\prime \prime}(r)}{r} \quad \text{for}\quad ...
Hyperon's user avatar
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0 votes

In Lagrangian mechanics, do we need to filter out impossible solutions after solving?

The way I see it: application of Hamilton's stationary action is a two stage process: Use stationary action to obtain an equation of motion Use the equation of motion and initial conditions to obtain ...
Cleonis's user avatar
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0 votes

Discrepance between gauge symmetry and Noether's first theorem

There's no conserved charge associated with the local symmetry. The reason for that is that gauge "symmetry" is not a physical symmetry. The current for the EM Lagrangian is $J^\mu=\partial_\...
Lucky Charms's user avatar
0 votes

IPhO2014 T1 with Lagrange multipliers

To solve for the unknows $~\ddot x~,\ddot y,\ddot d~,\lambda~$ you need additional equation which come from the constraint equation $$ g=(x-d)^2+y^2-R^2=0$$ from here $$\ddot g=(x-d)\ddot x+y\,\ddot y-...
Eli's user avatar
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1 vote
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Derivative interactions in the Wilsonian renormalisation Group

I will try to answer your questions one by one. Please comment if you need any clarifications. Before I do that, let me briefly review the idea of RG flow which might make the answers to (1) and (2) ...
Nandagopal Manoj's user avatar
4 votes
Accepted

Different representations of the Yukawa interaction

Your second expression is meaningless. You need to saturate a row vector with a column vector, not to mention the spinor implicit index saturation, so it should be, instead, $$y_e \bar{e}_R H^\dagger ...
Cosmas Zachos's user avatar
3 votes

Four-divergence term in Lagrangian

It is known that adding a four divergence term, $\partial_\mu A^\mu$ does not affect the equations of motion. I am trying to reason this based on the Euler-Lagrange equation. But I want to show this ...
hft's user avatar
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8 votes
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How can a scalar field have components and how do I interpret these components?

The components of the electric (field) vector $\vec{E} \,$ "live" in ordinary three-dimensional space (where we also live in). If you place a small test charge $q$ at some point in space (...
Hyperon's user avatar
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1 vote

Oscillating inverted hemisphere Lagrangian mechanics problem

In any circumstance, we can account for the kinetic energy of a rotating object in one of two ways: $T = \frac12 I \omega^2$, where $I$ is the moment of inertia about the (instantaneous) pivot point ...
Michael Seifert's user avatar
0 votes

Oscillating inverted hemisphere Lagrangian mechanics problem

By considering a rolling full sphere (or a wheel for simplicty), its clear, that the full sphere center moves horizontally like an the axis of a wheel during oscillating by rolling without slip. The ...
Roland F's user avatar
1 vote

Oscillating inverted hemisphere Lagrangian mechanics problem

Probably simplest to take the moment of inertia about the hemisphere's centre of mass. Then if the centre of mass moves the hemisphere has translational kinetic energy and if the hemisphere rotates ...
gandalf61's user avatar
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2 votes

Equations of motion for Lagrangian of scalar QED

I have to determine the equations of motion for both the complex scalar field $\varphi$ and the electromagnetic field $A_\mu$ by using the Euler-Lagrange equations. Now I know, that because the ...
hft's user avatar
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1 vote

What is the physical meaning of the counterterms we add in Lagrangians?

There are two answers to this depending on what you mean. Based on your question it seems both levels of understanding would be useful. The second answer actually has more of a "physical meaning&...
JohnA.'s user avatar
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0 votes

Onsager-Machlup functional and the Boltzmann distribution

One can find the extremum trajectory and the quadratic fluctuations around it - by usual procedure of varying action. (If I am not mistaken, this leads directly to Newton equations with damping.) This ...
Roger V.'s user avatar
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2 votes

Why the choice of Configuration Space in Hamilton's Principle is $(q, \dot{q}, t)$?

Well, the function $f$ in the action integral is always identical to the Lagrangian in Hamilton's principle: $$\delta I=\delta\int_1^2L(q,\dot q,t)\;dt=0,\;\;\text{For Some Path in Configuration space}...
Albertus Magnus's user avatar
0 votes
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Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?

On the question whether it is possible to go from Newton's laws to Hamilton's stationary action: Yes, that is possible. (As pointed out by physics stackexchange contributor Kevin Zhou: in physics you ...
Cleonis's user avatar
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2 votes

Tychonian Universe

but I cannot seem to use these and a coordinate transformation to find: $$r_{ME}=-r_{ES}(t)+r_{MS}(t)$$ to plot a graph that resembles image [1] The problem seems to be that you are considering only ...
hft's user avatar
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0 votes

Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?

Yes it is typically the case for field theories. A simple construction involves many harmonic oscillators with large frequencies. When there are an infinite number, the frequencies can be unbounded ...
LPZ's user avatar
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0 votes

Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?

If you divide the action integral by the fixed time between the two fixed points in space, the variation principle states, that all paths, invariant with respect to the time mean of the diffenence ...
Roland F's user avatar
4 votes
Accepted

How to Relate the Functional Derivative to Infinitesimal Change in Noether's Theorem

I will be quoting a lot from the functional derivative page on wikipedia, since it provides a great reference. Recall that a function is an object which takes as input a function and has as output a ...
AccidentalTaylorExpansion's user avatar
5 votes

How to Relate the Functional Derivative to Infinitesimal Change in Noether's Theorem

One way to do a calculation similar to OP's without a notion of infinitesimal is to consider a one-parameter deformation of the field $\phi^\epsilon$ (with $\phi = \phi^0$) and define $\delta$ to mean ...
SolubleFish's user avatar
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1 vote

How to Relate the Functional Derivative to Infinitesimal Change in Noether's Theorem

It is not entirely clear what OP is trying to achieve. If OP is trying to avoid using infinitesimal variations and infinitesimals, one can in principle equivalently use vector fields and Lie ...
Qmechanic's user avatar
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2 votes

How does this term in the Majorana mass not vanish?

Remember that $\nu_L$ is a two-component column vector with Grassmann entries $\nu_L^1$ and $\nu_L^2$, so with $$ \sigma_2 =\left[\matrix{0&-i\cr i&0}\right], $$ we have $$ \nu_L^T \sigma_2 \...
mike stone's user avatar
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4 votes
Accepted

Why are 2-point functions Green's functions?

It follows from the Schwinger-Dyson equations. Let's prove it for the case you're interested in (though the equations are more general). I will do it for a scalar field and let you work out the ...
Prahar's user avatar
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1 vote
Accepted

Problem solving for Wilsonian Effective Action

I haven't checked the details of your integrals, but assuming that that's all done correctly, the only thing you're missing is log rules and a series expansion. Using $a_i$ to mean the corresponding ...
Rokas Veitas's user avatar
4 votes
Accepted

Hamiltonian of a complex scalar quantum field

In field theory texts, e.g. Peskin and Schroeder, one will often see the stress energy tensor written simply as: $$T^{\mu\nu}={\partial\mathcal L\over\partial(\partial_\mu\phi)}\partial^\nu\phi-\delta^...
Albertus Magnus's user avatar
11 votes

Hamiltonian of a complex scalar quantum field

Where did I go wrong? Seems like probably there are a number of different places you could have gone wrong. Since this is a homework-like question, I will answer in terms of classical field theory, ...
hft's user avatar
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0 votes

Conserved current transforming under adjoint

Charges are constructed by integrating currents, $$ Q^a = \int_\Sigma n^\mu j_\mu^a $$ $Q^a$ transforms in the adjoint, therefore $j^a_\mu(x)$ must transform in the adjoint.
Prahar's user avatar
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2 votes

Changing variables from $\dot{q}$ to $p$ in Lagrangian instead of Legendre Transformation

Obviously, you can write the Lagrangian $L(t,q,\dot{q})$ as a function of $t,q,p$ instead of $t,q,\dot{q}$, by just inverting the relation $$p= \frac{\partial L(t,q,\dot{q})}{\partial \dot{q}}.$$ ...
Valter Moretti's user avatar
1 vote

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

Why not? Those two forms are mathematically equivalent, since $a=b$ is same identity as $-a=-b$, so you can choose one of those representations which look nicer to you. For example, say we are ...
Agnius Vasiliauskas's user avatar
0 votes

Conserved current transforming under adjoint

It's straightforward, provided you appreciate the currents and their charges are the vectors (states) of the adjoint, not its transformation matrices/generators. Ignoring normalizations and ...
Cosmas Zachos's user avatar
0 votes

Any good resources for Lagrangian and Hamiltonian Dynamics?

Ok, so some of the other answers here seem like they may have forgotten what it's like to be in your position as a starting student (Goldstein as a starter book... really? Lagrange's original work?! ...
-1 votes

Directly integrating the Lagrangian for a simple harmonic oscillator

The action is the difference of time averaged kinetic and potential energy, that is, zero.
my2cts's user avatar
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0 votes

Directly integrating the Lagrangian for a simple harmonic oscillator

Indeed you have the option of actually performing the integration as stated in (1). $$ S = \int\left(\frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\right)dt.\tag{1} $$ That is to say: The only way to ...
Cleonis's user avatar
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1 vote

Directly integrating the Lagrangian for a simple harmonic oscillator

This question can be cast in a convenient way by using the initial conditions $$ x_a=x(t_a) \quad x_b=x(t_b), $$ and taking $$ x(t)=A(x_a,x_b)\cos(\omega t)+B(x_a,x_b)\sin(\omega t). $$ Then, you can ...
Jon's user avatar
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3 votes
Accepted

Directly integrating the Lagrangian for a simple harmonic oscillator

Well, if we know the classical solution $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ (which we do for the harmonic oscillator), we can plug it into the action functional $S[q]$ and obtain the on-shell action ...
Qmechanic's user avatar
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1 vote

Functional variation about metric

OP is right. In order to be able to vary wrt. the metric $g_{\mu\nu}$ it is important to know the hidden/implicit dependence of $g_{\mu\nu}$ in various quantities. Be aware that this is fundamentally ...
Qmechanic's user avatar
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0 votes

Functional variation about metric

I'm not sure how to format. I'm new here. I'm also very rusty. I'm helping because I'm not sure whether or not someone else will answer. This is my best thought. I feel like that term should be ...
Steven Dorsher's user avatar
1 vote

Difference between renormalizable and super-renormalizable theories

Qmechanic's answer is very good and tackles your main question well. I want to however add an important detail to your first question. Namely, we generally do not just consider $[\lambda]$ when ...
mika's user avatar
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2 votes
Accepted

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

If you were to prefer one over the other, perhaps you could argue for $$\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right) = 0.$$ since that's what ...
Señor O's user avatar
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0 votes

How to show the Gauss-Bonnet term is a total derivative?

The bulk of this post is correct for positive definite metrics, but the Lorentzian case is more subtle than I thought. See remarks at the end of the answer. What I find odd is that nobody has ...
Bence Racskó's user avatar
3 votes
Accepted

Difference between renormalizable and super-renormalizable theories

It should stressed be that Peskin & Schroeder are here using the old Dyson definitions of renormalizability. For a more general derivation of eq. (10.13), see e.g. this Phys.SE post, which also ...
Qmechanic's user avatar
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4 votes

Ghosts in QCD Lagrangian

Briefly speaking, OP's Lagrangian density (1) is the un-gauge-fixed original Lagrangian density ${\cal L}_0$ for QCD. ${\cal L}_0$ defines the classical theory. Now let us quantize the theory. In the ...
Qmechanic's user avatar
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1 vote

Derivation of the geodesic equation. Why do we start with the special relativistic action?

The action between 2 values $a$ and $b$ of a parameter $\lambda$ is: $$s_{ab} = \int_a^b \sqrt{\frac{\partial X^{\mu}}{\partial \lambda}\frac{\partial X^{\nu}}{\partial \lambda}g_{\mu\nu}} d\lambda$$ ...
Claudio Saspinski's user avatar
3 votes

Derivation of the geodesic equation. Why do we start with the special relativistic action?

The derivative $\frac{ds}{d\lambda}$ gives a vector field which is tangent to the world line at every point of the world line. This tangent vector does not depend on how space behaves away from the ...
paulina's user avatar
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8 votes
Accepted

Ghosts in QCD Lagrangian

It's conventional to specify the classical Lagrangian, which does not include ghost terms. (Ghosts only contribute at loop level). One reason not to write ghost terms, when one is speaking generically ...
Andrew's user avatar
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