Questions tagged [boundary-terms]
128 questions
29
votes
2 answers
Invariance of Lagrangian in Noether's theorem
Often in textbooks Noether's theorem is stated with the assumption that the Lagrangian needs to be invariant $\delta L=0$.
However, given a lagrangian $L$, we know that the Lagrangians $\alpha L$ (where $\alpha$ is any constant) and $L +…
Isaac
- 2,986
25
votes
2 answers
Explicit Variation of Gibbons-Hawking-York Boundary Term
Are there any references that present the explicit variation of the Hilbert-Einstein action plus the Hawking-Gibbons-York boundary term, and demonstrate the cancellation of the normal derivatives of metric variations? I have tried to read the…
Michael Shaw
- 497
13
votes
1 answer
Does the Gibbons-Hawking boundary action have an anomaly inflow interpretation?
The Einstein-Hilbert action on a manifold $M$ with boundary is
$$\frac{-1}{16\pi G}\int_M d^n x \sqrt{-g} R +\frac{1}{8\pi G} \int_{\partial M} d^{n-1}x \sqrt{|h|} K$$
where $K$ is the extrinsic curvature of $\partial M$ in the induced metric $h$.…
Dwagg
- 2,062
13
votes
2 answers
Necessity of the Gibbons-Hawking-York (GHY) boundary term
The fundamental point of my question is whether the GHY-boundary term in general relativity is even necessary at all, and if yes, then why is it so, and what is its physical significance.
Several points:
Despite the appearance of an integral, the…
Bence Racskó
- 11,869
11
votes
1 answer
When is numerical value of Lagrangian evaluated on-shell a full differential?
I noticed recently that for many field equations, Lagrangian evaluated on-shell (i.e. using equations of motions) is a full derivative- a divergence or something, or in other words a boundary term. This is true for Weyl, Dirac, Klein-Gordon (without…
Blazej
- 2,267
11
votes
1 answer
Evaluating the Einstein-Hilbert action
The Einstein-Hilbert action is given by,
$$I = \frac{1}{16\pi G} \int_{M} \mathrm{d}^d x \, \sqrt{-g} \, R \, \, + \, \, \frac{1}{8\pi G}\int_{\partial M} \mathrm{d}^{d-1}x \, \sqrt{-h} \, K$$
including the Gibbons-Hawking-York boundary term. A…
user2062542
- 739
10
votes
2 answers
How the boundary term in the variation of the action vanishes
In David Tong's QFT lecture notes (Quantum Field Theory: University of Cambridge Part III Mathematical Tripos, Lecture notes 2007, p.8), he states that
We can determine the equations of motion by the principle of least action. We vary the path,…
user12906
9
votes
3 answers
Surface terms for path integrals in field theory?
My question relates to something that I´ve seen in many books and appears in all its glory here: Ryder, pg 198
My question is about eq. 6.74. Which I repeat below:
$$i \int {\cal D}\phi \frac{\delta \hat{Z} [\phi] }{\delta \phi} \exp \left(i\int…
Forever_a_Newcomer
- 2,546
9
votes
2 answers
Variation of Action with time coordinate variations
I was trying to derive equation (65) in the review by László B. Szabados in Living Reviews in Relativity (2002, Article 4)
This slightly unusual then usual classical mechanics because it includes a variation of time also, $\delta t$.
Usually one…
user50482
- 93
8
votes
1 answer
Why the variation of a surface term is zero?
My original question is like:
Why are the Euler-Lagrange equations invariant if we add a surface term to the action?
And there is an answer by Javier:
https://physics.stackexchange.com/a/205585/
I have some questions about Javier's answer, but…
Hallow_Juan
- 95
- 5
8
votes
4 answers
Why can we neglect surface terms in Field Theory?
My teacher says that the integral $$\int _{-\infty }^{\infty }\frac{\partial J^{\mu }}{\partial x^{\mu}}d^4x$$ that we met in QFT can always be neglected since $$\int _{-\infty }^{\infty }\frac{\partial J^{\mu }}{\partial x^{\mu}…
ZHANG Juenjie
- 1,101
8
votes
2 answers
Why are these two definitions for symmetries in the Lagrangian equivalent?
I have heard the following two definitions for a symmetry of the Lagrangian:
If under a coordinate transformation the form of the Lagrangian
remains unchanged then there is a symmetry.
If $\delta \mathcal{L}=\partial_\mu F^\mu$, where $\mathcal{L}$…
Dargscisyhp
- 5,380
- 5
- 30
- 49
8
votes
1 answer
Boundary term in Einstein-Hilbert action
Why is the boundary term in the Einstein-Hilbert action, the Gibbons-Hawking-York term, generally "missing" in General Relativity courses, IMPORTANT from the variational viewpoint, geometrical setting and the needs of Black Hole Thermodynamics?…
riemannium
- 6,843
6
votes
3 answers
Something fishy with canonical momentum fixed at boundary in classical action
There's something fishy that I don't get clearly with the action principle of classical mechanics, and the endpoints that need to be fixed (boundary conditions). Please, take note that I'm not interested in answers with the hamiltonian, which isn't…
Cham
- 8,015
6
votes
1 answer
Why does a surface term have to vanish for an operator to be Hermitian?
In section $1.10$ of Shankar's Principles of Quantum Mechanics, it is stated that an operator $K$ satisfying $\langle x \vert K \vert x' \rangle = \langle x' \vert K \vert x \rangle ^*$ is not enough to guarantee that $\langle g \vert K \vert f…
weirdmath
- 141