11

The term vanished because we can translate this term to one making a statement about the fields at the boundary and assume that the fields themselves vanish in spatial and temporal infinity. By Stokes' Theorem, we can translate volume integrals into surface integrals. More specifically Gauss' Theorem states that the integral of a divergence of a field over ...


8

The last term is of the form $$ \int d^n x \partial_\mu X^\mu $$ and by Stoke's theorem, this equals a surface integral at $\infty$ $$ \int_\infty ds^\mu X_\mu $$ which vanishes provided $X^\mu\to 0$ sufficiently rapidly at infinity. Addendum. I realize that this is a rather terse answer, so let me add some detail. Notice that the oriented surface ...


6

Theorem: let $L$ be a homogeneous function of degree $k$; then the on-shell lagrangian is a total derivative. Proof: according to the Euler's homogeneous function theorem, $$ k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1} $$ On the other hand, because of the Euler-Lagrange equations, $$ (1)=q\frac{\partial L}{\...


6

It seems that OP is pondering the following. What happens in a field theory [in OP's case: GR] if spacetime $M$ has a non-empty boundary $\partial M\neq \emptyset$, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields $\phi^{\alpha}(x)$ [in OP's case: the metric tensor $g_{\mu\nu}(x)$]? I) Firstly, it should stressed that ...


6

This the Variational Problem with unknown end-time and one proceeds like this: $$\delta S= \int_{t_i}^{t_f+\delta t_f} L \left(q+\delta q,\dot{q}+\delta \dot{q},t\right) dt - \int_{t_i}^{t_f} L \left(q, \dot{q}, t\right) dt$$ After several transformations and integration by parts one finally gets the usual Euler-Lagrange diff eq plus a boundary condition ...


6

I) In general, for a given choice of boundary conditions, it is important to adjust the action with compatible boundary terms/total divergence terms in order to ensure the existence of the variational/functional derivative. As OP observes, the problem is (when deriving the Euler-Lagrange expression) that the usual integration by parts argument fails if the ...


6

Perhaps its a little clearer if you shorten the contents of the brackets (and lets drop the constants too): $$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$ $$ = \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} ...


5

It's not quite clear what the inner workings of your function are, because your $K$ does not match 't Hooft's, and you have not provided your working. However, there is a simpler test case which encapsulates the problem better. Consider the zero-order modified Bessel function $K_0(x)$, which has the integral representation $$\mathop{K_{0}}\left(x\right) = \...


5

Well, if you have a term like $\partial_\mu \mathcal{J}^\mu$, the divergence theorem lets you convert it into a surface term upon integrating to find the action, and since variations are assumed to vanish at the boundary, this term goes away. The Euler-Lagrange equations don't change because they come from setting the variation of the action to zero. ...


4

Requiring only first derivatives in an action is required because higher time derivatives generically lead to ghost instabilities in the field theory. This goes under the name "Ostragradsky instability," see the excellent reference by Richard Woodard discussing the kinds of issues that arise when this sort of instability is present. The main problem with ...


4

Two points. First, the variation of the GHY term is discussed in detail in this post: Explicit Variation of Gibbons-Hawking-York Boundary Term Second, the GHY term on its own is enough to render the boundary value problem well-defined, but it is not sufficient for a physically interesting variational principle. This requires additional surface terms in the ...


4

I am doing my Bachelors on something like that and we even chose to include double boundary terms in the variation. In addition to this paper, Variational principle and 1-point functions in 3-dimensional flat space Einstein gravity by Stephane Detournay et al., we get one further term: $$-\frac{(1-\alpha)}{16\pi G}\int_{\mathcal{\partial M}}d^3x\sqrt{-\...


3

Regarding the last question, the Palatini formulation doesn't require the boundary term whereas standard GR does because the two formulations are not strictly equivalent. If the action is just the Einstein-Hilbert action, they give the same field equations, true, but for more generic actions (e.g., involving combinations and powers of the Ricci scalar $\...


3

OP asks: Why the variation of a surface term is zero? Answer: Assume that the action is schematically of the form $$S=S_1+B_2,\tag{1}$$ where $S_1$ is a bulk term and $B_2$ is a boundary term (BT). (E.g. in GR $S_1$ is the EH action and $B_2$ is the GHY BT.) Then the variation of the bulk term is of the form $$\delta S_1~=~(\text{bulk term}) + \delta B_1,...


3

The physical meaning is that total derivative/divergence terms are just boundary terms in the action, and boundary conditions fixes the boundary, so they cannot enter actively into the stationary action principle nor alter the EL equations (assuming the variational problem is well-posed). See also this related Phys.SE post.


2

Note that adding boundary terms to the action does not change the equations of motion! It merely changes the boundary conditions which you have to impose in order for the variational principle to yield the equations of motion. That said, the role of the Gibbons-Hawking-York boundary term is not to cancel the all the surface integrals you obtain from varying ...


2

You want to take a look at Eric Poisson's Advanced General Relativity: https://www.physics.uoguelph.ca/poisson/research/agr.pdf The induced metric can be obtained by looking at the invariant interval $ds^2$, and then setting $dx_i=0$ for one of your coordinates. Note that you also forgot an $\varepsilon$ in your equation which is -1 for spacelike boundaries ...


2

I) Hint: Decompose the full infinitesimal variation $$ \tag{A} \delta q~=~\delta_0 q + \dot{q} \delta t $$ in a vertical infinitesimal variation $\delta_0 q$ and a horizontal infinitesimal variation $\delta t$. Similarly the full infinitesimal variation becomes $$ \tag{B} \delta I~=~\delta_0 I + \left[ L ~\delta t \right]_{t_1}^{t_2}, $$ where the ...


2

I got the answer reading a book from E. Poisson, what I was doing was indeed wrong, you have to start with the induced metric given by $$ h_{ab}= g_{\mu\nu}e^{\mu}_a e^{\nu}_b $$ where $$e^{\mu}_a=\frac{\partial x^{\mu}}{\partial y^a}$$ are the tangent vectors to curves of the hypersurface. Then, you just replace $g$ by $h$ in the usual relation $$\delta\...


2

The Gibbons-Hawking boundary term for a spacetime manifold is explicitly, $$S_{GH}=\frac{1}{8\pi G}\int_{\partial M} d^3x \, \sqrt{|h|} \, K$$ where $\partial M$ is the boundary of $M$, $K$ the extrinsic curvature, and $h$ the determinant of the metric on the boundary. Let us Wick rotate the Schwarzschild metric to, $$ds^2 = \left( 1-\frac{2GM}{r}\right)d\...


2

Let us assume that the new and old Lagrangian density $${\cal L} \quad\to\quad {\cal L} +\sum_{\mu=0}^3 d_{\mu} {\cal X}^{\mu}\tag{A}$$ does not depend explicitly on the spacetime point $x$. Then Noether's 1st theorem states that the canonical stress-energy-momentum (SEM) tensor density $$ {\cal T}^{\mu}{}_{\nu} ~:=~ \frac{\partial{\cal L}}{\partial( \...


2

OP is observing that in Minkowski space $g_{\mu\nu}=\eta_{\mu\nu}$, it doesn't matter whether we write $${\cal L} ~=~\sqrt{|g|}\partial\phi\partial\bar{\phi} \tag{1} $$ or $${\cal L}~=~-\sqrt{|g|}\phi\Box\bar{\phi}\tag{2} $$ for the Lagrangian density, if we don't care about total divergence terms. OP is pondering what happens in curved spacetime $(M,g)$? ...


2

The covariant divergence of a vector is $$\nabla_\mu V^\mu = \frac{\partial_\mu (V^\mu \sqrt{-g})}{\sqrt{-g}}$$ Meaning that adding a covariant divergence to the Lagrangian will result in the following change : $$\Delta S = \int d^4x \sqrt{-g} \nabla_\mu V^\mu = \int d^4x \partial_\mu (V^\mu \sqrt{-g})$$ which is once again easy to see that it vanishes ...


2

The derivative of $x f(x)$ is $x f'(x) + f(x)$, and here you're integrating $k \int_{-\infty}^\infty dx ~ x ~ f'(x)$ for some constant $k$, and some complicated function $f$. When you integrate this by parts, you raise $f' dx$ and lower $x$ to find:$$k \int_{-\infty}^\infty dx ~ x ~ f'(x) = k \left[x ~f(x)\right]_{-\infty}^{~\infty} - k \int_{-\infty}^\infty ...


2

Without some particular hypotheses on $J^\mu$ the statement is simply false. There are so many elementary examples...Take $$J^\mu = x^\mu$$ in Minkowskian coordinates in Minkowski spacetime, for instance. It is sufficient that $\partial_\mu J^\mu \geq 0$ with $\partial_\mu J^\mu >0$ in a region with non-vanishing four-volume to make false the statement. ...


2

So as to complement Nakahara's and Nogeira's proofs, when I did this calculation the most puzzling part was the origin of the $\frac{2}{3}$ in front of $A\wedge A\wedge A$, but it is easy to find it out: \begin{align*} Tr[F\wedge F] =&\ Tr\left[dA\wedge dA+dA\wedge A\wedge A +A\wedge A\wedge dA+A\wedge A\wedge A\wedge A\right] \\[...


2

In terms of the components $A=A_\mu dx^{\mu}$, we have$$ \\\ \frac{\theta}{2\pi}\mathrm{tr}\left[F\wedge F\right]=\frac{2\theta}{\pi}\mathrm{tr}\left[\varepsilon^{\mu\nu\rho\sigma}(\partial_{\mu} A_{\nu}+A_{\mu}A_{\nu})(\partial_{\rho} A_{\sigma}+A_{\rho}A_{\sigma})\right] \\\\ $$ And then $$ \frac{\theta}{2\pi}\mathrm{tr}\left[F\wedge F\right]=\frac{2\theta}...


2

I would like to add that in some cases, even if a term is a total derivative -- so a surface term, by Stokes's theorem -- we can't neglect it in QFT! This will also expand a little on Valter Moretti's point that we need to use some sort of boundary conditions. For this purpose, let us consider $$S_\theta = \int_M \operatorname{tr} F \wedge F.$$ Here $M$ is ...


2

In general the vanishing of the boundary term is what gives Lagrangian mechanics its really peculiar character in terms of how it predicts things. Newtonian and Hamiltonian mechanics are raring to go, "come on, give me the forces and the initial conditions and I will move the particle forwards a little bit, and then forwards from there a little bit more, ...


Only top voted, non community-wiki answers of a minimum length are eligible