CAF123 said:
Thanks for all the replies,
So let me write down two cases: 1) If we consider a field variation that is constrained to be a symmetry of the action then ##\delta S = 0##. If also the field configs are constrained to satisfy EL equations then we get $$\delta S = 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) d^4 x = \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) |_{\text{boundary}}$$ So this means the boundary term has to vanish in this case too right so that it agrees with the ##\delta S=0## on the lhs?
I just wondered about this because in the above posts, it was said that arbitrary variations of the fields vanish and in this case the field variation is not arbitrary but constrained so that ##\delta S=0##.
2) The second case is similar to the above. Let's suppose the field variation is not constrained so that ##\delta S \neq 0##, but again suppose field configs satisfy EL. Then $$\delta S \neq 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi \right) d^4 x $$ and so here the field variation should not vanish on the boundary in order to agree with the lhs?
I see that locally when we remove the integral signs 1) gives ##\partial_{\mu}J^{\mu} = 0## while the second case would give a source term (in accordance with symmetry of actions giving rise to conserved curents) but it's just the boundary terms I'm left to understand it seems with the above questions.
Thanks to all!
You need to understand the meaning of
symmetry transformations,
variation principle and
Noether theorem. So, pay attention to what I am going to say.
1.
Symmetry Transformations
Definition: given a
particular (infinitesimal) field transformation \delta_{\epsilon}\Phi, and a
differentiable action S[\Phi] = \int_{D} d^{4}x \ \mathcal{L}(\Phi , \partial\Phi), the integral of a local Lagrangian \mathcal{L} over
bounded and
arbitrarily contractible region D of space-time. We say that \delta_{\epsilon}\Phi is a symmetry transformation if (and only if),
without using the EOM, the action changes according to \delta S[\Phi , \delta_{\epsilon}\Phi] \equiv S[\Phi + \delta_{\epsilon}\Phi] - S[\Phi] = \int_{D} d^{4}x \ \partial_{\mu}\Lambda^{\mu}(\Phi , \delta_{\epsilon}\Phi) , \ \ \ (1) for some \Lambda^{\nu}. Eq(1) should be understood as a
restriction on the given transformation \delta_{\epsilon}\Phi. Also, since the equations of motion have not been used in deriving Eq(1), it must hold
for all \Phi.
So, if under some transformation \delta_{\eta}\Phi, you obtain \delta S[\Phi , \delta_{\eta}\Phi] = \int_{D} d^{4}x \ K(\Phi , \delta_{\eta}\Phi), \ \ \ \mbox{forall} \ \Phi(x) , \ \ \ \ (2) for some function K (\phi , \delta_{\eta}\Phi) \neq \partial_{\mu} F^{\mu}, then \delta_{\eta}\Phi
is not a symmetry transformation.
Advise: always pay attention and understand
the assumptions used in deriving mathematical expressions.
So, I will repeat: The function \delta_{\epsilon}\Phi is a symmetry transformation
if, and only if, it satisfies Eq(1)
for all \Phi(x).
Before we consider examples of (1) and (2), let us understand why Eq(1) is a good definition for symmetries: if \Lambda^{\mu} = 0, then S[\Phi + \delta_{\epsilon}\Phi] = S[\Phi]. Thus, if \varphi (x) solves the EOM, so will the transformed field \varphi (x;\epsilon) = \varphi (x) + \delta_{\epsilon}\varphi(x). This is also the case when \Lambda^{\mu} \neq 0, because \mathcal{L} and \mathcal{L} + \partial_{\mu}\Lambda^{\mu} lead to the same Euler-Lagrange equation.
1.1 Example of symmetry transformation:
Under space-time translation x \to x + \epsilon by the infinitesimal constant \epsilon, an arbitrary field \Phi changes according to \delta_{\epsilon}\Phi (x) = - \epsilon^{\mu}\partial_{\mu}\Phi . \ \ \ \ \ \ \ (3) This induces the following change in the action
\delta S[\Phi , \delta_{\epsilon}\Phi] = \int_{D} d^{4}x \left( \frac{\partial \mathcal{L}}{\partial \Phi} \delta_{\epsilon}\Phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\nu}\Phi)} \partial_{\nu}(\delta_{\epsilon}\Phi) \right) . Assuming that \mathcal{L} has no explicit dependence on x^{\mu}, and substituting Eq(3), we find that the change in the action is given by the following
non-vanishing integral of a
total divergence \delta S[\Phi , \delta_{\epsilon}\Phi] = \int_{D} d^{4}x \ \partial_{\mu}(- \epsilon^{\mu}\mathcal{L}) \ \ \ \forall \Phi(x) . Thus, according to Eq(1), spacetime translation is a symmetry with \Lambda^{\mu} (\Phi , \delta_{\epsilon}\Phi) = - \epsilon^{\mu} \mathcal{L}.2.
Arbitrary (algebraic) variation & The Noether Identity
An
arbitrary variation \delta\Phi of an
arbitrary field \Phi, induces the following variation on the action integral
\delta S[\Phi , \delta\Phi] = \int_{D} d^{4}x \left( E(\Phi) \delta\Phi + \partial_{\mu}\left( \Pi^{\mu} \delta\Phi \right) \right) , \ \ \ \ \ \ (4) where \Pi^{\mu} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} , is the symplectic momentum, and E(\Phi) \equiv \frac{\partial \mathcal{L}}{\partial \Phi} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\Phi)} \right) , is the Euler
derivative. If E(\psi) = 0, we say that \psi(x) is a solution to the EOM. So, keep in mind that E(\Phi) is an
operator acting on the
arbitrary field \Phi(x). This means that equation (4) is a valid expression
for all \Phi(x) and
for all the variation \delta\Phi. Thus, it must also be valid
for all \Phi and for the
symmetry variation \delta_{\epsilon}\Phi as defined in equation (1). So, if we set \delta \Phi = \delta_{\epsilon}\Phi in Eq(4) and subtract Eq(1) from it, we get \int_{D} d^{4}x \left( E(\Phi) \delta_{\epsilon}\Phi + \partial_{\mu}J^{\mu} \right) = 0 , \ \ \ \ \ \ \ (5) where J^{\mu} \equiv \Pi^{\mu}\delta_{\epsilon}\Phi - \Lambda^{\mu} . \ \ \ \ \ \ \ \ \ \ (6) Since D \subset \mathbb{R}^{4} is an
arbitrary contractible region, the integrand in Eq(5) must vanish
identically. Thus, for the symmetry transformation \delta_{\epsilon}\Phi, the following (Noether)
identity holds for
arbitrary field \Phi(x)
E(\Phi) \delta_{\epsilon}\Phi + \partial_{\mu}J^{\mu} = 0 , \ \ \ \forall \Phi(x) . \ \ \ \ \ (7)
Thus,
in presence of a continuous symmetry, a continuity equation \partial_{\mu}J^{\mu} = 0 is satisfied for the on-shell configuration E(\psi) = 0. And that is the statement of the (first) Noether theorem.
2.1 Example of a transformation that is not symmetry (explicit symmetry breaking):
Consider n fermion fields \psi_{i} with
different masses m_{i}. The Dirac lagrangian for such system can be written as \mathcal{L} = \bar{\Psi} \left( {\not} \partial + M \right) \Psi , where \Psi = (\psi_{1}, \psi_{2}, \cdots , \psi_{n})^{T} and M is n \times n mass matrix not proportional to the identity matrix I_{n}, i.e., there is
no mass degeneracy. Now, consider the finite SU(n) transformations \Psi \to U(\eta) \Psi , \ \ \ \bar{\Psi} \to \bar{\Psi}U^{\dagger}(\eta) , \ \ U(\eta) = \exp (\eta^{a}T_{a}) . Infinitesimally, we have \delta_{\eta}\Psi = \eta^{a}T_{a}\Psi , \ \ \ \delta_{\eta}\bar{\Psi} = - \eta^{a} \bar{\Psi}T_{a} . Transforming the fields in the Lagrangian, we get \mathcal{L} \to \mathcal{L} (\eta) = \bar{\Psi} \left( {\not} \partial + U^{\dagger}(\eta) M U(\eta) \right) \Psi . Expanding both sides to first order in \eta, we obtain
\mathcal{L} (\eta) = \mathcal{L} (0) + \eta^{a} \frac{d}{d \eta^{a}} \mathcal{L}(0) = \mathcal{L} (0) + \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi , where \mathcal{L}(0) = \mathcal{L} is the un-transformed Lagrangian. From that we get \delta_{\eta} \mathcal{L} = \eta^{a} \frac{d}{d \eta^{a}} \mathcal{L}(0) = \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi . \ \ \ \ (8)
Clearly, the RHS of (8)
cannot be written as
total divergence, also it
does not vanish since for our model M \neq m I_{n}. Thus, the SU(n) transformation U(\eta)
is not symmetry transformation.
Now, let us subject \mathcal{L} to an arbitrary variation and find
\delta \mathcal{L} = \delta \bar{\Psi} \left( {\not} \partial + M \right) \Psi + \bar{\Psi} \left( {\not} \partial + M \right) \delta\Psi . In terms of the Euler derivatives, E(\Psi) and E(\bar{\Psi}), we can rewrite the above as \delta \mathcal{L} = \delta \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta \Psi + \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} \delta \Psi \right) . To compare this equation with Eq(8), we set \delta = \delta_{\eta}
\delta_{\eta} \mathcal{L} = \delta_{\eta} \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta_{\eta} \Psi + \eta^{a} \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} T_{a} \Psi \right) . \ \ \ (9) Thus, from (8) and (9), we obtain the
would-be Noether identity
\delta_{\eta} \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta_{\eta} \Psi + \eta^{a} \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} T_{a} \Psi \right) = \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi . Finally, if we go for the
on-shell configuration E(\psi) = E(\bar{\psi}) = 0, we find
\partial_{\mu} j^{\mu}_{a} = \bar{\psi} [ M , T_{a} ] \psi , \ \mbox{where} \ j^{\mu}_{a} \equiv \bar{\psi} \gamma^{\mu}T_{a} \psi .
This example shows that Noether’s theorem is more intelligent than us. It has detected that our model is not invariant under SU(n), and calculated for us the divergence of the
would-be-conserved symmetry current.
3.
Boundary integrals: in the action principle and in Noether’s theorem
This is a tough business, and still is an active research area in gauge theory and GR. So, we will not discuss long rage forces (massless gauge fields) , but deal only with massive fields having
compact support, i.e., \Phi(t , r) \to 0 as r \to \infty.
The principle of least action states that the action must be
stationary, \delta S[\Phi] = 0, under
arbitrary variation of the fields \delta\Phi, and
fixed initial and final field configurations. That is to say that the variation satisfy \delta \Phi|_{t_{1}} = \delta \Phi|_{t_{2}} = 0 but
arbitrary inside the integration domain (
the bulk) D. Thus, in order for the “least action principle” to make sense, the action must be
differentiable, i.e., it must possesses well-defined functional derivatives. So, we must have \delta S[\Phi] = \int_{D} d^{4}x \ ( \mbox{Stuff} ) \ \delta \Phi , \ \ \ \ \ \ (3.1) with
no extra
non-vanishing terms. But we saw, Eq(4), that
\delta S[\Phi] = \int_{D} d^{4}x \ E(\Phi) \delta\Phi + \int_{D} d^{4}x \ \partial_{\mu} \left( \frac{ \partial \mathcal{L}}{ \partial ( \partial_{\mu}\Phi ) } \delta \Phi \right) . \ \ \ (3.2)
Thus, the second integral in Eq(3.2) must vanish in order to conclude that \delta S[\Phi] = 0 \ \Leftrightarrow \ E(\Phi) = 0 in D and that, by
fixing initial and final data, there exists a
unique solution to the evolution equation E(\Phi) = 0. So, given \delta \Phi|_{t_{1}} = \delta \Phi|_{t_{2}} = 0, and assuming compact support, that is \Phi(t,r) tends to zero sufficiently fast as r \to \infty, we need to show that \int_{D} d^{4}x \ \partial_{\mu}\left( \frac{ \partial \mathcal{L}}{ \partial ( \partial_{\mu}\Phi )} \delta \Phi \right) = 0 . We take D to be a “fat” world-tube containing the fields. Let \partial D be the
continuous,
orientable, and
piecewise smooth boundary of D. Let t = t_{1} and t = t_{2} be 3-dimensional cross-sections of the tube denoted by \partial D(t_{1}) and \partial D(t_{2}) respectively and \partial D(r = \infty) the “cylindrical surface” that connects the two (constant time) cross-sections at r = \infty. Thus, it is clear that \partial D = \partial D(t_{1}) \cup \partial D(t_{2}) \cup \partial D(r = \infty ) . \ \ \ \ \ \ \ \ \ \ (3.3) Applying Gauss’s theorem to the second integral in Eq(3.2), we get
\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta\Phi \right) = \int_{\partial D} d^{3}x \ n_{\mu} \ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta \Phi , \ \ \ \ \ (3.4) where n_{\mu} is the outer pointing unit normal on \partial D: d^{3}x \ n_{\mu}|_{\partial D(t_{2})} = - d^{3}x \ n_{\mu}|_{\partial D(t_{1})} = (d^{3} \vec{x} , \vec{0}), \ \ d^{3}x \ n_{\mu}|_{\partial D(r = \infty)} = r^{2} d\Omega dt \hat{r} . \ \ \ \ (3.5) Substituting (3.3) and (3.5) in RHS of (3.4), we find\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta \Phi \right) = \int_{t_{2}} d^{3}\vec{x} \frac{\partial \mathcal{L}}{ \partial ( \partial_{0} \Phi)} \delta \Phi - \int_{t_{1}} d^{3}\vec{x} \frac{\partial \mathcal{L}}{ \partial ( \partial_{0} \Phi ) } \delta \Phi + \int_{r = \infty} \ d\Omega \ dt \ r^{2} \frac{\partial \mathcal{L}}{\partial (\partial_{r}\Phi)} \delta \Phi . Now, the first two integrals on the RHS both vanish because of the fixed initial and final data \delta \Phi (t_{1}) = \delta \Phi (t_{2}) = 0. And in the
absence of massless gauge fields and other constraints, the third integral vanishes because of the compact support assumption: at large r, \frac{ \partial \mathcal{L} }{ \partial ( \partial_{r} \Phi ) } decays faster than the growth of r^{2}. Thus, in non-gauge field theories, the second integral in Eq(3.2) does indeed vanish leaving us with
\delta S[ \Phi ] = \int_{D} d^{4}x \ E(\Phi) \ \delta \Phi . Since \delta \Phi is an arbitrary function in the bulk D, then \delta S[ \Phi ] = 0 implies (and is implied by) E(\Phi) = 0 in D.
In gauge field theories and other constraint systems, boundary integrals require more careful analysis. Basically, the question of whether the boundary integral
vanishes or not, is expressed in terms of
asymptotic symmetries, i.e., the set of all symmetry transformations (having a
non-zero conserved Noether charge) that preserve the
asymptotic boundary conditions on the fields. For those who are interested I recommend the following paper and text[
1] T. Regge and C. Teitelboim, “Role of Surface Integrals in the Hamiltonian Formulation of General Relativity" Annals Phys.,
88, 286, 1974.
[
2] M. Henneaux and C. Teitelboim,
Quantization of Gauge Systems. Princeton University Press, 1992.
The last boundary integral, I leave for you as exercise. For the conserved symmetry current J^{\mu}, show that \int_{D} d^{4}x \ \partial_{\mu}J^{\mu}(t, \vec{x}) = 0 \ \Rightarrow \ Q(t_{1}) = Q(t_{2}) , where Q(t) \equiv \int d^{3}\vec{x} \ J^{0}(t , \vec{x}) is the Noether charge associated with the symmetry group.
I.e., are you sure you don't need a surface integral on the rhs? (You've written "evaluate at boundary", instead of "integrate over boundary surface".)
