Invariance of action ⇒ covariance of field equations? Is this statement true?
I have only seen examples of this, like the invariance of Electromagnetic action under Lorentz transformations.
How could we prove it?
The action is a scalar, S, so how I can't even transform it like U−1SU...
Answer
In this answer we formally show that a (quasi)symmetry of an action implies a corresponding symmetry of its EOM†. The answer does not discuss form covariance of EOM. For further relations between symmetries of action, EOM, and solutions of EOM, see e.g. this Phys.SE post.
Let us first recall the definition of a quasi-symmetry of the action
SV[ϕ] := ∫VL,L := L dnx.
It means that the action (1) changes by a boundary integral
SV′[ϕ′]+∫∂V′dn−1x (…) = SV[ϕ]+∫∂Vdn−1x (…)
under the transformation. In the following we will assume that the spacetime integration region V is arbitrary.
Theorem. If a local action functional SV[ϕ] has a quasi-symmetry transformation ϕα(x) ⟶ ϕ′α(x′),xμ ⟶ x′μ,
then the EOM eα(ϕ(x),∂ϕ(x),…;x) := δSV[ϕ]δϕα(x) ≈ 0must have a symmetry (wrt. the same transformation) eα(ϕ′(x′),∂′ϕ′(x′),…;x′) ≈ eα(ϕ(x),∂ϕ(x),…;x).
I) Formal finite proof: This works both for a discrete and a continuous quasi-symmetry.
eα(ϕ(x),∂ϕ(x),…;x) := δSV[ϕ]δϕα(x) (2)= δSV′[ϕ′]δϕα(x) ‡∼ ∫V′dnx′ δSV′[ϕ′]δϕ′α(x′)δϕ′α(x′)δϕα(x)
II) Formal infinitesimal proof: This only works for a continuous quasi-symmetry. From the infinitesimal transformation (3)
δϕα(x) := ϕ′α(x′)−ϕα(x),δxμ := x′μ−xμ,
we define a so-called vertical transformation
δ0ϕα(x) := ϕ′α(x)−ϕα(x) = δϕα(x)−δxμ dμϕα(x),dμ := ddxμ,
which transforms of the fields ϕα(x) without transforming the spacetime points xμ. The quasi-symmetry implies that the Lagrangian n-form L transforms with a total spacetime derivative
δL = dμfμ dnx,δ0L = dμ(fμ−L δxμ) dnx.
The EOM (4) are typically of second order, so let us assume this for simplicity. (This assumption is not necessary.) Then the infinitesimal transformation of EOM (4) reads
δeα(x) = δ0eα(x)+δxμ dμeα(x)⏟≈0 ≈ δ0eα(x)
In the very last step of eq. (10) we used that the infinitesimal variation
δ0SV[ϕ]+∫Vdnx dμ(L δxμ) = δSV[ϕ] = ∫∂Vdn−1x (…)
of the action is a boundary integral by assumption (2), so that its functional derivative (10) must vanish (if it exists).
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† Terminology and Notation: Equations of motion (EOM) means Euler-Lagrange equations (1). The words on-shell and off-shell refer to whether EOM are satisfied or not. The ≈ symbol means equality modulo EOM.
‡ Warning: This step is not always justified. The ∼ symbol indicates that we have formally integrated by part and ignored boundary contributions. Also we have assumed that the pertinent functional derivative is well-defined and exists. This caveat is the main shortcoming of the formal proof given here. The point is quite serious, e.g. in the case of a global (=x-independent) variation, which typically doesn't vanish on the boundary. So boundary contributions could in principle play a role.
However, instead of using functional derivatives and integrations, it is possible to prove eq. (10) x-locally
δ0eα(x) = … = Eα(0)dμ⏟=0(fμ(x)−L(x) δxμ)−∑k≥0dk(eβ(x)⏟≈0⋅Pα(k)δ0ϕβ(x))
using only higher partial field derivatives
Pα(k) := ∂∂ϕα(k),k ∈ Nn0,
and higher Euler operators
Eα(k) := ∑m≥k(mk)(−d)mPα(m),
that all refer to the same spacetime point x. This x-local approach circumvent the problem of un-accounted boundary contributions.
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