Scalar field Lagrangian
Let
where we abuse notation and invoke a
so that the action functional
Euler-Lagrange equations
Let
A field
Proof
Let
be a variation of πΌ : ( β π 0 , π 0 ) β πΆ πΌ ( π ) agreeing on the boundary. Then π β [ πΌ ( π ) ] = β« π β π Λ πΏ ( π , π | π , d π | π ) Δ π = β« π Δ π π₯ Λ πΏ ( π₯ , π | π₯ , d π | π₯ ) whence
πΏ β [ π , πΌ ] = d d π β£ π = 0 β« π Δ π π₯ Λ πΏ ( π₯ , πΌ ( π ; π₯ ) , d πΌ ( π ; π₯ ) ) = β« π Δ π π₯ d d π β£ π = 0 Λ πΏ ( π₯ , πΌ ( π ; π₯ ) , d πΌ ( π ; π₯ ) ) = β« π Δ π π₯ ( π Λ πΏ π π π πΌ π π ( πΌ ; π₯ ) + π Λ πΏ π ( d π π ) π ( d πΌ π ) π π ( 0 ; π‘ ) ) = β« π Δ π π₯ ( π Λ πΏ π π π πΌ π π ( πΌ ; π₯ ) + π Λ πΏ π ( d π π ) π 2 πΌ π π π₯ π π π ( 0 ; π‘ ) ) . Applying integration by parts we get
πΏ β [ π ; πΌ ] = β« π Δ π π₯ π πΌ π π ( πΌ ; π₯ ) ( π Λ πΏ π π β π π π₯ π π Λ πΏ π ( d π π ) ) so by the Fundamental lemma of variational calculus
0 = π Λ πΏ π π β π π π₯ π π Λ πΏ π ( d π π ) as claimed.
If
Proof
Since
, we have Λ πΏ = β | π | Β― πΏ 0 = β | π | π Β― πΏ π π β π π π₯ π β | π | π Β― πΏ π ( d π π ) so applying ^P1 we have
0 = β | π | π Β― πΏ π π β β | π | β π π Β― πΏ π ( d π π ) whence the claimed equation.