This and F[db. cl = 0 (which is the equation dF = d(dA) = 0) are the Maxwell equations for the source-free electromagnetic field. The energy-momentum tensor is

Example 3: A charged scalar field

This is really a combination of two real scalar fields ftx and ft2. These are combined into a complex scalar field ft = fti + ift2, which could represent, for example, n+ and mesons. The total Lagrangian of the scalar field and electromagnetic field is

L = -U^a + ieAuft^OP-.t-^rf)-^ Vc where c is a constant and ft is the complex conjugate of ft. Varying ft, xjr and Aa independently, one obtains wi2

ft-.abS^-jift+ieAa g°6( 2ft. b+i eAb ft) + ieAa.b g^ft = 0, and its complex conjugate, and

¿^c^-ic^jo-i^o^+ic^^ + ic^,,^) = 0. The energy-momentum tensor is

+ ^ J^bft-ft.bieAaft)+l. FacFbdtf* + e*AuAbftft+Lgub.

Example 4: An isentropic perfect fluid

The technique here is rather different. The fluid is described by a function p, called the density, and a congruence of timelike curves, called the flow lines. By a congruence of curves, is meant a family of curves, one through each point of If 3) is a sufficiently small compact region, one can represent a congruence by a diffeomorphism y: [a, 6] x where [a, 6] is some closed interval of R1 and Jf is some three-dimensional manifold with boundary. The curves are said to be timelike if their tangent vector W = (8/8t)y, te[a, 6], is timelike everywhere. The tangent vector V is defined by V = (— g(W, W) W, so g(V, V) = — 1, and the fluid current vector is defined by j = pV. It is required that this is conserved, i.e. ja._ a = 0. The behaviour of the fluid is determined by prescribing the elastic potential (or internal energy) e as a function of p. The Lagrangian is taken to be

and the action I is required to be stationary when the flow lines are varied and p is adjusted to keep ja conserved. A variation of the flow lines is a differentiable map y: ( — 8,8)* [a,b~\ x 3) such that and y(ti,[o,6],>") = y([o>6],>") on J(-3>, (ue(-8,8)).

Then it follows that AW = Z^W where the vector K is K = (8j8u)y. This vector may be thought of as representing the displacement, under the variation, of a point of the flow line. It follows that

Substituting for A Va and integrating along the flow lines, one finds

Therefore the variation of the action integral is

Was this article helpful?

## Post a comment