Mike

2003-10-26 15:54:37 UTC

Using Stoke's theorem, one can show that the closed line integral must be

conserved from one end of a world-tube to the other end in a scalar field.

If the line integral must be conserve from one closed curve to the next,

then does this result in the need of a geodesic surface from one end of the

tube to the other? I'm thinking in terms of classical string theory and

trying to justify the Euler-Lagrange equations. If an extremal surface can

be justified for other reasons, then perhaps the Euler-Lagrange equation can

be derived from that. I am told that the Euler-Lagrange vector is normal to

the surface and that it is zero for an extremal surface which must then be a

geodesic surface. But if the surface must be extremal for other reasons,

then perhaps we can derive the meaning of energy and momentum as some

geometric features on this surface. The surface may need to be extremal, for

example, because the most probable path is one along equi-potential paths

where the gradient in the direction of the normal is zero.

I would appreciate some comments, thank you.

Mike.

conserved from one end of a world-tube to the other end in a scalar field.

If the line integral must be conserve from one closed curve to the next,

then does this result in the need of a geodesic surface from one end of the

tube to the other? I'm thinking in terms of classical string theory and

trying to justify the Euler-Lagrange equations. If an extremal surface can

be justified for other reasons, then perhaps the Euler-Lagrange equation can

be derived from that. I am told that the Euler-Lagrange vector is normal to

the surface and that it is zero for an extremal surface which must then be a

geodesic surface. But if the surface must be extremal for other reasons,

then perhaps we can derive the meaning of energy and momentum as some

geometric features on this surface. The surface may need to be extremal, for

example, because the most probable path is one along equi-potential paths

where the gradient in the direction of the normal is zero.

I would appreciate some comments, thank you.

Mike.