Variational principle

Action

For a given trajectory of motion of a physical system the action, S, is a suitable covariant scalar integral along the trajectory, The variational principle is then formulated as follows: that trajectory, on which the action has an extremum, is the solution. Or: the solution is the stationary trajectory of the action. Or: the variation of the action vanishes on the solution,
δS=0.
(1)

Example: a free body

The action of a free body with mass m is given by the following covariant integral along the body's trajectory,
S=−m⌠
⌡ 		ds  .
(2)
To calculate the variation of the action we first vary the expression ds2=gabdxadxb,
2dsδds=δgabdxadxb+2gabδdxadxb
(3)
which gives the variation of ds,
δds= 1
2
 δgabuaubds+dδxa ua.
(4)
The second term should be integrated by parts using
dδxa ua = d(δxa ua) − δxcduc.
(5)
The full differential does not contribute to the variation, and we finally arrive at1
δS=−m⌠
⌡ 		dsδxc
 − duc
ds
 + 1
2
 gab,cuaub
 =0
(6)
Since the variation δx is arbitrary, it is the expression in brackets that should be equal zero, which gives the equation of motion
duc
ds
 − 1
2
 gab,cuaub=0.
(7)
which is equivalent to the geodesic equation (see the exercise).

Exercises

  1. Derive the Newton's equation of motion
    m
    d
    →
    v
     
    dt
    		 =−
    →
    ∇
     
    		 V
    of a non-relativistic body with mass m, coordinates [(r)\vec], and velocity [(v)\vec] in a potential V([(r)\vec]) using the action
    S=⌠
    ⌡

    m
    →
    v
     
    		 2
     
    2
    		 −V(
    →
    r
     
    		 )

    		 dt.
  2. In the Minkowski space derive the Lorentz force equation
    m dua
    ds
    		 =eFabub
    for a relativistic body with mass m and charge e moving in an electromagnetic field Fab=∂aAb−∂bAa using the action
    S=−mc⌠
    ⌡     		ds − e⌠
    ⌡     		Aadxa.
  3. In the Minkowski space derive the second Maxwell equation, ∂a∂aAb=4πjb, using the action
    S=− 1
    8π
    		 ⌠
    ⌡     		∂aAb∂aAb d4x− ⌠
    ⌡     		Aaja d4x.
    Show that with the Lorentz condition, ∂aAa=0, it is equivalent to ∂aFab=4πja.
  4. In a curved space derive the geodesic equation2
    dua
    ds
    		 +Γabcubuc=0,
    for the motion of a free body in a curvilinear frame using the action
    S=−mc⌠
    ⌡     		ds,
    where ds2=gabdxadxb.

Footnotes:

1 f,a ≡ [(∂f)/(∂xa)]
2  where Γabc=1/2(gab,c−gbc,a+gca,b) are the Christoffel symbols.

File translated from TEX by TTH, version 3.77.
On 5 Oct 2007, 22:36.