QuantumPhysics.dvi
wang
(Wang)
#1
The variation is then computed using standard chain rule,
δS[q] = S[q+δq]−S[q]
=
∫t 2
t 1
dt
(
L(q+δq,q ̇+δq ̇;t)−L(q,q ̇;t)
)
=
∫t 2
t 1
dt
∑
i
(∂L
∂qi
δqi+
∂L
∂q ̇i
δq ̇i
)
(6.5)
Using now the fact that
δq ̇i=
d
dt
δqi (6.6)
integrating by parts and using the vanishing boundary conditions, we are left with
δS[q] =
∫t 2
t 1
dt
∑
i
(
−
d
dt
∂L
∂q ̇i
+
∂L
∂qi
)
δqi (6.7)
Finally, requiring that a path be stationary amounts to requiring that a small variation
around the path leaves the action unchangedδS[q] = 0, which clearly is equivalent to the
Euler-Lagrange equations.
In the simplest examples, the Lagrangian is the difference between kinetic and potential
energyV of the generalized positions,
LV=
1
2
∑N
i=1
miq ̇i^2 −V(q 1 ,···,qN) (6.8)
for which the Euler-Lagrange equations are
miq ̈i=−
∂V
∂qi
(6.9)
In a slightly more complicated example, we have the Lagrangian of a charged particle in
3 space dimensions in the presence of an electricE(r,t) and magnetic fieldB(r,t), which
derive from an electric potential Φ(r,t) and vector fieldA(r,t) as follows,
E=−∇~Φ−
∂A
∂t
B=∇×~ A (6.10)
The associated Lagrangian for a particle with electric chargee, massm, and positionris,
LA=
1
2
mr ̇^2 −eΦ(r,t) +eA(r,t)·r ̇ (6.11)
Check that the Euler-lagrange equation coincides with the Lorentzforce law
m ̈r=eE(r,t) +er ̇×B(r,t) (6.12)