Download A Spring in Imaginary Time Part 3-Advanced Classical and Relativistic Mechanics-Lecture Handout and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity!
q˙(s) · q˙(s) + V (q(s))
]
ds,
we set δE = 0 and investigate the implications.
δE = δ
(∫ (^) s 1
s 0
[
k 2
q˙(s) · q˙(s) + V (q(s))
]
ds
= (^) ∂ε∂
(∫ (^) s 1
s 0
[
k 2
q˙(s) · q˙(s) + V (q(s))
]
ds
ε=
def of δ
∫ (^) s 1
s 0
k 2
∂ ∂ε [ ˙q(s)^ ·^ q˙(s)] +^
∂ ∂ε [V^ (q(s))]^ ds
ε=
linearity
∫ (^) s 1
s 0
k q˙ε(s) (^) ∂ε∂ [ ˙qε(s)] + ∇V (qε(s)) (^) ∂ε∂ [qε(s)] ds
ε=
chain rule
∫ (^) s 1
s 0
k q˙ε(s) (^) ∂t∂∂ε∂ [qε(s)] + ∇V (qε(s)) (^) ∂ε∂ [qε(s)] ds
ε=
mixed partials
∫ (^) s 1
s 0
−k
∂t q˙ε(s)
∂ε [qε(s)] +^ ∇V^ (qε(s))^
∂ ∂ε [qε(s)]^ ds
ε=
IBP
∫ (^) s 1
s 0
−kq¨ε(s) + ∇V (qε(s))
∂ ∂ε [qε(s)]^ ds
ε=
factoring
∫ (^) s 1
s 0
−kq¨(s) + ∇V (q(s))
∂ ∂ε [qε(s)]ε=0 ds^ letting^ ε^ = 0.
So if this is 0 for all allowable variations δq, we must have an integrand of 0, i.e., kq¨(s) = ∇V (q(s)).
- Suppose the spring is in a constant downwards gravitational field in R^3 , so that
V (x, y, z) = mgz, where m is the mass density of the spring and g is the acceleration of gravity (9. m/s^2 ). What sort of curve does the spring trace out, in equilibrium?
Apply the answer from (1), ∇V = kq¨(s), and compute ∇V (x, y, z) = ∇(mgz) = [0, 0 , mg]
1
Classical Mechanics
to obtain the system
q ¨ 1 (s) = 0 q¨ 2 (s) = 0 q¨ 3 (s) = mg k.
All equations may be solved directly by successive integrations; the first two yield linear functions, and the third gives a polynomial in z:
q 1 (s) = (b 1 − a 1 )s + a 1 q 2 (s) = (b 2 − a 2 )s + a 2 q 3 (s) = mg 2 k s^2 +
b 3 − a 3 − mg 2 k
s + a 3 ,
where the values of the constants are deduced by comparison to the components of q(s 0 ) = a, q(s 1 ) = b. Thus the spring traces out a parabola lying in the vertical plane whose intersection with the xy-plane is the straight line from (a 1 , a 2 ) to (b 1 , b 2 ).
- Using the energy, as given previously, replace the parameter s by it and show that up to a constant, the energy of the static string becomes the action for a particle moving in a potential.
E =
∫ (^) s 1
s 0
[
k 2
q˙(s) · q˙(s) + V (q(s))
]
ds
∫ (^) s 1
s 0
[
k 2
∂ ∂s q(s)^ ·^
∂ ∂s q(s) +^ V^ (q(s))
]
ds
∫ (^) t 1
t 0
[
k 2
∂ ∂t q(it)^ ·^
∂ ∂t q(it) +^ V^ (q(it))
]
d(it)
∫ (^) t 1
t 0
[
k 2
i q˙(it) · i q˙(it) + V (q(it))
]
i dt
∫ (^) t 1
t 0
[
k 2
q˙(it) · q˙(it) + V (q(it))
]
i dt
= −i
∫ (^) t 1
t 0
[
k 2
q˙(it) · q˙(it) − V (q(it))
]
dt
= −iS(q)
