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A Spring in Imaginary Time
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One of the stranger aspects of Lagrangian dynamics is how it turns into statics when we replace the time coordinate t by it — or in the jargon of physicists, when we ‘Wick rotate’ to ‘imaginary time’! People usually take advantage of this to do interesting things in the context of quantum mechanics, but the basic ideas are already visible in classical mechanics. So, let’s look at them!
- Suppose you have a spring in Rn^ whose ends are held fixed, tracing out a curve
q: [s 0 , s 1 ] → Rn
with endpoints q(s 0 ) = a, q(s 1 ) = b.
Suppose the spring is put into a potential
V : Rn^ → R
(perhaps due to gravity, but not necessarily). What curve will the spring trace out when it is in equilibrium?
Hint: Hooke’s law says that a stretched spring has energy proportional to the square of how much it is stretched. Here this is true of each little piece of the spring, so its total energy due to stretching will be k 2
∫ (^) s 1
s 0
q ˙(s) · q˙(s) ds
for some ‘spring constant’ k. But in addition, each little piece will acquire energy due to the potential V at that point, so the spring will also have potential energy ∫ (^) s 1
s 0
V (q(s)) ds.
The total energy of the spring is thus:
E =
∫ (^) s 1
s 0
k 2
q˙(s) · q˙(s) + V (q(s))
ds.
Our study of statics has taught us that in equilibrium, a static system minimizes its energy, or at least finds a critical point. So, set δE = 0
for all allowed variations δq of the path, and work out the differential equation this implies for q.
Answer: A variation δq(s) produces the variation
δE =
∫ (^) s 1
s 0
(k q˙(s) · δ q˙(s) + ∇V (q(s)) · δq(s)) ds
in E. The first term in the integrand can be integrated by parts, and, as δq(s 0 ) = δq(s 1 ) = 0, this yields
δE =
∫ (^) s 1
s 0
(−k q¨(s) · δq(s) + ∇V (q(s)) · δq(s)) ds =
∫ (^) s 1
s 0
(−k ¨q(s) + ∇V (q(s))) · δq(s) ds.
If δE is to vanish for all variations δq(s), we must have
−k q¨(s) + ∇V (q(s)) = 0,
which is the required differential equation for 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 (9.8 meters/second^2 ). What sort of curve does the spring trace out, in equilibrium?
Answer: As ∇V (x, y, z) = mgzˆ, the equations for x, y, and z are
¨x(s) = 0, y¨(s) = 0, and ¨z(s) =
mg k
The solutions are x(s) = uxs + vx y(s) = uys + vy
z(s) = uz s + vz + mg 2 k
s^2 ,
where ui and vi denote constants easily calculated from s 0 , s 1 , a, and b. These solutions paramet- rically define a parabola in R^3.
- The calculation in problem 1 should remind you strongly of the derivation of the Euler-Lagrange equations for a particle in a potential. To heighten this analogy, take the energy
E =
∫ (^) s 1
s 0
k 2
q˙(s) · q˙(s) + V (q(s))
ds.
and formally replace the parameter s by it, replacing the real interval [s 0 , s 1 ] ⊂ R by the imaginary interval [t 0 , t 1 ] ⊂ iR, where iti = si. Show that up to some constant factor, the energy of the static spring becomes the action for a particle moving in a potential.
Answer: Formally replacing s by it in our expression for E yields
E =
∫ (^) it 1
it 0
k 2
q˙(it) · q˙(it) + V (q(it))
d(it).
Using d dt q(it) = i q˙(it),
we can write E in the form
E =
∫ (^) it 1
it 0
k 2
d dt q(it) ·
d dt q(it) + V (q(it))
d(it) = −i
∫ (^) it 1
it 0
k 2
d dt q(it) ·
d dt q(it) − V (q(it))
d(t).
This quantity is proportional to the action of a particle of mass k moving in a potential V (q(it)).
- Fill in the blanks in this analogy:
