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particles of mass m move along the x-axis, subject to a potential
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9. Two point particles of mass m move along the x-axis, subject to a potential V (x 1
2 ) =
1 2 ax 2 1 + ax 2 2 + b(x 1 − x
2 ) 2 . The system is in thermal equilibrium with a thermostat at temperature T . Compute the canonical partition function and the average value of the energy. 10. A one-dimensional system is composed of N points of mass m constrained to move along the x-axis and subject to the potential V (x
1 , . . . , x N ) = v(x
1 ) + v(x
2 − x
1 ) + . . . + v(x N − x
N −1 ) + f x N , where f is a prescribed positive constant and v(x) = + ∞, if x < a, b(x
− a), if x > a, where a and b are two prescribed positive constants. Assume that the system is in thermal equilibrium with a thermostat at temperature T . Compute the canonical partition function, the heat capacity, the average length x N , the coefficient of thermal dilation and the elasticity module (1/ x N )(∂ x N /∂f ).
11. A point particle of mass m is constrained to move along the x-axis under the action of a conservative force field with potential energy V (x) = ⎧
⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1 2 mω 2 (x + a) 2 , if x ≤ −a, 0, if − a ≤ x ≤ a, 1 2 mω 2 (x − a) 2 , if x ≥ a,
658 Statistical mechanics: Gibbs sets 15.16 where a > 0. Assuming that the system is in contact with a thermostat at temperature T , compute the canonical partition function. Compute H and show its graph as a function of a > 0. 12. Consider a system of N point particles of mass m moving along the x-axis under the action of a conservative force field of potential energy V (x
1 , . . . , x N ) = v(x
1 ) + v(x
2 − x
1 ) + . . . + v(x N − x
N −1 ), where v(x) =
⎧ ⎨ ⎩ + ∞, if x < a, 1 2 mω 2 (x − a) 2 , if x ≥, a with a > 0. Assuming that the system is in contact with a thermostat at temperature T , compute the canonical partition function, H ,
x N (average length of the system) and (1/ x N )(∂/∂T ) x N (coefficient of thermal dilation), and show their graphs as functions of the temperature T . 13. Prove that the grand canonical partition function for a system confined in a region of volume V and with Hamiltonian H = N i =1 (p 2 i /2m + φ(q i )) is given by Z G (V, T, z) = exp(z Φ (V, T )), where Φ (V, T ) = (2πmkT ) 3/2 V exp( −βφ(Q)) dQ. Prove that for any external potential φ the state equation of this system is always the equation of a perfect gas. 14. (Thompson 1972) Prove that if the potential Φ (r) satisfies Φ ( |q l − q
m |) ≤
C/ |q l − q m | d +ε if |q l − q m | ≥ R, where d is the dimension of the space of configurations of each particle, C, ε are arbitrary positive constants, for every pair of subdomains D 1 and D
2 , of volume V 1 and V
2 and containing N 1 and N
2 particles, respectively, of a domain D of volume V with a distance at least equal to R between them, we have Z(V, N
1 + N
2 , T )
≥ Z(V 1 , N 1 , T )Z(V
2 , N
2 , T ) exp( −N 1
2 βC/R
d +ε ). 15. (Uhlenbeck and Ford 1963) Consider a system with partition function Z(N, V ) = V j=0
( V N−k ). Prove that the grand canonical partition function is Z G (V, z) = ∞ N =0 z N Z(N, V ) = (z V +1 − 1)/z − 1 (1 + z) V , and the grand canonical potential χ(z) in the thermodynamical limit is given by χ(z) =
log(1 + z), if |z| ≤ 1, log z(1 + z), if |z| > 1. Deduce p as a function of v from the relations βp = χ(z) and v −1 = z∂χ/∂z, eliminating the fugacity z. Prove that βp = log 2 if 2 3 ≤ v ≤ 2. 15.17 Statistical mechanics: Gibbs sets 659 16. (Huang 1987, Problem 9.5) Consider a system with grand canonical partition function Z(z, V ) = (1 + z) V (1 + z
αV ), where α > 0 is fixed. Write the state equation (naturally, in the thermodynamical limit), eliminating the fugacity z from the parametric form (15.115) and prove that there exists a phase transition. Find the specific volumes of the two phases. Find the zeros of the partition function Z(z, V ) in the z ∈ C plane (for fixed volume V ) and prove that if V → ∞, then the zeros approach the real axis at z = 1. 17. (One-dimensional Ising model) Consider a system made of a one- dimensional lattice such that to each site there corresponds a variable (spin) s i that can assume the values ±1. Each spin interacts with the two adjacent spins s i
and with an external magnetic field in such a way that the total energy of a configuration {s i
E( {s i }) = −J |i−j|=1
s i s j − H
i s i . The case J > 0 corresponds to a ferromagnetic model, while J < 0 is associated with the antiferromagnetic case. Assume that H = 0 and that the total number of spins is N . In the case that the points s 1 and s
N of the lattice are free, or the case when the lattice closes to form a ring, and hence that s 1 = s N +1 , prove that the canonical partition functions are given by Z f (x) = 2x −(N−1)/2
(x + 1) N −1 in the free case, Z r (x) = x −N/2
[(x + 1) N + (x − 1) N ] in the ring case, where x = e J/kT . Find the zeros x r and x
f of the two partition functions (answer: x r
−1 with multiplicity N − 1; x f = i((2n + 1)π)/2N , where i = √ −1 and
n = 0, 1, . . . , N − 1) and check that, setting x f = is, the density of the zeros µ(s) = (1/N ) dn/ds in the thermodynamical limit N → ∞ is given by µ(s) = 1
1 1 + s
2 . Recalling that the physically significant region corresponds to x real and positive (why?), do these models present phase transitions? 15.17
Additional remarks and bibliographical notes In the classical literature, the fact that the ergodic hypothesis has been formulated by Boltzmann assuming the existence of a trajectory passing through all points in the phase space accessible to the system (hence corresponding to a fixed value of the energy) is often discussed. Clearly this condition would be sufficient to
660 Statistical mechanics: Gibbs sets 15.17 ensure that temporal averages and set averages are interchangeable, but at the same time its impossibility is evident. Indeed, the phase trajectory S t X of a Hamiltonian flow is a regular curve of zero measure, and hence it can be dense at most on the constant energy surface. The reasoning of Boltzmann is, however, much richer and more complex (and maybe this is the reason why it was not appreciated by his contemporaries) and it deserves a brief discussion. In addition, through a modern exposition of his ideas, we can criticise the tendency, which emerges in some texts (including the treatise of Huang 1987) to consider ergodic theory as a mathematical discipline without (almost) any physical relevance. For the reader interested in going into more detail into the topics we are going to discuss below, we refer to the excellent article of Gallavotti Classical statistical mechanics, in the Enciclopedia delle scienze fisiche, published by the Istituto della Enciclopedia Italiana, from which we took most of the considerations that follow. Consider a system of N interacting particles described by the Hamiltonian (15.1) and contained in a finite volume V with perfect walls (isolating and against which the particles collide elastically). Instead of assuming, as usual, that the system can take a continuum of states in the space Γ , we subdivide the latter into small cells ∆ , each determining the position and velocity of each particle with the uncertainty unavoidable in every measurement process. This approach is due to Boltzmann himself, and it is deeply innovative, anticipating in a sense (though not intentionally) the criticism to the determinism of classical mechanics which came much later with the uncertainty principle and the development of quantum mechanics. If h denotes the uncertainty in the measurements of position and velocity, and hence if δqδp
≈ h, h 3N is the volume of a cell. The microscopic state space is then the set of the cells
∆ subdividing the space Γ .
t associated with (15.1) induces, in this context, a transformation S = S
τ which transforms the cells ∆ into one another: S ∆ = ∆ . Here τ is a ‘microscopic time’, very short with respect to the duration T of any macroscopic measurement of the system and on a scale in which the movement of the particles can be measured (accounting for the finite precision). Typical values for τ and T are of the order of 10 −12
seconds and one second, respectively. (A deeper discussion of this point is given by Gallavotti in the quoted article; the reader will notice how his arguments have many analogies with the typical arguments of kinetic theory.) 15.18 Statistical mechanics: Gibbs sets 661 By the theorem of Liouville, the map S is injective and surjective. Essentially S is the canonical linear map obtained by solving over a time interval τ Hamilton’s equations (15.1) linearised at the centre of the cell ∆ considered. The effect of S is therefore to permute the cells ∆ among them. Since the system under consideration is closed and isolated, its energy E is macroscopically fixed (and lies between E and E + ∆ E). Since the volume V accessible to the particles is finite, the number V( ∆ ) of cells representing the energetically possible states is very large, but finite. For example, if we assume for N , V and E the value of a mole of a perfect gas made by hydrogen molecules, with mass m = 2 × 10 −24
g, in standard conditions of pressure and temperature, and hence N = 6 × 10 23
× 10 10 erg, where h is the Planck constant, h = 6 × 10
−27 erg s and for ∆ E the value h/τ = 6 × 10 −15
erg, we find that
V( ∆ ) is of the order of 10 10 25 . The cells can therefore be numbered: ∆ 1 , . . . , ∆ V . The temporal average of a function f becomes ˆ f = lim j →∞ 1 j j −1 i =0 f ( S i ∆ ), (15.116)
where f ( ∆ ) is the value f takes on the cell ∆ . The set average is given by f = 1 V( ∆ ) V (∆) i =1 f ( ∆ i ). (15.117)
The ergodic hypothesis becomes ˆ f = f , (15.118) and it is clearly equivalent to assume that S acts as a one-cycle permutation: a given cell ∆ evolves successively into different cells until it returns to the initial state in a number of steps equal to the number V( ∆ ) of cells. It follows that, by numbering the cells appropriately, we have S ∆
= ∆ i +1 , i = 1, . . . , V − 1 (15.119)
and S ∆ V = ∆ 1 . The ergodic hypothesis is not necessary in the most general formulation. For statistical mechanics to be solidly based, it would be sufficient for (15.118) to be valid only for the few thermodynamical quantities of interest. It would be sufficient that, instead of satisfying (15.119), every cell in its evolution visited mainly those cells in which the observable quantities of interest take an approximately constant value, which as we saw (Section 15.6), in the Boltzmann interpretation are the majority of cells with fixed energy. 662 Statistical mechanics: Gibbs sets 15.18 15.18
Additional solved problems Problem 1 Consider a system of N equal homogeneous plane plates, with a centre of sym- metry and an axis x through their centres and orthogonal to the plates. The centres of mass G 1 , . . . , G N are fixed (for example, they are equidistant) and the plates can rotate without friction around the x-axis. At rest, the plates occupy configurations that can be obtained by translating one into another along the x- axis. When not in equilibrium, there is a torsion energy V = N −1 i =1 1 2 γ(ϕ
i −ϕ i −1 ) 2 where ϕ i is the rotation angle of the ith plate with respect to equilibrium (ϕ 0 = 0)
and γ is a positive constant. If the system is subject to energy fluctuations corresponding to the temperat- ure T , find: (i) the canonical partition function; (ii) the average values of energy, kinetic energy, torsion energy, and the average value of the relative rotation angle between two contiguous plates and of its square. Solution
If I denotes the moment of inertia of the plates with respect to the x-axis, the Hamiltonian of the system is H = N
=1 p 2 i 2I + N −1 i =1 1 2 γ (ϕ i +1 − ϕ i −1 ) 2 . The rotation angles vary between −∞ and +∞. To compute the partition func- tion, we must compute the integral R N e − 1 2 βγ Σ N−1 i=1
(ϕ i+1
−ϕ i ) 2 dϕ 1 . . . dϕ N . It is convenient to use the transformation ϕ i −ϕ i −1 = η i , i = 1, . . . , N , whose Jacobian is equal to 1. The integral is thus reduced to +∞ −∞ e − 1 2 βγη
2 dη N = (2π/βγ) N/ 2 . It is immediately obvious that the integral in the space of momenta is factorised as in +∞
e (−β/2I)p
2 dp N = (2πI/βγ) N/ 2 . Therefore the partition function is Z =
1 N !h
N 2π β N I γ N/ 2 , from which we find H = −∂/∂β log Z = NkT . The average p 2 i /2I
is given by +∞ −∞ p 2 i 2I e −βp 2 i / 2I dp i +∞ −∞ e −βp 2 i / 2I dp i −1 , 15.18 Statistical mechanics: Gibbs sets 663 because all other factors cancel out, and hence we can write p 2 i 2I = − ∂ ∂β log +∞ −∞ e −βp 2 i / 2I dp i = − ∂ ∂β log 2πI β 1/2 = 1 2 kT. It follows that the average kinetic energy is 1 2
is equipartitioned between the averages of the kinetic and torsion energy. As for the angle of relative rotation, η i , we clearly have η i = 0 and η 2 i
kT /γ, since N/2γ η 2 i = 1 2 N kT . Problem 2 Consider the system of N uncoupled harmonic oscillators with Hamiltonian 1 2 (p 2 i + ω 2 q 2 i ) contained in the cube of side 2A. Describe the corresponding canonical set at temperature T . Solution The partition function can be written as Z = 1
3N +∞ −∞ e −βp
2 / 2 dp 3N A −A e −βω 2 q 2 / 2 dq 3N . Setting f (α) = 1 √ π α −α e −y 2 dy, we have Z = 1 N !h 3N 2π βω 3N f 3N β 2 ωA . Therefore we immediately find the average energy H = 3N kT − 3N
ωA √ 2πβ e −(βω
2 / 2)A 2 f −1 β 2 ωA (in agreement with the equipartition theorem for A → +∞). We can then compute the Helmholtz free energy Ψ = − 1 β log Z and the pressure P = −∂ Ψ
3 , we can write P = 3N kT 1 24A 2 2e −βω 2 A 2 / 2 A −A e −βω 2 q 2 / 2 dq = N V kT 2Ae
−βω 2 A 2 / 2 A −A e −βω 2 q 2 / 2 dq . Note that for ω → 0 we obtain the perfect gas pressure P = (N/V )kT . The same happens, keeping the ratio N/V fixed, for small A. However for ω large or A large we see that P tends to zero (the increase in the attractive force or moving away the walls have asymptotically the effect of suppressing the pressure). 664 Statistical mechanics: Gibbs sets 15.18 Problem 3 Model a system of N biatomic particles by attributing to each pair the Hamiltonian h i
1 2m (p (i)2 1 + p (i)2 2 ) + 1 2 a(q (i) 1 − q (i) 2 ) 2 , i = 1, . . . , N. The system is contained in the cube defined by |q (i) j,k | ≤ A, with k = 1, 2, j = 1, 2, 3, i = 1, . . . , N . Using the Hamiltonian H = N i =1 h i and for prescribed temperature T , compute the canonical partition function, the average energy and the average square diameter |q 1 − q 2 | 2 . Solution We have a system of 2N particles in R 3 . Hence we write Z = 1 (2N !) 1 h 6N +∞ −∞ e −(β/2m)p 2 dp 6N A −A A −A e −(βa/2)(q 1 −q 2 ) 2 dq 1 dq 2 3N , where we denote by q 1 , q 2 two corresponding components of the vectors q (i) 1
(i) 2 , for generic i. The first integral is simply (2πm/β) 3N . To evaluate the second integral it is convenient to use the transformation q 1 + q 2 = √ 2ξ, −(q
1 − q
2 ) =
√ 2η,
with unit Jacobian. The double integral then becomes 2 √ 2A √ 2A − √ 2A e −βaη 2 dη = 8A 2 1 √ 2A √ 2A 0 e −βaη 2 dη. In summary, Z = V 2N (2N )! 1 h 6N 2πm
β 3N 2 3N 1 √ 2A √ 2A 0 e −βaη 2 dη 3N . Taking the logarithm, the principal terms, up to factors that render the variables dimensionless, are (N 1) log Z = 2N log V 2N − 3N log β + 3N log 1 √ 2A √ 2A 0 e −βaη
2 dη . 15.18 Statistical mechanics: Gibbs sets 665 We can now compute H = − ∂ ∂β log Z = 3N kT + kT A
2βa 0 y 2 e −y 2 dy A √ 2βa
0 e −y 2 dy def = 3N kT (1 + ϕ(A 2βa)).
In the case A √ 2βa 1 we have H ≈ 9 2 N kT , since ϕ(A √ 2βa)
≈ 1 2 . Regarding the average 2η 2 , we only need to compute 2η 2 = 2 √ 2A 0 η 2 e −βaη
2 dη √ 2A 0 e −βaη 2 dη = 2a β ϕ(A 2βa),
and consequently we have |q 1 − q 2 | 2 = 6
a β ϕ(A 2βa) ≈ 3
a β , for A 2βa
1. Problem 4 A cubic box of side l resting on the horizontal plane z = 0 contains a system of N particles subject to weight. Describe the canonical and microcanonical sets. Solution The Hamiltonian is H = P
2m + mg
N i =1 z i , where as usual P is the momentum vector in R 3N . We write the canonical partition function in the form (V = l 3 ) Z = 1 N !h 3N R 3N e −(β/2m)P
2 dP V
N 1 l l 0 e −βmgz dz N . Then
R 3N e −(β/2mP 2 ) dP = 2πm
β 3N/2
, from which it follows that Z = V
N !h 3N 2πm β 3N/2
1 − e
−βmgl βmgl
N .
666 Statistical mechanics: Gibbs sets 15.18 Taking N
1, and neglecting β-independent terms we get log Z =
− 3 2 N log β − N log(βmgl) + N log(1 − e −βmgl ),
H = − ∂ ∂β log Z =
3 2 N β + N β − N β βmgle
−βmgl 1 − e −βmgl = N kT
5 2 − xe −x 1 − e −x , with x = βmgl. For low values of β (high temperatures) we have H ≈ 3 2 N kT
(the influence of gravity is not felt). At low temperatures, however, it remains H ≈ 5 2 N kT . We now evaluate the average height: z = l 0 ze −βmgz
dz l 0 e −βmgz
dz = l x 1 − xe −x 1 − e −x , x = βmgl. At low temperatures (x → ∞) we have z ≈ 0, while at high temperatures (x → 0) we can easily verify that z ≈ 1 2 . Both results are consistent with physical intuition. We can also compute the free energy: − Ψ
1 β log Z = 1 β N log V N − 3 2 N β log β
− N β log(βmgl) + N β log(1 − e
−βmgl ) and the pressure: P = − ∂ Ψ ∂V = N βV + N β xe −x 1 − e −x 1 l dl dV − N β 1 3V = N V kT 1 + 1 3 xe −x 1 − e −x − 1 3 . At high temperatures (x → 0) we again find P ≈ N/V kT , while at low temperatures (x → ∞) the asymptotic value is P ≈ 2 3 N/V kT . We now try to describe the microcanonical set for the same system, consid- ering energies E > E 0 = N mgl (which do not admit states with zero global momentum P). We study the set {H ≤ E} in the space Γ . For prescribed values of the heights z i , we have for the norm of P the bound P 2 /2m
≤ E −mg N i =1 z i . Setting
N i =1 z i = N z G the momentum P varies in the ball of R 3N with radius √ 2m(E
−mgNz G ) 1/2 whose volume is χ 3N 1/3N [2m(E −mgNz G )] 3N/2 = v(E, z
G ). Hence the measure B(E, V ) of the set {H ≤ E} is B(E, V ) = l 2N l
a(z G )v(E, z G ) dz
G ,
15.18 Statistical mechanics: Gibbs sets 667 where a(z G ) is the measure of the (N − 1)-dimensional section of the cube in R N of side l with the hyperplane N i =1 z i = N z G . Passing to the dimensionless variables ξ i = z i /l ∈ (0, 1) we can factorise in B the coefficient l 3N = V
N , and
hence in the entropy we can single out the term N log(V /N ) (for N 1). Since l 0
G ) dz
G = l
N , when E
0 is negligible with respect to E we find the same result obtained for a perfect gas. Problem 5 In a monatomic gas of N particles confined in a square of side l at temper- ature T compute the probability that at least one particle has kinetic energy greater than: (i) α H /N = αε, with α ∈ (0, ∞); (ii) the sum of the energy of all other particles. Solution These probabilities can be computed as ratios between the ρ-measure in the space Γ
whole space, i.e. the function Z. In case (i) we must force a momentum p i to be greater than the absolute value of √ 2mαε; hence we must compute the ratio ∞ √ 2mαε 2πpe −βp
2 / 2m dp ∞ 0 2πpe −βp
2 2m dp = ∞ √ αβε ye −y 2 dy = e
−αβε . Since ε = 1/β (the system is plane), the sought probability is ν = e −α for a
specific particle. Naturally 1 − ν is the probability that a specific particle has energy less than αε. The probability that no particle has energy greater than αε is (1
− ν) N , and therefore the probability that at least one particle has energy greater than αε is 1 − (1 − ν) N .
N N −j ν j (1 − ν) N −j (note that the sum over j from 0 to N yields 1). Con- sidering j as a continuous variable, the value maximising this probability is j = νN = N e −α . To answer the second question, we must compute the ratio R 2(N−1) e −βP
∗2 / 2m P >P ∗ e −βP 2 / 2m dP dP ∗ R 2N e −βP
2 / 2m dP −1 , 668 Statistical mechanics: Gibbs sets 15.18 which gives the probability ν that the event is verified for a specific particle. In the first integral P ∗ is a momentum in R 2(N −1) . Hence we write ν = χ
∞ 0 P ∗2N−3 e −βP ∗2 / 2m ∞ P ∗ 2πP e −βP
2 / 2m dP dP ∗ χ 2N ∞ 0 P 2N −1 e −βP
2 / 2m dP = π
χ 2(N −1)
χ 2N ∞ 0 X 2N −3 e −2X
2 dX ∞ 0 X 2N −1 e −X 2 dX = 1 2 N −1 π χ 2(N −1) χ 2N Γ (N − 1)
Γ (N )
= 1 2 N −1 (for example for N = 2 we find trivially that one of the two particles has probability 1 2 of having greater energy than the other). The probability that any particle has energy greater than the rest of the system is N 1 2
−1 (the events referring to a single particle are mutually exclusive). Note that we could do all computations explicitly because we chose a two- dimensional system, but the same procedure applies in three dimensions. We suggest continuing the problem by finding the probability that sets of 2, 3, . . . , N < N particles have globally energy greater than the energy of the complementary system.
Problem 6 A cylinder of volume V and cross-section Σ contains N particles and the system is at a temperature T . (i) Find the average number of particles that pass through the generic section in unit time. (ii) Find the pressure as an average of the momentum transfer rate per collision with the walls, proving that this average is equal to 2 3 ( H /V ). Solution
The average displacement of a molecule in unit time, in the direction orthogonal to Σ (x-axis) with positive orientation, is s =
∞ 0 p 1,1 m e −βp 2 1,1 / 2m dp 1,1 ∞ 0 e −βp
2 1,1
/ 2m dp 1,1 −1 = 2 π 1 mβ . The global number of molecules which pass through Σ in the unit of time (in both directions) is then ν = 2
2 π kT m Σ V N (for m
10 −23
g, T = 300 K, Σ = 1 cm 2 , N/V = 10 18 cm
, we find ν 10 23 s −1 ). The momentum transfer per unit time on the unit surface normal to the x-axis is given by Π +
= N i =1 2 m 1 V (p i, 1 ) 2 + , 15.18 Statistical mechanics: Gibbs sets 669 where
the symbol
( ·) + indicates that
we only
consider the
positive components. Without any
computation we find Π + 1 . It is enough to note that 3 j =1 ( Π + j + Π − j ) = (4/V )H and passing to the averages (because of isotropy) we have
6 Π = 4 V H ; hence we can really identify Π with the pressure. It should be noted that the results of the last problem can be obtained using the formalism of kinetic theory, and the Maxwell–Boltzmann distribution. The procedure is identical, even formally.
16 LAGRANGIAN FORMALISM IN CONTINUUM MECHANICS 16.1 Brief summary of the fundamental laws of continuum mechanics The model of a continuum relies on the hypothesis that we can describe the distribution of mass through a density function ρ(P, t), in such a way that the mass of each measurable part D of the system under consideration is representable in the form M (D) =
D ρ(x, t) dx. (16.1) A continuum can be three-dimensional, two-dimensional (plates or membranes) or one-dimensional (strings and beams). The following is a way to represent the configurations of a continuum with respect to a frame S = (0, x 1 , x 2 , x
3 ). We choose a reference configuration C ∗ and
we denote by x ∗ 1 , x ∗ 2 , x ∗ 3 the coordinates of its points. Any other configuration C is then described by a diffeomorphism: x = x(x ∗
x ∗ ∈ C ∗ . (16.2) The coordinates x ∗ 1 , x ∗ 2 , x ∗ 3 play the role of Lagrangian coordinates. If the system is in motion, instead of (16.2) we have x = x(x ∗
x ∗ ∈ C ∗ (16.3)
describing the motion of every single point (typically x ∗ = x(x ∗ , 0)).
Expression (16.3) is the so-called Lagrangian description of the motion. Its inverse
x ∗ = x ∗ (x, t)
(16.4) provides, for every fixed x in the space, the Lagrangian coordinates of the points occupying the position x as time varies (Eulerian description). The fundamental law of the kinematics of continua is mass conservation, expressed by the continuity equation ∂ρ ∂t + div(ρv) = 0, (16.5)
which is just a particular case of the balance equation ∂ G ∂t + div j = γ 672 Lagrangian formalism in continuum mechanics 16.1 for a scalar quantity G, carried by a current density j (i.e. j · n = amount of G carried through the unit surface with normal n in unit time), where γ represents the source or sink (rate of production or absorption of G per unit of volume). The proof is very simple. Equation (16.5) is written in Eulerian form. Since the derivative along the motion (Lagrangian derivative) is d G
= ∂ G ∂t + v
· ∇G, the Lagrangian form of the continuity equation is dρ dt
(16.6) The dynamics of continua require appropriate modelling of forces. We split the forces acting on a part D of the continuum into two categories: (a) surface forces: forces that are manifested through contact with the boundary of D; (b) body forces: all other forces (a typical example is weight). The model for body forces can be constructed using the simple hypothesis that they are proportional to the mass element ρ(x, t) dx on which they act, through a coefficient f (x, t), called the specific mass force (dimensionally, an acceleration: g in the case of weight). To define surface forces, we consider an element dσ of the boundary of D with normal direction n external to D and we say that the force that the complementary set exerts on D through dσ is expressed by Φ (x, t; n) dσ, where Φ has the dimension of a pressure and it is called the specific stress ( Φ · n is
the compression stress if negative, and the tension stress if positive, while the component normal to n is called shear stress). The basic theorem of the dynamics of continua is due to Cauchy. T heorem 16.1 (Cauchy) For every unit vector n = 3 i =1 α i e i the specific stress has the following expression: Φ (x, t; n) = 3 i =1 α i Φ (x, t; e i ). We omit the proof. Cauchy’s theorem yields as a result that the products T ij (x, t) = Φ (x, t; e i ) · e j , (16.7) with e 1 , e 2 , e
3 an orthonormal triple in x, are the elements of a tensor T (the stress tensor ) which defines the stress state in (x, t). Knowledge of T ij yields the reconstruction of the stress relative to every unit vector n: Φ j (x, t; n) = 3 i =1 α i T ij (x, t), j = 1, 2, 3, (16.8)
with α 1 , α 2 , α
3 direction cosines of n. 16.1 Lagrangian formalism in continuum mechanics 673 Using the theorem of Cauchy it is possible to deduce that the first and second cardinal equations applied to every subset of a continuous system yield, respectively, the following equations: ρ(f − a) + div T = 0, (16.9) T ij = T ji , i = / j.
(16.10) In the former, by definition div T is the vector divT = 3
=1 ∂T ij ∂x i e j . (16.11) Equation (16.9) holds generically for all continua. The mechanical nature of the system must be specified through additional equations. Expression (16.10) represents the so-called stress symmetry. A special case of great interest is the case of fluids. D efinition 16.1 A fluid is a continuum for which the shear stresses at equilib- rium are zero. If this also happens in a dynamic situation, then the fluid is called perfect or ideal. For fluids we have an additional simplification of the stress tensor, as the diagonal elements are equal (the proof is left as an exercise). Moreover, since the fluid resists only compression, the common value of the diagonal elements of the stress tensor must be negative: T ij =
ij , where p > 0 is the pressure. The equilibrium equation of a fluid can now be written as ρf =
∇p (16.12)
(since div T = −∇p), and the equation of motion (for a perfect fluid) is ρ(f − a) = ∇p. (16.13) In both it is necessary to specify the relation between ρ and p: ρ = ρ(p), ρ (p)
≥ 0 (16.14)
(state equation). A fluid for which the relation (16.14) is known is called barotropic. For simplicity we only consider isothermal phenomena. For a barotropic fluid we can introduce the function P(p) =
dp ρ(p)
(16.15) (potential energy of the pressure), and hence (1/ρ) ∇ p
∇P(p). If in addition f =
∇u(x), equation (16.12) can be immediately integrated to give P(p(x)) = u(x) + constant 674 Lagrangian formalism in continuum mechanics 16.1 (the constant can be determined using the boundary condition p(x 0 ) = p
0 at some given point x 0 ), while after some manipulations (16.13) can be written as ∂v ∂t
× v = −∇B (16.16)
(Euler equation), where B =
1 2 v 2 − u + P
(16.17) is the Bernoulli trinomial. The Euler equation is invariant with respect to time reversal (t → −t, v → −v) and indeed it describes a non-dissipative phenomenon. While useful in many circumstances, it is not adequate to describe many phe- nomena of practical importance (for example the motion of objects in fluids). It is then necessary to construct a more sophisticated model of the fluid (the model of viscous fluids), which we do not discuss here. See for example Landau (1990). Example 16.1 : linear acoustics Consider a perfect fluid in equilibrium (neglecting gravity) at uniform pressure p 0
0 : ρ = ρ 0 + p − p 0 c 2 , with c = [ρ (p 0 )] −1/2 having the dimension of a velocity (ρ (p 0 ) is assumed to be positive and the fluid is said to be compressible). In the small perturbations approximation (the linear approximation) we consider p − p 0
− ρ 0 , v, etc. to be first-order perturbations, and we neglect higher-order terms such as curl v × v,
v · ∇ρ, etc. Considering the linearised version of the Euler equation: ∂v ∂t
1 ρ 0 ∇p = 0 and of the continuity equation: dρ dt
0 div v = 0, eliminating div v (after taking the divergence of the first equation) and writing ∂ρ/∂t = (1/c 2 )(∂p/∂t), we find that the pressure satisfies the wave equation (or the d’Alembert equation) ∇ 2 p − 1 c 2 ∂ 2 p ∂t 2 = 0.
(16.18) Particularly interesting solutions are the plane waves (depending on only one space coordinate, corresponding to the direction of propagation), which in the most common case can be represented as a superposition of progressive waves: p(x, t) = f (x − ct)
(16.19) 16.1 Lagrangian formalism in continuum mechanics 675 and regressive waves: p(x, t) = g(x + ct), (16.20)
which highlight the role of c as the velocity of propagation of the wave. If ψ(x, t) is a plane wave then a spherical wave can be constructed via the transformation ϕ(r, t) = 1 r ψ(r, t), (16.21)
where r is the distance from the centre of the wave. Plane waves can also appear as stationary waves: ψ(x, t) = A sin kx cos νt, (16.22)
where the wave number k and the frequency ν must be related by kc = ν. Example 16.2 : vibrating string Consider a perfectly flexible string kept straight with tension T at the two endpoints. The equilibrium configuration is straight. If we perturb either the configuration (plucked string) or the velocity (hammered string), or both, in such a way that the string oscillates with velocity approximately orthogonal to the string at rest, we can easily prove the following facts regarding the linearised motions:
(i) the tension is constant along the string; (ii) the equation of small shear vibrations is ∂ 2
∂x 2 − 1 c 2 ∂ 2 u ∂t 2 = 0, (16.23) where u(x, t) is the displacement from equilibrium and c = T /ρ, where ρ is the constant (linear) density of the string. The Cauchy problem for equation (16.23) with initial values u(x, 0) = ϕ(x), −∞ < x < +∞, (16.24)
∂u ∂t t =0 = ψ(x),
−∞ < x < +∞ (16.25)
has the d’Alembert solution u(x, t) = 1 2
1 2c x +ct x −ct ψ(ξ), dξ, (16.26)
where we recognise the progressive and regressive waves generated by the perturbation ϕ and by the perturbation ψ. 676 Lagrangian formalism in continuum mechanics 16.2 Example 16.3 : longitudinal vibrations of a rod Hooke’s law for elastic materials applied to a homogeneous cylindrical rod subject to tension or compression T 0 (t) (for unit cross-section) implies that T = E ∆ , (16.27) where
∆ / is the relative elongation and E is Young’s modulus. Neglecting shear deformations, denoting by u(x, t) the displacement from equilibrium and extrapolating Hooke’s law to T = E ∂u
, (16.28)
the equation of motion can be written as ∂ 2 u ∂x 2 − 1 c 2 ∂ 2 u ∂t 2 = 0, (16.29)
where c = E/ρ (ρ is the rod’s density). Concerning the historical aspects of the theory of wave propagation, we suggest Truesdell (1968) or Manacorda (1991). We refer the reader interested in the physics of musical instruments to Fletcher and Rossing (1991). 16.2 The passage from the discrete to the continuous model. The Lagrangian function We consider again the problem of longitudinal vibrations of an elastic homo- geneous rod (Example 16.3) and we aim to construct an approximation of this system using a discrete set of point particles. We denote by ρ the density and by S the area of the cross-section of the rod. We subdivide the rod into N equal parts and we replace them with a chain of point particles of mass m/N , where m is the mass of the entire rod (Fig. 16.1). To model the internal forces we assume that two consecutive points are connected by springs of negligible mass, with an elastic constant k which we specify later and with length at rest equal to ε = /N ( is the length of the rod). We denote by x (0)
1 , x
(0) 2 , . . . , x (0) n the x-coordinates of the point particles at equilibrium (x (0)
s = sε). Consider now the generic triple (P i −
, P i , P i +1 ) and denote by u i the displacement of P i from equilibrium (Fig. 16.2). The stretching of the spring between P i +1 and P i is u i +1 − u i , and hence the global potential energy is V = 1
k N − 1 i =1 (u i +1 − u i ) 2 , (16.30) 16.2 Lagrangian formalism in continuum mechanics 677
1
2
−1
n x Fig. 16.1 P (0)
i–1 P (0)
i+1 P i+1 P i–1 x P (0)
i u i P i Fig. 16.2 while the kinetic energy is T =
1 2 m N N i =1 ˙ u 2 i . (16.31) We can then write the Lagrangian of the system L(u, ˙ u) =
1 2 m N N i =1 ˙ u 2 i − k N − 1 i =1 (u i +1 − u i ) 2 (16.32) and finally obtain the equations of motion m N
u i − k[u i +1 − 2u i + u
i − 1 ] = 0, i = 1, 2, . . . , N (with u 0
m ¨ u i − kε
u i +1 − 2u i + u i − 1 ε 2 = 0, i = 1, . . . , N, (16.33)
where there appears the discretisation of the second derivative with respect to x and at the same time the product kε. It is obvious that m/l = ρS is the linear density of the rod. We must make precise the choice of k. Since the elastic tension force between two contiguous points can be written in the form kε u
+1 − u
i ε , (16.34) following Hooke’s law (16.27), we write kε = E (Young’s modulus). 678 Lagrangian formalism in continuum mechanics 16.3 We now recall that the system (16.33) is just the space discretisation of equation (16.29). It is known that if we construct a regular function u N (x, t) which takes the values u i at the points x i , then u
N converges for N → ∞ to the solution u of (16.29) (for the prescribed boundary conditions). It is most interesting to rewrite the Lagrangian (16.32) in the form L =
1 2 ρS N i =1 ˙ u 2 i − ES
N − 1 i =1 u i +1 − u i ε 2 · ε and to pass to the limit for N → ∞, obtaining the following integral expression: L =
0 1 2 ρS ∂u ∂t 2 − ES
∂u ∂x 2 dx. (16.35)
Therefore we discover that we can associate with the continuous model a Lagrangian and also define a Lagrangian density: L ∂u
, ∂u ∂t = 1 2 ρ ∂u ∂t 2 − 1 2 E ∂u ∂x 2 (16.36) (the factor S can be replaced in (16.35) by a double integral over cross-sections) such that L = C
(16.37) where
C is the configuration of the system under consideration. 16.3
Lagrangian formulation of continuum mechanics It is now natural to consider whether the equation of motion (16.29) can be obtained by imposing the condition that an action-type functional related to the Lagrangian (16.37) is stationary with respect to certain classes of perturbations. We consider the problem from a general point of view, assuming that with every continuum described by a field function u(x, t) (for example the displacement from the equilibrium configuration) we can associate a Lagrangian density with the necessary regularity with respect to its arguments: L u, ∂u
1 , ∂u ∂x 2 , ∂u ∂x 3 , ∂u ∂t , x 1 , x 2 , x
3 , t
(16.38) 16.3 Lagrangian formalism in continuum mechanics 679 and extend the validity of Hamilton’s principle. P ostulate 16.1 The natural motion of the system corresponds to a stationary point of the functional A =
t 1 t 0 C (t) Ldx dt (16.39)
(where C(t) is the configuration of the system at time t) with respect to the (regular) perturbations δu(x, t) which vanish ∀x ∈ C(t), when t = t 0 , t = t
1 , and
on the boundary ∂ C, for every t ∈ (t 0 , t
1 ). Hence we seek the conditions for a point to be stationary for the functional (16.39) in the specified class. Denoting by u ∗ (x, t) the value of the field for the natural motion, we introduce the field u = u
∗ + δu
and evaluate the first variation of A: δA =
t 1 t 0 C (t) ∂ L ∂u δu + 3 i =1 ∂ L ∂ξ i ∂δu ∂x i + ∂ L ∂ζ ∂δu ∂t dx dt, where ξ = ∇u, ζ = ∂u/∂t. Given the assumptions on δu, the divergence theorem yields t
t 0 C 3 i =1 ∂ L ∂ξ i ∂δu
∂x i dx dt = − t 1 t 0 C 3 i =1 ∂ ∂x i ∂ L ∂ξ i δu dx dt. In addition, using d dt C (t)
∂ L ∂ζ δu dx = C (t) ∂ ∂t ∂ L ∂ζ δu dx + ∂ C (t) ∂ L ∂ζ δu v
n dσ,
with v n being the normal velocity of the points of ∂ C(t), and remarking that the integral over ∂ C(t) is zero, since δu is zero on the boundary, we can rewrite the last term in δA as t 1
0 C (t) ∂ L ∂ζ ∂δu ∂t dx dt = − t 1 t 0 C (t) ∂ ∂t ∂ L ∂ζ δu dx dt, taking into account that δu = 0 on C(t 0
C(t 1 ). 680 Lagrangian formalism in continuum mechanics 16.4 An argument similar to the one used to prove the analogous theorem in the discrete case yields the following conclusion. T heorem 16.2 The characteristic condition for a point to be stationary for the functional (16.39) in the class of perturbations considered, is that ∂ ∂t ∂ L ∂ζ + 3 i =1 ∂ ∂x i ∂ L ∂ξ i − ∂ L ∂u = 0 (16.40)
(recall that ξ i = ∂u/∂x i , ζ = ∂u/∂t). Due to Postulate 16.1, equation (16.40) represents the equation of motion, naturally a partial differential equation. Remark 16.1 From equation (16.40) we deduce that the terms linear in ξ i and in ζ (with constant coefficients) in the expression for L are not essential. More generally, we can consider Lagrangian densities depending on scalar
functions u 1 , u 2 , . . . , u . The variational problem can be stated in a similar way, leading to an equation of the type (16.40) for every unknown function u k . 16.4 Applications of the Lagrangian formalism to continuum mechanics We now consider a few concrete examples illustrating the theory developed in the previous section. (A) Longitudinal vibrations of an elastic rod Using the Lagrangian density (16.36) in equation (16.40), we clearly find the d’Alembert equation (16.29). (B ) Linear acoustics To determine the Lagrangian density for ‘small perturbations’ of a perfect gas, neglecting the effect of the body forces, we note that in the Lagrangian density there must appear two contributions, due to the specific kinetic energy, and to the specific potential energy (which must be subtracted from the former). Denoting by u(x, t) the displacement vector, the kinetic energy of the unit of mass is 1
(∂u/∂t) 2 . To evaluate the potential energy V of the unit of mass we write the energy balance d V + p d
1 ρ = 0, (16.41) where p d(1/ρ) is the work done by the unit of mass of the gas for the variation d(1/ρ) of its volume. Recall that we are dealing with a barotropic fluid, and
16.4 Lagrangian formalism in continuum mechanics 681 hence that ρ = ρ(p) (in the case of sound vibrations one must consider adiabatic transformations, hence pρ − γ = constant). We then obtain from (16.41) that V = −
1/ρ 1/ρ
0 p dη,
where we have introduced the variable η = 1/ρ. Consider the linear approximation of p as a function of η around η 0 = 1/ρ
0 (henceforth, zero subscripts denote quantities at equilibrium): p = p
0 + (η
− η 0 ) dp dη η =η 0 . Computing the integral, we obtain V = − p
0 1 ρ − 1 ρ 0 + 1 2 dp dη η =η 0 1 ρ − 1 ρ 0 2 . (16.42) We now compute (dp/dη) η =η 0 , writing η = 1/ρ(p) and differentiating with respect to η: 1 =
− ρ (p)
ρ 2 (p) dp dη , from which dp dη η =η 0 = − ρ 2 0 ρ 0 · (16.43) We now express (1/ρ) − (1/ρ
0 ) through the variation relative to ρ, i.e. δ = (ρ − ρ
0 )/ρ:
1 ρ − 1 ρ 0 = − 1 ρ 0 δ, (16.44) and we substitute (16.43) and (16.44) into equation (16.42): V = p
ρ 0 δ + 1 2 1 ρ 0 δ 2 . (16.45) Recall that we set 1/ρ 0 = c 2 . The last step to obtain the Lagrangian density consists of expressing δ in terms of the displacement u. To this end, it is sufficient to write the linearised continuity equation ∂ρ ∂t
0 div
∂u ∂t = 0 682 Lagrangian formalism in continuum mechanics 16.4 in the form ∂δ ∂t + div ∂u ∂t = 0. (16.46) Integrating the latter expression (and denoting by γ the value of δ + div u for t = 0) we find the relation δ =
−div u + γ. (16.47)
This yields the sought Lagrangian density L =
1 2 ∂u ∂t 2 + p 0 ρ 0 + c
2 γ div u − 1 2 c 2 (div u) 2 , and, recalling Remark 16.1, we can suppress the linear term in div u, and arrive at the expression L =
1 2 ∂u ∂t 2 − 1 2 c 2 (div u)
2 . (16.48) It is a trivial exercise to check that (16.43) leads to ∇(∇ · u) − 1 c
∂ 2 u ∂t 2 = 0, from which, by taking the divergence, we obtain the wave equation for δ. (C ) Electromagnetic field The idea of deducing the field equations from a Lagrangian density is entirely general and can be applied to fields other than mechanics as well, although outside the conceptual framework of mechanics, there does not exist a general criterion to deduce the Lagrangian density. We now consider an example and illustrate how it is possible to derive the Maxwell equations for the electromagnetic field in a vacuum from a Lagrangian density. The unknown functions on which the Lagrangian density L depends are the scalar potential φ(x, t) and the vector potential A(x, t) (Section 4.7), through which we can express the electric field E and the magnetic induction field B: E = −∇φ −
1 c ∂A ∂t , (16.49) B = curl A. (16.50)
The equations div B = 0, (16.51) curl E +
1 c ∂B ∂t = 0
(16.52) are automatically satisfied thanks to (16.49) and (16.50). 16.4 Lagrangian formalism in continuum mechanics 683 The equations to be deduced from the Lagrangian formulation are therefore the remaining Maxwell equations: div E = 4πρ, (16.53) curl B
− 1 c ∂E ∂t = 1 c 4πj. (16.54) Let us check that a correct choice for L is L =
1 8π (E 2 − B
2 ) +
1 c j · A − ρφ, (16.55)
where E and B are given by (16.49), (16.50). The equation for φ is i ∂
i ∂ L ∂(∂φ/∂x i ) − ∂ L ∂φ = 0.
(16.56) Since
E 2 = i ∂φ ∂x i 2 + 2 c i ∂φ ∂x i ∂A i ∂t + 1 c 2 i ∂A i ∂t 2 , equation (16.56) takes the form 1 4π
φ + 1 c ∂ ∂t divA + ρ = 0, (16.57)
and hence it coincides with (16.53). We now write B 2
i ∂A i ∂x j − ∂A j ∂x i 2 and compute for example ∂ ∂t ∂ L ∂(∂A
1 /∂t)
= 1 4πc 2 ∂ 2 A 1 ∂t 2 + 1 4πc ∂ 2 φ ∂t∂x
1 = − 1 4πc
∂E 1 ∂t , − i ∂ ∂x i ∂ L ∂(∂A 1 /∂x
i ) = 1 4π ∂ ∂x 2 ∂A 1 ∂x 2 − ∂A 2 ∂x 1 + 1 4π ∂ ∂x 3 ∂A 1 ∂x 3 − ∂A 3 ∂x 1 = − 1 4π ∂B 2 ∂x 3 − ∂B 3 ∂x 2 , ∂ L ∂A 1 = 1 c j 1 . It is now immediate to verify that the equation ∂ ∂t ∂ L ∂(∂A
1 /∂t)
+ i ∂ ∂x i ∂ L ∂(∂A
1 /∂x
i ) − ∂ L ∂A 1 = 0
coincides with the first component of (16.54). The computation for the other components is similar. 684 Lagrangian formalism in continuum mechanics 16.5 16.5
Hamiltonian formalism For the mechanics of continua and for field theory, just as for the mechanics of systems of point particles, it is possible to develop a Hamiltonian formalism parallel to the Lagrangian formalism. If L (u, ∂u/∂x 1 , ∂u/∂x
2 , ∂u/∂x
3 , ∂u/∂t, x, t) is a Lagrangian density depending on an unknown vector u(x, t) ∈ R , we can define the vector π of the densities of momenta: π k = ∂ L ∂(∂u k /∂t) , k = 1, 2, . . . , . (16.58) To pass to the Hamiltonian formalism it is necessary for the system (16.58) to be invertible, and hence it must be possible to write ∂u k /∂t as functions of the vector π. We can then define the Hamiltonian density: H = π · ∂u
− L, (16.59)
with H = H u,
∂u ∂x 1 , ∂u ∂x 2 , ∂u ∂x 3 , π, x, t . (16.60)
Computing the derivatives ∂ H/∂u
k , ∂
H/∂π k using (16.58), (16.59) and (16.40) (written for each component u k ), we arrive at Hamilton’s equations: ∂ H ∂π k = ∂u k ∂t , ∂ H ∂u k = − ∂ L ∂u k = − ∂π k ∂t − i ∂ ∂x i ∂ L ∂(∂u
k /∂x
i ) , (16.61) and hence, noting that ∂ H/[∂(∂u k
i )] =
−∂L/[∂(∂u k /∂x i )], we have ∂ H
k = − ∂π k ∂t + i ∂ ∂x i ∂ H ∂(∂u
k /∂x
i ) . (16.62) The lack of formal symmetry between the two expressions (16.61) and (16.62) can be corrected by introducing the operators δ δu k = ∂ ∂u k − i ∂ ∂x i ∂ ∂(∂u k /∂x
i ) , δ δπ k = ∂ ∂π k − i ∂ ∂x i ∂ ∂(∂u
k /∂π
i ) = ∂ ∂π k , 16.6 Lagrangian formalism in continuum mechanics 685 through which we find the more familiar expressions ∂u k ∂t = δ H δπ k , (16.63) ∂π k ∂t = − δ H δu k , k = 1, 2, . . . , . (16.64) Example 16.4 Using the Lagrangian density (16.36) for the longitudinal vibrations of an elastic rod, we have π = ρ ∂u
, from which it follows that H = π
2ρ + 1 2 E ∂u ∂x 2 , and Hamilton’s equations are ∂u ∂t = π ρ , ∂π ∂t = E ∂ 2 u ∂x 2 . The first is the definition of π; the combination of the two yields the d’Alembert equation. 16.6
The equilibrium of continua as a variational problem. Suspended cables The stationary points of the functional (16.39) with respect to the perturbations described in Postulate 16.1 define the motion of the system. Similarly equilibrium is characterised by the stationary points of the functional V =
C V u,
∂u ∂x 1 , ∂u ∂x 2 , ∂u ∂x 3 , x 1 , x
2 , x
3 dx,
(16.65) where
V is obtained by suppressing the kinematic terms in L, with respect to the subclass of perturbations which vanish on the boundary. Clearly V can be interpreted as the total potential energy. The Euler equation is therefore i ∂ ∂x i ∂ V ∂ξ i − ∂ V ∂u = 0
(16.66) (as usual, ξ i = ∂u/∂x
i ), with which one needs to associate possible constraints. 686 Lagrangian formalism in continuum mechanics 16.6 As an application, consider the two problems relating to the equilibrium of suspended cables, where a ‘cable’ is a homogeneous string, perfectly flexible and of constant length. (A) Cable subject only to weight (catenary, Fig. 16.3) If y = f (x) is the equation for the generic configuration of the cable, the potential energy is proportional to the height of the centre of mass, defined by y G (f ) = x B 0 f (x)
1 + f 2 (x) dx. (16.67) This must be minimised by imposing the constraint that the cable has fixed length , and hence x B 0 1 + f
2 (x) dx = . (16.68)
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