A Physical Constants1
Constants in Mechanics
−11 3 −1 −2 Gravitational constant G =6.672 59(85) × 10 m kg s (1996) −11 3 −1 −2 =6.673(10) × 10 m kg s (2002) −2 Gravitational acceleration g =9.806 65 m s 30 Solar mass M =1.988 92(25) × 10 kg 8 Solar equatorial radius R =6.96 × 10 m 24 Earth mass M⊕ =5.973 70(76) × 10 kg 6 Earth equatorial radius R⊕ =6.378 140 × 10 m 22 Moon mass M =7.36 () × 10 kg 6 Moon radius R =1.738 × 10 m 9 Distance of Earth from Sun Rmax =0.152 1 × 10 m 9 Rmin =0.147 1 × 10 m 9 Rmean =0.149 6 × 10 m 8 Distance of Moon from Earth Rmean =0.380 × 10 m 7 Period of Earth w.r.t. Sun T⊕ = 365.25 d = 3.16 × 10 s 6 Period of Moon w.r.t. Earth T =27.3d=2.36 × 10 s
1 From Particle Data Group, American Institute of Physics 2002 http://pdg.lbl.gov/2002/contents http://physics.nist.gov/constants 326 A Physical Constants
Constants in Electromagnetism
def Velocity of light c =. 299, 792, 458 ms−1 def. −7 −2 Vacuum permeability μ0 =4π × 10 NA =12.566 370 614 ...× 10−7 NA−2 1 × −12 −1 Vacuum dielectric constant ε0 = 2 =8.854 187 817 ... 10 Fm μ0c Elementary charge e =1.602 177 33 (49) × 10−19 C e2 =1.439 × 10−9 eV m = 2.305 × 10−28 Jm 4πε0
Constants in Thermodynamics
−23 −1 Boltzmann constant kB =1.380 650 3 (24) × 10 JK =8.617 342 (15) × 105 eV K−1 23 −1 Avogadro number NA =6.022 136 7(36) × 10 mole Derived quantities : Gas constant R = NLkB Particle number N,molenumbernN= NLn
Constants in Quantum Mechanics
Planck’s constant h =6.626 068 76 (52) × 10−34 Js h = =1.054 571 596 (82) × 10−34 Js 2π =6.582 118 89 (26) × 10−16 eV s 2 4πε0 −10 Bohr radius a∞ = =0.529 177 208 3 (19) × 10 m me2 # $2 m e2 2 1 e2 Rydberg energy E0 = = 2 = 2 4πε0 2ma0 2 4πε0a0 1 2 2 = 2 mc α =13.605 691 72 (53) eV A Physical Constants 327
−31 Electron rest mass me =9.109 381 88 (72) × 10 kg =0.510 998 902 (21) MeVc−2 −28 Myon rest mass mμ =1.883 566() × 10 kg −27 Proton rest mass mp =1.672 621 58 (13) × 10 kg = 938.271 998(38) MeVc−2 −27 Neutron rest mass mn =1.674 954 3 () × 10 kg 1 g −27 Atomic mass unit (amu) mu = =1.660 538 73 (13) × 10 kg NA mole = 931.494 013(37) MeVc−2 e [c] × −5 −1 Bohr magneton μB = 2mc =5.788 381 749 (43) 10 eV T 2 Fine-structure constant α = e =1/137.035 999 76 (50) 4πε0 c
Conversion Factors
1eV= 1.602 176 462 (63) × 10−19 J B Scalars, Vectors, Tensors
In this appendix some relations are collected, which could be useful in non- relativistic mechanics.
B.1 Definitions and Simple Rules
B.1.1 Definitions
(Mathematical) definition: (i) Tensors (of a given rank) are elements of a vector space (i.e., they obey certain operation rules.1 (ii) Under transformation, a tensor of nth rank (with n indices) has, depending on the type of the index, the behavior of the position vectors. (Sloppy, physical) definition: A vector is a quantity with modulus and direc- tion. Comments: • A coordinate transformation (from unprimed to primed coordinates) of a vector a with the (cartesian) components ai is given by j ai = Ui aj. (B.1) j
Correspondingly, the behavior of a tensor nthr rank under a transforma- tion is given by j1 ··· jn Ti1...in = Ui1 Uin Tj1...jn . (B.2) j1...jn
1 See Sect. B.2.1 for vectors. 330 B Scalars, Vectors, Tensors
• Scalars and vectors are tensors of zeroth and first rank, respectively. • A scalar is a number. But not each number is a scalar: For example the x component of a vector is a number, but no scalar, because it changes with a coordinate transformation. • Tensors of second rank can be represented by matrices. But not each ma- trix is a tensor: For example the transformation matrix U of (B.1) with the j elements Ui is not a tensor, since it does not refer to a given coordinate system, but gives the relation between two different systems.
B.1.2BehaviorUnderInversion
Under inversion a tensor of even-numbered rank (in particular thus a scalar) transforms into itself; a tensor of odd-numbered rank (thus in particular a vector) transforms into its negative. Pseudo tensors are likewise elements of vector spaces, have but the wrong behavior under transformation: Under inversion a pseudo tensor of even- numbered rank (in particular a pseudo scalar) transforms into its negative; a pseudo tensor of odd-numbered rank (in particular a pseudo vector) trans- forms into itself. Notation: In order to discriminate, a vector is denoted alternatively as a polar vector and a pseudo vector also as an axial vector. Comments: • The cross product of two polar vectors (e.g., in the case of the angular momentum) is an axial vector (pseudo vector).2 • The tensor product (dyadic product) of two polar vectors is a tensor.3 • The scalar product of a vector with a pseudo vector (i.e. the triple scalar product of three vectors) is a pseudo scalar.4
B.2 Vectors
B.2.1 Rules for Vectors
One has the following axioms for vectors as elements of a vector space: A sum of vectors a ∈ Rd and b ∈ Rd is defined, which results in a vector c ∈ Rd, a + b = c ⇔ ai + bi = ci. A multiplication is defined of a vector a ∈ Rd withascalarα ∈ K;thespace is linear: d = αa ⇔ di = αai. 2 See Appendix B.2.3. 3 See Appendix B.2.2. 4 See Appendix B.2.4. B.2 Vectors 331
There is an inner product (scalar product), of two vectors resulting in a scalar s,
a · b = s.
B.2.2 Dyadic Product (Tensor Product) of Vectors
The dyadic product (tensor product)
ab≡ a ⊗ b
of two vectors a, b is a tensor T = ab= a ⊗ b with c · T = c · a ⊗ b = c · a b T · d = a ⊗ b · d = a b · d .
Comments: • The behavior of the tensor T under transformation is apparent. • In contrast to the scalar product a · b = b · a the tensor product is not commutative, ab= ba. • The symbol ⊗ is rarely used and is used here only for distinction from the dot product (Fig. B.1). • It is thus of importance, for the distinction from the tensor product to notate the product symbol (dot) in the case of the scalar product!! • The multiplication rule “row times column” for matrices holds likewise for the scalar product of vectors. For the scalar product this is a sum of products (of two terms), while for the tensor product each tensor element consists of just one product.
) *)* . . ... · . . ⊗ ...
Fig. B.1. The scalar product (left) and the tensor product (right) of two vectors with the rule “row times column”
B.2.3 Vector Product (Cross Product) of Vectors
The vector product (cross product) is restricted to vectors in R3;by
c = a × b 332 B Scalars, Vectors, Tensors
Fig. B.2. The outer product of two vectors
a third vector c is assigned to two vectors a, b with
a · c =0= b · c & & & &2& &2 & &2 &a × b& = &a · a& &b · b& − &a · b& a, b, and c = a × b forming a right-handed system (Fig. B.2). The resulting vector c is perpendicular to the vectors a and b; the modulus is equal to the area spanned by the vectors a and b, |a × b| = ab sin α, where α is the angle (in the mathematically positive sense from a to b)be- tween the vectors a and b. The cross product has the following properties: a × b = −b × a a × a =0 a × b =0 ⇒ a =0 or b =0 or a = αb.
B.2.4 Triple Scalar Product of Vectors Like the external product thus the triple scalar product is restricted to vectors 3 in R ;with a · b × c = a × b · c according to sections B.2.1 and B.2.2 a (pseudo) scalar is assigned to three (polar) vectors. The triple product is the volume of the parallelepiped spanned by the three vectors and positive if the three vectors form a right-handed system.
B.2.5 Multiple Products of Vectors The following relations for vectors in R3 are of use occasionally: a × b × c = a · c b − a · b c (B.3) a × b · c × d = a · c b · d − a · d b · c . (B.4) Comment: a × b × c = a × b × c. C Rectangular Coordinate Systems
C.1 Definitions
3 Let a point in R be represented by the (generally curvilinear) coordinates ξi (i =1, 2, 3).
Definition: The coordinate sheet fi is the sheet fi(ξi)=0(i =1, 2, 3).
Definition: The coordinate line si is the intersecting line of the sheets fj(ξj )=0andfk(ξk)=0(withi, j, k cyclic).
Definition: The unit vector ei is the vector tangential to the coordinate line si in the direction of increasing value of ξi with ei · ei =1.
C.2 Cartesian Coordinates
The position vector is r = x ex + y ey + z ez. The line element is (Fig. C.1)
ds =dx ex +dy ey +dz ez.
z dz
dy dx
y
Fig. C.1. The volume element in cartesian co- x ordinates 334 C Rectangular Coordinate Systems
The surface element is
da =dy dz ex +dz dx ey +dx dy ez. The volume element is (Fig. C.1) d3r =dx dy dz.
C.3 Spherical Polar Coordinates (Spherical Coordinates)
The transformation from cartesian coordinates to spherical polar coordinates is given by ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ er sin ϑ cos ϕ sin ϑ sin ϕ cos ϑ ex ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ eϑ = cos ϑ cos ϕ cos ϑ sin ϕ − sin ϑ ey eϕ − sin ϕ cos ϕ 0 ez
Fig. C.2. Spherical polar coordinates (“spherical coordinates”)
The position vector is (Fig. C.2)
r = r er. The line element is
ds =dr er + r dϑ eϑ + r sin ϑ dϕ eϕ. The surface element is 2 da = r sin ϑ dϑ dϕ er + r dr sin ϑ dϕ eϑ + r dr dϑ eϕ. The volume element is d3r = r2 dr sin ϑ dϑ dϕ = −r2 dr d(cos ϑ)dϕ. C.5 Plane Polar Coordinates 335 C.4 Cylindrical Coordinates
The coordinate transformation is ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ e cos ϕ sin ϕ 0 ex ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ eϕ = − sin ϕ cos ϕ 0 ey ez 001ez
Fig. C.3. Cylindrical coordinates
The position vector is (Fig. C.3)
r = e + z ez.
The line element is
ds =d e + dϕeϕ +dz ez.
The surface element is
da = dϕ dz e +d dz eϕ + d dϕ ez.
The volume element is d3r = d dϕ dz.
C.5 Plane Polar Coordinates
The plane polar coordinates one obtains from the cylindrical coordinates with z = 0 or from the spherical polar coordinates with ϑ = π/2, # $ # $# $ e cos ϕ sin ϕ e r = x . eϕ − sin ϕ cos ϕ ey 336 C Rectangular Coordinate Systems C.6 Inverse Relations
The (rarely needed) inverse of the relations between the unit vectors in short- hand notation is given by
e = T · e ⇔ e = TT · e with the transpose TT = T−1 of the matrix T.
Problems
C.1. Vectors and Coordinate Systems. (a) Write down examples for physical quantities which can be described as vectors. (b) Write down the decomposition of a vector in Cartesian, plane, and spher- ical polar coordinates. What is the difference between the Cartesian and the polar coordinates? What is the relation between spherical and plane polar coordinates? C.2. The Line Element in Spherical Coordinates. For a motion in R3 one can employ Cartesian coordinates r(t)={x(t),y(t),z(t)} or spherical coordinates {r(t),θ(t),ϕ(t)}. Determine the infinitesimal line element (ds)2 =(dx)2 +(dy)2 +(dz)2 in spherical coordinates, and write down the squared modulus of the velocity in these coordinates. C.3. Vector Functions, Trajectory. (a) What is a vector function? Write down examples. (b) The trajectory of a point mass is described by the time-dependent position vector r(t). Write down the decomposition of r(t) in Cartesian and polar coordinates. How are these decompositions different from each other with respect to the time dependence? (c) The set of all points through which r(t) runs for all times t one denotes as space curve or trajectory of the point mass. For the description of the space curve one uses a coordinate system (local trihedron), which is defined by following set of orthogonal unit vectors: dr(s) tangent unit vector: =: e (s) ds t where s is the arc length (along the space curve), & & & & det(s) &det(s)& principal normal: & & =: en(s) ds ds Problems 337
and binormal: eb := et × en & & & & dr(t) &dr(t)& Prove that et(s) can also be written as & &. dt dt
From the definition of en(s) follows: & & & & det &det & = κ en, where κ = & & , ds ds is denoted as the curvature. (d) Determine the unit vectors of the local trihedron for a circular orbit. Em- ploy the parameter representation r(ϕ)=(r cos ϕ, r sin ϕ, 0). Which meaning has κ in this case, and which relation exists to the plane polar coordinates? (e) Prove that de b ∼ e . ds n de One writes b = −τ e and denotes τ as torsion of the space curve. ds n
(f) Determine et, en and eb for the spiral line (helix) h r(ϕ)=(r cos ϕ, r sin ϕ, ϕ), 2π and write down κ and τ. Hint: First express the variable ϕ by the arc length. What does one get for the case h =0? Write down in particular κ and τ for this case. (g) Compare the radius of curvature 1/κ of the helix with that of the circular orbit. D Nabla (Del) Operator and Laplace Operator
Let f be a scalar field and g a vector field. Equivalent notations: ∇f ≡ grad f gradient of f ∇·g ≡ div g divergence of g ∇×g ≡ rot g rotation of g
D.1 Representations of the Nabla and Laplace Operators
Coordinate-Free Representation
1 ∇◦ = lim da ◦ Δ →0 V ΔV ∂ΔV 1 (n ×∇) ◦ = lim ds ◦, Δ →0 F ΔF ∂ΔF where n is the unit vector normal to the surface element ΔF ; ∂ΔV and ∂ΔF are the surface of the volume ΔV and the boundary of the sheet ΔF , respectively, with the abbreviation
◦ = ·v(r)or◦ = ×v(r)or◦ = f(r).
Representation in Cartesian Coordinates
∂ ∂ ∂ ∇ = e + e + e (D.1) x ∂x y ∂y z ∂z ∂2 ∂2 ∂2 ∇·∇ = + + . (D.2) ∂x2 ∂y2 ∂z2 340 D Nabla (Del) Operator and Laplace Operator
Representation in Spherical Polar Coordinates
∂ 1 ∂ 1 ∂ ∇f = er f + eϑ f + eϕ f (D.3) ∂r r ∂ϑ r sin ϑ ∂ϕ ∇· 1 ∂ 2 ∂ ∂ g = 2 sin ϑ (r gr)+r (sin ϑgϑ)+r gϕ (D.4) r sin ϑ & ∂r ∂ϑ& ∂ϕ & & & er r eϑ r sin ϕ eϕ & 1 & & ∇× g = & ∂ ∂ ∂ & (D.5) r2 sin ϑ & ∂r ∂ϑ ∂ϕ & gr rgϑ r sin ϑgϕ 1 ∂ ∂ 1 ∂2 2 ∂ 1 ∇·∇ = r2 + Λ(ϑ, ϕ)= + + Λ(ϑ, ϕ)(D.6) r2 ∂r ∂r r2 ∂r2 r2 ∂r r2 1 ∂ ∂ 1 ∂2 Λ(ϑ, ϕ)= sin ϑ + . (D.7) sin ϑ ∂ϑ ∂ϑ sin2 ϑ ∂ϕ2
Representation in Cylindrical Coordinates
∂ 1 ∂ ∂ ∇f = e f + e f + e f (D.8) ∂ ϕ ∂ϕ z ∂z 1 ∂ 1 ∂ ∂ ∇· g = (g)+ gϕ + gz (D.9) ∂& ∂ϕ& ∂z & & & e eϕ ez & 1 & & ∇× g = & ∂ ∂ ∂ & (D.10) & ∂ ∂ϕ ∂z & g gϕ gz 1 ∂ ∂ 1 ∂2 ∂2 ∇·∇ = + + . (D.11) ∂ ∂ 2 ∂ϕ2 ∂z2
D.2 Standard Relations
r (gradient) ∇r = e = (D.12) r r df (gradient) ∇f(r)= e (D.13) dr r (a =const) ∇(a · r)=a (D.14) (divergence) ∇· r = 3 (D.15) (a =const) ∇·(a × r) = 0 (D.16) (rotation) ∇× r = 0 (D.17) (a =const) ∇×(a × r)=2a (D.18) D.3 Rules 341
(vector gradient) ∇⊗r = 1 (∇α rβ = δαβ) (D.19) ∂ (formal notation) ∇ = (D.20) ∂r ∞ 1 (Taylor series) f(r + a)=ea·∇f(r):= (a ·∇)nf(r) (D.21) n! n=0 1 Δ = −4πδ(r − r0) (D.22) |r − r0|
Comment: According to the last equation
− 1 1 4π |r − r0|
is the Green function to the Laplace operator.
D.3 Rules
One uses the three-dimensional generalization of the rules as known from one dimension. The formal notation
∇≡ ∂ ∂r is of use (which is not defined via the limit of a difference quotient). One must pay attention to the order of the quantities, in particular in the case of the cross product.
Chain and Product Rules
df ∇[f(g(r))] = ∇g(r) (D.23) dg ∇ (fg)=g ∇f + f ∇g (D.24) ∇·(fu)=f ∇·u + u ·∇f (D.25) ∇ (u · v)=u ·∇v + v ·∇u +u × (∇×v)+v × (∇×u) (D.26) ∇·(v × w)=w ·∇×v − v ·∇×w (D.27) ∇×sv = s ∇×v − v ×∇s (D.28) ∇×(v × w)=v ∇·w + w ·∇v − w ∇·v − v ·∇w (D.29) 342 D Nabla (Del) Operator and Laplace Operator
Fig. D.1. Theorem of Gauß: The field E with its source ∇·E (left) outside and (right) inside of the volume V enclosed by the area ∂V ;thepointswhereE cuts the surface ∂V are marked by points; the source of the field (right)ismarkedbya small circle
Multiple Differentiation
∇·∇f =Δf (D.30) ∇×∇s = 0 (D.31) ∇·∇×v = 0 (D.32) ∇×(∇×v)=∇∇·v −∇·∇v (D.33) Δ(fg)=fΔg + gΔf +2(∇f) · (∇g). (D.34)
D.4 Integral Theorems
Analogously to the integration of differentials of functions on R1 one can integrate multi-dimensional integrals once: Abbreviation:
◦ = ·v(r)or◦ = ×v(r)or◦ = f(r).
D.4.1 The Theorem of Gauß
Let f(r), ∇f(r)andv(r), ∇·v(r), respectively be continuous in a volume V . Then one has the theorem of Gauß d3r ∇◦= da ◦ . (D.35) V ∂V
Comment: The most frequent examples are of the form d3r ∇·E = da · E, V ∂V with the electric field E(r), see Figs. D.1 and D.2. D.4 Integral Theorems 343
Fig. D.2. Theorem of Gauß: The two-dimensional analogue of Fig. D.1
Fig. D.3. Theorem of Stokes: The field B along a surface with its source ∇×B (left) outside and (right) inside of the (in this example: plane) area F enclosed by the curve ∂F; the point where ∇×B cuts the area F is marked by the open point
D.4.2 Stokes’ Theorem
Let f(r), ∇f(r)andv(r), ∇·v(r), respectively be continuous on a sheet F . Then one has Stokes’ theorem da ×∇◦= ds ◦ (D.36) F ∂F
Comment: The most frequent examples are of the form da ∇×B = da × B, F ∂F with the magnetic induction field B(r), see Fig. D.3.
D.4.3 The Theorem of Green
As a special application of the theorem of Gauß with
◦ = f(r)∇g(r) − g(r)∇f(r) 344 D Nabla (Del) Operator and Laplace Operator
one obtains the theorem of Green d3r [f(r)∇·∇g(r) − g(r)∇·∇f(r)] V (D.37) = da · [f(r)∇g(r) − g(r)∇f(r)] . ∂V
Problems
D.1. Nabla Operator. Gradient, divergence and curl: Determine
∇·r, ∇r, ∇(a · r), ∇f(r), ∇×r, ∇·(a × r). Vector gradient (a) With the help of
α × (β × γ)=(α · γ)β − (α · β)γ
derive a relation for ∇×(a × b). (b) With the help of this procedure understand the operation of a vector gradient. Now determine ∇r.
D.2. Vector Fields.
(a) Sketch the vector fields A(r)=r and A(r)=B × r, by placing arrows with appropriate direction and magnitude of the field on the lines of con- stant field strength |A(r)| =const.B is perpendicular to the drawing plane. (b) A rotational field is described by
v(r)=ω × r.
Determine the vorticity ∇ × v(r). A point charge causes a field of the form r G(r)=γ . r3 Determine the source strength ∇ · G(r). (c) Determine the effect of divergence and rotation on a gradient field A = ∇Φ(r), where Φ(r) is assumed to be twice continuously differentiable but arbitrary otherwise. Problems 345
D.3. Line Integrals.
A. Nonconservative forces: Given be a force field F of the form
F =(3x2 +2y,−9yz, 8xz2). · Determine the work c A dr done an the system along the path (a) C1: straight line from (0,0,0) to (1,1,1) and (b) C2: parabolic arc from (0,0,0) to (1,1,1)
B. Conservative forces: Given be a force F = r/r3. (c) What is special of this force F in comparison to the force field from problem (c) of problem D.2? What conclusion is to be drawn for the work done on the system? · (d) Determine c F dr along the paths (d1) C1:alongthex-axis from (∞;0) to (1;0) (d2) C2: along the bisecting line from (∞; ∞) to (1;1) (d3) C3: along of the arc of the unit circle in the first quadrant from (1;0) to (0;1) E Variational Method
E.1 Functions and Functionals
Definition: A function y = f(x)mapsavalue x to a value y,
x → y.
A functional Y = F [x(t)] maps a function x(t)toavalueY ,
x(t) → Y.
Comments: • In the context of the Hamilton principle of Sect. 3.10, a trajectory q(t)is mapped to the action W . • Here and in the following the functional is a derivative and an integral. Example: (function) y = f(x)= 1+x2 t2 (functional) Y = F [x]= 1+x ˙2 dt. t1 The function is extremal for a value x with df df dy = dx =0 ⇒ =0. dx dx The functional is extremal for a trajectory x(t)with
δY =0. 348 E Variational Method
Here,
f(x + ε dx) − f(x) df dy = lim = dx ε→0 ε dx F [x(t)+εδx(t)] − F [x(t)] δF δY = lim =: δx ε→0 ε δx often: δF = F [x + δx] − F [x]
Here δx is called the variation of the trajectory x(t)andδF is called the variation (or also the differential) of the functional F .
E.2 Variational Problem and Euler Equation
The Hamilton principle treated in Sect. 3.10 is a special case of an integral extremum principle. The extremum principle can be treated in general and shall be touched upon here rudimentary. Let x(t) be a function; the derivatives shall be abbreviated byx ˙ =dx/dt, etc. Furthermore, let g(x(t), x˙(t),t) be a function of multiple variables (u = x, v =˙x and t). Finally, let t2 Y = F [x]= g(x, x,˙ t)dt t1
be a functional. (In this case the functional is a time derivative and an inte- gral.) Under the constraint of fixed boundary points x(t1)andx(t2)or
δx(t1)=0=δx(t2) (E.1)
Y is extremal (“stationary”) only if the Euler equation holds, # $ ∂g d ∂g δY =0 ⇔ − =0. (E.2) ∂x dt ∂x˙
This relation is basic to many optimizations in science and technology. Proof: With the varied trajectory x(t)+δx(t) one has t2 δY = [g(x + δx, x˙ + δx,˙ t) − g(x, x,˙ t)] dt t1 # $ t2 ∂g ∂g = δx + δx˙ dt + O(δ∈). t1 ∂x ∂x˙ E.2 Variational Problem and Euler Equation 349
Terms of higher order are neglected in the following, since they vanish in the limit of small variations (δx → 0). The second term can be integrated in parts, & # $ t2 &t2 t2 ∂g ∂g & d ∂g δx˙ dt = δx& − δx dt, t1 ∂x˙ ∂x˙ t1 t1 dt ∂x˙ where the integrated (first) term vanishes because of the boundary condition δx(ti) = 0. Then one is left with # $ t2 ∂g d ∂g δY = − δxdt. t1 ∂x dt ∂x˙
Since this has to hold for arbitrary times t1 (and t2), the integrand (the product of the contents of the square brackets and δx) must vanish, and since this has to hold for arbitrary variations δx, the contents of the square brackets itself must vanish. Comments: • For
F [x]= g({xi}, {x˙ i},t)dt
one has # $ t2 ∂g ∂g δF[x]= δxi + δx˙ i dt ∂xi ∂x˙ i t1 i & # $ &t2 t2 ∂g & ∂g d ∂g = δxi& + − δxi dt. ∂x˙ ∂x dt ∂x˙ i i t1 i t1 i i
With δxi(t1)=0=δxi(t2) and the independence upon the δxi(t)one obtains in generalization of (E.2) the Euler equations # $ ∂g d ∂g δY =0 ⇔ − =0 ∀i. (E.3) ∂xi dt ∂x˙ i
• For F [x]= g(x, x,˙ x,...¨ )dt
one has # $ t2 ∂g ∂g ∂g δF[x]= δx + δx˙ + δx¨ + ... dt ∂x ∂x˙ ∂x¨ t1 # $ & & &t2 &t2 ∂g d ∂g & ∂g & = − + ... δx& + + ... δx˙& + ... ∂x˙ dt ∂x¨ ∂x¨ # $ t1 # $ t1 t2 2 ∂g − d ∂g d ∂g + + 2 + ... δxdt. t1 ∂x dt ∂x˙ dt ∂x¨ 350 E Variational Method
One mustrequire correspondingly more boundary conditions (δx˙(ti)=0, etc.). Then the integrated terms vanish, and the Euler equation is the van- ishing of the square brackets under the integral. More details in Goldstein Sect. 2-2.
Problems
E.1. Dido’s Problem. Given the integral x2 I = F [y(x),y(x),x]dx. x1 The problem of the variational method is to find such a curve y(x)forwhich the integral I is extremal. This leads to to the Euler–Lagrangian equation d ∂F ∂F − =0 dx ∂y ∂y as a necessary condition for the extremum of the integral I. (a) Often the Integrand F [y,y] does not explicitly depend upon x.Provethat one integration of the Euler–Lagrangian equation in this case leads to ∂F F − y =const. ∂y
(b) Using the result of part (a) solve Dido’s problem which reads: Among all curves y(x)oflengthl, determine that curve y(x) connecting two given points P1 and P2 which, together with the straight line connect- ing these points, is a boundary of the area of largest size. F Linear Differential Equations with Constant Coefficients
The general solution xi(t) of a system of N coupled linear inhomogeneous equations (of nth order)