Allgemeine Mechanik Musterlösung 6.

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1 llgemeine Mechanik Musterlösung 6. HS 2017 Prof. Thomas Gehrmann Übung 1. Brachistochrone Ein Teilchen rutscht reibungslos auf einer Kurve im Gravitationsfeld. Start- und Endpunkte der Kurve sind gegeben und von der Höhendifferenz b getrennt. Die horizontale Distanz zwischen den Punkten ist a. Finden Sie die Kurve welche die kürzeste Laufzeit liefert. Zeichnen Sie die Kurve für die drei Fälle a) a/b < π/2; b) a/b = π/2; c)a/b > π/2. Hinweis. In der Euler-Lagrange Gleichung ist der Integrand von x unabhängig. Kontrollieren Sie, dass die Zykloide, parametrisiert als x(τ) = (τ sin τ) and y(τ) = b (1 cos τ) die differentielle Gleichung erfüllt. y(x) b g a x bbildung 1: Beispiel für Brachistochrone Lösung. The time T required to get from point (x 1, y 1 ) = (0, b) to point (x 2, y 2 ) = (a, 0) can be expressed as a functional of the function y(x). n element of the path is = 1 + y 2 dx. From energy conservation mv 2 /2 = mg(b y) we obtain the velocity v = /dt. Then we have 2 J[y] = T = v = 1 a 1 + y dx (x) 2 (L.1) 2g b y This functional is of the following form J[y] = 1 2g x2 1 x 1 dxf (y, y ) with F (y, y ) = Since F x = 0, the Euler - Lagrange equation gives us y 2 b y (L.2) F y (y, y )y F (y, y ) = const (L.3) This is an analogue to Exercise 4 of Sheet 5. In our case it implies For x(τ) = (τ sin τ) and y(τ) = b (1 cos τ) we have (b y)(1 + y 2 ) = const (L.4) dx dτ = (1 cos τ), dy 1 dτ = sin τ (L.5)

2 and thus y = dy dx = dy dτ dτ dx = sin τ 1 cos τ 1 + y 2 = 1 + sin 2 τ (1 cos τ) 2 = 2 1 cos τ = 2 b y (L.6) which in turn gives us 2 = const. The curve is determined by the initial point (0, b), as well as the final point (a, 0) which specifies the integration constant. The 3 cases are plotted in bbildung 1. For a/b = π/2 the curve has zero derivative in the last point. Note: The curve of fastest descent is called brachystochrone. It describes e.g. the (flipped) trajectory of a given point on the wheel of a moving car. Übung 2. Fermat-Prinzip Eine Lichtwelle, die sich von Punkt zum Punkt B ausbreitet, folgt genau dem Weg, welcher der minimalen Zeit entspricht (Fermat-Prinzip). a) Die Punkte und B befinden sich in einem Medium, in dem der Brechungsindex von der Position abhängt. Formulieren Sie das Fermat-Prinzip als ein Pfadintegral, analog zum Lagrangeformalismus. b) Verwenden Sie dieses Prinzip, um die Brechungs- und Reflexionsgesetze herzuleiten. Hinweis. Die Lichtgeschwindigkeit im Medium mit Brechungsindex n ist c/n. Lösung. a) b) Refraction: dt = c( r) dt = c( r) T = B dt = B, δt = 0 (L.7) c( r) T = C B c + C c/n = 1 c [ C + n CB ] = 1 y c [ 2 + x n y 2 + ( x x 1 ) 2 ] (L.8) Since ẋ 1 is cyclic, the Euler - Lagrange equation can be reduced to T x 1 = 0. T = 1 x 1 c [ x 1 x x n 1 y 2 + x 2 1 y 2 + ( x x 1 ) ] = = x 1 CB n( x x 1 ) C CB C (L.9) (L.10) Thus we obtain Reflection: x 1 C = n x x 1 CB sin α sin β = n (L.11) T = C B c + C c = 1 c [ C + CB ] = 1 y c [ 2 + x y 2 + ( x x 1 ) 2 ] (L.12) 2

3 bbildung 2: 3 cases for different combinations of a and b for Brachistochrone question 3

4 bbildung 3: Fermat-Prinzip Similarly as in the previous case, we require x 1 B c = 1 c [ x 1 C x x 1 CB ] = 0 x 1 C = x x 1 CB sin α = sin β (L.13) Übung 3. Geodäte 1 Finden Sie die Kurve, die den kürzesten Pfad zwischen zwei Punkten (θ 1, φ 1 ) und (θ 2, φ 2 ) auf der Oberfläche einer Kugel mit Radius R beschreibt. Drücken Sie das Resultat als x(s), y(s), z(s) aus und passen Sie es an die Randbedingungen an. Hinweis. Beginnen Sie mit den llgemeinen geodäsischen Gleichungen dxi, mit der entsprechenden Zwangsbedingung g(x, y, z) = 0. Lösung. By definition the geodesic lines minimise the path between the two points J[x, y, z] = t2 The minimisation is understood in our case under the constraint t 1 dt ẋ 2 + ẏ 2 + ż 2. (L.14) g(x, y, z) = x 2 + y 2 + z 2 r 2 = 0. (L.15) Thus we write the Euler-Lagrange equations for the functional F (x, y, z, ẋ, ẏ, ż, t) = ẋ 2 + ẏ 2 + ż 2 λ(t)g(x, y, z) (L.16) 4

5 that with the definition v = ẋ 2 + ẏ 2 + ż 2 read d ẋ dt v = λg x d ẏ dt v = λg y d ż dt v = λg z To parametrise the curve with the path length s we write ẋ = dx(s) dt d dt = v d = dx dt = vx s and using the product rule d ẋ dt v = vx ss Thus the Euler-Lagrange equations in the new parametrisation become x ss = 2xλ/v (L.17) (L.18) (L.19) (L.20) (L.21) (L.22) (L.23) (L.24) and similar for y and z. Since the path must satisfy the constraint we can use its equation to simplify and solve the Euler-Lagrange. We differentiate it twice with respect to s. x 2 s + y 2 s + z 2 s + xx ss + yy ss + zz ss = 0 (L.25) Now we substitute the second derivatives from L.24 and observe that x 2 s + y 2 s + z 2 s = (dx 2 + dy 2 + dz 2 )/ 2 = 1 (L.26) which together with the constraint equation lea to 2λ/v = 1/(r 2 ) (L.27) With this the Euler-Lagrange equations become x ss = x r 2 (L.28) y ss = y r 2 (L.29) z ss = z r 2 (L.30) each of which has the solution sin(s/r) + B cos(s/r), where the constants should be chosen to match the positions of the start and end points. This solution describes big circles on the sphere (circles of perimeter 2πr) 5

6 Lösung. lternative solution: The geodesics can be found by parametrizing the curve in terms of a single parameter θ and minimizing the arc length. For the Cartesian coordinates on a sphere we have x = R sin φ cos Θ(φ) (L.31) s : y = R sin φ sin Θ(φ) (L.32) z = R cos φ, for φ 0 φ φ 1 (L.33) for some suitable function Θ(φ). Note that this choice of the variables names assumes Θ is the azimuthal and φ is the polar angle which is not the standard convention! The length L of any parametric curve s is given as L = φ1 φ 0 (dx ) 2 + dφ We may rewrite this as ( ) dy 2 + dφ L(Θ) = ( ) dz 2 φ1 dφ = R 1 + Θ dφ (φ) 2 sin 2 φ dφ (L.34) φ 0 φ1 φ 0 F (φ, Θ (φ)) dφ (L.35) The problem can now be reduced to the problem of finding the function Θ = Θ(φ) which minimizes the length functional L(Θ) subject to the constraints (L.31). Since F is independent of Θ, the Euler - Lagrange equation is reduced to F Θ (φ, Θ (φ)) = C (L.36) In combination with (L.34) this gives us dθ dφ = sin φ sin 2 φ 2 (L.37) for some constant related to C. It is convenient to replace with = sin α so the differential equation becomes dθ dφ = sin α sin φ (L.38) sin 2 φ sin 2 α This equation can be integrated through the substitution tan φ = 1 u (L.39) which lea to By integrating we obtain where β is a constant of integration. Hence, we find that dθ du = tan α 1 u 2 tan 2 α Θ + β = cos 1 (u tan α) Θ(φ) = β + cos 1 ( tan α tan φ ) (L.40) (L.41) (L.42) where α and β can be chosen so that the resulting extremum curve connects the given two points on the sphere. It is convenient in this case to transform the extremum relation back into Cartesian coordinates. We write (L.42) as cos[θ(φ) + β] = tan α (L.43) tan φ 6

7 and then multiply both sides of the equation by R sin φ. In this way we easily find that x cos β y sin β = z tan α (L.44) which must hold for any point on the geodesic curve connecting our two points. It is an equation of a plane passing through the origin in 3-space. Hence, the geodesic curves are arcs of great circles on the sphere. Übung 4. Geodäte 2 Bestimmen Sie die Geodäten auf einem Kegelmantel mit der Gleichung z = λ r (in Zylinderkoordinaten r, ϕ, z). Überprüfen Sie Ihr Ergebnis für den Fall λ 0. Hinweis. Die gesuchten Kurven sollen die Form r = r(ϕ, r 0, ϕ 0 ) haben. In cylinder coordinate, one have the conversion relation for a cone in Cartesian coor- Lösung. dinate as: x = rsin(ϕ) dx = sin(ϕ)dr + rcos(ϕ)dϕ y = rcos(ϕ) dy = cos(ϕ)dr rsin(ϕ)dϕ z = λr dz = λdr (L.45) ny point on the cone can be described by (r, ϕ) The length of any curve on the cone is = dx 2 + dy 2 + dz 2 = ( ) dr 2 dr 2 + r 2 dϕ 2 + λ 2 dr 2 = dϕ r 2 + (1 + λ 2 ) (L.46) dϕ To find the geodesic between two fixed points on the cone, we calculate the length between the fixed points (r 0, ϕ 0 ) and (r 1, ϕ 1 ). L = (r1,ϕ 1 ) (r 0,ϕ 0 ) = (r1,ϕ 1 ) (r 0,ϕ 0 ) dϕ r 2 + (1 + λ 2 )ṙ 2 = (r1,ϕ 1 ) (r 0,ϕ 0 ) f(r, ṙ)dϕ (L.47) Here we assume r is a function of ϕ, thus r = r(ϕ), on the geodesic. We also have ṙ = dr/dϕ. s f(r, ṙ) does not contain ϕ explicitly, the condition for L having extreme vlaue is when f(r, ṙ) ṙ f ṙ = c 0. (L.48) Here c 0 is a constant to be determined by boundaries conditions. Put in the defination of f(r, ṙ), one have: r 2 r 2 + (1 + λ 2 )ṙ = c 2 0 ṙ = r (L.49) r 2 /c λ 2 Define x = r/c 0 and rewerite the above equation as dr r r 2 /c = dϕ 1 + λ 2 d(r/c 0 ) (r/c 0 ) (r/c 0 ) 2 1 = dx x x 2 1 = dϕ 1 + λ 2 (L.50) 7

8 Integrate both side and use arcsec(x) = arcsec( R dx, x x2 1 r ϕ + ϕc )= c0 1 + λ2 one has r= c 0 c cos ϕ+ϕ 1+λ2 (L.51) Here ϕc is a contant to be determined by boundary conditions. To understand above equation into more details, one could alwasy rotate the cylinder coordinate by some angle ϕ ϕ (ϕc +ϕ0 ) such that the fixed point (r0, ϕ0 ) becomes (r0, ϕc ). In the rotated cylinder coordinate, the geodesic now pass the fixed point (r0, ϕc ) thus, r( ϕc ) = r0 = c = c0. 0 c +ϕc cos ϕ 1+λ2 (L.52) To summary, now one has r(ϕ) = r. 0 ϕ ϕ 0 cos 1+λ2 (L.53) Note that above equation is the formular for a general geodesic on a cone. It passes (r0, ϕ0 ) but the value of r0 and ϕ0 could be arbitrary. To apply such formula on two fixed points (r1, ϕ1 ) and (r2, ϕ2 ), one nee to use geodesic equation to solve r0 = r0 (r1, r2, ϕ1, ϕ2 ) and ϕ0 = ϕ0 (r1, r2, ϕ1, ϕ2 ). When λ = 0, the above geodesic function becomes r(ϕ)cos(ϕ ϕ0 ) = r0. (L.54) cone with λ = 0 is actually a cone with infinite base radius. In such case, the geodesic is indepent of z axis and correspon to a straight line. With proper rotation of ϕ axis, the geodesic is a straight line with x = r0. See figure below: 8