when viewed from the top, the objects should move as if interacting gravitationally

Transcription

1 2 Elastic Space

2 2 Elastic Space The dynamics and apparent interactions of massive balls rolling on a stretched horizontal membrane are often used to illustrate gravitation. Investigate the system further. Is it possible to define and measure the apparent 2

3 when viewed from the top, the objects should move as if interacting gravitationally 3

4 The Correct Shape rubber sheet 1 2 mv2 mgh(r) h(r): membrane height at radius r both systems: E k E p 1 h r = κ m M 1 r gravity 2 mv2 κ mm r = φ = E p m grav. potential F g = κ Mm r 2 4

5 Is the Shape Correct?? C 1 r 5

6 Is the Shape Correct?? C 1 is it only an experimental problem r (i.e. imperfect membrane), or is it something fundamental? 6

7 PHYSICS OF A RUBBER SHEET What Does Rubber Do?

8 Description of Rubber Sheet Curvature 13 based on Feynman, R. P.: The Feynman Lectures On Physics, Vol. 2, Ch. 12: 2 h = g k ρ valid for small deformations only; Laplace operator derivation in Appendix h: membrane height at radius r ρ: distribution of mass k: membrane stiffness g: gravitational acceleration 8

9 Rubber Sheet vs. Gravity rubber sheet shape gravitational potential 2 h = g k ρ 9

10 Rubber Sheet vs. Gravity rubber sheet shape gravitational potential 2 h = g k ρ 2 φ = 4πκρ Laplace operator (derivations in Appendix) h: membrane height ρ: distribution of mass k: membrane stiffness g: gravitational acceleration ρ: distribution of mass φ: gravitational potential κ: gravitational constant 10

11 Rubber Sheet vs. Gravity rubber sheet shape gravitational potential? C 1 r 11

12 rubber sheet shape 3D 2 h = g k ρ gravitational potential 3D 2 φ = 4πκρ the rubber sheet model is a 2D problem! 12

13 GRAVITY IN 2 DIMENSIONS Rubber Sheet and Gravity Equivalence 13

14 Gravitation why F g 1 r 2 φ 1 r? depends on the dimensionality of the world informal reasoning: F g intensity intensity in 3D 1 r 2 intensity in 2D 1 r F g 1 r (formal derivation based on Divergence Theorem in Appendix) 14

15 2D F g 1 r 15

16 2D : F g 1 r φ log r R 0 16

17 2D : F g 1 r φ log r R 0 h = C ln r R 0 C = 16 mm R 0 = 3 mm 17

18 2D : F g 1 r φ log r R 0 h = C ln r R 0 2D rubber sheet assumes C = 16 the mm correct shape to model 2D gravity R 0 governed = 3 mm by the same equations! 18

19 FINDING THE GRAVITATIONAL CONSTANT 19

20 Where to get κ m? from the membrane shape (potential): φ r = 2κ m M ln r R 0 (by solving 2 h = g ρ using k h(r) = C ln r R 0 κ m = C 2M 20

21 Experiment: changing mass, keeping diameter 13 small steel ball strong magnet additional weights 21

22 κ m [m/kg] 13 Results u(r) = C ln r R 0 κ m = C 2M 0,06 0,04 0, ,1 0,2 0,3 0,4 0,5 M [kg] κ m = 5.3 ± m kg 1 22

23 κ m [m/kg] 13 Results u(r) = C ln r R 0 κ m = C 2M 0,06 0,04 0, ,1 0,2 0,3 0,4 0,5 M [kg] κ m = 5.3 ± 0.2 too stretched 10 2 m kg 1 23

24 DYNAMICS 24

25 Dynamics the shape is correct approx. works but: energy losses (friction / rolling resistance) conservation of mechanical energy + elasticity 25

26 m = 1 g, d = 12 mm m = 2.2 g, d = 16 mm 13 y [m] t [s] y [m] t [s] x [m] x [m] Σ E p E k Σ E p E k M = 1.8 kg 26

27 Ellipses? Bertrand's theorem: stable, closed orbits can only exist if φ 1 r or φ r2 no closed orbits here pericenter trajectory M = 1.8 kg m = 9.7 g r init = 13.4 mm Johnson, Porter Wear (2010). Classical Mechanics With Applications 27

28 Laws 1 st Law 2 nd Law 3 rd Law no elliptical orbits 28

29 Laws 1 st Law 2 nd Law 3 rd Law no elliptical orbits theoretically: (conservation of momentum) experiment: (energy losses friction) 29

30 Laws 1 st Law 2 nd Law 3 rd Law (for circular orbits) a 3 T 2 = const. for orbiting same mass no elliptical orbits theoretically: (conservation of momentum) experiment: (energy losses friction) force equilibrium (F g = mω 2 a) a 2 T 2 a T = const. 30

31 Conclusion Thank you for your attention! h = C ln r/r 0 dynamics: same motion shape = potential description of sheet curvature same equations as gravity, but 2D F = κ m Mm r ; Σ E p E k κ m = 0.053kg 1 m 2 s 2 Laws: th., exp a 3 T 2 = C a T = C

32 Thank you for your attention! h = C ln r/r 0 dynamics: same motion shape = potential description of sheet curvature same equations as gravity, but 2D F = κ m Mm r ; κ m = m2 kg s 2 Σ E p E k Laws: th., exp a 3 T 2 = C a T = C

33 APPENDIX gravity, rubber sheet: equations gravity in N-D derivation of 3 rd Law small slopes approximation 33

34 φ: Poisson intensity g = F : gradient of potential m g(r) = φ(r) Gauss s Theorem: g(r) = 4πκρ(r) together: Poisson equation φ r = 4πκρ r Δφ(r) = 4πκρ(r) 34

35 φ: Boundary Conditions, ρ(r) Δφ(r) = 4πκρ(r) φ = const., therefore n φ S = 0 ρ: point mass in our measurements: ρ r = m δ r, where δ r is -function ( δ x dx = 1, x R 0 : δ x = 0) 35

36 h: Poisson net force causing the rubber sheet to bend: F = k y sin θ 2 k y sin θ 1 ΔF = k y(sin θ 2 sin θ 1 ) h for θ 1 : sin θ tan θ h x, then F = k y h 2 x h 1 x = k y 2 h x 2 x k: 36

37 h: Poisson deformation is caused by gravity: F = g ρ x y substitute ΔF = k y 2 h x: x 2 2 h x 2 = F k x y = ρg k generalization (vector fields): h(r) = g k σ(r) 37

38 h: Boundary Conditions, ρ(r) Δφ(r) = g k ρ(r) φ = const., therefore n h = 0 ρ: area density negligible in comparison to mass of ball m, therefore approximated by - function: ρ r = m δ r 38

39 Rubber Sheet vs. Gravity two differential equations of the same type, in the same region von Neumann boundary conditions: nf r = 0 in a closed region such differential equations have at most one solution (Dirichlet unique solution theorem) character of solutions for φ and h will be the same φ h 39

40 The Form of Newton s Law of Gravitation 13 why F g 1 r 2? Gauss : V g da = 4πκM (gravitational flux through any closed surface is proportional to the enclosed mass) special case: spherical symmetry, point mass: g A s = 4πκM ; A s = 4πr 2 g = 4πκM A s = κm r 2 40

41 The Form of Newton s Law of Gravitation 13 why F g 1 r 2? Gauss : V g da = 4πκM (gravitational flux through any closed surface is proportional to the enclosed mass) special case: spherical symmetry, point mass: holds in N dimensions, too g = 4πκM A s ; A s r N 1 g κm r (N 1) 41

42 Gravitational Force in 2D g = 2κM r potential φ r = r g dr = g log r + C 42

43 Derivation of the 3 rd Law standard - 3D κ m Mm r 2 = Mm mω2 r κ m r substitute ω = 2π : T r 3 T 2 = κ mm 4π 2 r 3 T 2 = 1 4π 2 κm F g = F cf 2D rubber sheet = mω2 r r 2 T 2 = κ mm 4π 2 r T = 1 2π κ mm 43

44 Large slopes do not work well if the slope is too big, the projected force will not be monotonous + part of E k we will work only with small slopes 44

45 we will work only with small slopes 45

46 we will work only with small slopes we can assume: uniform tension sin θ θ tan φ = Δh Δx cos θ 1 experiment: no inelastic (permanent) deformation of membrane deformation) 46