Circular orbits on a warped spandex fabric

Prof. Chad A. Middleton CMU Physics Seminar

October 24, 2013

To appear in the American Journal of Physics

•  describes the curvature of

spacetime •  describes the matter &

energy in spacetime

Einstein’s theory of general relativity

Sean M. Carrol, Spacetime and Geometry: An Introduction to Einstein’s General Relativity (Addison Wesley, 2004)

Matter tells space

how to curve, space tells matter

how to move.

Gµ� =8�G

c4Tµ�

Gµ⌫

Tµ⌫

Consider a spherically-symmetric, non-rotating massive object… Embedding diagram (t = t0 , θ = π/2).. •  2D equatorial ‘slice’ of the 3D space at one

moment in time

Einstein’s theory of general relativity

Sean M. Carrol, Spacetime and Geometry: An Introduction to Einstein’s General Relativity (Addison Wesley, 2004)

z(r) = 2p

2M (r � 2M)

Is there a warped 2D surface that will yield the orbits of planetary motion?

•  Newton’s 2nd Law..

•  using the relation..

•  yields Kepler’s 3rd Law..

Notice: •  Kepler’s 3rd Law is independent of m!

Kepler’s 3rd Law for planetary motion

M

m GmM

r2=

mv2

r~F

~v

~r

v = 2⇡r/T

T 2 =

✓4⇡2

G

◆· r

3

M

Outline •  A marble rolling on a cylindrically-symmetric surface

(Lagrangian dynamics)

•  The shape of the spandex fabric (Calculus of Variations)

•  Small curvature regime o  Kepler-like expression o  Experimentation

•  Large curvature regime o  Kepler-like expression o  Direct measurement of the modulus of elasticity o  Experimentation

•  Circular orbits in GR

A marble rolling on a cylindrically-symmetric surface •  is described by a Lagrangian of the form.. •  Now, for the marble..

Notice: •  The marble is constrained to reside on the fabric..

z = z(r)

I =

2

5

mR2and !2

= v2/R2so

1

2

I!2=

1

5

mv2

L =1

2m(r2 + r2�2 + z2) +

1

2I!2 �mgz

for the radial-coordinate.. •  yields the equation of motion for the marble.. *

•  compare to the equation of motion for planetary orbits..

L.Q. English and A. Mareno, “Trajectories of rolling marbles on various funnels”, Am. J. Phys. 80 (11), 996-1000 (2012)

* will NOT yield Newtonian-like orbits of planetary motion for a marble on ANY cylindrically-symmetric surface!

The Lagrange equation of motion

@L

@r� d

dt

@L

@r= 0

(1 + z02)r + z0z00r2 � r�2 +5

7gz0 = 0

r � r�2 +GM

r2= 0

For the equation of motion for the marble..

•  setting for circular orbits, we obtain..

Notice: •  we used the relation •  depends linearly on the slope of the spandex fabric.

r = r = 0

Circular orbits on a cylindrically-symmetric surface

(1 + z02)r + z0z00r2 � r�2 +5

7gz0 = 0

4⇡2r

T 2=

5

7g · z0(r)

v = r� = 2⇡r/T

Technique: 1.  Construct potential energy (PE) integral functional of

spandex fabric. i.  Elastic PE of the spandex. ii.  Gravitational PE of the spandex. iii.  Gravitational PE of the central mass.

2.  Apply Calculus of Variations.

⇒  The elastic fabric-mass system will assume the shape which minimizes the total PE of the system.

i. and iii. first considered in “Comment on “The shape of ‘the Spandex’ and orbits upon its surface”, Am. J. Phys. 70 (10), 1056-1058 (2002)

The shape of the spandex fabric

Elastic PE of a differential concentric ring of the fabric of unstretched width dr… •  Define the modulus of elasticity, E .. •  Integrating the differential segment over the whole fabric,

the total elastic PE of the fabric is…

Elastic PE of the spandex fabric

dUe =1

2(p

dr2 + dz2 � dr)2

E =dr

2⇡r

dr

dr

dzp dr

2 + dz2

Ue =

Z R

0⇡E · r(

p1 + z02 � 1)2 dr

Gravitational PE of a differential concentric ring of the fabric..

•  The mass of the differential ring is a constant under stretching..

where , are the unstretched, variable areal mass densities.

•  Integrating the differential segment over the whole fabric, the total gravitational PE of the fabric is..

Gravitational PE of the spandex fabric

dUg,s = dmsg · z

Ug,s =

Z R

02⇡�0g · rz dr

dms = �0 · 2⇡rdr = �(z0) · 2⇡rpdr2 + dz2

�0 �(z0)

Notice: •  we approximate the central mass as being point-like.

Gravitational PE of the central mass

Ug,M = Mg · z(0) = �Z R

0Mg · z0(r) dr

where we defined the functional.. To minimize the total PE, subject to the Euler-Lagrange eqn..

The total PE of the spandex-central mass system

f(z, z0; r) ⌘ ⇡E · r(p

1 + z02 � 1)2 + 2⇡�0g · rz �Mg · z0

@f

@z� d

dr

@f

@z0= 0

U = Ue + Ug,s + Ug,M =

Z R

0f(z, z0; r) dr

The Euler-Lagrange equation takes the form..

•  which can be integrated..

•  where we defined the parameter..

The shape equation for the elastic fabric

d

dr

rz0

1� 1p

1 + z02

�� Mg

2⇡E

�=

�0g

E· r

rz01� 1p

1 + z02

�= ↵(M + ⇡�0r

2)

↵ ⌘ g

2⇡E

The circular equation of motion & the shape equation

rz01� 1p

1 + z02

�= ↵(M + ⇡�0r

2)

•  Circular equation of motion

•  The shape equation

4⇡2r

T 2=

5

7g · z0(r)

•  Small curvature regime.. so •  Large curvature regime..

so

z0(r) ⌧ 11p

1 + z

02= 1� 1

2z

02 + o(z04)

z0(r) � 1 1p1 + z

02=

1

z

01p

1 + 1/z02=

1

z

0 (1�1

2z02+ o(1/z04))

T 3 / r/p�0

When … •  expanding the shape equation and inserting into the circular eqn of

motion

Notice: •  * when •  when •  Two competing terms on equal footing when

*Gary D. White and Michael Walker, “The shape of ‘the Spandex’ and orbits upon its surface”, Am. J. Phys. 70 (1), 48-52 (2002).

The small curvature regime z0(r) ⌧ 1

T 3 =

✓28⇡2

5g

◆3/21p2↵

· r2

(M + ⇡�0r2)1/2

T 3 / r2/pM M � ⇡�0r

2

M ⌧ ⇡�0r2

M ' ⇡�0r2 ⇠ 0.10 kg

•  4 ft. diameter trampoline frame o  styrofoam insert for zero pre-stretch o  truck tie down around perimeter

•  Camera mounted directly above, ramp mounted on frame

•  Most circular video clip (0f ~12) imported into Tracker

•  Position determined every 1/30 s and average radius, rave , calculated per revolution

•  Shift by 1/8 revolution for subsequent data point

The experiment in the small curvature regime

Plot of T3 vs rave2/(M+πσ0rave

2)1/2… Notice: •  orbit for zero central mass! •  slope = 45.6 kg1/2 s3/m2 w/ R2 = 0.994 The slope yields a value for α and E..

The experiment in the small curvature regime

0.1 0.2 0.3 0.4 0.5 0.6rave 2 !"M!"#0rave 2 #1!2 "m2 !kg1!2 #

5

10

15

20

25

T3 "s3 #0.601 kg0.535 kg0.199 kg0.067 kg0 kg

Central Mass

↵ =

1

2

✓28⇡2

5g

◆3

· 1

slope

2 ' 0.043 m/kg

E =g

2⇡↵' 36 N/m

When … •  expanding the shape equation and inserting into the circular eqn of

motion

Notice: •  * when •  When , , so above equation invalid! •  Using , we get poor results!

* corresponds to the solution of the 2D Laplace equation with cylindrical-symmetry

The large curvature regime z0(r) � 1

T =

✓28⇡2

5g

◆1/2

· r

(M↵+ r)1/2

T / r/pM r ⌧ M↵

r � M↵ z0(r) ' 1

↵ = 0.043 m/kg

The shape equation in the large curvature regime is.. •  integrating yields..

Notice: •  Plot of z(M)/ln(RB) vs (M - M0) yields

the slope, which is the value of α!

Direct measurement of the modulus of elasticity, E

Top 10 diamonds: M = 0.274kg - 1.174kg in 0.1 kg intervals Bottom 14 diamonds: M = 1.274kg - 7.774kg in 0.5 kg intervals

z0(r) ' M↵

r+ 1

z(M)

ln(RB)= (M �M0)↵

Plot of z(M)/ln(RB) vs (M - M0) yields the slope, which is the value of α! Notice: •  M = 0.274 – 0.674 kg regime •  M = 5.274 – 7.774 kg regime

Direct measurement of the modulus of elasticity, E

0 1 2 3 4 5 6 7M!M0 !kg"

0

0.02

0.04

0.06

0.08

z#ln!RB"!m"

5.274!7.774 kg0.774!4.774 kg0.274!0.674 kg

Central Mass

↵ ' 0.030 m/kg

↵ ' 0.006 m/kg

Plot of T vs rave/(Mα+rave)1/2.. •  •  Notice: •  ~10% error for α = 0.006 kg/m, which compares

with ~94% error when α = 0.043 kg/m.

The experiment in the large curvature regime

0.05 0.10 0.15 0.20rave !"M!"rave #1!2 "m1!2 #

0.1

0.2

0.3

0.4

0.5

T"s#7.774 kg7.274 kg6.274 kg5.274 kg

Central Mass

slopeth =

✓28⇡2

5g

◆1/2

= 2.37 s/m

1/2

slope

exp

= 2.62 s/m

1/2

The metric exterior to a spherically-symmetric object of mass M, in the presence of a cosmological constant (or vacuum energy), Λ..

where Notice.. •  , Schwarzschild – de Sitter spacetime •  , Schwarzschild – Anti-de Sitter spacetime

•  , Schwarzschild solution

Circular orbits in GR

⇤ > 0

⇤ < 0

⇤ = 0

⇢vac =⇤

8⇡G

ds2 = �✓1� 2GM

r� ⇤

3r2◆dt2 +

✓1� 2GM

r� ⇤

3r2◆�1

dr2 + r2(d✓2 + sin2 ✓d�2)

By normalizing the four-velocity and employing conservation of energy and angular momentum.. •  The radial equation of motion..

where For circular orbits, set.. • 

• 

Circular orbits in GR

E =1

2

✓dr

d⌧

◆2

+ Veff (r) Veff (r) = �GM

r+

`2

2r2� GM`2

r3� ⇤

6(`2 + r2)

d

drVeff (r) = 0

E = Veff (r)

One arrives at an exact Kepler-like expression of the form.. *

•  Kepler’s 3rd Law when Λ -> 0.

Compare to the Kepler-like relation for a marble on the warped spandex fabric in the small curvature regime.. •  Areal mass density, σ0, plays the role of a negative cosmological constant, Λ.

*N. Cruz, M. Olivares, and J. Villanueva, “The geodesic structure of the Schwarzschild Anti-de Sitter black hole”, Classical and Quantum Gravity 22, 1167 (2005)

Circular orbits in GR

T 2 / r3

(M � ⇤3Gr3)

T 3 / r2

(M + ⇡�0r2)1/2

•  The mass of the spandex fabric interior to the orbit of a marble matters. •  The modulus of elasticity, E, describing the spandex fabric is not constant

and is a function of the stretch. •  Areal mass density, σ0, plays the role of a negative cosmological constant, Λ.

Conclusion