Unit 2-3 Assignments for Physics 5103 - Reading in · 2019. 11. 6. · Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG!Name_____ !1 Assignment 8

!1Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

Assignment 8 Oct 23 , 2019 Due Wednesday Oct 30: Based on Unit 2 Chapter 1-3 and Unit 3 Chapter 1-3.

Well-known Coordinates (OCC) NOTE: Save copy of solution to this Ex.1(b) for next Assignment 9. 1. Find Jacobian, Kajobian, Em, Em, metric tensors gmn and gmn for OCC (a) and (b). (You may do (b) then reduce to (a).)

(a) Cylindrical coordinates {q1=ρ, q2=φ, q3=z}: x=x1=ρ cosφ, y=x2=ρ sinφ, z=x3. (b) Spherical coordinates: {q1=r, q2=θ, q3=φ }: x=x1=rsinθ cosφ, y=x2=rsinθ sinφ, z=x3=rcosθ.

"Plopped" Parabolic Coordinates (GCC) (In attached figure) 2. Consider the GCC(Cartesian) definition: q1 = (x)2 + y q2 = (y)2 - x (a) Does an analytic Cartesian coordinate definition x j = x j(qm) exist? If so, show. (b) Derive the Jacobian, Kajobian, unitary vectors Em, Em, and metric tensors for this GCC. (c) On the appropriate graph on attached pages sketch the unitary vectors at the point (x=1, y=1) (Arrow) and at the point (x=1, y=0). Where, if anywhere, is the grid an OCC however briefly? Indicate loci on graph. (d) Find and indicate where, if anywhere, are there Jacobian or Kajobian singularities of this GCC. Show on graph.

"Sliding" Parabolic Coordinates (GCC) (In attached figure) 3. Consider the Cartesian(GCC) definition: x = 0.4 (q1)2 - q2 , y = q1 - 0.4 (q2)2 (a) Does an analytic GCC coordinate definition qm = qm(x j) exist? If so, show. (b) Derive the Jacobian, Kajobian, unitary vectors Em, Em, and metric tensors for this GCC. (c) On the appropriate graph on attached pages sketch the unitary vectors near point (x=1, y=1) (Arrow) and near point (x=1, y=0). Where, if anywhere, is the grid an OCC however briefly? Indicate loci on graph. (d) Find and indicate where, if anywhere, are there Jacobian or Kajobian singularities of this GCC. Show on graph.

4. Covariant vs Contravariant Geometry (In attached figure)GCC components of a vector V in attached figure are realized by line segments OA, BV, etc. Give each segment length by single terms of the form Vm or Vm times (√gmm)+1, (√gmm)-1, (√gmm)+1, or (√gmm)-1 with the correct m=1 or 2. Also label each unitary vector as E1 , E1 , E2, or E2, whichever it is. You should be able to do this quickly without looking at the text figures.

Extra Credit 3D problem: ”Unprofessional" Paraboloidal Coordinates (GCC) (In attached figure) 5. The surface (See xyz-plot) introduces 3D partial derivative chain rules. It is the (q3=0)-surface in a 3D GCC coordinate grid q1=x, q2=y, q3 . It contains a projection of an orthogonal (x,y) Cartesian coordinate grid on the surface that is obviously not orthogonal most places. a. Derive the 3-by-3 Jacobian J(x,y,z) and Kajobian K(x,y,z) for (q3=0). b. Extract covariant ! and contravariant ! vectors represented in Cartesian (x,y,z) basis. c. Derive the 3-by-3 covariant metric gυν(x,y) and contravariant metric gυν(x,y) for (q3=0) and tell which if any points on the surface have grids that are locally orthogonal and which if any are locally orthonormal. (Larger graph provided separately for Ex.5d and Ex.5e. d. Calculate and sketch covariant on (q3=0) surface where (x=4,y=-2) and where (x=3,y=+2). e. Calculate and sketch contravariant on (q3=0) surface where (x=4,y=+2) and where (x=0,y=+4).

z = f (x, y) =21 x2 + y2

=21x2+y2− z

E1 ,E2 ,E3{ } E

1 ,E2 ,E3{ }

E1 ,E2 ,E3{ }

E1 ,E2 ,E3{ }


� Above is 2D plot for Ex. 4Below is 3D (xyz) plot for Extra credit Ex.5.

!  

E

E

EE

q1=69

q1=70

q2=666

q2=667

VVVVVV

VV ((√√gg ))

VV ((√√gg ))

VV ((√√gg ))

O

A

B

segment BV

segment OB

VV ((√√gg ))segment AVsegment OA

D

E

F

VV ((√√gg ))segment OE

VV ((√√gg ))

segment EVH

VV ((√√gg ))segment OF

VV ((√√gg ))segment FV

V=V1E1+V2E

2=V1E1+V2E2


"Plopped" and"Sliding" Parabolic Coordinates are 2D (xy) plots for Ex.2 and Ex.3 (despite 3D appearance (only) of latter.)

! "Plopped" Parabolic Coordinates for Ex.2

! "Sliding" Parabolic Coordinates for Ex.3 


! Assignment 8 (contd.) - Extra credit Ex. 5 "Unprofessional" Paraboloidal Coordinates


Assignment 8 Solutions Ex.1 Compute Jacobian, Kajobian, Em, Em, metric tensors gmn and gmn for the following OCC.

(c) Cylindrical coordinates {q1=ρ, q2=φ}: x=x1=ρ cosφ, y=x2=ρ sinφ. Spherical coordinates: {q1=r, q2=θ, q3=φ }: x=x1=rsinθ cosφ, y=x2=rsinθ sinφ, z=x3=rcosθ. 3.6.1 Jacobian, Kajobian, Em, Em, metric gmn and gmn for spherical coordinates and cylindrical coordinates Spherical coordinates: {q1=r, q2=θ, q3=φ }: x=x1=rsinθ cosφ, y=x2=rsinθ sinφ, z=x3=rcosθ, reduce to cylindrical coordinates {q1=ρ, q2=φ}: x=x1=ρ cosφ, y=x2=ρ sinφ for ρ=r and θ=π/2: (So spherical coordinates are detailed first below.)Jacobian matrices and determinants:

�

“Kajobian” matrix inverses of J.

�

Covariant metric tensor gµν is matrix product g=JT·J of Jacobian and its transpose. OCC g’s are diagonal.�

Assignment 8 Solutions (contd.) Ex.2 "Plopped" Parabolic Coordinate solutions Consider the GCC(Cartesian) definition: q1 = (x)2 + y , q 2 = (y)2 - x (a) Does an analytic Cartesian coordinate definition x j = x j(m) exist? Not a very useful one.(b) Derive the Jacobian, Kajobian, unitary vectors Em, Em, and metric tensors for this GCC.(c) On the appropriate graph on the following page sketch the unitary vectors at the point (x=1, y=1) (Arrow) and at the point (x=1, y=0). Where, if anywhere, are they OCC?(d) Find and indicate where, if anywhere, are the singularities of this GCC.

Inverting:

Er Eθ Eφ

J =

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

=sinθ cosφ r cosθ cosφ −r sinθ sinφsinθ sinφ r cosθ sinφ r sinθ cosφ

cosθ −r sinθ 0

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

θ=π /2r=ρ

⎯ →⎯⎯cosφ 0 −ρ sinφsinφ 0 ρ cosφ

0 −ρ 0

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

det J = det J T =∂{xyz}∂{rθφ}

= r2 sinθ θ=π /2r=ρ

⎯ →⎯⎯ ρ2

K = J−1 =

∂r∂x

∂r∂y

∂r∂z

∂θ∂x

∂θ∂y

∂θ∂z

∂φ∂x

∂φ∂y

∂φ∂z

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

=

r cosθ sinφ r sinθ cosφ−r sinθ 0

−r cosθ cosφ −r sinθ sinφ−r sinθ 0

r cosθ cosφ −r sinθ sinφr cosθ sinφ r sinθ cosφ

−sinθ sinφ r sinθ cosφcosθ 0

sinθ cosφ −r sinθ sinφcosθ 0

−sinθ cosφ −r sinθ sinφsinθ sinφ r sinθ cosφ

sinθ sinφ r cosθ sinφcosθ −r sinθ

−sinθ cosφ r cosθ cosφcosθ −r sinθ

sinθ cosφ r cosθ cosφsinθ sinφ r cosθ sinφ

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

r2 sinθ=Er

Eθ

Eφ

sinθ cosφ sinθ sinφ cosθcosθ cosφ

rcosθ sinφ

r−sinθr

−sinφr sinθ

cosφr sinθ

0

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

θ=π / 2r=ρ

⎯ →⎯⎯

cosφ sinφ 0

0 0 − 1ρ

−sinφρ

cosφρ

0

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

Covariant: grr = 1, gθθ = r2 , gφφ = r

2 sin2 θ , Contravariant: grr = 1, gθθ = 1 / r 2 , gφφ = 1 / r 2 sin2 θ ,

∂q 1

∂x∂q 1

∂y∂q 2

∂x∂q 2

∂y

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

=E1

E2⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

=2x 1−1 2y

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

∂x∂q 1

∂x∂q 2

∂y∂q 1

∂y∂q 2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

= E1

E2( ) =

2y −11 2x

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1 + 4xy

detJ = 1 + 4xy = 0 when: xy =−1

4

g11 = E1iE1 = 2x 1( )i 2x 1( ) = 4x 2 + 1g12 = E1iE2 = 2x 1( )i −1 2y( ) = 2(y − x)g 22 = E2iE2 = −1 2y( )i −1 2y( ) = 1 + 4y 2detg = g

11g

22− g

12g

12= (detJ)2 = (1 + 4xy)2

OCC where: g12 = 0 = 2(y − x) or: y = x

g11

= E1iE

1=

2y1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟i2y1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1| g |

=4y 2 + 1

1 + 4xy( )2

g12

= E1iE

2=

2y1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟i−12x

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1| g |

=2(x −y)

1 + 4xy( )2

g22

= E2iE

2=−12x

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟i−12x

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

1| g |

=1 + 4x 2

1 + 4xy( )2


""Plopped" Parabolic Coordinates "Sliding" Parabolic Coordinates

Assignment 8 (contd.) - solutionsEx.3 "Sliding" Parabolic Coordinates Cartesian(GCC) definition: x = 0.4 (q1)2 - q2 , y = q1 - 0.4 (q2)2

(a) Does an analytic GCC coordinate definition qm = qm(x j) exist?

�Not practical to solve quartic equation for q1 or q2.

(b) Derive the Jacobian, Kajobian, unitary vectors Em, Em, and metric tensors for this GCC.

Inverting:

q1=2

q1=-2 -1 0 +1 +2+2+1 0 -1

q2=+2

+10

-1

-2

-1

0

+1

q1=2

q2=+2 q2=+1 q2=0

q1=1

q1=0

q2=-1q2=0

q2=-2

E2

E2

E1

E1E2

E2

E1

E1

E1E2

Jacobiansingularityloci

orthogonalityloci

q1=-q2

x=y

q1 = const.⇒ y = q1 − 0.4(x − 0.4(q1)2)2

q2 = const.⇒ x = −q2 − 0.4(y + 0.4(q2)2)2

∂x∂q1

∂x∂q 2

∂y∂q1

∂y∂q 2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

= E1

E2( ) =

45q1 −1

1 −45q 2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

detJ = 1−1625

q1q 2 = 0 when: q1q 2 =2516

∂q1

∂x∂q1

∂y∂q 2

∂x∂q 2

∂y

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

=E1

E2⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

=

−45q 2 +1

−1 +45q1

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

1−1625

q1q 2

g11

= E1iE

1=

45q1

1

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟i

45q1

1

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟=

1625

q1( )2 + 1

g12

= E1iE

2=

45q1

1

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟i−1

−45q 2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟=−45

q1 + q 2( )

g22

= E2iE

2=−1

−45q 2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟i−1

−45q 2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟= 1 +

1625

q 2( )2

detg = g11g

22− g

12g

12= (detJ)2 = 1−

1625

q1q 2⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

g11 = E1iE1 =−

45q 2 1

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟i −

45q 2 1

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

detg=

+1625

q 2( )2 + 1

1−1625

q1q 2⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

g12

= E1iE2 =−

45q 2 1

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟i −1 +

45q1

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

detg=

45

q1 + q 2( )

1−1625

q1q 2⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2

g 22 = E2iE2 =−1 +

45q1

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟i −1 +

45q1

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

detg=

1 +1625

q1( )2

1−1625

q1q 2⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟

2


Assignment 8 solution to Ex.4 GCC Coordinate diagram.

�

Differential � or approximation � shows how to scale

covariant vectors � ( � ) and � ( � ). As sketched above, vectors � and �

approximately frame a “unit” parallelogram-grid-cell between points � , � , � , and � separated by unit GCC difference � . Of course the vectors would be better approximations of a smaller cell, say, a nano unit cell with � . Any consistent scale may be applied to draw Em-vectors since they have different units than the GCC qm-coordinates themselves. But, then the contravariant

Em-vectors must scale inversely so that and .

Assignment 8 solution to Extra credit Ex.5 3D-GCC Coordinates 4 The surface is (q3=0) part of a 3D GCC coordinate grid q1=x, q2=y, q3 containing a projection of orthogonal (x,y) Cartesian coordinate grid. (That grid on the surface is obviously not orthogonal most places. ) a. Derive Jacobian J(x,y) and Kajobian K(x,y) for (q3=0). b. Extract and in (x,y,z) basis.

Kajobian is easiest and derived first: Inverse is Jacobian. It happens to be identical to Kajobian here!

c. Derive the 3-by-3 covariant metric gυν(x,y) and contravariant metric gυν(x,y) for (q3=0) and tell which if any points on the surface have grids that are locally orthogonal and which if any are locally orthonormal. The covariant and contravariant metrics are not identical. Only origin has orthogonality or orthonomality.

E

E

EE

q1=69

q1=70

q2=666

q2=667

VVVVVV

VV ((√√gg ))

VV ((√√gg ))

VV ((√√gg ))

O

A

B

segment BV

segment OB

VV ((√√gg ))segment AVsegment OA

D

E

F

VV ((√√gg ))segment OE

VV ((√√gg ))

segment EVH

VV ((√√gg ))segment OF

VV ((√√gg ))segment FV

11 1

1 1 1

11 1

1 1 1 2 2 222 2

2 2 2

22 2

+1-1

+1

+1

-1

-1-1

+1

2

21

1

V=V1E1+V2E

2=V1E1+V2E2

V1=V•E1V2=V•E2

V1=V•E1V2=V•E2

dr = ∂r∂q1

dq1 + ∂r∂q2

dq2 = E1dq1 +E2dq

2 Δr ! E1Δq

1 +E2Δq2

E1 Δq1 = 1, Δq2 = 0 E2 Δq

1 = 0, Δq2 = 1 E1 E2(q1 = 69,q2 = 666) (q1 = 70,q2 = 666)

(q1 = 69,q2 = 667) (q1 = 70,q2 = 667) Δqm = 1

Δqm = 10−9

Em iEm = 1 Em iE

n = δmn

z = f (x, y) =21 x2 + y2 =2

1 x2 + y2 − z

E1 ,E2 ,E3{ } E

1 ,E2 ,E3{ }

∂x∂q1

∂x∂q2

∂x∂q3

∂y∂q1

∂y∂q2

∂y∂q3

∂z∂q1

∂z∂q2

∂z∂q3

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

=

E1=∇q1 E2=∇q2 E3=∇q3

1 0 x0 1 2y0 0 −1

∂q1∂x

∂q1∂y

∂q1∂z

∂q2∂x

∂q2∂y

∂q2∂z

∂q3∂x

∂q3∂y

∂q3∂z

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

=

1 0 x E1= ∂q1∂r

0 1 2y E2= ∂q2∂r

0 0 −1 E3= ∂q3∂r


!

d. Calculate and sketch covariant on (q3=0) surface at (x=4,y=-2) and (x=3,y=+2).

e. Calculate and sketch contravariant on (q3=0) surface at (x=4,y=+2) and (x=0,y=+4).Assignment X-solutions

"Unprofessional" Paraboloidal Coordinates (contd.)

!

g11 = E1iE1

= 1g12 = E1iE2

= 0g13 = E1iE3

= +x

0g22 = E2iE2

= 1g23 = E2iE3

= +2y

+x +2yg33 = E3iE3

= 1+x2+4y2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⋅

g11 = E1iE1= 1+ x2

g12 = E1iE2= 2xy

g13 = E1iE3= −x

2xy

g22 = E2iE2= 1+ 4y2

g23 = E2iE3= −2y

−x −2yg33 = E3iE3

= 1

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

=1 0 00 1 00 0 1

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

E1 ,E2 ,E3{ }

E1 ,E2 ,E3{ }

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Unit 2-3 Assignments for Physics 5103 - Reading in · 2019. 11. 6. · Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG!Name_____ !1 Assignment 8