hw3sol - Physics 570 Homework No 3 Solutions due Wednesday...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 570 Homework No. 3 Solutions: due Wednesday, 10 February, 2010 1. Let { x,y,z } be local coordinates on a 3-dimensional manifold, and define the following four 1-forms: φ ∼ ≡ yz dx + dz , ψ ∼ ≡ (sin z ) dx + (cos z ) dy , ξ ∼ ≡ dy + z dz , α ∼ ≡ y dx- xdy x 2 + y 2 . and also define the following tangent vector: e u = ∂ x + ∂ y . a. Calculate the exterior differential of each of the four 1-forms. b. Calculate the 3-form ψ ∼ ∧ dψ ∼ . c. Find the 1-form ( φ ∼ ∧ ξ ∼ )( e u, · ), where the · inside the parentheses means that the 2-form is still “waiting” for an additional vector in order to determine the appropriate scalar, and therefore can be considered as a 1-form. d. Lastly, since you did show that the exterior derivative of the 1-form α ∼ above was zero, there must be some function f such that α ∼ = df . Please present such a function, but be assured that the function f is not unique. ........................................................................................... a. The calculations are straightforward, I hope: dφ ∼ = y dz ∧ dx + z dy ∧ dx , dψ ∼ = cos z dz ∧ dx- sin z dz ∧ dy ,dξ ∼ = 0 , d α ∼ = 2 dy ∧ dx x 2 + y 2- 2 ( xdx + y dy ) ∧ ( y dx- xdy ) ( x 2 + y 2 ) 2 = 2 dy ∧ dx x 2 + y 2- 2 ( x 2 + y 2 ) dy ∧ dx ( x 2 + y 2 ) 2 = 0 . b. Already having dψ ∼ again we can simply follow through the manipulations: ψ ∼ ∧ dψ ∼ = (sin z dx + cos z dy ) ∧ (cos z dz ∧ dx- sin z dz ∧ dy ) = sin 2 z dx ∧ dy ∧ dz + cos 2 z dx ∧ dy ∧ dz = dx ∧ dy ∧ dz . c. We first need to determine the desired 2-form: φ ∼ ∧ ξ ∼ = yz dx ∧ dy + dz ∧ dy + yz 2 dx ∧ dz . Then when we are letting a 2-form act on a vector, we need to remember that it is skew- symmetric. A simple example is that ( dx ∧ dy )( ∂ x + ∂ y , · ) = ( dx ⊗ dy- dy ⊗ dx )( ∂ x + ∂ y , · ) = dy- dx . Therefore, for our current case we have ( yz dx ∧ dy + dz ∧ dy + yz 2 dx ∧ dz )( ∂ x + ∂ y , · ) ≡ ( ∂ x + ∂ y ) c ( yz dx ∧ dy + dz ∧ dy + yz 2 dx ∧ dz ) = yz dy + yz 2 dz- yz dx- dz =- yz ( dx- dy ) + ( yz 2- 1) dz ....
View Full Document

{[ snackBarMessage ]}

Page1 / 6

hw3sol - Physics 570 Homework No 3 Solutions due Wednesday...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online