H01-Newtons-Laws-w-Sols

# H01-Newtons-Laws-w-Sols - Homework 1 Solutions Phys. 105,...

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Unformatted text preview: Homework 1 Solutions Phys. 105, UCSC, F2010 Due Oct. 5, Tuesday. Each problem is worth 5 points, except problems 5, 6, 7 (10 points each). Problem 1 Assume that ↔ O 1 and ↔ O 2 are arbitrary orthogonal matrices of the same dimensions. For each of the following statements, prove it if it is true, or provide one counter-example (using 2 × 2 matrices is good enough) if it is not true. (a) ↔ O 1 + ↔ O 2 is also orthogonal. Solution This is not true. ↔ O 1 = ( 1 0 0 1 ) and ↔ O 2 = ( 1 0 − 1 ) are orthogonal, but ↔ O 1 + ↔ O 2 = ( 2 0 0 0 ) is not. (b) ↔ O 1 ↔ O 2 is also orthogonal. Solution This is true. ↔ O 1 ↔ O 2 ( ↔ O 1 ↔ O 2 ) t = ↔ O 1 ↔ O 2 ↔ O t 2 ↔ O t 1 = ↔ O 1 ↔ O t 1 = 1 . (c) r ↔ O 1 is also orthogonal, where r is an arbitrary real number. Solution Definitely not. For any orthogonal matrix, r ↔ O 1 is not orthogonal, if | r | negationslash = 1 , since then the magnitude of each column/row vector is | r | , not 1. (d) ↔ O t 1 is also orthogonal. Solution Yes. Because ↔ O t 1 ↔ O 1 = 1 , ↔ O 1 is the inverse of ↔ O t 1 . This means that the inverse of ↔ O t 1 is its transpose ( ↔ O t 1 ) t = ↔ O 1 . Problem 2 Given an arbitrary orthogonal coordinate transformation ↔ O : → r ′ = ↔ O → r prove following statements. You can use the fact that the scalar product is a scalar, without proof, if necessary. You can consider the 3D space, specifically, but these statements are valid for any dimensions. (a) → v = ˙ → r is a vector. [Trivial, but show it. Stated in class, without really showing.] Solution → r ′ = ↔ O → r . Differentiating both sides and assuming that ↔ O contains time independent quantities only, we get ˙ → r ′ = ↔ O ˙ → r . (b) The gradient operator → ∇ def = ∑ 3 i =1 ˆ i ∂ ∂x i is a vector, where x i is the i-th Cartesian component of the position vector → r ( x 1 = x, x 2 = y, x 3 = z , in 3D) and ˆ i is the i-th unit vector in the Cartesian coordinate system ( ˆ 1 = ˆ x, ˆ 2 = ˆ y, ˆ 3 = ˆ z , in 3D). (cf. Section 1.6 of the text.) Solution ↔ O → r = → r ′ , → r = ↔ O t → r ′ . In component form, x i ′ = ∑ j O ij x j , x i = ∑ j O ji x j ′ . By chain rule, ∂ ∂x i ′ = ∑ j ∂x j ∂x ′ i ∂ ∂x j = ∑ j O ij ∂ ∂x j . QED. 1 Homework 1 Solutions Phys. 105, UCSC, F2010 (c) The distance between two positions → r 1 and → r 2 is a scalar. Solution Distance = radicalbig ( → r 1 − → r 2 ) · ( → r 1 − → r 2 ) . So, this is a scalar, since a scalar product is a scalar. (d) The magnitude of the relative velocity → v 1 − → v 2 is a scalar. Solution Same as (c). Change r to v ....
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H01-Newtons-Laws-w-Sols - Homework 1 Solutions Phys. 105,...

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