Physics 451
Fall 2004 Homework Assignment #1 — Solutions
Textbook problems: Ch. 1: 1.1.5, 1.3.3, 1.4.7, 1.5.5, 1.5.6 Ch. 3: 3.2.4, 3.2.19, 3.2.27 Chapter 1 1.1.5 A sailboat sails for 1 hr at 4 km/hr (relative to the water) on a steady compass heading of 40◦ east of north. The saiboat is simultaneously carried along by a current. At the end of the hour the boat is 6.12 km from its starting point., The line from its starting point to its location lies 60◦ east of north. Find the x (easterly) and y (northerly) components of the water velocity. This is a straightforward relative velocity (vector addition) problem. Let ~vbl denote the velocity of the boat with respect to land, ~vbw the velocity of the boat with respect to the water and ~vwl the velocity of the water with respect to land. Then ~vbl = ~vbw + ~vwl where
~vbw = 4 km/hr @ 50◦ = (2.57ˆ x + 3.06ˆ y ) km/hr ◦ ~vbl = 6.12 km/hr @ 30 = (5.3ˆ x + 3.06ˆ y ) km/hr
Thus ~vwl = ~vbl − ~vbw = 2.73ˆ x km/hr 1.3.3 The vector ~r, starting at the origin, terminates at and specifies the point in space (x, y, z). Find the surface swept out by the tip of ~r if (a) (~r − ~a ) · ~a = 0 The vanishing of the dot product indicates that the vector ~r − ~a is perpendicular to the constant vector ~a. As a result, ~r − ~a must lie in a plane perpendicular to ~a. This means ~r itself must lie in a plane passing through the tip of ~a and perpendicular to ~a
r−a r a
(b) (~r − ~a ) · ~r = 0 This time the vector ~r − ~a has to be perpendicular to the position vector ~r itself. It is perhaps harder to see what this is in three dimensions. However, for two dimensions, we find
r−a a r
which gives a circle. In three dimensions, this is a sphere. Note that we can also complete the square to obtain (~r − ~a ) · ~r = |~r − 12 ~a |2 − | 12 ~a |2 Hence we end up with the equation for a circle of radius |~a |/2 centered at the point ~a/2 |~r − 12 ~a |2 = | 12 ~a |2 ~ × B) ~ · (A ~ × B) ~ = (AB)2 − (A ~·B ~ )2 . 1.4.7 Prove that (A This can be shown just by a straightforward computation. Since ~×B ~ = (Ay Bz − Az By )ˆ A x + (Az Bx − Ax Bz )ˆ y + (Ax By − Ay Bx )ˆ z we find ~×B ~ |2 = (Ay Bz − Az By )2 + (Az Bx − Ax Bz )2 + (Ax By − Ay Bx )2 |A = A2x By2 + A2x Bz2 + A2y Bx2 + A2y Bz2 + A2z Bx2 + A2z By2 − 2Ax Bx Ay By − 2Ax Bx Az Bz − 2Ay By Az Bz = (A2x + A2y + A2z )(Bx2 + By2 + Bz2 ) − (Ax Bx + Ay By + Az Bz )2 where we had to add and subtract A2x Bx2 +A2y By2 +A2z Bz2 and do some factorization to obtain the last line. However, there is a more elegant approach to this problem. Recall that cross products are related to sin θ and dot products are related to cos θ. Then ~×B ~ |2 = (AB sin θ)2 = (AB)2 (1 − cos2 θ) = (AB)2 − (AB cos θ)2 |A ~·B ~ )2 = (AB)2 − (A
~ of a particle is given by L ~ = ~r × p~ = m~r ×~v where p~ 1.5.5 The orbital angular momentum L is the linear momentum. With linear and angular velocity related by ~v = ω ~ × ~r, show that ~ = mr2 [~ L ω − rˆ(ˆ r·ω ~ )] Here, rˆ is a unit vector in the ~r direction. ~ = m~r × ~v and ~v = ω Using L ~ × ~r, we find ~ = m~r × (~ L ω × ~r ) Because of the double cross product, this is the perfect opportunity to use the ~ × (B ~ × C) ~ = B( ~ A ~ · C) ~ − C( ~ A ~ · B) ~ “BAC–CAB” rule: A ~ = m[~ L ω (~r · ~r ) − ~r(~r · ω ~ )] = m[~ ω r2 − ~r(~r · ω ~ )] Using ~r = r rˆ, and factoring out r2 , we then obtain ~ = mr2 [~ L ω − rˆ(ˆ r·ω ~ )]
(1)
1.5.6 The kinetic energy of a single particle is given by T = 21 mv 2 . For rotational motion this becomes 21 m(~ ω × ~r )2 . Show that ~ )2 ] T = 12 m[r2 ω 2 − (~r · ω We can use the result of problem 1.4.7: T = 21 m(~ ω × ~r )2 = 12 m[(ωr)2 − (~ ω · ~r )2 ] = 12 m[r2 ω 2 − (~r · ω ~ )2 ] Note that we could have written this in terms of unit vectors r·ω ~ )2 ] T = 21 mr2 [ω 2 − (ˆ Comparing this with (1) above, we find that ~ ·ω T = 21 L ~ which is not a coincidence.
Chapter 3 3.2.4 (a) Complex numbers, a + ib, with a and b real, may be represented by (or are isomorphic with) 2 × 2 matrices: a b a + ib ↔ −b a Show that this matrix representation is valid for (i) addition and (ii) multiplication. Let us start with addition. For complex numbers, we have (straightforwardly) (a + ib) + (c + id) = (a + c) + i(b + d) whereas, if we used matrices we would get a b c d (a + c) + = −b a −d c −(b + d)
(b + d) (a + c)
which shows that the sum of matrices yields the proper representation of the complex number (a + c) + i(b + d). We now handle multiplication in the same manner. First, we have (a + ib)(c + id) = (ac − bd) + i(ad + bc) while matrix multiplication gives a b c d (ac − bd) = −b a −d c −(ad + bc)
(ad + bc) (ac − bd)
which is again the correct result. (b) Find the matrix corresponding to (a + ib)−1 . We can find the matrix in two ways. We first do standard complex arithmetic (a + ib)−1 =
a − ib 1 1 = = 2 (a − ib) a + ib (a + ib)(a − ib) a + b2
This corresponds to the 2 × 2 matrix −1
(a + ib)
↔
1 2 a + b2
a −b b a
Alternatively, we first convert to a matrix representation, and then find the inverse matrix −1 1 a −b a b −1 (a + ib) ↔ = 2 −b a a + b2 b a Either way, we obtain the same result. 3.2.19 An operator P~ commutes with Jx and Jy , the x and y components of an angular momentum operator. Show that P~ commutes with the third component of angular momentum; that is, [P~ , Jz ] = 0 We begin with the statement that P~ commutes with Jx and Jy . This may be expressed as [P~ , Jx ] = 0 and [P~ , Jy ] = 0 or equivalently as P~ Jx = Jx P~ and P~ Jy = Jy P~ . We also take the hint into account and note that Jx and Jy satisfy the commutation relation [Jx , Jy ] = iJz or equivalently Jz = −i[Jx , Jy ]. Substituting this in for Jz , we find the double commutator [P~ , Jz ] = [P~ , −i[Jx , Jy ]] = −i[P~ , [Jx , Jy ]] Note that we are able to pull the −i factor out of the commutator. From here, we may expand all the commutators to find [P~ , [Jx , Jy ]] = P~ Jx Jy − P~ Jy Jx − Jx Jy P~ + Jy Jx P~ = Jx P~ Jy − Jy P~ Jx − Jx P~ Jy + Jy P~ Jx =0 To get from the first to the second line, we commuted P~ past either Jx or Jy as appropriate. Of course, a quicker way to do this problem is to use the Jacobi identity [A, [B, C]] = [B, [A, C]] − [C, [A, B]] to obtain [P~ , [Jx , Jy ]] = [Jx , [P~ , Jy ]] − [Jy , [P~ , Jx ]] The right hand side clearly vanishes, since P~ commutes with both Jx and Jy . 3.2.27 (a) The operator Tr replaces a matrix A by its trace; that is Tr (a) = trace(A) =
X i
Show that Tr is a linear operator.
aii
Recall that to show that Tr is linear we may prove that Tr (αA+βB) = α Tr (A)+ β Tr (B) where α and β are numbers. However, this is a simple property of arithmetic X X X Tr (αA + βB) = (αaii + βbii ) = α aii + β bii = α Tr (A) + β Tr (B) i
i
i
(b) The operator det replaces a matrix A by its determinant; that is det(A) = determinant of A Show that det is not a linear operator. In this case all we need to do is to find a single counterexample. For example, for an n × n matrix, the properties of the determinant yields det(αA) = αn det(A) This is not linear unless n = 1 (in which case A is really a single number and not a matrix). There are of course many other examples that one could come up with to show that det is not a linear operator.
