# Mathematical Methods For Physicists Solutions Ch. 2, Webber and Arfken

October 19, 2017 | Author: Josh Brewer | Category: Tensor, Euclidean Vector, Theoretical Physics, Mathematical Analysis, Differential Geometry

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Ch. 2: 2.5.11, 2.6.5, 2.9.6, 2.9.12, 2.10.6, 2.10.11, 2.10.12...

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Physics 451

Fall 2004 Homework Assignment #4 — Solutions

Textbook problems: Ch. 2: 2.5.11, 2.6.5, 2.9.6, 2.9.12, 2.10.6, 2.10.11, 2.10.12 Chapter 2 2.5.11 A particle m moves in response to a central force according to Newton’s second law m~¨r = rˆ f (~r ) Show that ~r × ~r˙ = ~c, a constant, and that the geometric interpretation of this leads to Kepler’s second law. ~ = ~r × p~ = m~r × ~r˙ . To show Actually, ~r × ~r˙ is basically the angular momentum, L ~ is constant, we can take its time derivative that L ~˙ = d (m~r × ~r˙ ) = m~r˙ × ~r˙ + m~r × ~¨r L dt The first cross-product vanishes. So, by using Newton’s second law, we end up with ~˙ = ~r × rˆ f (~r ) = (~r × ~r ) f (~r ) = 0 L r ~ is a constant in time (ie that it is This indicates that the angular momentum L ~ conserved). The constant vector ~c of the problem is just L/m. Note that this proof works for any central force, not just the inverse square force law. For the geometric interpretation, consider the orbit of the particle m dr r

The amount of area swept out by the particle is given by the area of the triangle dA = 12 |~r × d~r | So the area swept out in a given time dt is simply d~r 1 dA 1 = 2 ~r × = 2 |~r × ~r˙ | dt dt Since this is a constant, we find that equal areas are swept out in equal times. This is just Kepler’s second law (which is also the law of conservation of angular momentum).

2.6.5 The four-dimensional, fourth-rank Riemann-Christoffel curvature tensor of general relativity Riklm satisfies the symmetry relations Riklm = −Rikml = −Rkilm With the indices running from 0 to 3, show that the number of independent components is reduced from 256 to 36 and that the condition Riklm = Rlmik further reduces the number of independent components to 21. Finally, if the components satisfy an identity Riklm + Rilmk + Rimkl = 0, show that the number of independent components is reduced to 20. Here we just have to do some counting. For a general rank-4 tensor in four dimensions, since each index can take any of four possible values, the number of independent components is simply independent components = 44 = 256 Taking into account the first symmetry relation, the first part Riklm = −Rikml indicates that the Riemann tensor is antisymmetric when the last pair of indices is switched. Thinking of the last pair of indices as specifying a 4×4 antisymmetric matrix, this means instead of having 42 = 16 independent elements, we actually only have 21 (4)(3) = 6 independent choices for the last index pair (this is the number of elements in an antisymmetric 4 × 4 matrix). Similarly, the second part of the first symmetry relation Riklm = −Rkilm indicates that the Riemann tensor is antisymmetric in the first pair of indices. As a result, the same argument gives only 6 independent choices for the first index pair. This accounts for independent components = 6 · 6 = 36 We are now able to handle the second condition Riklm = Rlmik By now, it should be obvious that this statement indicates that the Riemann tensor is symmetric when the first index pair is interchanged with the second

index pair. The counting of independent components is then the same as that for a 6 × 6 symmetric matrix. This gives independent components = 21 (6)(7) = 21 Finally, the last identity is perhaps the trickiest to deal with. As indicated in the note, this only gives additional information when all four indices are different. Setting iklm to be 0123, this gives R0123 + R0231 + R0312 = 0

(1)

As a result, this can be used to remove one more component, leading to independent components = 21 − 1 = 20 We can, of course, worry that a different combination of iklm (say 1302 or something like that) will give further relations that can be used to remove additional components. However, this is not the case, as can be seen by applying the first to relations. Note that it is an interesting exercise to count the number of independent components in the Riemann tensor in d dimensions. The result is independent components for d dimensions =

1 2 2 12 d (d

− 1)

Putting in d = 4 yields the expected 20. However, it is fun to note that putting in d = 1 gives 0 (you cannot have curvature in only one dimension) and putting in d = 2 gives 1 (there is exactly one independent measure of curvature in two dimensions). 2.9.6

a) Show that the inertia tensor (matrix) of Section 3.5 may be written Iij = m(r2 δij − xi xj )

[typo corrected!]

for a particle of mass m at (x1 , x2 , x3 ). Note that, for a single particle, the inertia tensor of Section 3.5 is specified as Ixx = m(r2 − x2 ),

Ixy = −mxy,

Using i = 1, 2, 3 notation, this is the same as indicating  Iij =

m(r2 − x2i ) i = j −mxi xj i= 6 j

etc

We can enforce the condition i = j by using the Kronecker delta, δij . Similarly, the condition i 6= j can be enforced by the ‘opposite’ expression 1 − δij . This means we can write Iij = m(r2 − x2i )δij − mxi xj (1 − δij )

(no sum)

distributing the factors out, and noting that it is safe to set xi xj δij = x2i δij , we end up with Iij = mr2 δij − mx2i δij − mxi xj + mx2i δij = m(r2 δij − xi xj ) Note that there is a typo in the book’s version of the homework exercise! b) Show that Iij = −Mil Mlj = −milk xk ljm xm where Mil = m1/2 ilk xk . This is the contraction of two second-rank tensors and is identical with the matrix product of Section 3.2. We may calculate −Mil Mlj = −milk xk ljm xm = −mlki ljm xk xm Note that the product of two epsilons can be re-expressed as lki ljm = δkj δim − δkm δij This is actually the BAC–CAB rule in index notation. Hence −Mil Mlj = −m(δkj δim − δkm δij )xk xm = −m(δkj xk δim xm − δkm xk xm δij ) = −m(xj xi − xk xk δij ) = m(r2 δij − xi xj ) = Iij Note that we have used the fact that xk xk = x21 + x22 + x23 = r2 . 2.9.12 Given Ak = 21 ijk Bij with Bij = −Bji , antisymmetric, show that Bmn = mnk Ak Given Ak = 21 ijk Bij , we compute mnk Ak = 12 mnk kij Bij = 12 kmn kij Bij = 12 (δmi δnj − δmj δni )Bij = 12 (Bmn − Bnm ) = Bmn We have used the antisymmetric nature of Bij in the last step.

(2)

2.10.6 Derive the covariant and contravariant metric tensors for circular cylindrical coordinates. There are several ways to derive the metric. For example, we may use the relation between Cartesian and cylindrical coordinates x = ρ cos ϕ,

y = ρ sin ϕ,

z=z

(3)

to compute the differentials dx = dρ cos ϕ − ρ sin ϕ dϕ,

dy = dρ sin ϕ + ρ cos ϕ dϕ,

dz = dz

The line element is then ds2 = dx2 + dy 2 + dz 2 = (dρ cos ϕ − ρ sin ϕ dϕ)2 + (dρ sin ϕ + ρ cos ϕ dϕ)2 + dz 2 = dρ2 + ρ2 dϕ2 + dz 2 Since ds2 = gij dxi dxj [where (x1 , x2 , x3 ) metric tensor (matrix) as  1 gij =  0 0

= (ρ, ϕ, z)] we may write the covariant 0 ρ2 0

 0 0 1

(4)

Alternatively, the metric is given by gij = ~ei · ~ej where the basis vectors are ~ei =

∂~r ∂xi

Taking partial derivatives of (3), we obtain ~eρ = x ˆ cos ϕ + yˆ sin ϕ ~eϕ = ρ(−ˆ x sin ϕ + yˆ cos ϕ) ~ez = zˆ Then gρρ = ~eρ · ~eρ = (ˆ x cos ϕ + yˆ sin ϕ) · (ˆ x cos ϕ + yˆ sin ϕ) = cos2 ϕ + sin2 ϕ = 1 gρϕ = ~eρ · ~eϕ = (ˆ x cos ϕ + yˆ sin ϕ) · ρ(−ˆ x sin ϕ + yˆ cos ϕ) = ρ(− cos ϕ sin ϕ + sin ϕ cos ϕ) = 0 etc . . . The result is the same as (4). The contravariant components of the metric is  1 0 g ij =  0 ρ−2 0 0

given by the matrix inverse of (4)  0 0 (5) 1

2.10.11 From the circular cylindrical metric tensor gij calculate the Γk ij for circular cylindrical coordinates. We may compute the Christoffel components using the expression Γijk = 21 (∂k gij + ∂j gik − ∂i gjk ) However, instead of working out all the components one at a time, it is more efficient to examine the metric (4) and to note that the only non-vanishing derivative is ∂ρ gϕϕ = 2ρ This indicates that the only non-vanishing Christoffel symbols Γijk are the ones where the three indices ijk are some permutation of ρϕϕ. It is then easy to see that Γρϕϕ = −ρ, Γϕρϕ = Γϕϕρ = ρ Finally, raising the first index using the inverse metric (5) yields 1 ρ Note that the Christoffel symbols are symmetric in the last two indices. Γρ ϕϕ = −ρ,

Γϕ ρϕ = Γϕ ϕρ =

(6)

2.10.12 Using the Γk ij from Exercise 2.10.11, write out the covariant derivatives V i ;j of a ~ in circular cylindrical coordinates. vector V Recall that the covariant derivative of a contravariant vector is given by V i ;j = V i ,j + Γi jk V k where the semi-colon indicates covariant differentiation and the comma indicates ordinary partial differentiation. To work out the covariant derivative, we just have to use (6) for the non-vanishing Christoffel connections. The result is V ρ ;ρ = V ρ ,ρ + Γρ ρk V k = V ρ ,ρ 1 V ϕ ;ρ = V ϕ ,ρ + Γϕ ρk V k = V ϕ ,ρ + Γϕ ρϕ V ϕ = V ϕ ,ρ + V ϕ ρ z z z k z V ;ρ = V ,ρ + Γ ρk V = V ,ρ V ρ ;ϕ = V ρ ,ϕ + Γρ ϕk V k = V ρ ,ϕ + Γρ ϕϕ V ϕ = V ρ ,ϕ − ρV ϕ 1 V ϕ ;ϕ = V ϕ ,ϕ + Γϕ ϕk V k = V ϕ ,ϕ + Γϕ ϕρ V ρ = V ϕ ,ϕ + V ρ ρ V z ;ϕ = V z ,ϕ + Γz ϕk V k = V z ,ϕ V ρ ;z = V ρ ,z + Γρ zk V k = V ρ ,z V ϕ ;z = V ϕ ,z + Γϕ zk V k = V ϕ ,z V z ;z = V z ,z + Γz zk V k = V z ,z Note that, corresponding to the three non-vanishing Christoffel symbols, only three of the expressions are modified in the covariant derivative.