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 = ˆr f (~r )
Show that ~r × ˙~r = ~c, a constant, and that the geometric interpretation of this leads to Kepler’s second law.
Actually, ~r × ˙~r is basically the angular momentum, ~L = ~r × ~p = m~r × ˙~r. To show that ~L is constant, we can take its time derivative
˙~L = d
dt(m~r × ˙~r ) = m ˙~r × ˙~r + m~r × ¨~r
The first cross-product vanishes. So, by using Newton’s second law, we end up with
˙~L = ~r × ˆrf(~r ) = (~r × ~r )f(~r ) r = 0
This indicates that the angular momentum ~L is a constant in time (ie that it is 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 dA dt = 1 2 ~r ×d~r dt = 12|~r × ˙~r |
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 compo-nents is reduced from 256 to 36 and that the condition
Riklm = Rlmik
further reduces the number of independent components to 21. Finally, if the com-ponents 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 12(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 = 12(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 some-thing 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 com-ponents in the Riemann tensor in d dimensions. The result is
independent components for d dimensions = 121 d2(d2− 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(r2δij − xixj) [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, etc
Using i = 1, 2, 3 notation, this is the same as indicating
Iij =
m(r2− x2
i) i = j
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 − mxixj(1 − δij) (no sum)
distributing the factors out, and noting that it is safe to set xixjδij = x2iδij, we
end up with
Iij = mr2δij − mx2iδij− mxixj + mx2iδij = m(r2δij− xixj)
Note that there is a typo in the book’s version of the homework exercise! b) Show that
Iij = −MilMlj = −milkxkljmxm
where Mil = m1/2ilkxk. This is the contraction of two second-rank tensors and
is identical with the matrix product of Section 3.2. We may calculate
−MilMlj = −milkxkljmxm = −mlkiljmxkxm
Note that the product of two epsilons can be re-expressed as
lkiljm = δkjδim− δkmδij (2)
This is actually the BAC–CAB rule in index notation. Hence
−MilMlj = −m(δkjδim− δkmδij)xkxm= −m(δkjxkδimxm− δkmxkxmδij)
= −m(xjxi− xkxkδij) = m(r2δij − xixj) = Iij
Note that we have used the fact that xkxk = x21+ x22+ x23 = r2.
2.9.12 Given Ak = 12ijkBij with Bij = −Bji, antisymmetric, show that
Bmn= mnkAk
Given Ak = 12ijkBij, we compute
mnkAk = 12mnkkijBij = 12kmnkijBij = 12(δmiδnj − δmjδni)Bij
= 12(Bmn− Bnm) = Bmn
2.10.6 Derive the covariant and contravariant metric tensors for circular cylindrical coordi-nates.
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 + dy2+ dz2 = (dρ cos ϕ − ρ sin ϕ dϕ)2+ (dρ sin ϕ + ρ cos ϕ dϕ)2+ dz2 = dρ2+ ρ2dϕ2+ dz2
Since ds2 = g
ijdxidxj [where (x1, x2, x3) = (ρ, ϕ, z)] we may write the covariant
metric tensor (matrix) as
gij = 1 0 0 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 = ˆx cos ϕ + ˆy sin ϕ ~ eϕ = ρ(−ˆx sin ϕ + ˆy cos ϕ) ~ez = ˆz Then
gρρ = ~eρ· ~eρ = (ˆx cos ϕ + ˆy sin ϕ) · (ˆx cos ϕ + ˆy sin ϕ) = cos2ϕ + sin2ϕ = 1
gρϕ = ~eρ· ~eϕ = (ˆx cos ϕ + ˆy sin ϕ) · ρ(−ˆx sin ϕ + ˆy 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 given by the matrix inverse of (4)
gij = 1 0 0 0 ρ−2 0 0 0 1 (5)
2.10.11 From the circular cylindrical metric tensor gij calculate the Γkij for circular cylindrical
coordinates.
We may compute the Christoffel components using the expression Γijk = 12(∂kgij + ∂jgik− ∂igjk)
However, instead of working out all the components one at a time, it is more effi-cient to examine the metric (4) and to note that the only non-vanishing derivative is
∂ρgϕϕ = 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 Γρϕϕ = −ρ, Γϕρϕ= Γϕϕρ=
1
ρ (6)
Note that the Christoffel symbols are symmetric in the last two indices. 2.10.12 Using the Γk
ij from Exercise 2.10.11, write out the covariant derivatives Vi;j of a
vector ~V in circular cylindrical coordinates.
Recall that the covariant derivative of a contravariant vector is given by Vi;j = Vi,j + ΓijkVk
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
Vρ;ρ= Vρ,ρ+ ΓρρkVk = Vρ,ρ Vϕ;ρ= Vϕ,ρ+ ΓϕρkVk = Vϕ,ρ+ ΓϕρϕVϕ = Vϕ,ρ+ 1 ρV ϕ Vz;ρ = Vz,ρ+ ΓzρkVk = Vz,ρ Vρ;ϕ = Vρ,ϕ+ ΓρϕkVk = Vρ,ϕ+ ΓρϕϕVϕ = Vρ,ϕ− ρVϕ Vϕ;ϕ = Vϕ,ϕ+ ΓϕϕkVk = Vϕ,ϕ+ ΓϕϕρVρ = Vϕ,ϕ+ 1 ρV ρ Vz;ϕ = Vz,ϕ+ ΓzϕkVk = Vz,ϕ Vρ;z = Vρ,z + ΓρzkVk = Vρ,z Vϕ;z = Vϕ,z + ΓϕzkVk = Vϕ,z Vz;z = Vz,z + ΓzzkVk = Vz,z
Note that, corresponding to the three non-vanishing Christoffel symbols, only three of the expressions are modified in the covariant derivative.
