Questions for PHZ 6607, Special and General Relativity I
1) Consider an arbitrary point (t, x, y, z) and a nearby point (t + ∆t, x + ∆x, y, z).
a) Write down an expression for the invariant separation between the two points (to second
order): indicate how to tell when the separation is spacelike and when it is timelike;
We find:
∆l2 = −c2 ∆t2 + ∆x2
(
>0
=0
<0
spacelike,
lightlike,
timelike.
Now consider a description of the two points with respect to a frame moving in the xdirection with speed v.
b) Find expressions for ∆t′ and ∆x′ , the coordinate difference between the two points in
the moving frame, and repeat part a) in terms of these primed quantities; and
We have:
∆t′ = γ(∆t − v∆x/c2 )
∆x′
Thus:
= γ(∆x − v∆t),
where
γ = 1/
q
1 − v 2 /c2 .
∆l′2 = −c2 ∆t′2 + ∆x′2
2v∆x∆t v 2 ∆x2
2
2 2
2
2 2
2
= −c γ ∆t −
∆x
−
2v∆x∆t
+
v
∆t
+
γ
+
c2
c4
v2 2 2
= γ2 1 − 2
−c ∆t + ∆x2
c
= −c2 ∆t2 + ∆x2 = ∆l2 .
c) On the basis of your answers above, comment on an implied property of the Lorentz
transformation.
Lorentz transformations preserve ∆l2 , whether it is spacelike, lightlike or timelike.
2a) Write out in full the non-relativistic equations for the motion of a free particle in
spherical polar coordinates.
We have:
Hence:
1 L = m ṙ 2 + r 2 θ̇ 2 + r 2 sin2 θ φ̇2 .
2
r̈ − r θ̇ 2 − r sin2 θ φ̇2 = 0,
r 2 θ̈ + 2r ṙ θ̇ − r sin θ cos θ φ̇2 = 0,
r 2 sin2 θ φ̈ + 2r sin2 θ ṙ φ̇ + 2r 2 sin θ cos θ θ̇ φ̇
and
= 0.
b) Read off the connection coefficients which underlie the identification of these equations
of motion as the equations for a geodesic parameterized by Newtonian time, and explain exactly
how you do this.
From these we can identify and read off:
1
1
φ
φ
Γrθθ = −r, Γrφφ = −r sin2 θ, Γθrθ = , Γrφ = , Γθφφ = − sin θ cos θ, and Γθφ = cot θ,
r
r
all other being zero, where we have used: ẍi + Γijk ẋj ẋk = 0, and the symmetry: Γijk = Γi(jk) .
c) Find as many independent constants of the motion as you can (and compare with the
situation in quantum mechanics).
It is obvious that Lz = mr 2 sin2 θ φ̇ is conserved. Not so obvious is the fact that Lx and Ly are
also conserved. Even less obvious is the fact that px , py and pz are all conserved too.
3a) Show for a differentiable function ϕ defined over a manifold that the set {∂ϕ/∂xa}
transforms as do the components of a tensor of rank one; what type of tensor has this construction
defined?
We find:
∂xa ∂ϕ
∂ϕ
,
=
∂xa′
∂xa′ ∂xa
which is the transformation law for a covector.
b) Write out the components of this tensor in polar coordinates explicitly in terms of its
components in cartesian coordinates.
We calculate, ∇ϕ = ϕ,r , ϕ,θ , ϕ,φ , where:
ϕ,r = x,r ϕ,x + y,r ϕ,y + z,r ϕ,z
= sin θ cos φ ϕ,x + sin θ sin φ ϕ,y + cos θ ϕ,z
ϕ,θ = x,θ ϕ,x + y,θ ϕ,y + z,θ ϕ,z
= r cos θ cos φ ϕ,x + r cos θ sin φ ϕ,y − r sin θ ϕ,z
ϕ,φ = x,φ ϕ,x + y,φ ϕ,y + z,φ ϕ,z
= −r sin θ sin φ ϕ,x + sin θ cos φ ϕ,y
c) Compare this with the expression you have previously used for the gradient of a function
in spherical polar coordinates, and comment on any differences you see.
The usual result quoted in Electromagnetic contexts (e.g., in Griffiths, Jackson, etc.) would be:
1
1
,
∇ϕ = ϕ,r , ϕ,θ ,
ϕ
r
r sin θ ,φ
for which the magnitude squared would be calculated using an orthonormal metric, not the metric
in spherical polar coordinates.
4) Maxwell’s equations for electromagnetism may be written in a geometrically covariant
form as:
∇i B i = 0, ∇i E i = 4πρ, εijk ∇j Ek = −Ḃ i , εijk ∇j Bk = Ė i + 4πJ i ,
2
where ∇i is a derivative operator (generally not ∂/∂xi ), and E i is not to be confused with Ei .
j
In accord with these equations, the electromagnetic stress tensor Ti may be defined according to
the formula:
1 j
j
4πTi = −Ei E j − Bi B j + δi (Ek E k + Bk B k ).
2
a) Try to show that, in a vacuum, it follows as a consequence of Maxwell’s equations that
j
Ṗi + ∇j Ti = 0,
where Pi is the electromagnetic energy flux vector defined by
4πPi = εijk E jB k .
Indicate as precisely as you can what extra information you would need to complete the proof.
Substituting, we have:
j
j
j
j
= − ∇ j E i E − E i ∇ j E − ∇ j B i B − B i ∇j B +
i
1h
k
k
k
k
,
(∇i Ek ) E + Ek ∇i E + (∇i Bk ) B + Bk ∇i B
2
4π Ṗi = εijk Ė jB k + εijk E jḂ k
= εijk εjlm ∇l Bm B k − εijk E j εklm ∇l Em
k
l
m
l
m
l
m
l
m
= δk δi − δi δk (∇l Bm ) B − δi δj − δj δi E j (∇l Em )
= (∇k Bi ) B k − (∇i Bk ) B k − E j ∇i Ej + E j ∇j Ei .
j
4π∇j Ti
and
Adding, we find:
4π
j
Ṗi + ∇j Ti
i
1h
k
k
k
k
= − (∇i Ek ) E − Ek ∇i E + (∇i Bk ) B − Bk ∇i B
.
2
To complete the proof, the first two terms and the last two terms would have to separately cancel.
b) Suppose there exists a symmetric second rank tensor gij which allows you to relate the
covector Ei to the contravariant vector E i via Ei = gij E j . If this is to be enough extra information
to prove the above conservation equations, what can you deduce about ∇k gij ?
Substituting, we compute ∇i Ek = ∇i gkj E j +gkj ∇i E j , and ∇i Bk = ∇i gkj B j +gkj ∇i B j ,
and hence:
j k
1
j
j
k
4π Ṗi + ∇j Ti = − ∇i gkj E E + B B ,
2
where we have used the symmetry of gkj to replace gkj E k byEj , etc. before canceling.
3
Questions for PHZ 6607, Special and General Relativity I
5a) Using the Lagrangian
1
£ = mgµν x˙µ x˙ν ,
2
in cartesian (spatial) coordinates, find expressions for the conserved linear and angular canonical
momenta px , py , pz , Lx , Ly and Lz , in terms of the ‘velocities’.
We have £ = 12 m(ẋ2 + ẏ 2 + ż 2 ), and find:
∂£
= m ẋ,
∂ ẋ
∂£
= m ẏ,
=
∂ ẏ
∂£
=
= m ż, and, by definition,
∂ ż
= ypz − zpy = m(y ż − z ẏ),
px =
py
pz
Lx
Ly = zpx − xpz = m(z ẋ − xż),
Lz = xpy − ypx = m(xẏ − y ẋ).
b) Similarly, starting with the Lagrangian written in spherical polar coordinates, find expressions for the canonical momenta pr , pθ and pφ in terms of the ‘velocities’ which arise for these
coordinates.
We have £ = 21 m(ṙ 2 + r 2 θ̇ 2 + r 2 sin2 θ φ̇2 ), and find:
∂£
= m ṙ,
∂ ṙ
∂£
pθ =
= mr 2 θ̇, and
∂ θ̇
∂£
pφ =
= mr 2 sin2 θ φ̇.
∂ φ̇
pr =
c) Use the coordinate relations:
x = r sin θ cos φ,
y = r sin θ sin φ, and
z = r cos θ,
write down the relations giving the cartesian velocities in terms of the polar velocities.
We calculate:
ẋ = sin θ cos φ ṙ + r cos θ cos φ θ̇ − r sin θ sin φ φ̇,
ẏ = sin θ sin φ ṙ + r cos θ sin φ θ̇ + r sin θ cos φ φ̇,
ż = cos θ ṙ − r sin θ θ̇.
d) Hence, obtain expressions for all the conserved momenta listed in part a) in terms of the
canonical momenta for polar coordinates.
4
For the linear momenta, we find simply:
px = m(sin θ cos φ ṙ + r cos θ cos φ θ̇ − r sin θ sin φ φ̇),
py = m(sin θ sin φ ṙ + r cos θ sin φ θ̇ + r sin θ cos φ φ̇),
pz = m(cos θ ṙ − r sin θ θ̇).
For the angular momenta, we find:
Lx = mr sin θ sin φ(cos θ ṙ − r sin θ θ̇) − mr cos θ(sin θ sin φ ṙ + r cos θ sin φ θ̇ + r sin θ cos φ φ̇)
= −mr 2 (sin φ θ̇ + sin θ cos θ cos φ φ̇),
Ly = mr cos θ(sin θ cos φ ṙ + r cos θ cos φ θ̇ − r sin θ sin φ φ̇) − mr sin θ cos φ(cos θ ṙ − r sin θ θ̇),
= mr 2 (cos φ θ̇ − sin θ cos θ sin φ φ̇),
Lz = mr sin θ cos φ(sin θ sin φ ṙ + r cos θ sin φ θ̇ + r sin θ cos φ φ̇)
− mr sin θ sin φ(sin θ cos φ ṙ + r cos θ cos φ θ̇ − r sin θ sin φ φ̇)
= mr 2 sin2 θ φ̇,
The last of which is exactly equal to pφ as computed in b) above.
e) What type of tensor are the velocities and what type the momenta? (This difference
explains why it is dangerous to go directly from the velocities in one set of coordinates to the
momenta in another. It also explains why p·x is always a scalar.)
The velocities are tensors of rank (1, 0) and the momenta are tensors of rank (0, 1).
6) Consider the four-dimensional spacetime metric which in {t, r, θ, φ} coordinates has diagonal components {−(1 − rs /r), 1/(1 − rs/r), r 2, r 2 sin2 θ}, all other components being zero. Notice
that in the usual sense, the parameter rs has the dimensions of length.
a) Write out expressions for the canonical momenta of a particle whose Lagrangian is given
by:
1
£ = mgµν x˙µ x˙ν , in which ˙ = d/dλ,
2
in which the spacetime geometry is given by this metric.
Calculating (and restoring c2 ), we find:
rs pt = −mc2 1 −
ṫ,
r
m ṙ
,
pr =
1 − rrs
pθ = mr 2 θ̇,
pφ = mr 2 sin2 θ φ̇.
b) Write out Hamilton’s equations of motion for this particle, and indicate which canonical
momenta are conserved.
5
Solving for the velocities and substituting, we obtain:
1 µν
g pµ pν
2m
p2t
1
1
rs 2
1
1 +
p2θ +
=−
p2 .
1−
pr +
r
2
2
s
2
2m
r
2mc 1 − r
2mr
2mr sin2 θ φ
H = pµ ẋµ = L =
Hamilton’s equations give:
∂H
,
∂xµ
∂H
ẋµ = +
.
∂pµ
ṗµ = −
Since the metric is independent of both t and φ, both pt = −E and pφ = Lz are conserved. Thus,
the non-trivial equations become:
p2t
1 2
1
rs
rs 2
pr +
pθ +
ṗr = −
p2φ ,
2 −
2
2
2
3
2
3
r
s
2mr c 1 −
2mr
mr
mr sin θ
r
cos θ
ṗθ =
p2φ .
3
2
mr sin θ
c) By explicit calculation, show that all components of the angular momentum which were
defined in Question 5) are in fact conserved for the particle moving in this non-flat geometry.
Hence, or otherwise, argue whether or not this spacetime is spherically symmetric.
From above we find:
Lx = − sin φpθ − cot θ cos φpφ ,
Ly = − cos φpθ + cot θ sin φpφ ,
Lz = pφ .
Note: L2 = p2θ + p2φ / sin2 θ. Computing, we find:
p2φ
cos φ pθ pφ
L̇x = − cos φ 2 2 − sin φ 2 3 +
+ cot θ sin φ 2 2 = 0,
mr sin θ
mr sin θ sin2 θ mr 2
mr sin θ
2
p2φ
cos θpφ
pφ pθ
sin φ pθ pφ
L̇y = + sin φ 2 2 − cos φ 2 3 −
+ cot θ cos φ 2 2 = 0.
mr sin θ
mr sin θ sin2 θ mr 2
mr sin θ
pφ pθ
cos θp2φ
Spherical symmetry follows from the form of the metric; it also follows from the conservation of
all three angular momenta.
d) Find an expression for dr/dφ, assuming the motion takes place in the equatorial plane.
Obtain the corresponding equation for a particle of mass m moving under Newtonian gravity
around a central body of mass M.
In the equatotial plane, sin θ = 1 and L = Lz . To find an expression for dr/dφ it is simplest to
start from the ”conservation of momentum” pµ pν g µν = −m2 c2 , which gives:
m2 ṙ 2
L2
E2
= −m2 c2 .
+
+
− 2
rs
rs
2
r
c 1− r
1− r
6
We can rearrange this to find ṙ, via:
ṙ 2
E2
L2
rs 2
= 2 2 − 1−
c + 2 2 .
r
m c
m r
Using φ̇ = L/mr 2 (recall sin θ = 1 and Lz = L), expanding and dividing, we finally obtain:
2
rs c2
rs L 2
m2 r 4 E 2
L2
dr
2
−c +
=
− 2 2+ 2 3 .
dφ
r
L2
m2 c2
m r
m r
In the Newtonian case we would have correspondingly:
1
GM m
L2
2
ETOT =
−
mṙ +
,
2
2
r
mr
and
φ̇ =
L
,
mr 2
where ˙ now means d/dt rather than d/dτ . Solving for ṙ and dividing, we obtain:
L2
dr 2 m2 r 4 2ETOT 2GM
=
+
− 2 2 .
dφ
m
r
L2
m r
e) Using your knowledge of the difference between Newtonian physics and relativistic physics,
and by assuming that the motion takes place in a region where both |v|/c the ratio rs /r are small,
indicate how the resulting motion in this spacetime can be compared with motion in a Newtonian
gravitational potential, and find the implied correspondence between the parameter rs and the
mass causing the potential: if you have used c = 1, restore it in the final relation you obtain.
In the relativistic case, let us replace E by mc2 + ∆E and assume ∆E/mc2 ≪ 1. Then, we have:
∆E 2
L2
dr 2 m2 r 4 2∆E
rs c2
rs L 2
+ 2 2+
− 2 2+ 2 3 .
=
dφ
m
r
L2
m c
m r
m r
Now, the first term here corresponds almost directly to the first term in the Newtonian case, and
the second term here represents a small relativistic correction to it. The third term here can be
compared with the second term in the Newtonian case since they both have a 1/r dependance.
These terms are exactly equal if rs = GM/c2 . The fourth term here corresponds exactly to the
third term in the Newtonian case, and the fifth term here is a small relativistic correction to the
fourth term, since now rs /r ∼ 2(|v|/c)2, and both ratios are assumed small.
7