18.152 PSET #8
Jack Spilecki
Problem I.
First, assume that L and L are null-vectors normalized by m(L, L) = m(L, L) =
0 and m(L, L) = −2. Supplement L and L by vectors e(1) , e(2) , . . . , e(n−1) to form
a basis of R1+n , N , as in the hint. We then have:
1
2
T (L, L) = (∂α φLα ) ∂β φLβ − m(L, L)(m−1 )γδ ∂γ φ∂δ φ = (∂α φLα ) ≥ 0.
2
Similarly,
1
2
T (L, L) = (∂α φLα ) ∂β φLβ − m(L, L)(m−1 )γδ ∂γ φ∂δ φ = (∂α φLα ) ≥ 0.
2
Lastly,
1
T (L, L) = (∂α Lα ) ∂β Lβ − m(L, L)(m−1 )γδ ∂γ φ∂δ φ
2
β
= (∂α φLα ) ∂β φL + 2(m−1 )γδ ∂γ φ∂δ φ
1
1
mγδ ∂γ φ∂δ φ
= (∂α φLα ) ∂β φLβ − (∂γ φLγ )(∂δ φLδ ) − (∂δ φLδ )(∂γ φLγ ) + 2
2
=
mαβ ∂α φ∂β φ.
Then we have T (L, L) ≥ 0, since m is positive-definite on span(e(1) , . . . , e(n−1) ),
and vanishes on span(L, L).
Suppose T (L, L) = 0. Then ∂α φ, a vector, can only have nonzero components
in the L and L directions, in the basis N . If both T (L, L) and T (L, L) are zero,
then even these components are zero, implying φ is constant, which contradicts the
hypothesis |∇t,x φ| =
6 0. So, at least one of these terms, T (L, L), T (L, L) or T (L, L),
must be nonzero, hence, positive, since each is nonnegative.
For the general case of X, Y future-directed time-like vectors, we use the result
from Problem V of last week’s PSET, and write X = aL + bL and Y = cL + dL
for null-vectors L and L satisfying the above conditions, and a, b, c, d positive real
numbers. Then
T (X, Y ) = acT (L, L) + bdT (L, L) + (ad + bc)T (L, L) > 0,
since at least one of T (L, L), T (L, L) and T (L, L) is positive, and a, b, c, d are all
positive.
1
Problem II.
P
0
µ
(a) Firstly, K = 1 + µ (K )2 ≥ 1 > 0, so K is future-directed. Furthermore,
P3
j
if |x|2 := j=1 (xj )2 = 0, then, K = 0, for j = 1, 2, 3, and so K is time-like. If
|x|2 > 0, then
m(K, K) = 4t2 |x|2 − (1 + t2 + |x|2 )2
< 4(1 + t2 )|x|2 − (1 + t2 + |x|2 )2
= −(1 + t2 − |x|2 )2
< 0,
and so K is time-like in this case too.
0
j
(b) First note that K 0 = −K , and K j = K for j = 1, 2, 3. Then, if µ, ν ∈
{1, 2, 3},
∂µ K ν + ∂ν K µ = 2tδµν + 2tδνµ = 4tmµν .
If one of the coordinates is spatial, and the other time, then by symmetry of m it
suffices to show the result for µ ∈ {1, 2, 3}, and ν = 0. In this case, we have
∂µ K 0 + ∂0 K µ = −2xµ + 2xµ = 0 = 4tmµ0 .
Lastly, if µ = ν = 0, we have
∂0K 0 + ∂0K 0 = −2t − 2t = 4tm00 .
(c) First, note that mαβ (m−1 )βγ = δβγ , i.e. 1 if β = γ, and 0 otherwise.
This is because mαβ (m−1 )βγ = (mm−1 )αγ = Iαγ . Furthermore, mαβ (m−1 )αβ =
P
α,β Iαβ = 4. Then we have, using these identities, and symmetry in the indices
of m and m−1 , to contract over repeated indices:
1
mµν T µν = mµν (m−1 )µα (m−1 )νβ ∂α φ∂β φ − mαβ (m−1 )γδ ∂γ φ∂δ φ
2
1
−1 αβ
−1 γδ
= (m )
∂α φ∂β φ − mαβ (m ) ∂γ φ∂δ φ
2
= (m−1 )αβ ∂α φ∂β φ − 2(m−1 )γδ ∂γ φ∂δ φ
= −(m−1 )αβ ∂α φ∂β φ.
In the last step we relabelled the summed-over indices γ and δ to α and β respectively, which is allowed since they are dummy indices.
(d) From Corollary 13.1.0.3 (replace J µ with −J µ ), and parts (b) and (c),
1
∂µ(K) J µ = − T αβ (δα K β + δβ K α ) = −2mαβ T αβ = 2(m−1 )αβ ∂α φ∂β φ.
2
2
(e) We have:
∂µ −2tφ(m−1 )µα ∂α φ + φ2 (m−1 )µα ∂α t = −2δµ tφ(m−1 )µα ∂α φ − 2t∂µ φ∂α φ(m−1 )µα
+ 2φ∂µ φ(m−1 )µα ∂α t
= −2t∂µ φ∂α (m−1 )µα .
Relabel µ to β and use symmetry to swap α and β. Then, by adding this expression
(K)
to the expression for ∂µ J µ found in (d), we find:
∂µ Jfµ = 0.
(f.1) From the expansions given (specifically 0.0.11), we have
(
)
i
h
i
h
i
1 h
2
2
2
2
(K) 0
1 + (r + t) T (L, L) + 1 + (r − t) T (L, L) + 2 1 + t + r T (L, L) ,
J =
4
so it suffices to find T (L, L), T (L, L) and T (L, L).
P3
Since L = (1, xr ), we have ∂µ φLµ = ∂t φ + i=1 xri ∂i φ = ∂t φ + r̂ · ∇φ = ∂t φ + ∂r φ.
Since m(L, L) = 0, we then have:
1
T (L, L) = (∂α Lα ) ∂β Lβ − m(L, L)(m−1 )γδ ∂γ φ∂δ φ
2
= (∂t φ + ∂r φ)2
= (∇L φ)2
Similarly, L = (1, − xr ), so ∂µ φLµ = ∂t φ − ∂r φ, and since m(L, L) = 0 as well, we
have:
T (L, L) = (∂t φ − ∂r φ)2 = (∇L φ)2 .
Finally, we have (m−1 )αβ ∂α φ∂β φ = −∂α Lα ∂β Lβ + mαβ ∂α φ∂β φ, and so, using
m(L, L) = −2, we have:
1
Tαβ Lα Lβ = ∂α φLα ∂β φLβ − m(L, L)(m−1 )αβ ∂α φ∂β φ
2
= ∂α φLα ∂β φLβ + (m−1 )αβ ∂α φ∂β φ
=
mαβ ∂α φ∂β φ.
Plugging in the results for T (L, L), T (L, L) and T (L, L), we arrive at the desired
expression for
(K)
J 0.
(f.2) From (0.0.7), we have:
Je0 =(K) J 0 − 2tφ(m−1 )0α ∂α φ + φ2 (m−1 )0α ∂α t
=(K) J 0 + 2tφ∂t φ − φ2 .
Combining this with the result from (f.1) gives (0.0.12).
3
(g) Let BR (0) be the ball of radius R centred at the origin. Then, using integration by parts in spherical coordinates:
Z
Z
φ2 dx =
φ2 (∂r r)r2 sin(θ)drdθdϕ
BR (0)
BR (0)
1
=
3
Z
2
φ r sin(θ)dθdϕ −
3
∂BR (0)
2 3
Z
rφ∂r φdx.
BR (0)
Since φ is of compact support, the boundary integral vanishes for sufficiently large
R, and we get:
Z
φ2 dx =
−
R3
2
3
Z
rφ∂r φdx.
R3
(h) Define Q := [0, t] × BR (0), a region of spacetime in 1+3 dimensions. We
integrate Jeµ over Q and use the divergence theorem. We have
Z
N̂α Jeα [φ(σ)]dσ =
Z
∂µ Jeµ [φ(t, x)] dtdx = 0,
Q
∂Q
since ∂µ Jeµ = 0 from (e). Note that ∂Q consists of three components: L := {0} ×
BR (0) (oriented “backwards in time”), U := {t} × BR (0) (oriented “forwards in
time”), and S := (0, 1) × ∂BR (0). We may take R sufficiently large such that
φ(τ, x) = 0 for all (τ, x) ∈ S, by assumption. Then the only contributions to the
surface integral come from U and L. So, for sufficiently large R we have:
Z
Je0 [φ(σ)]dσ = 0,
U ∪(−L)
since N̂ is in the t direction.
Write Je0 = 41 E + 2tφ∂t φ − φ2 = 14 (E + 8φ2 ) + 2tφ∂t φ − 3φ2 , where E denotes
the expression contained in curly braces and where we have added and subtracted
2φ2 in the second equality. Then, for sufficiently large R, we may use the equality
from (g):
Z
1
2
2
(E + 8φ ) + 2tφ∂t φ − 3φ dx
{t}×BR (0) 4
Z
1
=
(E + 8φ2 ) + 2tφ∂t φ + 2rφ∂r φ dx.
{t}×BR (0) 4
Je0 [φ(σ)]dσ =
{t}×BR (0)
Z
But ∇L φ = ∂t φ + ∂r φ and ∇L φ = ∂t φ − ∂r φ, so we have:
2tφ∂t φ + 2rφ∂r φ = tφ(∇L φ + ∇L φ) + rφ(∇L φ − ∇L φ)
= φ(t + r)∇L φ + φ(t − r)∇L φ.
4
If we substitute this in to our previous integral, and evaluate at U and −L, taking
R → ∞, we get the desired equality:
Z 1
2
(E(t) + 8φ ) + φ(t + r)∇L φ + φ(t − r)∇L φ dx
R3 4
Z 1
2
=
(E(0) + 8φ ) + φr∇L φ − φr∇L φ dx.
R3 4
To show this is really the desired equality, we’ll expand out the first integrand and
show it matches the right hand side of (0.0.14) (the second follows from the first
upon substituting t = 0):
1
(E(t) + 8φ2 ) + φ(t + r)∇L φ + φ(t − r)∇L φ
4
(
i
i
i
h
h
1 h
=
1 + (r + t)2 (∇L φ)2 + 1 + (r − t)2 (∇L )2 + 2 1 + t2 + r2 mαβ ∂α φ∂β φ
4
)
+ 8φ2 + 4φ(t + r)∇L φ + 4φ(t − r)∇L φ
)
(
h
i
2
1
2
αβ
2
2
2
2
m ∂α φ∂β φ .
=
(∇L φ) + [(r + t)∇L φ + 2φ] + (∇L φ) + (t − r)∇L φ + 2φ + 2 1 + t + r 4
5