PACS numbers:
I.
PROBLEM 13 DISCUSSION
Problem 13 considered a transformation in 2d flat space:
λ
t(λ) = a sinh( )
a
λ
x(λ) = a cosh( )
a
ds2 = −dt2 + dx2
dxa
= (cosh(λ/a), sinh(λ/a))
u=
dλ
(I.1)
(I.2)
(I.3)
(I.4)
from this we can calculate the acceleration according to the equation previously derived
dua
Dua
=
+ Γabc ub uc
Dλ
dλ
1
= (sinh(λ/a), cosh(λ/a))
a
1
1
2
a b
a = a a gab = 2 (− sinh2 + cosh2 ) = 2
a
a
(I.5)
(I.6)
(I.7)
Note that we have a path that has constant acceleration but it is NOT a parabola – it’s a hyperbola.
II.
SCALAR FIELD ACTION PROBLEM DISCUSSION
The action of a scalar field φ is
1
S=
2
Z
√
g g ab ∇a φ ∇b φ d4 x
We obtain the equations of motion by varying w.r.t φ:
Z
√ ab
δS =
g g ∇a φ ∇b (δφ) d4 x
Z
√ ab
gg ∂a φ ∂b (δφ) d4 x
=
Z
Z
√
√
= − ∂b ( g g ab ∇a φ)δφ d4 x + ∂b ( g g ab ∇a φ δφ) d4 x
(II.1)
(II.2)
(II.3)
(II.4)
ignoring the boundary term for now, we have
δS = 0 = −
Z
∂b √
√
√ g(g ab ∇a φ)δφ gd4 x
g
⇒ ∇b g ab ∇a φ = 0
(II.5)
(II.6)
which are the equations of motion for a scalar field. The boundary term is generally dealt with by the initial conditions.
2
III.
CONTINUATION OF CRISTOFFEL SYMBOLS
Recall from last time the definition of the Cristoffel symbols, we’ll derive the relation shown in class in more detail.
1
(III.1)
Γa bc = g ad (gdc,b + gbd,c − gbc,d )
2
1
Γa ac = g ad (gdc,a + gad,c − gac,d )
(III.2)
2
1
(III.3)
= g ad ∂c gad
2
1
=
∂c g
(III.4)
2g
1
= ∂c ln g
(III.5)
2
√
(III.6)
= ∂c ln g
IV.
SYMMETRIES/ISOMETRIES
Take, for example, flat space, where ds2 = dx2 + dy 2 . Rotation in x-y space leaves the space unchanged – the metric
is form invariant.
As another example, consider 2-d Minkovski space, ds2 = −dt2 + dx2 . Now let
x → x cosh φ + t sinh φ
t → t cosh φ + x sinh φ
(IV.1)
(IV.2)
x = ρ cosh τ, dx = dρ cosh τ + ρ sinh τ dτ
y = ρ sinh τ, dy = dρ sinh τ + ρ cosh τ dτ
(IV.3)
(IV.4)
we can easily see that the metric remains unchanged. This looks a bit like a rotation in x-t space. To clarify this,
let’s attempt the coordinate change
Using this, we arrive at ds2 = dρ2 − ρ2 dτ 2 . This looks like polar coordinates, and we see that τ → τ + δτ leaves the
distance unchanged. This change of coordinates only maps x > t. However, we can be properly clever and adjsut the
map (e.g. by allowing complex τ ) to join it with x and −x. We further observe that the proper time is τρ . Lines of
constant ρ are hyperbolae, and lines of constant τ are the t = 0 lines for Lorentz-boosted frames.
V.
VECTORS AND DIFFERENTIALS
We’ve shown that we can use a differential to describe a tangent vector space
df
∂xa ∂f
=
dx
∂λ ∂xa
∂xa ∂
d
=
dλ
∂λ ∂xa
In 2-space, let’s consider the vectorial object
d
dθ .
(V.1)
(V.2)
In polar coordinates, this is
d
∂
∂
= (0 ·
+1·
)
dθ
∂r
∂θ
(
VI.
d a
) = (0, 1)
dθ
(V.3)
(V.4)
LOCAL INFINITESIMAL TRANSFORMATIONS
′
Let x′a = xa + ǫξ a (xb ). Then what is gab
(xa )? Consider the constant ds2 ,
ds2 = gab dxa dxb
=
′
gab
dx
′a
dx
′b
(VI.1)
(VI.2)
3
We expand this in the coordinate transformation above,
′
ds2 = gab
(dxa + ǫ∂c ξ a dxc )(dxb + ǫ∂d ξ b dxd )
2
ds =
′
gab
(x′ )(dxa
b
dx ) +
′
ǫ(gcb
b
c
a
dx ∂a ξ dx +
(VI.3)
′
gac
′
gab
(x′ ) = gab − ǫ(gcb ∂a ξ c + gac ∂b ξ c )
′
gab
(x′ ) +
∂gab c
∂gab
′
δx = gab
(xc ) c ǫξ c
∂xc
∂x
∴
′
gab
(xc )
c
a
c
b
dx ∂b ξ dx )
(VI.4)
(VI.5)
(VI.6)
(VI.7)
(VI.8)
c
c
c
= gab (x ) − ǫ [gcb ∂a ξ + gac ∂b ξ + ξ ∂c gab ]
(VI.9)
(VI.10)
and for the metric to be invariant under the transformation, that is for there to exist a symmetry, the quantity in the
brackets must be 0.