Solution to the exam in

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Solution to the exam in

FY3452 GRAVITATION OG COSMOLOGY

Friday June 3rd, 2011 This solution consists of 7 pages.

Problem 1. Lorentz transformations

r

= (x, 0, 0)

R θ

(t) = (0, 0, Vt)

x

z

An object is moving with constant velocityV along thez-axis, i.e. along the curve

r(t) =V tˆez. (1)

It is observed by detectors at rest at the positionR=xˆex. You may assumexto be non-negative. You may also choose to use units where the speed of lightc= 1.

First assume the object is emitting light (photons) in all directions. Seen from a coordinate system where the object is a rest this light is monochromatic, i.e. all photons have the same energy~ω0.

a) A photon emitted from the object is observed at positionRat timet. At which timet0 was it emitted?

Verify your solution by checking the special cases (i)x= 0, and (ii)t0= 0.

At timet0the object was at positionV t0eˆz, at a distance`=p

V²t²₀+x² fromR. I.e., the photon needs a timet−t0=`/cto travel from emission to detection,

c²(t−t0)²=V²t²₀+x². (2) Solving fort0 gives

t0= 1 1−(V /c)²

t−1

c

p(V t)²+ [1−(V /c)²]x²

. (3)

(i) Forx= 0 the photon needs the timet−t0 to travel a distanceV t0. It follows that we must havet₀=t/[1 + (V /c)], which agrees with (3) since [1−(V /c)]/

1−(V /c)²

= 1/[1 + (V /c)].

(ii) Fort₀ = 0 the photon was emitted from the origin (z= 0). I.e., the detection must occur at timet=x/c. Inserting this value fortinto (3) indeed gives t₀= 0.

Solution FY3452 Gravitation og cosmology, 03.06.2011 Page 2 of 7 b) From which position on thez-axis was the photon emitted? In which directionnˆ= cosθeˆz+ sinθeˆxis the photon observed to move? Express this by finding the quantity cotθ, withθas given in the figure above.

The emission occured from positionz=V t0, witht0 given by (3). The photon is travelling from (0,0, z) to (x,0,0), i.e. at an angleθsuch that

cotθ=−z

x =−V t0

x = (V /x) 1−(V /c)²

1 c

p(V t)²+ [ 1−(V /c)²]x²−t

. (4)

c) What is the egenvelocityu^0νof the object in a coordinate system where it is at rest? What is the eigenvelocity u^µof the object in our coordinate system (where the detectors are at rest)?

The egenvelocity is defined byu^µ= _dτ^dx^µ= _dτ^dt _dt^d (ct, x(t), y(t), z(t)), with the eigentimeτ choosen so that u^µu_µ=c². Hence we find

u^0ν = (c,0,0,0), (5)

and

u^µ= 1

p1−(V /c)²(c,0,0, V). (6)

d) Which energy~ωis the photon observed to have? Express your answer by~ω0, the angleθ, andV.

In the rest system of the object the photon is specified by the four-momentum p^0ν = ω₀

c (1,sinθ⁰cosφ⁰,sinθ⁰sinφ⁰,cosθ⁰), (7) withω0 independent of direction. In our system it is specified by the four-momentum

p^µ=ω

c (1,sinθcosφ,sinθsinφ,cosθ), (8) withω=ω(θ, φ) depending on the direction of the photon.

One simple method to find the connection is to use the invariance of scalar products, implying that u⁰_νp^0ν =u_µp^µ,

u⁰_νp^0ν =ω₀=u_µp^µ =1−(V /c) cosθ p1−(V /c)² ω.

I.e.,

~ω=

p1−(V /c)²

1−(V /c) cosθ ~ω0. (9)

By writingV =ctanhη we find that the photon is blueshifted by a factor e^η forθ= 0 (early detection times), redshifted by a factor e^−η for θ=π(late detection times), and unshifted when cosθ= tanhη/2.

A seemingly different method is to use the transformation formula between four-vectors in the two frames (withV =ctanhη),







 p⁰⁰ p^0x p^0y p^0z









=









coshη 0 0 −sinhη

0 1 0 0

0 0 1 0

−sinhη 0 0 coshη















 p⁰ p^x p^y p^z









. (10)

This gives

p⁰⁰= coshη p⁰−sinhη p^z= coshη(1−tanhη cosθ)p⁰, which is equivalent to (9).

Solution FY3452 Gravitation og cosmology, 03.06.2011 Page 3 of 7

Now assume instead that the object is a point chargeQ, so that it is surrounded by a rotation symmetric electric field in the coordinate system where it is at rest (at the origin)

E⁰(t⁰,x⁰) = Qx⁰

4πε0|x⁰|³, B⁰(t⁰,x⁰) =0. (11) This may also be expessed by the four-potential

A^0µ(t⁰,x⁰) = Q

4πε0|x⁰| (1,0). (12)

e) Express the quantity|x⁰|in our coordinates (t,x) (where the detectors are at rest).

Hint: Some general transformation formulae is included at the end of the problemset.

The two coordinate systems are related by the Lorentz transformation (with tanhη=V /c),







 ct⁰

x⁰ y⁰ z⁰

















coshη 0 0 −sinhη

0 1 0 0

0 0 1 0

−sinhη 0 0 coshη















 ct

x y z







 ,

so that

|x⁰|=p

x⁰²+y⁰²+z⁰²=p

x²+y²+ (coshη x−sinhη ct)²

= q

x²+y²+ cosh²η(z−V t)², (13)

where coshη= 1/p

1−(V /c)² (this quantity is commonly calledγ).

f ) Calculate the fieldE(t, x,0,0) observed by the detectors at positionR.

You may choose to use the transformation formula for the electromagnetic field tensor, or assume that the four-potential transform as a vector, and next compute theE-field from the transformed potentialA^ν(t,x).

The transformation formula for the electromagnetic field tensor reads

F^µν(x) = Λ^µ_αΛ^ν_βF^0αβ(x⁰), (14) here with Λ^µ_ν the inverse of the transformation matrix used in (10). We find

E^x(t,x) =F^0x(x) = Λ⁰₀Λ^x_xF^00x(x⁰) = coshη F^00x(x⁰) = Qcoshη x 4πε₀|x⁰|³,

since Λ^x_x= 1 is the only matrix element with an upper index x, and x⁰=x. For the same reasons,

E^y(t,x) =F^0y(x) = Λ⁰₀Λ^y_y F^00y(x⁰) = coshη F^00y(x⁰) =Qcoshη y 4πε₀|x⁰|³. Finally

E^z(t,x) =F^0z(x) = Λ⁰₀Λ^z_z−Λ⁰_zΛ^z₀

F^00z(x⁰) =F^00z(x⁰) =Qcoshη(z−V t) 4πε0|x⁰|³ . Here we have used thatF^0x0=−F^00x, that Λ⁰₀Λ^z_z−Λ⁰_zΛ^z₀

= cosh²η−sinh²η= 1, and finally thatz⁰= coshηz−sinhη ct= coshη(z−V t).

Alternatively we could first transform the four-potential according to the formula

A^µ(x) = Λ^µ_νA^0ν(x⁰), (15)

with Λ^µ_ν the inverse of the transformation matrix used in (10). This gives A^µ(x) = Q(coshη,0,0,sinhη)

4πε₀|x⁰| , (16)

Solution FY3452 Gravitation og cosmology, 03.06.2011 Page 4 of 7 from which we find, using |x⁰|=

q

x²+y²+ cosh²η(z−V t)², E^x=−

∂A^x c ∂t +∂A⁰

∂x

= Qcoshη x 4πε0|x⁰|³, E^y=−

∂A^y c ∂t +∂A⁰

∂y

=Qcoshη y 4πε0|x⁰|³, E^z=−

∂A^z c ∂t +∂A⁰

∂z

= Qcoshη(z−V t) 4πε0|x⁰|³ .

The computation ofE^x andE^y is very simple since A^x=A^y = 0. ForE^z the numerator combine as

−sinhη cosh²η tanhη+ coshη cosh²η

(z−V t) = coshη(z−V t), where we have used thatV /c= tanhη.

By either method we find

E^x(t, x,0,0) = Qxcoshη 4πε0

h

x²+ (coshη V t)²i^3/2,

E^y(t, x,0,0) = 0, (17)

E^z(t, x, ,0,0) = −QV tcoshη 4πε0

h

x²+ (coshη V t)²i^3/2.

g) Compare the observed direction of the field, cotϑ≡E^z/E^x, with the observed direction cotθof the photons in pointb).

We find cotϑ=−V t/xcompared to cotθ=−V t0/x. I.e., the electric field points away from thecurrent (in our coordinates) position of the object, while the photon momentum points away from the position of the object at the time of emission.

Problem 2. The Friedmann-Lemaˆıtre-Robertson-Walker universe

The Friedmann-Lemaˆıtre-Robertson-Walker metric is defined by the line element ds²=−dt²+a(t)²

dr²

1−kr² +r² dθ²+ sin²θdφ²

, (18)

where we units where the speed of lightc= 1, and wherek∈ {−1,0,1}.

a) Write down (i) the metric tensorgµν, and (ii) the inverse metric tensorg^µν for the universe defined by the line element (18). Assume that the metric tensor has signature (−,+,+,+).

We compare equation (18) with the general expression ds²=g_µνdx^µdx^ν to find

gµν =









−1 0 0 0

0 _1−kr^a(t)22 0 0

0 0 a(t)²r² 0

0 0 0 a(t)²r²sin²θ









. (19)

The inverse metric becomes

g^µν =











−1 0 0 0

0 ^1−kr_a(t)2² 0 0 0 0 _a(t)¹2r² 0 0 0 0 _a(t)2r¹²sin²θ











. (20)

Solution FY3452 Gravitation og cosmology, 03.06.2011 Page 5 of 7 b) Calculate the integration measure√

−gfor the universe defined by the line element (18).

Computing the determinant of (19) gives

√−g= a(t)³r²

√1−kr² sinθ. (21) c) Explain in qualitative terms what is meant by (i) covariant derivative, (ii) connection coefficients, (iii) Riemann tensor, (iv) Ricci tensor, (v) Einstein tensor, and (vi) scalar curvature. Explain briefly how you would compute these quantities from the line element (18). You need not perform any explicit computations here, but you should indicate the index structure of the quantities and relations involved.

(i) The covariant derivativeD_µ=∂_µ+Γ_µ is the correct differentiation operator on vectors (and tensors), taking into account that components of such quantities are with respect

to a basis which may be changing with position.

(ii) The components of the matricesΓµ makes up the connection coeffiients Γ^α_βµ. For a metric connection they can f.i. be found from the geodesic equations following from the metric gµν, by applying Hamiltons principle to the actionS=−¹₂R

dτ gµνx˙^µx˙^ν and comparing with the general form

¨

x^α+ Γ^α_βµx˙^βx˙^µ= 0. (22) Or they may be computed from the general expression

Γ^α_βµ=1

2g^αγ(gγβ,µ+gγµ,β−gβµ,γ). (23) (iii) The Riemann tensor con be defined as the commutator [D_µ, D_ν] ≡ R_µν. For each µν-combination this is a matrix with componentsR^α_βµν. In four space-time dimensions there are alltogether 4⁴= 256 components, out of which 20 are algebraically independent.

(iv) The Ricci tensor is obtained contraction of the Riemann tensor. Usually

Rβν =R^α_βαν, (24)

but sometimes with the opposite sign.

(v) The Einstein tensor is defined by

G_µν =R_µν−1

2g_µνg^αβR_αβ. (25)

It is distinguished by automatically being covariantly conserved,

T^µν_;ν= 0. (26)

(vi) The scalar curvature is defined as the trace of the Ricci tensor,

R=g^αβRαβ. (27)

As a scalar it is a natural candidate for a Lagrangian for the gravity field.

d) Assume that the matter content of this universe can be modelled by a scalar fieldϕand the associated action Smatter=−

Z d⁴x√

−g 1

2g^µν∂µϕ ∂νϕ+V(ϕ)

, (28)

whereV is some differentiable function of its argument. Assume that the field ϕonly depends on time, ϕ=ϕ(t).

Simplify the actionSmatterfor this case, and find the Euler-Lagrange equation for the fieldϕ.

Solution FY3452 Gravitation og cosmology, 03.06.2011 Page 6 of 7 The action can be written as,

Smatter= Z

dt a³ 1

2ϕ˙²−V(ϕ) Z

V

r²drsinθdθdφ

√

1−kr² , (29)

where the intergral over spatial coordinates may be treated as a constantV.

The Euler-Lagrange equations becomes d dt a³ϕ˙

=−a³V⁰(ϕ), which simplifies to

¨ ϕ+3 ˙a

a ϕ˙+V⁰(ϕ) = 0. (30)

I.e., the expansion of the universe induces a “friction force” in the equations of motion.

e) The Hilbert-Einstein action for the gravity part of the action is SHE= 1

16πGN

Z d⁴x√

−g R, (31)

whereGN is Newtons constant of gravity, andRis the scalar curvature. For the length element (18) the scalar curvature is

R= 6

a² a¨a+ ˙a²+k

. (32)

Simplify the actionSHE to the case that all dynamic quantities only depend ont(like you did forSmatterin the previous point), and use Hamiltons principle for the total action,Stotal=SHE+Smatter, to find the equation of motion forϕ.

We can write

SHE= 3 8π GN

Z

dt a²¨a+aa˙²+ka Z

V

r²drsinθdθdφ

√

1−kr² . (33)

We may perform a partial integration in the first term to bring this into a more familiar form involving at most first order derivatives,

S¯HE= 3 8π G_N

Z

dt −aa˙²+ka Z

V

r²drsinθdθdφ

√1−kr² ,

but it is also possible to apply Hamiltons principle to the original action. We obtain a total action

S¯total= Z

dt

−aa˙²+ka

+8πGN

3 a³ 1

2ϕ˙²−V(ϕ) 3V 8πGN

, (34)

form which we find the Euler-Lagrange equation fora, d

dt(−2aa) =˙

−˙a²+k

+ 8πGN

1

2ϕ˙²−V(ϕ)

, or simplified as

2¨a a +a˙²

a² + k a²

=−8πGN

1

2ϕ˙²−V(ϕ)

. (35)

f ) The action S_totalis invariant under time translations,ϕ(t)→ϕ(t+ε) anda(t)→a(t+ε). Use N¨others theorem to find the corresponding conserved quantity.

In this case N¨others theorem says that the conserved quantity is E= ∂L

∂a˙δa+∂L

∂ϕ˙δϕ−L,

Solution FY3452 Gravitation og cosmology, 03.06.2011 Page 7 of 7 withδa= ˙a,δϕ= ˙ϕ, andLgiven by the integrand in (34), except that we will ignore the constant 3V/(8πGN). With∂L/∂a˙ =−2aa˙ and∂L/∂ϕ˙ =a³ϕ˙ we find

E=− aa˙²+ka

+8πGN

3 a³ 1

2ϕ˙²+V(ϕ)

. (36)

Remark: The analysis above do not lead to the complete set of Einstein equations. A full analysis gives the additional constraint thatE= 0, equivalent to the condition

˙ a² a² + k

a² = 8π 3 GN

1

2ϕ˙²+V(φ)

, (37)

which is known as thefirst Friedmann equation.