Solution Manual For Relativity and Cosmology 1th Edition By From First Principles to Interpretations

“Relativity and Cosmology From First Principles to Interpretations” Balša Terzić Chapter 2 1. 𝑆 and 𝑆′ reference frames which are moving relative to one another with a velocity 𝑉 in the 𝑥-direction, measure the velocities of an object, also moving along the 𝑥-direction as 𝑣 and 𝑣′, respectively. Use the Lorentz transformations to derive the relationship between the velocities 𝑣 and 𝑣′.

One way to look at this is a composition of velocities: Imagine another frame 𝑆′′ which moves with a velocity (𝑣!" , 0, 0), with respect to frame 𝑆 " . With which velocity (𝑣! , 0, 0) does the frame 𝑆′′ move with respect to frame 𝑆?

From 𝑺 to 𝑺′: Using the general Lorentz transformation in Eqns. (2.20)-(2.21) in the textbook: 𝐪" = L#$ 𝐪 𝛾 −𝛾𝛽 0 0 −𝛾𝛽 𝛾 0 0 L#$ = 1 6 0 0 1 0 0 0 0 1 where 𝛽 = 𝑉/𝑐, 𝛾 = 1/91 − 𝑉 $ /𝑐 $ , derived from the relative motion between the frames 𝑆 and 𝑆′, which is 𝑉 along the 𝑥-axis.

From 𝑺′ to 𝑺′′: Using the general Lorentz transformation in Eqns. (2.20)-(2.21) in the textbook: 𝐪"" = L$% 𝐪′ 𝛾′ L$% = 1−𝛾′𝛽′ 0 0

−𝛾′𝛽′ 𝛾′ 0 0

0 0 0 06 1 0 0 1

where 𝛽′ = 𝑣!" /𝑐, 𝛾′ = 1/:1 − 𝑣!" $ /𝑐 $ , derived from the relative motion between the frames 𝑆′ and 𝑆′′, which is 𝑣!" along the 𝑥-axis.

From 𝑺 to 𝑺′′: Therefore,

𝐪"" = L$% 𝐪" = L$% (L#$ 𝐪) = (L$% L#$ )𝐪

(1 + 𝛽𝛽 " )𝛾𝛾′ −(𝛽 + 𝛽 " )𝛾𝛾′ 0 0 𝛾 "" −𝛾 "" 𝛽 "" 0 0 "" "" " " 𝛾 "" 0 06 L$% L#$ = 1−(𝛽 + 𝛽 )𝛾𝛾′ (1 + 𝛽𝛽 )𝛾𝛾′ 0 0 6 ≡ 1−𝛾 𝛽 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 $ $ where 𝛽′′ = 𝑣! /𝑐, 𝛾′′ = 1/91 − 𝑣! /𝑐 , derived from the relative motion between the frames 𝑆′′ and 𝑆, which is 𝑣! along the 𝑥-axis. The matrix L$% L#$ above is itself a Lorentz transformation matrix, hence the form on the far right.

Equating terms in the two matrices in the equation above yields (1 + 𝛽𝛽 " )𝛾𝛾 " = 𝛾 "" This is one equation for the one unknown (𝛾 "" ), from which we can compute the velocity 𝑣! (using 𝛾′′ = 1/91 − 𝛽 ""$ = 1/91 − 𝑣!$ /𝑐 $ ) between frames 𝑆′′ and 𝑆, which is what the problem asks. From here on out, it’s all algebra.

Use 𝛾′ = 1/91 − 𝛽 "$ , 𝛾′′ = 1/91 − 𝛽 ""$ and 𝛾 = 1/91 − 𝛽 $ : (1 + 𝛽𝛽 " ) 1 = 91 − 𝛽 $ 91 − 𝛽′$ 91 − 𝛽 ""$ 91 − 𝛽 $ 91 − 𝛽′$ (1 + 𝛽𝛽 " ) = square both sides 91 − 𝛽 ""$ (1 − 𝛽 $ )I1 − 𝛽 " $ J " $ (1 + 𝛽𝛽 ) = 1 − 𝛽 ""$ Solve for 𝛽′′: (1 − 𝛽 $ )I1 − 𝛽 " $ J ""$ 1−𝛽 = (1 + 𝛽𝛽 " )$ $ (1 − 𝛽 $ )I1 − 𝛽 " J (1 + 𝛽𝛽 " )$ − (1 − 𝛽 $ )I1 − 𝛽 " $ J ""$ 𝛽 =1− = (1 + 𝛽𝛽 " )$ (1 + 𝛽𝛽 " )$ (𝛽 + 𝛽 " )$ 1 + 2𝛽𝛽 " + 𝛽 $ 𝛽 "$ − 1 + 𝛽 "$ + 𝛽 $ − 𝛽 $ 𝛽 "$ 2𝛽𝛽 " + 𝛽 "$ + 𝛽 $ 𝛽 ""$ = = = (1 + 𝛽𝛽 " )$ (1 + 𝛽𝛽 " )$ (1 + 𝛽𝛽 " )$ " 𝑉 𝑣! " " 𝛽 + 𝛽 𝑣 ! 𝑐 + 𝑐 → 𝑣 = 𝑉 + 𝑣! 𝛽 "" = → = ! 𝑉 𝑣" 𝑉𝑣 " 1 + 𝛽𝛽 " 𝑐 1 + 𝑐 𝑐!

1 + $! 𝑐 2. Flat Euclidean 2D space can be represented in terms of Cartesian coordinates (𝑥, 𝑦), as well as in terms of polar coordinates (𝑟, 𝜃): 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃 a. Write down the line element in Cartesian coordinates and the metric 𝑔.

1 0 𝑑𝑠 $ = 𝑑𝑥 $ + 𝑑𝑦 $ 𝑔=U V 0 1 b. Write down the line element in polar coordinates and the metric 𝑔.

1 0 𝑑𝑠 $ = 𝑑𝑟 $ + 𝑟 $ 𝑑𝜃 $ 𝑔=U V 0 𝑟$

3. Consider 3D spherical coordinates. a. Write down the line element and the Euclidean metric 𝑔 1 0 0 $ $ $ $ $ $ $ $ 𝑑𝑠 = 𝑑𝑟 + 𝑟 𝑑𝜃 + 𝑟 sin 𝜃 𝑑𝜙 𝑔&' = X0 𝑟 Y 0 0 0 𝑟 $ sin$ 𝜃 b. Given the contravariant vector 𝑥 & = (1, 𝑟, 0)( , find 𝑥& . 𝑥& = 𝑔&' 𝑥 ' 𝑥) = 𝑔)) 𝑥 ) + 𝑔)# 𝑥 # + 𝑔)$ 𝑥 $ = 1 ⋅ 1 + 0 ⋅ 𝑟 + 0 ⋅ 0 = 1 𝑥# = 𝑔#) 𝑥 ) + 𝑔## 𝑥 # + 𝑔#$ 𝑥 $ = 0 ⋅ 1 + 𝑟 $ ⋅ 𝑟 + 0 ⋅ 0 = 𝑟 % 𝑥$ = 𝑔$) 𝑥 ) + 𝑔$# 𝑥 # + 𝑔$$ 𝑥 $ = 0 ⋅ 1 + 0 ⋅ 𝑟 + 𝑟 $ sin$ 𝜃 ⋅ 0 = 0 𝑥& = (1, 𝑟 % , 0) c. Given the covariant vector 𝑦& = (1, −𝑟 $ , cos $ 𝜃), find 𝑦 & . 𝑦 & = 𝑔&' 𝑦' 1

0 1 ⎛0 &' 𝑔 =⎜ 𝑟$ 0 0 ⎝

0 0 1

⎞ ⎟

𝑟 $ sin$ 𝜃⎠

𝑦 ) = 𝑔)) 𝑦) + 𝑔)# 𝑦# + 𝑔)$ 𝑦$ = 1 ⋅ 1 + 0 ⋅ (−𝑟 $ ) + 0 ⋅ cos $ 𝜃 = 1 1 𝑦# = 𝑔#) 𝑦) + 𝑔## 𝑦# + 𝑔#$ 𝑦$ = 0 ⋅ 1 + $ ⋅ (−𝑟 $ ) + 0 ⋅ cos $ 𝜃 = −1 𝑟 1 𝑦 $ = 𝑔$) 𝑦) + 𝑔$# 𝑦# + 𝑔$$ 𝑦$ = 0 ⋅ 1 + 0 ⋅ (−𝑟 $ ) + $ $ ⋅ cos $ 𝜃 𝑟 sin 𝜃 1 = $ cot $ 𝜃 𝑟 ( 1 $ 𝑦 = a1, −1, $ cot 𝜃b 𝑟 &

4. Suppose that a particle of mass 𝑚# and 3-velocity 𝒗# collides with another particle at rest with mass 𝑚$ and is absorbed by it. a. Use conservation of 4-momentum to compute the 3-velocity 𝒗 of the composite particle. Recover Newtonian limit for 𝑣# ≪ 𝑐.

Without loss of generality, and to simplify our computations, we can align the velocity vector to be along the, say, 𝑥-axis: 𝒗# = (𝑣# , 0, 0) Before the collision: 𝐩# = 𝑚# 𝛾# (𝑐, 𝒗# ) = 𝑚# 𝛾# (𝑐, 𝑣# , 0, 0)( Where 𝛾# = 1/91 − 𝑣#$ /𝑐 $

𝐩$ = (𝑚$ 𝑐, 0, 0, 0)(

After the collision: Aligning the incoming velocity vector to be along the, say, 𝑥-axis: 𝒗# = (𝑣# , 0, 0) also made the recoil velocity 𝒗$ be along the 𝑥-axis: 𝒗$ = (𝑣$ , 0, 0). " 𝐩#$ = 𝑚𝛾$ (𝑐, 𝒗$ ) = 𝑚𝛾$ (𝑐, 𝑣$ , 0,0) (1D momentum conservation: no transverse momenta) Conservation of 4-momentum: " 𝐩# + 𝐩$ = 𝐩#$ (𝑚# 𝛾# 𝑐, 𝑚# 𝛾# 𝑣# , 0, 0)( + (𝑚$ 𝑐, 0, 0, 0)( = (𝑚𝛾$ 𝑐, 𝑚𝛾$ 𝑣$ , 0, 0)( 0-term equation: 𝑚# 𝛾# 𝑐 + 𝑚$ 𝑐 = 𝑚𝛾$ 𝑐 → 𝑚# 𝛾# + 𝑚$ = 𝑚𝛾$ 1-term equation: 𝑚# 𝛾# 𝑣# = 𝑚𝛾$ 𝑣$ These are 2 equations for 2 unknowns: 𝑚 and 𝑣$ (recall 𝛾$ = 1/91 − 𝑣$$ /𝑐 $ ) Solve the 2nd equation for 𝑚𝛾$ = 𝑚# 𝛾# 𝑣# /𝑣$ and plug it into the first: 𝑚# 𝛾# 𝑣# 𝑚# 𝛾# + 𝑚$ = 𝑣$ Solve for 𝑣$ : 𝑚# 𝛾# 𝑣# 𝑣$ = 𝑚# 𝛾# + 𝑚$ b. From the result in part a. recover Newtonian limit for 𝑣# ≪ 𝑐.

In Newtonian mechanics, momentum conservation gives: 𝑚# 𝑚# 𝑣# = (𝑚# + 𝑚$ )𝑣$ → 𝑣$ = 𝑣 𝑚# + 𝑚$ # * The limit of the relativistic solution in part when 𝑣# ≪ 𝑐, or +! ≪ 1, or 𝛾 → 1: 𝑚# 𝛾# 𝑣# 𝑚# = 𝑣 .! →# 𝑚# 𝛾# + 𝑚$ 𝑚# + 𝑚$ #

lim 𝑣$ = lim 𝑣$ = lim

*! /+→)

.! →#

c. Use conservation of 4-momentum to compute the mass 𝑚 of the composite particle (Hint: 𝑚 ≠ 𝑚# + 𝑚$ . Conservation of 4-momentum demands that some of the 'energy of motion' be converted into rest mass energy) Substitute the solution from part a: 𝑚# 𝛾# 𝑣# 𝑣$ = 𝑚# 𝛾# + 𝑚$ into the second equation above: 𝑚# 𝛾# 𝑣# = 𝑚𝛾$ 𝑣$ to get 𝑚# 𝛾# 𝑣# 𝑚# 𝛾# 𝑣# = 𝑚𝛾$ 𝑚# 𝛾# + 𝑚$ Simplify to 𝑚𝛾$ 1= 𝑚# 𝛾# + 𝑚$ Solve for 𝑚 to get

𝑚=

𝑚# 𝛾# + 𝑚$ 𝛾$

ALTERNATIVE SOLUTION: Square both sides of

" 𝐩# + 𝐩$ = 𝐩#$ (𝑚# 𝛾# 𝑐, 𝑚# 𝛾# 𝑣# , 0, 0)( + (𝑚$ 𝑐, 0, 0, 0)( = (𝑚𝛾$ 𝑐, 𝑚𝛾$ 𝑣$ , 0, 0)( (𝐩# + 𝐩$ )𝟐 = 𝐩# ⋅ 𝐩# + 2𝐩# ⋅ 𝐩$ + 𝐩$ ⋅ 𝐩$ = −𝑚#$ 𝑐 $ −𝑚# 𝑚$ 𝛾# 𝑐 $ − 𝑚$$ 𝑐 $ " " 𝐩#$ ⋅ 𝐩#$ = −𝑚$ 𝑐 $ −𝑚$ 𝑐 $ = −𝑚#$ 𝑐 $ −𝑚# 𝑚$ 𝛾# 𝑐 $ − 𝑚$$ 𝑐 $ 𝑚$ = 𝑚#$ + 𝑚# 𝑚$ 𝛾# + 𝑚$$

𝑚 = :𝑚#$ + 𝑚# 𝑚$ 𝛾# + 𝑚$$ Hmmmmm… This is different from what we found with the first method…

But is it really? Let’s check.

First, substitute the solution for 𝑣$ found in part a: 𝑚# 𝛾# 𝑣# 𝑣$ = 𝑚# 𝛾# + 𝑚$ 𝑚# 𝛾# 𝑣# 𝛽$ = (𝑚# 𝛾# + 𝑚$ )𝑐 1 1 1 𝛾$ = = = 91 − 𝛽$$ 𝑚#$ 𝛾#$ 𝑣#$ (𝑚 𝛾 + 𝑚$ )𝑐 $ − 𝑚#$ 𝛾#$ 𝑣#$ k1 − k # # $ $ (𝑚# 𝛾# + 𝑚$ ) 𝑐 (𝑚# 𝛾# + 𝑚$ )$ 𝑐 $ 𝛾$ =

(𝑚# 𝛾# + 𝑚$ )𝑐 9(𝑚# 𝛾# + 𝑚$ )$ 𝑐 $ − 𝑚#$ 𝛾#$ 𝑣#$ 𝛾$ =

=

9(𝑚#$ 𝛾#$ + 2𝑚# 𝑚$ 𝛾# + 𝑚$$ )𝑐 $ − 𝑚#$ 𝛾#$ 𝑣#$ (𝑚# 𝛾# + 𝑚$ )𝑐

k𝑚#$ 𝛾#$ 𝑐 $ a1 − =

(𝑚# 𝛾# + 𝑚$ )𝑐

𝑣#$ b + 2𝑚# 𝑚$ 𝛾# 𝑐 $ + 𝑚$$ 𝑐 $ 𝑐$ (𝑚# 𝛾# + 𝑚$ )𝑐

9𝑚#$ 𝑐 $ + 2𝑚# 𝑚$ 𝛾# 𝑐 $ + 𝑚$$ 𝑐 $ Now substitute in the solution we found previously: 𝑚# 𝛾# + 𝑚$ 𝑚= 𝛾$ $ $ (𝑚# 𝛾# + 𝑚$ )9𝑚# 𝑐 + 2𝑚# 𝑚$ 𝛾# 𝑐 $ + 𝑚$$ 𝑐 $ 𝑚= = :𝑚#$ + 2𝑚# 𝑚$ 𝛾# + 𝑚$$ (𝑚# 𝛾# + 𝑚$ )𝑐 𝑚 = :𝑚#$ + 2𝑚# 𝑚$ 𝛾# + 𝑚$$ It checks: the “alternative” solution is the same as the one we found earlier (as it must!), only in disguise. This “alternative” solution is more insightful, since it

shows that the final mass is independent of the final velocity. The final velocity is a uniquely determined by the initial masses and the initial velocity. Therefore, upon substituting the final velocity in terms of these given quantities in our previous solution, that dependence “went away”.

5. A photon of wavelength 𝜆 travelling to the left along the 𝑥-axis hits a stationary electron of mass 𝑚0 and scatters off with a wavelength 𝜆" at an angle 𝜃 in (𝑥, 𝑧) plane (as measured in the frame of reference in which the electron was originally at rest). Use conservation of 4-momentum to compute the scattering wavelength 𝜆" .

Recall that a photon with frequency 𝜈 has a measured energy 𝐸 = ℎ𝜈 and spatial momentum of magnitude |𝐩| = ℎ𝑣/𝑐.

Hint: Start with the equation of 4-momentum conservation written in the form 𝐩0 + 𝐩1 = 𝐩"0 + 𝐩"1 and recall that 𝐩 ⋅ 𝐩 = −𝑚$ 𝑐 $ .

Before the collision: 𝐩0 = 𝑚0 (𝑐, 0, 0, 0)(

𝐩2 = a

( ℎ𝜈 ℎ𝜈 , − , 0,0b 𝑐 𝑐

After the collision: ( 𝐩"0 = I𝐸0 ′, 𝑝!" , 𝑝3" , 𝑝4" J

𝐩"1 = s

ℎ𝜈 " ℎ𝜈 " ℎ𝜈 " , cos 𝜃 , 0, sin 𝜃t 𝑐 𝑐 𝑐

(

𝐩0 + 𝐩1 = 𝐩"0 + 𝐩"1 𝐩"0 = 𝐩0 + 𝐩1 − 𝐩"1 Square both sides of the equation to get: 𝐩"0 ⋅ 𝐩"0 = (𝐩0 + 𝐩1 − 𝐩"1 )(𝐩0 + 𝐩1 − 𝐩"1 ) = 𝐩0 ⋅ 𝐩0 + 𝐩0 ⋅ 𝐩1 − 𝐩0 ⋅ 𝐩"1 + 𝐩1 ⋅ 𝐩0 + 𝐩1 ⋅ 𝐩1 − 𝐩1 ⋅ 𝐩"1 − 𝐩"1 ⋅ 𝐩0 − 𝐩"1 ⋅ 𝐩1 + 𝐩"1 ⋅ 𝐩"1 All of these dot products can be computed: 𝐩1 ⋅ 𝐩1 = 𝐩"1 ⋅ 𝐩"1 = 0 𝐩0 ⋅ 𝐩1 = −𝑚0 ℎ𝜈 𝐩0 ⋅ 𝐩"1 = −𝑚0 ℎ𝜈′ ℎ$ 𝜈𝜈 " ℎ$ 𝜈𝜈 " ℎ$ 𝜈𝜈 " 𝐩"1 ⋅ 𝐩1 = − $ − $ cos 𝜃 = − $ (1 + cos 𝜃) 𝑐 𝑐 𝑐 𝐩0 ⋅ 𝐩0 = −𝑚0$ 𝑐 $ 𝐩"0 ⋅ 𝐩"0 = −𝑚0$ 𝑐 $ Cancelling zero terms and regrouping the squared equation above 𝐩"0 ⋅ 𝐩"0 − 𝐩0 ⋅ 𝐩0 = 2(𝐩0 ⋅ 𝐩1 ) − 2(𝐩0 ⋅ 𝐩"1 ) − 2(𝐩1 ⋅ 𝐩"1 ) ℎ$ 𝜈𝜈 " 0 = 2(−𝑚0 ℎ𝜈) − 2(−𝑚0 ℎ𝜈′) − 2 s− $ (1 + cos 𝜃)t 𝑐

ℎ$ 𝜈𝜈 " 𝜈′ − 𝜈 = − (1 + cos 𝜃) 𝑚0 ℎ𝑐 $ ℎ𝜈𝜈 " 𝜈′ − 𝜈 = (1 + cos 𝜃) 𝑚0 𝑐 $ 𝜈′ − 𝜈 ℎ = (1 + cos 𝜃) " 𝜈𝜈 𝑚0 𝑐 $ 1 1 ℎ − = (1 + cos 𝜃) 𝜈′ 𝜈 𝑚0 𝑐 $ 𝑐 𝑐 ℎ − = (1 + cos 𝜃) 𝜈′ 𝜈 𝑚0 𝑐 ℎ 𝜆′ − 𝜆 = (1 + cos 𝜃) 𝑚0 𝑐 𝜆" = 𝜆 +

ℎ (1 + cos 𝜃) 𝑚0 𝑐

6. Find another frame of reference in Example 1 and compute what the observer observes for 𝐸.

How about a frame of reference in which the particle moves with the velocity 𝒗 in the −𝑥-direction and the observer is at rest and? Then, for the observer 𝐩 = (𝛾𝑚) 𝑐, −𝛾𝑚) 𝑣, 0,0)( , Then

𝐮 = (𝑐, 0,0,0)( .

𝐸 = −𝑝 ⋅ 𝑢 = −𝑔56 𝑝5 𝑢6 = −𝑔)) (𝛾𝑚) 𝑐)(𝑐) = 𝛾𝑚) 𝑐 $ .

This is the same results we found earlier. Therefore, the choice of reference frames is not important, as it should not be.

7. Consider Fig. 2.4. a. What does the path of a particle following the red path look like in space?

Particle does not move for 𝑡 = 4 m/𝑐. b. What does the path of a particle following the blue path look like in space? How fast is the particle moving and in which direction? What happens at the intermediate point C?

Particle moves to the right at 𝑣 = 𝑐/2 for 𝑡 = 2 m/𝑐, and then to the left at 𝑣 = 𝑐/2 for 𝑡 = 2 m/𝑐. C is the turning point in spacetime. c. What does the path of a particle following the purple path look like in space?

About how fast is the particle moving and in which direction?

Particle moves to the right at 𝑣 ≈ 𝑐 (just a little bit slower) for 𝑡 = 2 m/𝑐, and then to the left at 𝑣 ≈ 𝑐 for 𝑡 = 2 m/𝑐. d. Plot the path of a particle moving from the origin to the left at 𝑣 = 𝑐/3 for 𝑐𝑡 = 2 m and then stops for 𝑐𝑡 = 1 m and then to the right at 𝑣 = 2𝑐/3 for 𝑐𝑡 = 1 m. Compute the proper time 𝑑𝜏 measured by an experimenter traveling that path.

Plot of the path is shown below in thick, dashed red line.

Trip to the right at 𝑣 = 2𝑐/3: 𝑑𝑠 $ = −(𝑐𝑑𝑡)$ + 𝑑𝑥 $ = −4 + 3 = −1 → 𝑐𝑑𝜏 = 9−𝑠 $ = 1 Trip to the left at 𝑣 = 2𝑐/3: 𝑑𝑠 $ = −(𝑐𝑑𝑡)$ + 𝑑𝑥 $ = −4 + 3 = −1 → 𝑐𝑑𝜏 = 9−𝑠 $ = 1 Total: 𝑐𝑑𝜏 = 2 The higher the speed, the smaller the proper time 𝑐𝑑𝜏.

Chapter 4 1. Demonstrate explicitly that the quantity 𝐀 ⋅ 𝐁 = 𝐴5 𝐵5 transforms as a scalar, i.e. it is invariant under Lorentz transformation.

Using the Eqns. (4.7) and (4.5) from the textbook: 𝜕𝑥 7 𝜕𝑥 "5 8 𝐴"5 = "5 𝐴7 𝐵"5 = 𝐵 𝜕𝑥 𝜕𝑥 8 𝜕𝑥 7 𝜕𝑥 "5 8 𝜕𝑥 7 𝜕𝑥 "5 𝜕𝑥 7 8 𝐴"5 𝐵"5 = "5 𝐴7 𝐵 = 𝐴 𝐵 = 𝐴 𝐵8 = 𝛿78 𝐴7 𝐵8 = 𝐴8 𝐵8 𝜕𝑥 𝜕𝑥 8 𝜕𝑥 "5 𝜕𝑥 8 7 𝜕𝑥 8 7 = 𝐴5 𝐵5 = 𝑐𝑜𝑛𝑠𝑡.

2. Find the geodesic equation in cylindrical coordinates.

The coordinates are 𝑥 & = (𝑥 # , 𝑥 $ , 𝑥 % ) = (𝑟, 𝜙, 𝑧). The line element is 𝑑𝑠 $ = 𝑑𝑟 $ + 𝑟 $ 𝑑𝜙 $ + 𝑑𝑧 $ The metric is 1 0 0 1 0 0 # &' $ 𝑔&' = X0 𝑟 0Y and 𝑔 = „0 9 " 0…

0 0 1 0 0 1 One can do this problem in two ways: either use the definition of the geodesic equation from the textbook Eq. (4.87): 8 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : or solve the Lagrange equations (which is how we derived the geodesic equation).

1 𝐿 = 𝑔56 𝑥̇ 5 𝑥̇ 6 2 𝜕𝐿 𝑑 𝜕𝐿 − =0 5 𝜕𝑥 𝑑𝑠 𝜕𝑥̇ 5 Let’s do the second way.

1 1 1 1 𝐿 = 𝑔56 𝑥̇ 5 𝑥̇ 6 = 𝑟̇ $ + 𝑟 $ 𝜙̇ $ + 𝑧̇ $ 2 2 2 2 𝜕𝐿 𝑑 𝜕𝐿 − =0 𝜕𝑥 5 𝑑𝑠 𝜕𝑥̇ 5

𝑟-coordinate:

𝑑 (𝑟̇ ) = 0 → 𝑟𝜙̇ $ − 𝑟̈ = 0 → 𝑟̈ = 𝑟𝜙̇ $ 𝑑𝑠 We can extract, as we did in class and textbook, Christoffel symbols by comparing this differential equation with the definition of the geodesic equation from the textbook Eq. (4.87). Then 8 # ̇$ 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : and 𝜈 = 1 → 𝑟̈ = −Γ$$ 𝜙 𝑟̈ = 𝑟𝜙̇ $ # → Γ$$ = −𝑟 𝑟𝜙̇ $ −

𝜙-coordinate: 𝑑 2𝑟̇ 2𝑟̇ 0 − I𝑟 $ 𝜙̇J = 0 → − 2𝑟𝑟̇ 𝜙̇ − 𝑟 $ 𝜙̈ = 0 → 𝜙̈ + 𝜙̇ = 0 → 𝜙̈ = − 𝜙̇ 𝑑𝑠 𝑟 𝑟 Again, extracting Christoffel symbols: 8 $ $ ̇ $ $ 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : and 𝜈 = 2 → 𝜙̈ = −Γ#$ 𝑟̇ 𝜙̇ − Γ$# 𝜙𝑟̇ = −(Γ#$ + Γ$# )𝑟̇ 𝜙̇ 2 𝜙̈ = − 𝑟̇ 𝜙̇ 𝑟 $ $ → Γ#$ + Γ$# = #

2 𝑟

$ $ $ $ But Γ#$ = Γ$# , so Γ#$ = Γ$# = 9. 𝑧-coordinate: ; 0 − ;< (𝑧̇ ) = 0 → 𝑧̈ = 0 → 𝑧 = 𝑎𝑠 + 𝑏 (straight line in 𝑧) % All Christoffel symbols Γ78 are 0.

3. Consider polar coordinates 𝑑𝑠 $ = 𝑑𝑟 $ + 𝑟 $ 𝑑𝜃 $ . Find the covariant derivative 𝑉;55 of 𝑽 = 𝑟 $ cos 𝜃 𝒆𝒓 − sin 𝜃 𝒆𝜽 The summation over repeated indices is assumed here.

The coordinates are 𝑥 & = (𝑥 # , 𝑥 $ ) = (𝑟, 𝜃).

1 0 1 0 &' # t.

The metric is 𝑔&' = U $ V and 𝑔 = s0 0 𝑟 9" 𝑉;55 = 𝑉;## + 𝑉;$$ From the definition of the covariant derivative (textbook Eq. (4.69)): 5 8 𝑉;65 = 𝑉,65 + Γ68 𝑉 or, in our case, # 8 $ 8 𝑉;55 = (𝑉,## + Γ#8 𝑉 ) + (𝑉,$$ + Γ$8 𝑉 ) Also, 𝑽 = 𝑟 $ cos 𝜃 𝒆𝒓 − sin 𝜃 𝒆𝜽 = 𝑉 # 𝒆𝟏 + 𝑉 $ 𝒆𝟐 , so 𝑉 # = 𝑟 $ cos 𝜃 𝑉 $ = − sin 𝜃 It appears that we need to compute the Christoffel symbols for polar coordinates. You are welcome to do so using the definition of Christoffel symbols given in the textbook Eq. (4.68). However, we can use the results from the previous problem. Polar coordinates are cylindrical coordinates minus the 𝑧-coordinate! Aren’t you glad we computed the Christoffel symbols for the previous problem as a pedagogical by-product of our derivation? What a fortuitous coincidence J The only non-zero Christoffel symbols are: # Γ$$ = −𝑟

$ $ Γ#$ = Γ$# =

Then,

1 𝑟

# 8 $ 8 𝑉;55 = (𝑉,## + Γ#8 𝑉 ) + (𝑉,$$ + Γ$8 𝑉 ) # # # # $ $ $ # $ $ 5 𝑉;5 = (𝑉,# + Γ## 𝑉 + Γ#$ 𝑉 ) + (𝑉,$ + Γ$# 𝑉 + Γ$$ 𝑉 ) # # # # $ $ $ # $ $ 5 𝑉;5 = (𝑉,# + Γ## 𝑉 + Γ#$ 𝑉 ) + (𝑉,$ + Γ$# 𝑉 + Γ$$ 𝑉 ) terms in orange are 0 $ # 𝑉;55 = 𝑉,## + 𝑉,$$ + Γ$# 𝑉 𝜕 𝜕 1 𝑉;55 = (𝑟 $ cos 𝜃) + (−sin 𝜃) + (𝑟 $ cos 𝜃) 𝜕𝑟 𝜕𝜃 𝑟 𝑉;55 = 2𝑟 cos 𝜃 − cos 𝜃 + 𝑟 cos 𝜃 = (3𝑟 − 1) cos 𝜃

𝑉;55 = (3𝑟 − 1) cos 𝜃 4. Using the definition of the covariant derivative in Eq. (4.69) in the textbook, show that . . . 𝐴;56 − 𝐴;65 = −𝑅:56 𝐴: 5 5 5 8 5 8 5 Eq. (5.20): 𝑅6.: ≡ Γ6:,. − Γ.6,: + Γ6: Γ8. − Γ6. Γ8: 𝛼 → 𝛾, 𝛽 → 𝛿, 𝛾 → 𝛼, 𝛿→𝛽 . . .

8 .

8 . 𝑅:56 ≡ Γ:6,5 − Γ5:,6 + Γ:6 Γ85 − Γ:5 Γ86 8 Eq. (4.69): 𝐴5;6 ≡ 𝐴5,6 − Γ56 𝐴8 5 8 𝐴5;6 ≡ 𝐴5,6 + Γ68 𝐴 . . . : 𝐴;5 ≡ 𝐴,5 + Γ5: 𝐴 . . . 𝐴;6 ≡ 𝐴,6 + Γ6: 𝐴: . . . . .

7 7 𝐴;56 = (𝐴;5 );6 = (𝐴,5 + Γ5: 𝐴: ),6 + Γ67 (𝐴,5 + Γ5: 𝐴: ) .

.

.

.

7

.

7

= 𝐴,56 + Γ5:,6 𝐴: + Γ5: 𝐴:,6 + Γ67 𝐴,5 + Γ67 Γ5: 𝐴: Relabel dummy indices (ones over which Einstein’s summation is carried out) in the fourth term: 𝜇 → 𝛿, and factorize: . . . . . . 7 𝐴;56 = 𝐴,56 + Γ5:,6 𝐴: + Γ5: 𝐴:,6 + Γ6: 𝐴:,5 + Γ67 Γ5: 𝐴: .

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.

.

7

.

.

.

.

.

.

7

.

.

𝐴;56 = 𝐴,56 + (Γ5:,6 + Γ67 Γ5: )𝐴: + Γ5: 𝐴:,6 + Γ6: 𝐴:,5 . .

After finding 𝐴;56 computing 𝐴;65 is trivial: just replace dummy indices: 𝛼 → 𝛽, 𝛽 → 𝛼 𝐴;65 = 𝐴,65 + (Γ6:,5 + Γ57 Γ6: )𝐴: + Γ6: 𝐴:,5 + Γ5: 𝐴:,6

Therefore, .

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.

7

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7

.

.

𝐴;56 − 𝐴;65 = 𝐴,56 − 𝐴,65 + UΓ5:,6 + Γ67 Γ5: − Γ6:,5 − Γ57 Γ6: V 𝐴: + Γ5: 𝐴:,6 + Γ6: 𝐴:,5 .

.

− Γ6: 𝐴:,5 − Γ5: 𝐴:,6 Ordinary derivatives commute – the order of taking ordinary derivatives does not . . matter, so 𝐴,56 = 𝐴,65 . Also, the last 4 terms above cancel. Therefore, .

.

.

.

7

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.

7

𝐴;56 − 𝐴;65 = UΓ5:,6 + Γ67 Γ5: − Γ6:,5 − Γ57 Γ6: V 𝐴: Replacing dummy variables: 𝜈 → 𝜇 in the rewritten Eq. (5.20) above: . . .

8 .

8 . 𝑅:56 ≡ Γ:6,5 − Γ5:,6 + Γ:6 Γ85 − Γ:5 Γ86 . . . ,7 .

7 . 𝑅:56 ≡ Γ:6,5 − Γ5:,6 + Γ:6 Γ75 − Γ:5 Γ76

7

7

and recognizing that the lower indices of the Christoffel symbols commute: Γ:5 = Γ5: , we finally obtain . . . 𝐴;56 − 𝐴;65 = −𝑅:56 𝐴: 5. Starting with the definition of the Christoffel symbol in Eq. (4.68) in the textbook, show 8 8 that Γ56 = Γ65 . #

8 Eq. (4.68): Γ56 ≡ $ 𝑔8. I𝑔5.,6 + 𝑔.6,5 − 𝑔56,. J 8 8 After defining Γ56 we can easily find Γ65 by switching dummy indices 𝛼 ↔ 𝛽: 1 8 Γ65 ≡ 𝑔8. (𝑔6.,5 + 𝑔.5,6 − 𝑔65,. ) 2 Metric tensor is symmetric, so 𝑔56 = 𝑔65 , and so are their derivatives. Therefore, the last terms in the two definitions above are the same. So are the first term in the first equation and the second in the second equation; and the second term in the first equation and the first term in the second equation.

6. Compute the components of the Riemann tensor for a unit 2-sphere where 𝑑𝑠 $ = 𝑑𝜃 $ + sin$ 𝜃 𝑑𝜙 $ The coordinates are 𝑥 & = (𝑥 # , 𝑥 $ ) = (𝜃, 𝜙). The metric is 1 0 1 0 &' # t 𝑔&' = U $ V and 𝑔 = s0 0 sin 𝜃 BCD" E There are 2 coordinates, to each index (Greek or Latin) will only have 2 values: 1 or 2.

The definition of the Christoffel symbol is (textbook Eq.(4.68)): 1 F Γ56 = 𝑔8. (𝑔5.,6 + 𝑔.6,5 − 𝑔56,. ) 2 There are 2% = 8 Christoffel symbols in all, because it has 3 indices with 2 values. # # # Γ## = $ 𝑔## I𝑔##,# + 𝑔##,# − 𝑔##,# J + $ 𝑔#$ I𝑔#$,# + 𝑔$#,# − 𝑔##,$ J = 0 because 𝑔##,& = 0 (𝑔## = 1) and 𝑔#$ = 𝑔$# = 𝑔$# = 0 # # # Γ#$ = $ 𝑔## I𝑔##,$ + 𝑔#$,# − 𝑔#$,# J + $ 𝑔#$ I𝑔#$,$ + 𝑔$$,# − 𝑔#$,$ J = 0 because 𝑔##,& = 0 (𝑔## = 1) and 𝑔#$ = 𝑔$# = 𝑔$# = 0 # # Γ#$ = Γ$# = 0 because of the symmetry of the Christoffel symbols # # # Γ$$ = $ 𝑔## I𝑔$#,$ + 𝑔#$,$ − 𝑔$$,# J + $ 𝑔#$ I𝑔$$,$ + 𝑔$$,$ − 𝑔$$,$ J #

#

= $ (1)I−𝑔$$,# J = − $ 2 sin 𝜃 cos 𝜃 = −sin 𝜃 cos 𝜃 #

#

$ Γ## = $ 𝑔$# I𝑔##,# + 𝑔##,# − 𝑔##,# J + $ 𝑔$$ I𝑔#$,# + 𝑔$#,# − 𝑔##,$ J = 0 because 𝑔##,& = 0 (𝑔## = 1) and 𝑔#$ = 𝑔$# = 𝑔$# = 0 # # $ Γ#$ = $ 𝑔$# I𝑔##,$ + 𝑔#$,# − 𝑔#$,# J + $ 𝑔$$ I𝑔#$,$ + 𝑔$$,# − 𝑔#$,$ J = #

#

#

= $ 𝑔$$ 𝑔$$,# = $ BCD" E 2 sin 𝜃 cos 𝜃 = cot 𝜃 $ $ Γ$# = Γ#$ = cot 𝜃 because of the symmetry of the Christoffel symbols # $# # $ Γ$$ = $ 𝑔 I𝑔$#,$ + 𝑔#$,$ − 𝑔$$,# J + $ 𝑔$$ I𝑔$$,$ + 𝑔$$,$ − 𝑔$$,$ J = 0 because 𝑔##,& = 0 (𝑔## = 1) and 𝑔#$ = 𝑔$# = 𝑔$# = 0

# $ $ The only 3 non-zero terms Γ$$ , Γ#$ and Γ$# are highlighted in green.

From the definition of the Riemann tensor in the textbook Eq. (5.1): 5 5 5 8 5 8 5 𝑅6.: ≡ Γ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8: 1 8 7 8 7 𝑅56.: ≡ I𝑔6.,5: + 𝑔5:,6. − 𝑔6:,5. − 𝑔5.,6: J + 𝑔78 Γ5. Γ6: − 𝑔78 Γ5: Γ6.

2 G There are 2 = 16 components of the Riemann tensor in all, because it has 4 indices with 2 values. # $ $ The only 3 non-zero Christoffel symbols are Γ$$ , Γ#$ , and Γ$# only non-zero derivatives of the metric are 𝑔$$,## so: # 8 7 8 7 𝑅#### = $ I𝑔##,## + 𝑔##,## − 𝑔##,## − 𝑔##,## J + 𝑔78 Γ## Γ## − 𝑔78 Γ## Γ## = 0 #

7

7

8 8 𝑅###$ = $ I𝑔##,#$ + 𝑔#$,## − 𝑔#$,## − 𝑔##,#$ J + 𝑔78 Γ#$ Γ## − 𝑔78 Γ## Γ#$ = 0 𝑅##$# = −𝑅###$ = 0 skew symmetry in textbook Eq. (5.4) 𝑅#$## = 𝑅###$ = 0 interchange symmetry in textbook Eq. (5.5) 𝑅$### = −𝑅#$## = 0 skew symmetry in textbook Eq. (5.4) # 8 7 8 7 𝑅##$$ = $ I𝑔#$,#$ + 𝑔#$,#$ − 𝑔#$,#$ − 𝑔#$,#$ J + 𝑔78 Γ#$ Γ#$ − 𝑔78 Γ#$ Γ#$ = 0 𝑅$$## = 𝑅##$$ = 0 interchange symmetry in textbook Eq. (5.5) # 8 7 8 7 𝑅$$#$ = $ I𝑔$$,$# + 𝑔$#,$$ − 𝑔$#,$$ − 𝑔$$,$# J + 𝑔78 Γ$# Γ$$ − 𝑔78 Γ$$ Γ$# = 0 𝑅$$$# = −𝑅$$#$ = 0 skew symmetry in textbook Eq. (5.4) 𝑅$#$$ = 𝑅$$$# = 0 interchange symmetry in textbook Eq. (5.5) 𝑅$#$$ = 𝑅$$$# = 0 interchange symmetry in textbook Eq. (5.5) # 8 7 8 7 𝑅$$$$ = $ I𝑔$$,$$ + 𝑔$$,$$ − 𝑔$$,$$ − 𝑔$$,$$ J + 𝑔78 Γ$$ Γ$$ − 𝑔78 Γ$$ Γ$$ = 0 1 8 7 8 7 𝑅56.: ≡ I𝑔6.,5: + 𝑔5:,6. − 𝑔6:,5. − 𝑔5.,6: J + 𝑔78 Γ5: Γ6. − 𝑔78 Γ5. Γ6: 2 # 8 7 8 7 𝑅#$#$ = $ I𝑔$#,#$ + 𝑔#$,$# − 𝑔$$,## − 𝑔##,$$ J + 𝑔78 Γ#$ Γ$# − 𝑔78 Γ## Γ$$ #

# H"

$ $ = $ I−𝑔$$,## J + 𝑔$$ Γ#$ Γ$# = − $ HE" (sin$ 𝜃) + sin$ 𝜃 cot $ 𝜃 = − cos $ 𝜃 + sin$ 𝜃 + cos $ 𝜃 = sin$ 𝜃 𝑅#$$# = −𝑅#$#$ = − sin$ 𝜃 skew symmetry in textbook Eq. (5.23) 𝑅$##$ = −𝑅#$#$ = − sin$ 𝜃 skew symmetry in textbook Eq. (5.23) 𝑅$#$# = −𝑅$##$ = −(−𝑅#$#$ ) = 𝑅#$#$ = sin$ 𝜃 skew symmetry in textbook Eq. (5.23) applied twice # 8 7 8 7 𝑅#$$$ = $ I𝑔$$,#$ + 𝑔#$,$$ − 𝑔#$,$$ − 𝑔$$,#$ J + 𝑔78 Γ#$ Γ$$ − 𝑔78 Γ#$ Γ$$ = 0

Alternatively, use the mixed version of the Riemann tensor (textbook Eq. (5.20)): 5 5 5 8 5 8 5 𝑅6.: ≡ Γ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8: # $ $ The only 2 non-zero Christoffel symbols are Γ$$ , Γ#$ , and Γ$# only non-zero derivatives of the metric are 𝑔$$,## so: # # # 8 # 8 # 𝑅### = Γ##,# − Γ##,# + Γ## Γ8# − Γ## Γ8# = 0 # # # 8 # 8 # 𝑅##$ = Γ#$,# − Γ##,$ + Γ#$ Γ8# − Γ## Γ8$ = 0 # # # 8 # 8 # 𝑅#$# = Γ##,$ − Γ#$,# + Γ## Γ8$ − Γ#$ Γ8$ = 0

𝜕 # Γ − cot 𝜃 (−sin 𝜃 cos 𝜃) 𝜕𝜃 $$ H = HE (− sin 𝜃 cos 𝜃) − cot 𝜃 (−sin 𝜃 cos 𝜃) = − cos $ 𝜃 + sin$ 𝜃 + cos $ 𝜃 = sin$ 𝜃 … and like this ad nauseum… # # # 8 # 8 # # $ # 𝑅$#$ = Γ$$,# − Γ$#,$ + Γ$$ Γ8# − Γ$# Γ8$ = Γ$$,# − Γ$# Γ$$ =

Chapter 5 1. Use Eq. (5.20) in the textbook to express the Ricci tensor in terms of Christoffel symbols.

5 5 5 8 5 8 5 Eq. (5.20): 𝑅6.: ≡ Γ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8: Set 𝛼 = 𝛾 in Eq. (5.20): . . .

8 .

8 . 𝑅6: ≡ 𝑅6.: = Γ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8: 2. Derive Eq. (5.22) from Eq. (5.20) and Eq. (5.21) in the textbook.

5 5 5 8 5 8 5 Eq. (5.20): 𝑅6.: ≡ Γ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8: #

8 Eq. (5.21): Γ56 ≡ $ 𝑔8. I𝑔5.,6 + 𝑔.6,5 − 𝑔56,. J #

7

7

8 8 Eq. (5.22): 𝑅56.: ≡ $ I𝑔6.,5: + 𝑔5:,6. − 𝑔6:,5. − 𝑔5.,6: J + 𝑔57 Γ6: Γ8. − 𝑔57 Γ6. Γ8: 7

7

7

7

7

8 8 𝑅56.: = 𝑔57 𝑅6.: = 𝑔57 UΓ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8: V 7

7

7

7

8 8 𝑅56.: = 𝑔57 Γ6:,. − 𝑔57 Γ6.,: + 𝑔57 Γ6: Γ8. − 𝑔57 Γ6. Γ8: 7 7 8 8 𝑅56.: = Γ56:,. − Γ56.,: + 𝑔57 Γ6: Γ8. − 𝑔57 Γ6. Γ8: 8 8 𝑅56.: = Γ56:,. − Γ56.,: + Γ6: Γ58. − Γ6. Γ58: #

#

I

8 Γ56: = 𝑔58 Γ6: = 𝑔58 $ 𝑔8I I𝑔6I,: + 𝑔I:,6 − 𝑔6:,I J = $ 𝛿5 I𝑔6I,: + 𝑔I:,6 − 𝑔6:,I J #

Γ56: = $ I𝑔65,: + 𝑔5:,6 − 𝑔6:,5 J #

#

#

Γ56:,. = $ 𝑔65,:. + $ 𝑔5:,6. − $ 𝑔6:,5. #

#

8 Γ56. = 𝑔58 Γ6. = 𝑔58 $ 𝑔8: I𝑔6:,. + 𝑔:.,6 − 𝑔6.,: J = $ 𝛿5: I𝑔6:,. + 𝑔:.,6 − 𝑔6.,: J #

Γ56. = $ I𝑔65,. + 𝑔5.,6 − 𝑔6.,5 J #

#

#

Γ56.,: = $ 𝑔65,.: + $ 𝑔5.,6: − $ 𝑔6.,5: 7

7

8 8 𝑅56.: = Γ56:,. − Γ56.,: + 𝑔57 Γ6: Γ8. − 𝑔57 Γ6. Γ8:

1 1 1 1 1 1 = 𝑔65,:. + 𝑔5:,6. − 𝑔6:,5. − 𝑔65,.: − 𝑔5.,6: + 𝑔6.,5: 2 2 2 2 2 2 8 7 8 7 +𝑔57 Γ6: Γ8. − 𝑔57 Γ6. Γ8: #

7

7

8 8 𝑅56.: = $ I𝑔6.,5: + 𝑔5:,6. − 𝑔6:,5. − 𝑔5.,6: J + 𝑔57 Γ6: Γ8. − 𝑔57 Γ6. Γ8:

3. Derive Eq. (5.25) in the textbook.

Eq. (5.25): 𝑅56.: + 𝑅5.:6 + 𝑅5:6. = 0 1 8 7 8 7 𝑅56.: ≡ I𝑔6.,5: + 𝑔5:,6. − 𝑔6:,5. − 𝑔5.,6: J + 𝑔78 Γ5: Γ6. − 𝑔78 Γ5. Γ6: 2 1 8 7 8 7 𝑅5.:6 ≡ I𝑔.:,56 + 𝑔56,.: − 𝑔.6,5: − 𝑔5:,.6 J + 𝑔78 Γ56 Γ.: − 𝑔78 Γ5: Γ.6 2 1 8 7 8 7 𝑅5:6. ≡ I𝑔:6,5. + 𝑔5.,:6 − 𝑔:.,56 − 𝑔56,:. J + 𝑔78 Γ5. Γ:6 − 𝑔78 Γ56 Γ:.

2

Recall symmetries of the metric and the Christoffel symbols: 𝑔78 = 𝑔87 𝑔78,56 = 𝑔78,65 8 8 Γ56 = Γ65 𝑅56.: + 𝑅5.:6 + 𝑅5:6.

1 = I𝑔6.,5: − 𝑔.6,5: + 𝑔5:,6. − 𝑔5:,.6 − 𝑔6:,5. + 𝑔:6,5. − 𝑔5.,6: 2 + 𝑔5.,:6 + 𝑔.:,56 − 𝑔:.,56 + 𝑔56,.: − 𝑔56,:. J 7

7

7

7

7

7

8 8 8 8 8 8 + 𝑔78 UΓ5: Γ6. − Γ5: Γ.6 − Γ5. Γ6: + Γ5. Γ:6 + Γ56 Γ.: − Γ56 Γ:. V = 0

After recalling all the symmetries listed above, all of these terms cancel in pairs as listed.

To wrap my head around all those indices, I wrote a simple Python code. You can try to do the same in any of the programming languages/scripts/math software packages you are familiar with.

4. Derive Eq. (5.31) from Eqns. (5.29) and (5.30) in the textbook.

Eq. (5.27): 𝑅 ≡ 𝑔56 𝑅56 = 𝑅55 56

#

Eq. (5.29): 𝑅;5 = $ 𝑔56 𝑅;5 #

Eq. (5.30): 𝐺56 ≡ 𝑅56 − $ 𝑔56 𝑅

1 1 𝐺56;5 ≡ 𝑅56;5 − 𝑔56;5 𝑅 − 𝑔56 𝑅;5 2 2 # 5 56 56 From Eq. (5.29): 𝑅;5 = $ 𝛿J 𝑔 𝑅;J 1 56 5 𝑅.;5 = 𝑔6. 𝑅;5 = 𝑔6. 𝛿J5 𝑔56 𝑅;J 2 1 1 .

1 5 𝑅.;5 = 𝛿J5 𝛿.5 𝑅;J = 𝛿J 𝑅;J = 𝑅;.

2 2 2 Dummy variable renamed: 𝜖 → 𝛼 1 .

5 𝑅.;5 = 𝛿5 𝑅;5 2 5 5 𝑅.;5 = 𝛿I5 𝑅.;I 1 .

5 𝑔58 𝑅.;5 = 𝑔58 𝛿5 𝑅;5 2 5 5 𝑔58 𝑅.;5 = 𝑔58 𝛿I5 𝑅.;I = 𝛿I8 𝑅8.;I = 𝑅8.;8 1 1 . 𝑔58 𝛿5 𝑅;5 = 𝑔.8 𝑅;.

2 2 1 1 .

1 5 𝑅.;5 = 𝛿J5 𝛿.5 𝑅;J = 𝛿J 𝑅;J = 𝑅;.

2 2 2 1 5 𝑔I. 𝑅.;5 = 𝑔58 𝑅;.

2

56

#

56

#

#

5 𝑅;5 = $ 𝑔56 𝑅;5 → 𝑔6. 𝑅;5 = $ 𝑔6. 𝑔56 𝑅;5 → 𝑅.;5 = $ 𝛿.5 𝑅;5 #

#

#

5 𝑔56 𝑅.;5 = $ 𝑔56 𝛿.5 𝑅;5 → 𝑅6.;6 = $ 𝑔6. 𝑅;6 → 𝑅56;5 = $ 𝑔56 𝑅;5 1 5 𝑔58 𝑅.;5 = 𝑔58 𝛿.5 𝑅;5 2 5 5 𝑅.;5 = 𝛿I5 𝑅.;I 5 5 𝑔58 𝑅.;5 = 𝑔58 𝛿I5 𝑅.;I = 𝛿I8 𝑅8.;I = 𝑅8.;8 1 1 𝑔58 𝛿.5 𝑅;5 = 𝑔.8 𝑅;.

2 2 # # # 5 𝑔 𝛿 𝑅 → 𝑅 = 𝑔 𝑅 → 𝑅 = 𝑔 𝑅 .8;.

56;5 $ 58 . ;5 $ .8 ;. $ 56 ;5

Lecture slides, GR, slide 8: 𝑔56;5 = 0

1 1 1 1 𝐺56;5 ≡ 𝑅56;5 − 𝑔56;5 𝑅 − 𝑔56 𝑅;5 = 𝑔56 𝑅;5 − 0 − 𝑔56 𝑅;5 = 0.

2 2 2 2 5. Compute the Ricci scalar 𝑅 for the unit 2-sphere. (You can use the results of Exercise 6 in Chapter 4.)

From Exercise 6, Chapter 4: The coordinates are 𝑥 & = (𝑥 # , 𝑥 $ ) = (𝜃, 𝜙). The metric is 1 0 1 0 &' # 𝑔&' = U V and 𝑔 = s 0 BCD" Et 0 sin$ 𝜃 𝑅#$#$ = sin$ 𝜃 𝑅#$$# = −𝑅#$#$ = − sin$ 𝜃 skew symmetry in Lecture Notes Eq. (5.4) 𝑅$##$ = −𝑅#$#$ = − sin$ 𝜃 skew symmetry in Lecture Notes Eq. (5.4) 𝑅$#$# = −𝑅$##$ = −(−𝑅#$#$ ) = 𝑅#$#$ = sin$ 𝜃 skew symmetry in the Lecture Notes Eq. (5.4) applied twice .

Textbook Eq. (5.26), Ricci tensor: 𝑅56 ≡ 𝑅5.6 ≡ 𝑔:. 𝑅:5.6 𝑅56 = 𝑔## 𝑅#5#6 + 𝑔$$ 𝑅$5$6 1 1 𝑅## = 𝑅 = sin$ 𝜃 = 1 $#$# sin$ 𝜃 sin$ 𝜃 𝑅$$ = 1 × 𝑅#$#$ = 1 × sin$ 𝜃 = sin$ 𝜃 Textbook Eq. (5.27), Ricci scalar: 𝑅 ≡ 𝑔56 𝑅56 = 𝑅55 1 𝑅 ≡ 𝑔56 𝑅56 = 𝑔## 𝑅## + 𝑔$$ 𝑅$$ = 1 + $ sin$ 𝜃 = 1 + 1 = 2 sin 𝜃 𝑅=2

Chapter 6 1. Derive Eq. (6.21) in the textbook.

−𝑒 L 𝑔56 = 1 0 0 0

0 𝑒M 0 0

0 0 𝑟$ 0

0 0 6 0 𝑟 $ sin$ 𝜃

#

7

7

8 8 Eq. (5.22): 𝑅56.: ≡ $ I𝑔6.,5: + 𝑔5:,6. − 𝑔6:,5. − 𝑔5.,6: J + 𝑔78 Γ5: Γ6. − 𝑔78 Γ5. Γ6: .

Eq. (5.26): 𝑅56 ≡ 𝑅5.6 ≡ 𝑔:. 𝑅:5.6 Eq. (5.27): 𝑅 ≡ 𝑔56 𝑅56 = 𝑅55 #

Eq. (5.30): 𝐺56 ≡ 𝑅56 − $ 𝑔56 𝑅 #

) ) ) Eq. (6.11): Γ)# = Γ#) = $ 𝑁′, otherwise Γ56 =0 #

#

# # # # Eq. (6.14): Γ)) = $ 𝑁 " 𝑒 LNM , Γ## = $ 𝑃" , Γ$$ = −𝑒 NM 𝑟, Γ%% = −𝑒 NM 𝑟 sin$ 𝜃, ) otherwise Γ56 =0 #

#

$ $ $ $ Eq. (6.17): Γ#$ = Γ$# = 9 , Γ%% = − $ sin 2𝜃, otherwise Γ56 =0 #

% % % % % Eq. (6.20): Γ#% = Γ%# = 9 , Γ$% = Γ%$ = cot 𝜃, otherwise Γ56 =0 Coordinates are (𝑡, 𝑟, 𝜃, 𝜙). All non-zero Christoffel symbols depend only on 𝑟, 𝜃, so all . . derivatives with respect to 𝑡, 𝜙 (0th and 3rd coordinates) vanish: Γ6:,) = 0, Γ6:,% = 0 𝑅)) is symmetric, so we need to find only 10 of them.

From Problem 1:

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.

.

.

.

8 8 𝑅6: = Γ6:,. − Γ6.,: + Γ6: Γ8. − Γ6. Γ8:

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.

.

8 8 # # 8 ) 8 # 𝑅)) = Γ)),. − Γ).,) + Γ)) Γ8. − Γ). Γ8) = Γ)),# + Γ)) Γ#. − Γ)) Γ8) − Γ)# Γ8) # # ) # $ % # ) ) # = Γ)),# + Γ)) (Γ#) + Γ## + Γ#$ + Γ#% ) − Γ)) Γ#) − Γ)# Γ)) # # ) # $ % ) # = Γ)),# + Γ)) (Γ#) + Γ## + Γ#$ + Γ#% ) − 2Γ)# Γ)) ;

#

#

#

#

#

#

#

#

= ;9 U$ 𝑁 " 𝑒 LNM V + $ 𝑁 " 𝑒 LNM U$ 𝑁 " + $ 𝑃" + 9 + 9 V − 2 $ 𝑁′ $ 𝑁 " 𝑒 LNM 1 1 1 1 1 = 𝑁 "" 𝑒 LNM + (𝑁 " )$ 𝑒 LNM − 𝑁 " 𝑃" 𝑒 LNM + (𝑁 " )$ 𝑒 LNM + 𝑁 " 𝑃" 𝑒 LNM 2 2 2 4 4 1 " LNM 1 " $ LNM + 𝑁𝑒 − (𝑁 ) 𝑒 𝑟 2 1 "" LNM 1 " " LNM 1 " $ LNM = 𝑁 𝑒 − 𝑁𝑃𝑒 + (𝑁 ) 𝑒 + 𝑁 " 𝑒 LNM 2 4 4 1 1 1 1 = 𝑒 LNM a 𝑁 "" − 𝑁 " 𝑃" + (𝑁 " )$ + 𝑁 " b 2 4 4 𝑟 # " " # # " LNM # "" " $ 𝑅)) = 𝑒 U$ 𝑁 − G 𝑁 𝑃 + G (𝑁 ) + 9 𝑁 V

1 1 1 1 𝑔)) 𝑅)) = −𝑒 NM a 𝑁 "" − 𝑁 " 𝑃" + (𝑁 " )$ + 𝑁 " b 2 4 4 𝑟 .

.

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.

8 8 # ) 8 ) 8 # 𝑅)# = Γ)#,. − Γ).,# + Γ)# Γ8. − Γ). Γ8# = − Γ)#,# + Γ)# Γ). − Γ)) Γ8# − Γ)# Γ8# = 0 . .

8 .

8 .

8 . # ) ) # 𝑅)$ = Γ)$,. − Γ).,$ + Γ)$ Γ8. − Γ). Γ8$ = − Γ). Γ8$ = − Γ)) Γ#$ − Γ)# Γ)$ = 0 . .

8 .

8 .

8 . # ) ) # 𝑅)% = Γ)%,. − Γ).,% + Γ)% Γ8. − Γ). Γ8% = − Γ). Γ8% = − Γ)) Γ#% − Γ)# Γ)% = 0 . . . .

8 8 𝑅## = Γ##,. − Γ#.,# + Γ## Γ8. − Γ#. Γ8# # ) # $ % # . ) ) $ $ % % = Γ##,# − Γ#),# − Γ##,# − Γ#$,# − Γ#%,# + Γ## Γ#. − Γ#) Γ)# − Γ#$ Γ$# − Γ#% Γ%# ) $ % # (Γ ) # $ %) = − Γ#),# − Γ#$,# − Γ#%,# + Γ## #) + Γ## + Γ#$ + Γ#% ) ) # # $ $ % % − Γ#) Γ)# − Γ## Γ## − Γ#$ Γ$# − Γ#% Γ%# ) $ % # (Γ ) $ % ) ) ) $ $ % % = − Γ#),# − Γ#$,# − Γ#%,# + Γ## #) + Γ#$ + Γ#% − Γ#) Γ)# − Γ#$ Γ$# − Γ#% Γ%# #

#

#

#

#

$

#

#

#

#

#

M#

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#

$

#

% $ Γ#% = Γ#$ =9

= − $ 𝑁 "" + 9 " + 9 " + $ 𝑃" U$ 𝑁 " + 9V − G (𝑁 " )$ − 9 " $

#

= − $ 𝑁 "" + $ 𝑃" U$ 𝑁 " + 9 V − G (𝑁 " )$ #

𝑅## = − $ 𝑁 "" + G 𝑃" 𝑁 " + 9 − G (𝑁 " )$ 1 1 𝑃" 1 𝑔## 𝑅## = 𝑒 NM s− 𝑁 "" + 𝑃" 𝑁 " + − (𝑁 " )$ t 2 4 𝑟 4 #

#

8 8 $ % % % 𝑅#$ = Γ#$,. − Γ#.,$ + Γ#$ Γ8. − Γ#. Γ8$ = Γ#$ Γ$% − Γ#% Γ%$ = 9 cot 𝜃 − 9 cot 𝜃 = 0 8 8 𝑅#% = Γ#%,. − Γ#.,% + Γ#% Γ8. − Γ#. Γ8% = 0 . . .

8 8 . 𝑅$$ = Γ$$,. − Γ$.,$ + Γ$$ Γ8. − Γ$. Γ8$ # % # . # . $ . % . = Γ$$,# − Γ$%,$ + Γ$$ Γ#. − Γ$. Γ#$ − Γ$. Γ$$ − Γ$. Γ%$ # % # ) # $ % # $ $ # % % = Γ$$,# − Γ$%,$ + Γ$$ (Γ#) +Γ## +Γ#$ +Γ#% ) − Γ$$ Γ#$ − Γ$# Γ$$ − Γ$% Γ%$ # # % # ) # % % % $ = Γ$$,# − Γ$%,$ + Γ$$ (Γ#) +Γ## ) − Γ$% Γ%$ Γ#% = Γ#$ =9 #

#

= −𝑒 NM + 𝑃" 𝑒 NM 𝑟 + csc $ 𝜃 − 𝑒 NM 𝑟 U$ 𝑁 " + $ 𝑃" V − cot $ 𝜃 #

#

= −𝑒 NM + 𝑃" 𝑒 NM 𝑟 + 1 − 𝑒 NM 𝑟 U$ 𝑁 " + $ 𝑃" V #

#

csc $ 𝜃 − cot $ 𝜃 = 1

#

#

= 1 − 𝑒 NM U1 − 𝑃" 𝑟 + $ 𝑁 " 𝑟 + $ 𝑃" 𝑟V = 1 − 𝑒 NM U1 − $ 𝑃" 𝑟 + $ 𝑁 " 𝑟V = #

#

𝑅$$ = 𝑒 NM U𝑒 M − 1 + $ 𝑃" 𝑟 − $ 𝑁 " 𝑟V 1 1 1 𝑔$$ 𝑅$$ = $ 𝑒 NM a𝑒 M − 1 + 𝑃" 𝑟 − 𝑁 " 𝑟b 𝑟 2 2 . .

8 .

8 . 𝑅$% = Γ$%,. − Γ$.,% + Γ$% Γ8. − Γ$. Γ8% = 0 . .

8 .

8 . 𝑅%% = Γ%%,. − Γ%.,% + Γ%% Γ8. − Γ%. Γ8% # $ # . $ . # . $ . % . = Γ%%,# + Γ%%,$ + Γ%% Γ#. + Γ%% Γ$. − Γ%. Γ#% − Γ%. Γ$% − Γ%. Γ%% # $ # ) # $ % $ % = Γ%%,# + Γ%%,$ + Γ%% (Γ#) + Γ## + Γ#$ + Γ#% ) + Γ%% Γ$% # % $ % % # % $ − Γ%% Γ#% − Γ%% Γ$% − Γ%# Γ%% − Γ%$ Γ%% # $ # ) # $ % # % $ = Γ%%,# + Γ%%,$ + Γ%% (Γ#) + Γ## + Γ#$ ) − Γ%# Γ%% − Γ%$ Γ%% #

#

#

= −𝑒 NM sin$ 𝜃 + 𝑃" 𝑒 NM 𝑟 sin$ 𝜃 − cos 2𝜃 − 𝑒 NM 𝑟 sin$ 𝜃 U$ 𝑁 " + $ 𝑃" + 9 V

#

#

− 9 (−𝑒 NM 𝑟 sin$ 𝜃) + $ cot 𝜃 sin 2𝜃 #

#

= 𝑒 NM sin$ 𝜃 U−1 + 𝑃" 𝑟 − $ 𝑁 " 𝑟 − $ 𝑃" 𝑟 − 1 + 1 + 𝑒 M V #

#

𝑅%% = sin$ 𝜃 𝑒 NM U−1 + $ 𝑃" 𝑟 − $ 𝑁 " 𝑟 + 𝑒 M V 𝑅%% = sin$ 𝜃 𝑅$$ 1 1 𝑔%% 𝑅%% = $ $ sin$ 𝜃 𝑅$$ = $ 𝑅$$ 𝑟 sin 𝜃 𝑟 1 NM 1 " 1 " %% 𝑔 𝑅%% = $ 𝑒 a−1 + 𝑃 𝑟 − 𝑁 𝑟 + 𝑒 M b 𝑟 2 2 All the cross-terms vanish. 𝑅 = 𝑔)) 𝑅)) + 𝑔## 𝑅## + 𝑔$$ 𝑅$$ + 𝑔%% 𝑅%% #

#

#

#

#

#

M#

#

𝑅 = −𝑒 NM U$ 𝑁 "" − G 𝑁 " 𝑃" + G (𝑁 " )$ + 9 𝑁 " V + 𝑒 NM U− $ 𝑁 "" + G 𝑃" 𝑁 " + 9 − G (𝑁 " )$ V 1 NM M 1 1 1 1 1 𝑒 a𝑒 − 1 + 𝑃" 𝑟 − 𝑁 " 𝑟b + $ 𝑒 NM a−1 + 𝑃" 𝑟 − 𝑁 " 𝑟 + 𝑒 M b $ 𝑟 2 2 𝑟 2 2 # " " # # " # "" # " " M# # NM # "" " $ NM 𝑅 = −𝑒 U$ 𝑁 − G 𝑁 𝑃 + G (𝑁 ) + 9 𝑁 V + 𝑒 U− $ 𝑁 + G 𝑃 𝑁 + 9 − G (𝑁 " )$ V +

2 NM M 1 1 𝑒 a𝑒 − 1 + 𝑃" 𝑟 − 𝑁 " 𝑟b $ 𝑟 2 2 1 1 1 1 1 1 𝑃" 1 2 𝑅 = 𝑒 NM s− 𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ − 𝑁 " − 𝑁 "" + 𝑃" 𝑁 " + − (𝑁 " )$ + $ 𝑒 M 2 4 4 𝑟 2 4 𝑟 4 𝑟 " " 2 𝑃 𝑁 − $+ − t 𝑟 𝑟 𝑟 1 1 2𝑁 " 2𝑃" 2 M 2 𝑅 = 𝑒 NM s−𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ − + + $ 𝑒 − $t 2 2 𝑟 𝑟 𝑟 𝑟 +

00-term:

#

1 𝑅56 − 𝑔56 𝑅 = 0 2 1 𝑅)) − 𝑔)) 𝑅 = 0 2 # # # LNM # "" 𝑅)) = 𝑒 U$ 𝑁 − G 𝑁 " 𝑃" + G (𝑁 " )$ + 9 𝑁 " V #

#

#

#

𝑅)) − $ 𝑔)) = 𝑒 LNM U$ 𝑁 "" − G 𝑁 " 𝑃" + G (𝑁 " )$ + 9 𝑁 " V 1 1 1 2𝑁 " 2𝑃" 2 M 2 − (−𝑒 L )𝑒 NM s−𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ − + + $ 𝑒 − $t = 0 2 2 2 𝑟 𝑟 𝑟 𝑟

1 1 "" 1 " " 1 " $ 1 " 1 "" 1 " " 1 " $ 𝑁 " 𝑃" LNM 𝑅)) − 𝑔)) 𝑅 = 𝑒 s 𝑁 − 𝑁 𝑃 + (𝑁 ) + 𝑁 − 𝑁 + 𝑁 𝑃 − (𝑁 ) − + 2 2 4 4 𝑟 2 4 4 𝑟 𝑟 𝑒M 1 + $ − $t = 0 𝑟 𝑟 1 𝑃" 𝑒 M 1 𝑅)) − 𝑔)) 𝑅 = 𝑒 LNM s + $ − $ t = 0 2 𝑟 𝑟 𝑟 LNM L 𝑒 1 𝑒 a𝑃" − b + $ = 0 𝑟 𝑟 𝑟 −

𝑒 LNM " 1 𝑒L a𝑃 − b − $ = 0 𝑟 𝑟 𝑟

11-term:

1 𝑅## − 𝑔## 𝑅 = 0 2 1 "" 1 " " 𝑃" 1 " $ 𝑅## = − 𝑁 + 𝑃 𝑁 + − (𝑁 ) 2 4 𝑟 4 1 1 "" 1 " " 𝑃" 1 " $ 𝑅## − 𝑔## 𝑅 = − 𝑁 + 𝑃 𝑁 + − (𝑁 ) 2 2 4 𝑟 4 1 M NM 1 1 2𝑁 " 2𝑃" 2 M 2 − 𝑒 𝑒 s−𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ − + + $ 𝑒 − $t = 0 2 2 2 𝑟 𝑟 𝑟 𝑟 " 1 1 1 𝑃 1 1 1 1 𝑁" 𝑅## − 𝑔## 𝑅 = − 𝑁 "" + 𝑃" 𝑁 " + − (𝑁 " )$ + 𝑁 "" − 𝑁 " 𝑃" + (𝑁 " )$ + 2 2 4 𝑟 4 2 4 4 𝑟 𝑃" 1 M 1 − − $𝑒 + $ = 0 𝑟 𝑟 𝑟 1 𝑁" 1 M 1 𝑅## − 𝑔## 𝑅 = − $𝑒 + $ = 0 2 𝑟 𝑟 𝑟 " 𝑁 1 + $ (1 − 𝑒 M ) = 0 𝑟 𝑟 − 22-term:

𝑁" 1 − $ (1 − 𝑒 M ) = 0 𝑟 𝑟

1 𝑅$$ − 𝑔$$ 𝑅 = 0 2 1 1 𝑅$$ = 𝑒 NM a𝑒 M − 1 + 𝑃" 𝑟 − 𝑁 " 𝑟b 2 2 1 1 1 𝑅$$ − 𝑔$$ 𝑅 = 𝑒 NM a𝑒 M − 1 + 𝑃" 𝑟 − 𝑁 " 𝑟b 2 2 2 𝑟 $ NM 1 1 2𝑁 " 2𝑃" 2 M 2 − 𝑒 s−𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ − + + $ 𝑒 − $t = 0 2 2 2 𝑟 𝑟 𝑟 𝑟

1 1 " 1 " 𝑟 $ "" 𝑟 $ " " 𝑟 $ " $ NM M 𝑅$$ − 𝑔$$ 𝑅 = 𝑒 s𝑒 − 1 + 𝑃 𝑟 − 𝑁 𝑟 + 𝑁 − 𝑁 𝑃 + (𝑁 ) + 𝑟𝑁 " − 𝑟𝑃′ 2 2 2 2 4 4 − 𝑒 M + 1t = 0 1 1 1 𝑟$ 𝑟$ 𝑟$ 𝑅$$ − 𝑔$$ 𝑅 = 𝑒 NM s− 𝑃" 𝑟 + 𝑁 " 𝑟 + 𝑁 "" − 𝑁 " 𝑃" + (𝑁 " )$ t = 0 2 2 2 2 4 4 1 $ NM 𝑃" 𝑁 " 1 1 − 𝑟 𝑒 s − − 𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ t = 0 2 𝑟 𝑟 2 2 1 $ NM 𝑃" − 𝑁′ 1 1 − 𝑟 𝑒 s − 𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ t = 0 2 𝑟 2 2 1 1 1 𝑁′ − 𝑃′ − 𝑟 $ 𝑒 NM s−𝑁 "" + 𝑁 " 𝑃" − (𝑁 " )$ − t=0 2 2 2 𝑟 1 1 1 𝑁′ − 𝑃′ − 𝑟 $ 𝑒 NM s𝑁 "" − 𝑁 " 𝑃" + (𝑁 " )$ + t=0 2 2 2 𝑟

33-term:

1 𝑅%% − 𝑔%% 𝑅 = 0 2 𝑅%% = sin$ 𝜃 𝑅$$ 𝑔%% = sin$ 𝜃 𝑔$$ 1 1 1 𝑅%% − 𝑔%% 𝑅 = sin$ 𝜃 𝑅$$ − sin$ 𝜃 𝑔$$ 𝑅 = sin$ 𝜃 a𝑅$$ − 𝑔$$ 𝑅b = 0 2 2 2

This gives the same equation as the one for the 22-term.

2. Derive Eq. (6.22) in the textbook. 𝑒 LNM " 1 𝑒L − a𝑃 − b − $ = 0 𝑟 𝑟 𝑟 " 𝑁 1 − − $ (1 − 𝑒 M ) = 0 𝑟 𝑟 1 $ NM "" 1 " " 1 " $ 𝑁 " − 𝑃" − 𝑟 𝑒 š𝑁 − 𝑃 𝑁 + (𝑁 ) + ›=0 2 2 2 𝑟 𝑒 LNM " 1 𝑒L 1 𝑒M 1 MNL " − a𝑃 − b − $ = 0 | × −𝑟 𝑒 → 𝑃 − + = 0 → 𝑃" = (1 − 𝑒 M ) 𝑟 𝑟 𝑟 𝑟 𝑟 𝑟 𝑁" 1 1 1 − − $ (1 − 𝑒 M ) = 0 | × 𝑟 → −𝑁 " = (1 − 𝑒 M ) → 𝑁 " = − (1 − 𝑒 M ) 𝑟 𝑟 𝑟 𝑟 𝑑𝑃 𝑑𝑁 1 → =− = (1 − 𝑒 M ) 𝑑𝑟 𝑑𝑟 𝑟

3. Derive Eqs. (6.37)-(6.38).

The non-zero metric elements of the Schwarzschild metric are 𝑔)) = − a1 −

2𝐺𝑀 𝑟< b = − U1 − V 𝑐$𝑟 𝑟

2𝐺𝑀 N# 𝑟< N# 𝑔## = a1 − $ b = U1 − V 𝑐 𝑟 𝑟 𝑔%% = 𝑟 $ and the coordinates are (𝑐𝑡, 𝑟, 𝜃, 𝜙).

Solving the Lagrange equation (6.35) for 𝑐𝑡 coordinate yields 𝜕𝐿 𝑚𝑐 𝑟< 𝑟< =− 𝑡̇ U1 − V 𝑐 $ = −𝑚𝑐 $ U1 − V 𝑡̇ = 𝑐𝑜𝑛𝑠𝑡. = −𝐸 𝜕𝑡̇ 𝑟 𝑟 9−𝑣5 𝑣5 so 𝑡̇ =

𝐸

𝑟 𝑚𝑐 $ U1 − 𝑟< V

Similarly, solving the Lagrange equation (6.35) for 𝜙 coordinate yields: 𝜙̇ ≡

𝐿O 𝑙O 𝑑𝜙 = = 𝑑𝜏 𝑚𝑟 $ 𝑟 $

where 𝐿O is the angular momentum and 𝑙O ≡ 𝐿O /𝑚 the angular momentum per unit P"

mass. The units of [𝐿O ] = kg B £ and [𝑙O ] =

P" B

£.

The dot product in Eq. (6.35) becomes −𝑘𝑚$ 𝑐 $ = 𝐩 ⋅ 𝐩 = 𝑝5 𝑝5 = 𝑔56 𝑝6 𝑝5 = 𝑚$

𝑑𝑥 5 𝑑𝑥 5 𝑑𝜏 𝑑𝜏

𝑟< 𝑟̇ $ $ ̇$ = 𝑚$ ¥− U1 − V 𝑐 $ 𝑡̇ $ + 𝑟< + 𝑟 𝜙 ¦ 𝑟 U1 − 𝑟 V 𝑟< −𝑘𝑚 𝑐 = −𝑚 ¥U1 − V 𝑐 $ 𝑟 $ $

$

$ 𝑙O − − ¦ 𝑟 $ U1 − 𝑟< V 𝑟 $ 𝑚$ 𝑐 G U1 − < V 𝑟 𝑟

𝐸$

𝑟̇ $

$ 𝑙O 𝑘𝑐 = 𝑟 − 𝑟 − $ 𝑚$ 𝑐 $ U1 − 𝑟< V U1 − 𝑟< V 𝑟 $

𝐸$

𝑟̇ $

$ 𝑙O 𝑘= 𝑟 − 𝑟 − $ $ 𝑚$ 𝑐 G U1 − 𝑟< V 𝑐 $ U1 − 𝑟< V 𝑐 𝑟

𝐸$

𝑟̇ $

$ 𝑙O 𝑟< 𝐸$ 𝑟̇ $ 𝑟< 𝑘 U1 − V = $ G − $ − $ $ U1 − V 𝑟 𝑚 𝑐 𝑐 𝑐 𝑟 𝑟 $ 𝑙O 𝐸$ 𝑟̇ $ 𝑟< 𝑟< = $ + 𝑘 U1 − V + $ $ U1 − V $ G 𝑚 𝑐 𝑐 𝑟 𝑐 𝑟 𝑟 $ 𝑙O 𝐸$ 𝑟̇ $ 𝑟< = $ + s𝑘 + $ $ t U1 − V $ G 𝑚 𝑐 𝑐 𝑐 𝑟 𝑟 $ 𝑙O 𝐸$ 𝑟< $ $ = 𝑟̇ + 𝑐 s𝑘 + $ $ t U1 − V $ $ 𝑚 𝑐 𝑐 𝑟 𝑟

Define the relativistic effective potential per unit mass, as in Eq. (6.38): 𝑉QRR (𝑟) ≡ 𝑐 $ s𝑘 +

$ 𝑙O 𝑟< t U1 − V. $ $ 𝑐 𝑟 𝑟

Then Eq. (6.37) is recovered: 𝐸$ 𝑑𝑟 $ $ − 𝑉 (𝑟) = 𝑟̇ ≡ a b QRR 𝑚$ 𝑐 $ 𝑑𝜏 4. (Numerical) Compute orbits of particles around an isolated massive object 𝑀 in Schwarzschild geometry by numerically integrating Eqs. (6.37)-(6.38) for various initial conditions (𝑟, 𝜙) and values of the integrals of motion 𝐸 and 𝑙𝜙. Plot these orbits.

Chapter 7 1. Starting with Friedmann’s first and second equations with the cosmological constant: 𝑎̇ $ 8𝜋𝐺 𝑘 Λ a b = 𝜌− $+ 𝑎 3 𝑎 3 𝑎̈ 4𝜋𝐺 Λ (𝜌 + 3𝑃) + =− 𝑎 3 3 Find the corresponding Friedmann’s third equation.

Lecture 7, slide 21, but with Λ. 𝑎̇ $ 8𝜋𝐺 𝑘 Λ 8𝜋𝐺 $ Λ a b = 𝜌− $+ → 𝑎̇ $ = 𝜌𝑎 − 𝑘 + 𝑎$ 𝑎 3 𝑎 3 3 3 8𝜋𝐺 8𝜋𝐺 $ Λ → 2𝑎̇ 𝑎̈ = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3

|

𝑑 𝑑𝑡

𝑎̈ 4𝜋𝐺 Λ 4𝜋𝐺 Λ (𝜌 + 3𝑃) + =− → 𝑎̈ = − 𝑎(𝜌 + 3𝑃) + 𝑎 𝑎 3 3 3 3 → 2𝑎̇ a−

4𝜋𝐺 Λ 8𝜋𝐺 8𝜋𝐺 $ Λ 𝑎(𝜌 + 3𝑃) + 𝑎b = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3 3 3

4𝜋𝐺 Λ 8𝜋𝐺 8𝜋𝐺 $ Λ 𝑎(𝜌 + 3𝑃) + 𝑎b = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3 3 3 8𝜋𝐺 2Λ 8𝜋𝐺 8𝜋𝐺 $ Λ − 𝑎𝑎̇ (𝜌 + 3𝑃) + 𝑎𝑎̇ = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3 3 3 8𝜋𝐺 8𝜋𝐺 8𝜋𝐺 $ − 𝑎𝑎̇ (𝜌 + 3𝑃) − 𝜌(2𝑎𝑎̇ ) = 𝜌̇ 𝑎 3 3 3 −𝑎𝑎̇ (𝜌 + 3𝑃 + 2𝜌) = 𝜌̇ 𝑎$

2𝑎̇ a−

𝑎̇ − (3𝜌 + 3𝑃) = 𝜌̇ 𝑎 𝑎̇ 𝜌̇ = −3 (𝜌 + 𝑃) 𝑎 2. Starting from the thermodynamic argument 𝑑𝑈 = −𝑃𝑑𝑉, and the adiabatic condition 𝑈 = 𝜌𝑐2𝑉, derive Friedmann’s third equation. The solution will essentially restate the energy conservation derived for the Friedmann’s first equation.

From thermodynamics: • Adiabatic (no net heat flow in/out of the system) compression /expansion is ;S ;U 𝑑𝑈 = −𝑃𝑑𝑉 → ;T = −𝑃 ;T Homogeneous and isotropic Universe is adiabatic.

From the definition: 𝑈 = 𝑚𝑐 $ = 𝜌𝑉𝑐 $ ;S

;VWU+ " X

Taking the time derivative: ;T = ;T Combining the two expressions for 𝑑𝑈/𝑑𝑡: 𝑑(𝜌𝑉𝑐 $ ) 𝑑𝑉 𝑑𝜌 $ 𝑑𝑉 $ 𝑑𝑉 𝑉̇ = −𝑃 → 𝑉𝑐 + 𝜌 𝑐 = −𝑃 → 𝜌̇ 𝑐 $ = − (𝜌𝑐 $ + 𝑃) 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑉 G ;U U̇ Ż % $ ̇ 𝑉 = % 𝜋𝑎 , ;T = 𝑉 = 4𝜋𝑎 𝑎̇ → U = 3 Z Ż

Ż

𝜌̇ 𝑐 $ = −3 Z (𝜌𝑐 $ + 𝑃) 𝜌̇ = −3 Z (𝜌 + 𝑃) Third Friedmann’s Equation Energy conservation again.

3. Find a geodesic equation for the expanding, flat Universe for which the spacetime line element is given in Eq. (7.14) in the lecture notes: 𝑑𝑟 $ 𝑑𝑠 $ = −𝑐 $ 𝑑𝑡 $ + 𝑎$ (𝑡) s + 𝑟 $ 𝑑𝜃 $ + 𝑟 $ sin$ 𝜃 𝑑𝜙 $ t 1 − 𝑘𝑟 $ by completing the following steps: a. Identify coordinates (𝑥 ) , 𝑥 # , 𝑥 $ , 𝑥 % ).

The coordinates are 𝑥 & = (𝑥 ) , 𝑥 # , 𝑥 $ , 𝑥 % ) = (𝑐𝑡, 𝑟, 𝜃, 𝜙) b. Write down the space metric 𝑔&' . −1 0 0 $( ) 𝑎 𝑡 ⎛0 0 𝑔&' = ⎜ 1 − 𝑘𝑟 $ 0 0 𝑎$ (𝑡)𝑟 $ ⎝0 0 0

0 0

⎞ ⎟

0 𝑎 𝑡 𝑟 sin$ 𝜃⎠ $( ) $

c. Write down the (alternate) Lagrangian which minimizes the line element and the Lagrange equations. # # # Z " (T) # # 𝐿 = $ 𝑔56 𝑥̇ 5 𝑥̇ 6 = − $ 𝑐 $ 𝑡̇ $ + $ #N]9 " 𝑟̇ $ + $ 𝑎$ (𝑡)𝑟 $ 𝜃̇ $ + $ 𝑎$ (𝑡)𝑟 $ sin$ 𝜃 𝜙̇ $ 𝜕𝐿 𝑑 𝜕𝐿 − =0 5 𝜕𝑥 𝑑𝜆 𝜕𝑥̇ 5 Each coordinate is parametrized by the parameter 𝜆, so ̇ ≡ 𝑑/𝑑𝜆.

We let ′ ≡ 𝑑/𝑑𝑡. ; ;T ; Also, ;1 I𝑎(𝑡)J = ;1 ;T I𝑎(𝑡)J = 𝑡̇𝑎" .

As a sanity check, we should recover the results for a geodesic on a static 3-D sphere, which we did in Appendix B in the Lecture Notes. To recover those results, we set 𝑎 = 1 and 𝑘 = 0. d. Solve the Lagrange equations for each coordinate 𝑥 8 to obtain a differential equation with 𝑥̈ 8 on the left-hand side of the equation. 𝒄𝒕-coordinate (𝝂 = 𝟎):

𝜕𝐿 𝑑 𝜕𝐿 1 𝜕𝐿 𝑑 𝜕𝐿 𝜕𝐿 𝑑 𝜕𝐿 − =0 → a − b=0 → − =0 𝜕𝑐𝑡 𝑑𝜆 𝜕𝑐𝑡̇ 𝑐 𝜕𝑡 𝑑𝜆 𝜕𝑡̇ 𝜕𝑡 𝑑𝜆 𝜕𝑡̇ 𝜕𝐿 𝑑 − (−𝑐 $ 𝑡̇) = 0 𝜕𝑡 𝑑𝜆 𝑎𝑎′ 𝑟̇ $ + 𝑎𝑎" 𝑟 $ 𝜃̇ $ + 𝑎𝑎" 𝑟 $ sin$ 𝜃 𝜙̇ $ + 𝑐 $ 𝑡̈ = 0 1 − 𝑘𝑟 $ 𝑎𝑎" 𝑐 $ 𝑡̈ = − 𝑟̇ $ − 𝑎𝑎" 𝑟 $ 𝜃̇ $ − 𝑎𝑎" 𝑟 $ sin$ 𝜃 𝜙̇ $ 1 − 𝑘𝑟 $ 8 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : and 𝜈 = 0 ) $ ) ̇$ ) ̇$ → 𝑡̈ = −Γ## 𝑟̇ − Γ$$ 𝜃 − Γ%% 𝜙 " 𝑎𝑎 ) ) ) Γ## = , Γ$$ = 𝑎𝑎" 𝑟 $ , Γ%% = 𝑎𝑎" 𝑟 $ sin$ 𝜃 1 − 𝑘𝑟 $ ) All other Christoffel symbols Γ.: are 0. 𝒓-coordinate (𝝂 = 𝟏):

𝜕𝐿 𝑑 𝜕𝐿 − =0 𝜕𝑟 𝑑𝜆 𝜕𝑟̇ 𝜕𝐿 𝑑 𝑎$ − s 𝑟̇ t = 0 𝜕𝑟 𝑑𝜆 1 − 𝑘𝑟 $ 𝑎$ 𝑎$ 𝑎$ 2𝑎𝑎′𝑡̇ $ $ ̇$ $ $ $ $ ̇ 𝑘𝑟 𝑟̇ + 𝑎 𝑟𝜃 + 𝑎 𝑟 sin 𝜃 𝜙 − 𝑟̈ − 2𝑘𝑟 𝑟̇ − 𝑟̇ = 0 (1 − 𝑘𝑟 $ )$ (1 − 𝑘𝑟 $ )$ 1 − 𝑘𝑟 $ 1 − 𝑘𝑟 $ 1 − 𝑘𝑟 $ 𝑎$ 2𝑎𝑎′ $ $ ̇$ $ $ ̇$ − 𝑟̈ = s−𝑘𝑟 𝑟̇ + 𝑎 𝑟𝜃 + 𝑎 𝑟 sin 𝜃 𝜙 𝑟̇ 𝑡̇ t (1 − 𝑘𝑟 $ )$ 𝑎$ 1 − 𝑘𝑟 $ 𝑘𝑟 2𝑎′ $ $ ̇$ $) $ ̇$ − 𝑟̈ = − 𝑟̇ + 𝑟(1 − 𝑘𝑟 )𝜃 + 𝑟(1 − 𝑘𝑟 sin 𝜃 𝜙 𝑟̇ 𝑡̇ 1 − 𝑘𝑟 $ 𝑎 8 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : and 𝜈 = 1 # # # $ # ̇$ # ̇$ → 𝑟̈ = −(Γ)# + Γ#) )𝑟̇ 𝑡̇ − Γ## 𝑟̇ − Γ$$ 𝜃 − Γ%% 𝜙 " 𝑎 𝑘𝑟 # # # Γ)# = Γ#) = , Γ## = , 𝑎 1 − 𝑘𝑟 $ # # Γ$$ = −𝑟(1 − 𝑘𝑟 $ ), Γ%% = −𝑟(1 − 𝑘𝑟 $ ) sin$ 𝜃, # All other Christoffel symbols Γ.: are 0. 𝜽-coordinate (𝝂 = 𝟐):

𝜕𝐿 𝑑 𝜕𝐿 − =0 𝜕𝜃 𝑑𝜆 𝜕𝜃̇ 𝜕𝐿 𝑑 𝜕𝐿 − =0 𝜕𝜃 𝑑𝜆 𝜕𝜃̇ 𝑎$ 𝑟 $ sin 𝜃 cos 𝜃 𝜙̇ $ − 𝑎$ 𝑟 $ 𝜃̈ − 𝑎$ 2𝑟𝑟̇ 𝜃̇ − 2𝑎𝑎" 𝑡̇𝑟 $ 𝜃̇ = 0 2 2𝑎′ 𝜃̈ = sin 𝜃 cos 𝜃 𝜙̇ $ − 𝑟̇ 𝜃̇ − 𝑡̇𝜃̇ 𝑟 𝑎 8 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : and 𝜈 = 2 $ ̇$ $ $ )𝑟̇ ̇ $ $ )𝑡̇ ̇ → 𝜃̈ = −Γ%% 𝜙 − (Γ#$ + Γ$# 𝜃 − (Γ)$ + Γ$) 𝜃

1 𝑎′ $ $ $ $ Γ#$ = Γ$# = , Γ)$ = Γ$) = 𝑟 𝑎 $ All other Christoffel symbols Γ.: are 0. $ Γ%% = − sin 𝜃 cos 𝜃,

𝝓-coordinate (𝝂 = 𝟑):

𝜕𝐿 𝑑 𝜕𝐿 − =0 𝜕𝜙 𝑑𝜆 𝜕𝜙̇

𝑑 $ $ $ I𝑎 𝑟 sin 𝜃 𝜙̇ J = 0 𝑑𝜆 −2𝑎$ 𝑟𝑟̇ sin$ 𝜃 𝜙̇ − 2𝑎$ 𝑟 $ sin 𝜃 cos 𝜃 𝜃̇𝜙̇ − 𝑎$ 𝑟 $ sin$ 𝜃 𝜙̈ − 2𝑎𝑎̇ 𝑡̇𝑟 $ sin$ 𝜃 𝜙̇ = 0 2 2 cos 𝜃 2𝑎′ 𝜙̈ = − 𝑟̇ 𝜙̇ − 𝜃̇𝜙̇ − 𝑡̇𝜙̇ 𝑟 sin 𝜃 𝑎 8 𝑥̈ 8 = −Γ.: 𝑥̇ . 𝑥̇ : and 𝜈 = 3 % % )𝑟̇ ̇ % % % % → 𝜙̈ = −(Γ#% + Γ%# 𝜙 − (Γ$% + Γ%$ )𝜃̇𝜙̇ − (Γ)% + Γ%) )𝑡̇𝜙̇ 𝑎′ 1 cos 𝜃 % % % % % % Γ)% = Γ)% = , Γ#% = Γ#% = , Γ$% = Γ%$ = = cot 𝜃 𝑎 𝑟 sin 𝜃 % All other Christoffel symbols Γ.: are 0. −

It all checks with the Appendix B, as it should.

8 e. Identify all Christoffel symbols Γ.: by comparing the differential equations in part d. with the geodesic equation from the notes: 8 . : 𝑥̈ 8 = −Γ.: 𝑥̇ 𝑥̇ Color coded in part d.

Chapter 8 1. (Numerical) Use your favorite numerical tool to reproduce curves in the Figure 8.2.

Python code included in the distribution.

2. For the flat Friedmann’s Universe, compute when the transition between the radiationdominated and matter-dominated Universe happens.

Appendix G 3. Convert 𝜌^_C`,) = 3𝐻)$ /8𝜋𝐺 into: e. [5 pts] g cmN% f. [5 pts] eV cmN% g. [5 pts] protons cmN% h. [5 pts] 𝑀⨀ Mpc N% Use 𝐻) = 100ℎ km s N# Mpc N# . Give all answers in terms of the dimensionless parameter ℎ first and then use ℎ ≈ 0.73 for a numerical result. 𝐻) = 100ℎ km s N# Mpc N# ℎ ≈ 0.73 N## % 𝐺 = 6.67 × 10 m kg N# s N$ = 6.67 × 10Nb cm% g N# s N$ 1 year = 365 × 24 × 60 × 60 s = 3.15 × 10c s 1 pc = 3.26 ly 1 Mpc = 3.26 × 1 year × 30,000,000,000 cm s N# = 3.1 × 10$G cm 𝐻) = 100ℎ km s N# Mpc N# = 100ℎ × (1000 m) s N# (3.1 × 10$G cm)N# 𝐻) = 3.23 × 10N#b ℎ s N# 1 𝐻) = 2.26 × 10N#b s N# ≈ = 1/(13.7 × 10d years) 𝑡) 𝜌^_C`,) = a.

3𝐻)$ 3(3.23 × 10N#b ℎ s N# )$ = = 1.87 × 10N$d ℎ$ g cmN% 8𝜋𝐺 8𝜋(6.67 × 10Nb m% kg N# s N$ )

𝜌^_C`,) = 1.87 × 10N$d ℎ$ g cmN% ≈ 0.91 × 10N$d g cmN% 𝐸 = 𝑚𝑐 $ = (1 g)(3 × 10#) cm/s)$ = 9 × 10$) g cm$ s N$ 𝐸 = 9 × 10#% kg m$ s N$ = 9 × 10#% J 1 J = 6.24 × 10#b eV 1 g → 𝐸 = 9 × 10#% (6.24 × 10#b ) eV = 5.6 × 10%$ eV 𝜌^_C`,) = 1.87 × 10N$d ℎ$ g cmN% = 1.05 × 10G ℎ$ eV cmN%

b. 𝜌^_C`,) = 1.05 × 10G ℎ$ eV cmN% ≈ 5 × 10% eV cmN% 1 proton = 938 MeV = 9.38 × 10b eV 1 eV = (9.38 × 10b )N# protons 𝜌^_C`,) = 1.05 × 10G ℎ$ (9.38 × 10b )N# protons cmN% c. 𝜌^_C`,) = 1.1 × 10Ne ℎ$ protons cmN% ≈ 5.5 × 10Nf protons cmN% d.

or 𝜌^_C`,) ≈ 5.5 protons mN% 𝑀⊙ = 1.99 × 10%) kg = 1.99 × 10%% g → 1 g = (1.99 × 10%% )N# 𝑀⊙ 1 Mpc = 3.1 × 10$G cm → 1 cm = (3.1 × 10$G )N# Mpc 𝜌^_C`,) = 1.87 × 10N$d ℎ$ g cmN% 𝜌^_C`,) = 1.87 × 10N$d ℎ$ (1.99 × 10%% )N# (3.1 × 10$G )% ℎ$ 𝑀⊙ Mpc N% 𝜌^_C`,) = 2.8 × 10## ℎ$ 𝑀⊙ Mpc N% 𝜌^_C`,) = 1.37 × 10## 𝑀⊙ Mpc N%

Chapter 9 1. Derive Eq. (9.2) in the lecture notes using results from Table 8.2 and Eq. (8.8).

Eq. (8.8): −𝑘 = 𝑎$ 𝐻$ (1 − 2𝑞) W Eq. (8.10): 𝑞 = $W $%

#

#

#N$i&

&

#

Eq. (9.2): 𝑡) = h ¿

+ #N$i &

Eq. (8.20):

GjkW& %

#

−

i&

' (#N$i& )"

i&

' (#N$i& )"

#Ni

coshN# U i &V &

#Ni&

cos N# U i V

i

&

#

for 𝑞) < $ #

for 𝑞) ≥ $

#

= 𝐻)$ 𝑞) = $i &N# → 𝐻)$ = $i N# & & # # 1 → 𝐻) = (2𝑞) − 1)N$ → = (2𝑞) − 1)$ 𝐻)

𝑘 = +1, 𝑞) ≥ $

%h "

Eq. (8.6): 𝜌 = Gjk 𝑞 Eq. (8.8): −𝑘 = 𝑎$ 𝐻$ (1 − 2𝑞) For 𝑘 = +1 and now 𝑎 = 𝑎) = 1, 𝐻 = 𝐻) : −1 = 𝑎)$ 𝐻)$ (1 − 2𝑞) ) → 𝐻)$ = From Eq. (8.31):

# 1 1 → = (2𝑞) − 1)$ 2𝑞) − 1 𝐻)

𝑞) 𝑎(2𝑞) − 1) (1 − cos 𝜂) → 1 − = cos 𝜂 → 2𝑞) − 1 𝑞) 𝑎(2𝑞) − 1) 𝜂 = cos N# s1 − t 𝑞) 𝑞) (𝜂 − sin 𝜂) 𝑡= 2𝑞) − 1 Today 𝑎 = 𝑎) = 1, 𝜂 = 𝜂) and 𝑡 = 𝑡) so 𝑎(2𝑞) − 1) 2𝑞) − 1 1 − 𝑞) 𝜂) = cos N# s1 − t = cos N# a1 − b = cos N# a b 𝑞) 𝑞) 𝑞) 𝑞) (𝜂 − sin 𝜂) ) 𝑡) = 2𝑞) − 1 ) 𝑞) 1 − 𝑞) 1 − 𝑞) 𝑡) = acos N# a b − sin acos N# a bbb 2𝑞) − 1 𝑞) 𝑞) 𝑎=

From old books or by brute force: sin (cos N# 𝑥) = 9(1 − 𝑥)(1 + 𝑥) = 91 − 𝑥 $ so sin acos N# a

1 − 𝑞) 1 − 𝑞) $ 𝑞$ − 1 + 2𝑞) − 𝑞)$ 92𝑞) − 1 bb = k1 − a b =k ) = 𝑞) 𝑞) 𝑞) 𝑞)$ 𝑞) 1 − 𝑞) 92𝑞) − 1 scos N# a b− t 2𝑞) − 1 𝑞) 𝑞) 𝑞) 1 − 𝑞) 92𝑞) − 1 N# = 92𝑞) − 1 šcos a b − › % 𝑞) 𝑞) (2𝑞) − 1)$

𝑡) =

=

1 𝑞) 1 − 𝑞) 𝑞) 92𝑞) − 1 N# ¥ cos a b − ¦ % % 𝐻) (2𝑞 𝑞) 𝑞) $ $ (2𝑞 − 1) − 1) ) )

𝑡) =

1 𝑞) 1 − 𝑞) 1 ¥ cos N# a b− ¦ % 𝐻) (2𝑞 𝑞) 2𝑞) − 1 $ − 1) )

𝑡) =

1 1 𝑞) 1 − 𝑞) ¥ + cos N# a b¦ % 𝐻) 1 − 2𝑞) (2𝑞 𝑞) $ ) − 1)

#

𝑘 = −1, 𝑞) < $

%h "

Eq. (8.6): 𝜌 = Gjk 𝑞 Eq. (8.8): −𝑘 = 𝑎$ 𝐻$ (1 − 2𝑞) For 𝑘 = −1 and now 𝑎 = 𝑎) = 1, 𝐻 = 𝐻) : 1 = 𝑎)$ 𝐻)$ (1 − 2𝑞) ) → 𝐻)$ = From Eq. (8.42): 𝑎=

# 1 1 → = (1 − 2𝑞) )$ 1 − 2𝑞) 𝐻)

𝑞) 1 − 2𝑞) (cosh 𝜂 − 1) → 𝑎 + 1 = cosh 𝜂 → 1 − 2𝑞) 𝑞) 1 − 2𝑞) 𝜂 = coshN# a 𝑎 + 1b 𝑞) 𝑞) (sinh 𝜂 − 𝜂) 𝑡= 1 − 2𝑞)

Today 𝑎 = 𝑎) = 1, 𝜂 = 𝜂) and 𝑡 = 𝑡) so 1 − 2𝑞) 1 − 2𝑞) + 𝑞) 1 − 𝑞) 𝜂) = coshN# a 𝑎) + 1b = coshN# a b = coshN# a b 𝑞) 𝑞) 𝑞)

𝑞) (sinh 𝜂) − 𝜂) ) 1 − 2𝑞) 𝑞) 1 − 𝑞) 1 − 𝑞) 𝑡) = Äsinh acoshN# a bb − coshN# a bÅ 1 − 2𝑞) 𝑞) 𝑞) From old books or by brute force: sinh(coshN# 𝑥) = 9(𝑥 − 1)(𝑥 + 1) so 𝑡) =

1 − 𝑞) 1 − 𝑞) 1 − 𝑞) 1 − 2𝑞) 1 sinh acoshN# a bb = ka − 1b a + 1b = ka ba b 𝑞) 𝑞) 𝑞) 𝑞) 𝑞) 91 − 2𝑞) 𝑞) 𝑞) 1 − 𝑞) 91 − 2𝑞) 𝑡) = š − coshN# a b› 1 − 2𝑞) 𝑞) 𝑞) 𝑞) 1 − 𝑞) 91 − 2𝑞) N# = 91 − 2𝑞) š − cosh a b› % 𝑞) 𝑞) (1 − 2𝑞) )$

=

1 𝑞) 𝑞) 1 − 𝑞) 91 − 2𝑞) N# ¥ − cosh a b¦ % % 𝐻) (1 𝑞) 𝑞) (1 − 2𝑞) )$ − 2𝑞) )$

=

𝑡) =

1 1 𝑞) 1 − 𝑞) ¥ − coshN# a b¦ % 𝐻) 1 − 2𝑞) (1 𝑞) − 2𝑞) )$

2. Verify that for 𝑞 = 1/2, Eq. (8.20) is recovered from Eq. (9.2).

From Eq. (8.18): 2 1 𝑡) = 3 𝐻) # So, we need to show that the brackets in the solution for 𝑞) ≥ $ we derived in part A #

reduce to 2/3 for 𝑞) = $: 𝑡) =

1 1 𝑞) 1 − 𝑞) N# ¥ + cos a b¦ % 𝐻) 1 − 2𝑞) (2𝑞 𝑞) $ − 1) )

[] = ¥−

1 𝑞) 1 − 𝑞) + cos N# a b¦ % 2𝑞) − 1 (2𝑞 𝑞) $ ) − 1) #

Obviously, direct substitution of 𝑞) = $ into it would lead to mathematical problems. cos

N# (𝑥)

% # (1 − 𝑥)$ ≈ √2(1 − 𝑥)$ +

6√2

around 𝑥 = 1

𝑥≡

1 − 𝑞) 2𝑞) − 1 → 1−𝑥 = 𝑞) 𝑞) #

→ cos

N#

%

1 − 𝑞) 2𝑞) − 1 $ 1 2𝑞) − 1 $ a b ≈ √2 a b + a b 𝑞) 𝑞) 𝑞) 6√2 #

%

1 𝑞) 2𝑞) − 1 $ 1 2𝑞) − 1 $ [ ] = ¥− + „√2 a b + a b …¦ % 2𝑞) − 1 (2𝑞 𝑞) 𝑞) 6√2 $ ) − 1) [ ] = Ç−

1 1 1 1 1 92𝑞) 92𝑞) − 1 + + È = Ç + È 2𝑞) − 1 2𝑞) − 1 6√2 𝑞#/$ 2𝑞) − 1 6√2 𝑞#/$ )

=Ç

[ ]=Ç

92𝑞) − 1 I92𝑞) − 1JI92𝑞) + 1J 1

92𝑞) + 1

+

1

1

È

+

)

1

1

6√2 𝑞)#/$

6√2 𝑞)#/$ i& l#/$

=

È

1 1 1 2 √2 + = + = 2 6√2 2 6 3

3. For the flat Friedmann’s Universe with a cosmological constant, compute when the transition between the matter-dominated and dark energy-dominated Universe happens. 𝑎̇ $ 8𝜋𝐺 𝑘 Λ a b = 𝜌− $+ 𝑎 3 𝑎 3 Flat, radiation-dominated Universe: 𝑘 = 0, 𝜌 = 𝜌P = 𝜌P,) 𝑎N% 𝑎̇ $ 8𝜋𝐺 Λ 8𝜋𝐺 a b = 𝜌P,) 𝑎N% + = (𝜌P,) 𝑎N% + 𝜌m,) ) 𝑎 3 3 3 In Eq. (8.9), we defined the critical density 3𝐻$ 𝜌^_ = .

8𝜋𝐺 The mixture of dark matter and energy is given by Eq. (9.11): 𝜌 = 𝜌P + 𝜌m = 𝜌P,) 𝑎N% + 𝜌m Eq. (9.10): 𝜌P,) 8𝜋𝐺 ΩP,) ≡ → ΩP,) 𝐻)$ = 𝜌 𝜌^_,) 3 P,) 𝜌m,) 8𝜋𝐺 Ωm,) ≡ → Ωm,) 𝐻)$ = 𝜌 𝜌^_,) 3 m,) Flat Universe: Ω( = ΩP + ΩnQ = 1 $ 𝑎̇ 8𝜋𝐺 Λ a b = 𝜌P,) 𝑎N% + = 𝐻)$ (ΩP,) 𝑎N% + Ωm,) ) 𝑎 3 3 The transition will take place when the two energy density terms are equal: ΩP,) 𝑎N% = Ωm,) Solving for the scale of the Universe 𝑎(𝑡):

#/%

ΩP,) 𝑎=s t Ωm,) For the values of parameters mentioned in class: ΩP,) = 0.28 and Ωm,) = 0.72, the size of the Universe at the time of transition to dark energy-dominated Universe was #/% ΩP,) 0.28 #/% 𝑎Qo$ = s t =a b ≈ 0.73 Ωm,) 0.72 From Lecture 9, slide 21: pm T

Today, 𝑎(𝑡) ) = 𝑎) = 1, so

𝑎(𝑡) ∝ 𝑒 %

pmT

or 𝑎(𝑡) = 𝑎% 𝑒 % pm(TNT& )

𝑎(𝑡) = 𝑒 % From Lecture 9, slide 22:

Ωm,) ≡

Λ Λ k = 𝐻) 9Ωm,) $ → 3 3𝐻)

𝑎(𝑡) = 𝑒 qr(,& h& (TNT& ) At the transition, 𝑎(𝑡) = 𝑒 qr(,& h& (TNT&) = 𝑎Qo$ We find the time (the age of the Universe) at the matter-to-dark energy dominance transition by solving the equation above for 𝑡 𝑒 qr(,& h& (TNT&) = 𝑎Qo$ 𝑒 qr(,& h& T = 𝑒 qr(,& h& T& 𝑎Qo$ = 𝑒 qr(,& h& T& 𝑒 stu Z*+" = 𝑒 qr(,& h& T& vstu Z*+" 9Ωm,) 𝐻) 𝑡 = 9Ωm,) 𝐻) 𝑡) + log 𝑎Qo$ log 𝑎Qo$ 𝐻) 𝑡 = 𝐻) 𝑡) + 9Ωm,) Age of the Universe is about 13.7 billion years, so that 𝐻) 𝑡) = 1. Cast in this way, the second term will show us when the transition occurred in terms of this value, or in the fraction of the age of the Universe. log 𝑎Qo$ log 0.73 = ≈ −0.37 √0.72 9Ωm,) That means that the transition time was in about 0.63 𝑡) = 8.6 billion years.

4. Starting with Friedmann’s first and second equations with the cosmological constant: 𝑎̇ $ 8𝜋𝐺 𝑘 Λ a b = 𝜌− $+ 𝑎 3 𝑎 3 𝑎̈ 4𝜋𝐺 Λ (𝜌 + 3𝑃) + =− 𝑎 3 3

Find the corresponding Friedmann’s third equation.

Lecture 7, slide 21, but with Λ. 𝑎̇ $ 8𝜋𝐺 𝑘 Λ 8𝜋𝐺 $ Λ a b = 𝜌− $+ → 𝑎̇ $ = 𝜌𝑎 − 𝑘 + 𝑎$ 𝑎 3 𝑎 3 3 3 8𝜋𝐺 8𝜋𝐺 $ Λ → 2𝑎̇ 𝑎̈ = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3

|

𝑑 𝑑𝑡

𝑎̈ 4𝜋𝐺 Λ 4𝜋𝐺 Λ (𝜌 + 3𝑃) + =− → 𝑎̈ = − 𝑎(𝜌 + 3𝑃) + 𝑎 𝑎 3 3 3 3 → 2𝑎̇ a−

4𝜋𝐺 Λ 8𝜋𝐺 8𝜋𝐺 $ Λ 𝑎(𝜌 + 3𝑃) + 𝑎b = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3 3 3

4𝜋𝐺 Λ 8𝜋𝐺 8𝜋𝐺 $ Λ 𝑎(𝜌 + 3𝑃) + 𝑎b = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3 3 3 8𝜋𝐺 2Λ 8𝜋𝐺 8𝜋𝐺 $ Λ − 𝑎𝑎̇ (𝜌 + 3𝑃) + 𝑎𝑎̇ = 𝜌(2𝑎𝑎̇ ) + 𝜌̇ 𝑎 + 2𝑎𝑎̇ 3 3 3 3 3 8𝜋𝐺 8𝜋𝐺 8𝜋𝐺 $ − 𝑎𝑎̇ (𝜌 + 3𝑃) − 𝜌(2𝑎𝑎̇ ) = 𝜌̇ 𝑎 3 3 3 −𝑎𝑎̇ (𝜌 + 3𝑃 + 2𝜌) = 𝜌̇ 𝑎$

2𝑎̇ a−

𝑎̇ − (3𝜌 + 3𝑃) = 𝜌̇ 𝑎 𝑎̇ 𝜌̇ = −3 (𝜌 + 𝑃) 𝑎

Chapter 10 1. Consider the size of the particle horizon estimated for a flat, matter-dominated Universe in Eq. (10.23). Compute by how much the estimate would change if we accounted for radiation-dominated early period from 𝑡 = 0 to 𝑡 = 𝑡) /200,000.

Radiation-dominated Universe, Eq. (8.15): 𝑎(𝑡) = (2𝐻) )#/$ 𝑡 #/$ ≡ 𝑎# 𝑡 #/$ Matter-dominated Universe, Eq. (8.17): 𝑎(𝑡) = (3𝐻) /2)$/% 𝑡 $/% ≡ 𝑎$ 𝑡 $/% Eq. (10.18): T& 𝑑𝑡 " 𝜂(𝑡) ) ≡ 𝜂) = Ë " ) 𝑎(𝑡 ) T $/%

T $/%

Matter-dominated: 𝑎P (𝑡) = 𝑎) UT V

= UT V

&

&

𝜂(𝑡) ) ≡ 𝜂) = Ë T #/$

Radiation-dominated: 𝑎_ (𝑡) = 𝑎) UT V &

)

T*+!

)

𝑑𝑡 "

)

𝑡 " #/$

𝜂) = Ë

U𝑡 V

𝑑𝑡 "

T*+!

𝑡 " $/%

)

T #/$

U𝑡 V )

&

during 𝑡 ∈ Í0, 𝑡Qo# Î

T& 𝑑𝑡 𝑑𝑡 " +Ë " 𝑎_ (𝑡 " ) T*+! 𝑎P (𝑡 )

T&

+Ë

𝑑𝑡 " = 3𝑡) 𝑎P (𝑡 " )

= UT V

𝜂(𝑡) ) ≡ 𝜂) = Ë T*+!

T&

"

#/$ #/$ $/% #/% #/% 𝑡Qo# + 𝑡) 3(𝑡) − 𝑡Qo# )

= 𝑡)

#/$ #/$ $/% #/% 𝑡Qo# − 3𝑡) 𝑡Qo#

𝜂) = 3𝑡) + 𝑡)

The difference between accounting for the early (short) radiation-dominated Universe is shown in green in the equation above. Because 𝑡) ≫ 𝑡Qo# , (from the lecture notes 𝑡Qo# = 𝑡) /200,000), each of the green terms will be much smaller than the leading term (in red).

2. Compute by how much the estimate in the previous exercise would change if we accounted for the dark energy-dominated late period.

Dark energy-dominated: 𝑎m (𝑡) = 𝑒 h& (TNT& ) during 𝑡 ∈ Í𝑡Qo$ , 𝑡) Î T*+" T& 𝑑𝑡 " 𝑑𝑡 " ) 𝜂(𝑡) ≡ 𝜂) = Ë +Ë " 𝑎P (𝑡 " ) ) T*+" 𝑎m (𝑡 ) T*+" T& 𝑑𝑡 " 𝑑𝑡 " 𝜂) = Ë + Ë $ h& (T # NT& ) ) T*+" 𝑒 𝑡" % U𝑡 V )

$/% #/% 𝜂) = 3𝑡) 𝑡Qo$ −

1 Nh (T NT ) I𝑒 & & & − 𝑒 Nh&VT*+" NT&X J 𝐻)

1 I1 − 𝑒 Nh& VT*+" NT&X J 𝐻) Because 𝐻) (𝑡Qo$ − 𝑡) ) ≪ 1, we can expand the last term 1 1 I1 − 𝑒 Nh& VT*+" NT&X J ≈ U1 − 1 + 𝐻) I𝑡Qo$ − 𝑡) JV = 𝑡Qo$ − 𝑡) 𝐻) 𝐻) $/% #/% $/% #/% 𝜂) = 3𝑡) 𝑡Qo$ − I𝑡Qo$ − 𝑡) J = 3𝑡) 𝑡Qo$ + 𝑡) − 𝑡Qo$ Let Δ ≡ 𝑡) − 𝑡Qo$ > 0, but (𝑡) − 𝑡Qo$ )/𝑡) ≪ 1, so 𝑡Qo$ = 𝑡) − Δ $/% $/% #/% 𝜂) = 3𝑡) (𝑡) − Δ)#/% + Δ = 3𝑡) 𝑡) (1 − Δ/𝑡) )#/% + Δ 1 Δ 1 Δ $ #/% 𝜂) = 3𝑡) (1 − Δ/𝑡) ) + Δ ≈ 3𝑡) s1 − a b − a b t + Δ 3 𝑡) 9 𝑡) $ Δ 𝜂) ≈ 3𝑡) − Δ − +Δ 3𝑡) Δ$ 𝜂) ≈ 3𝑡) − 3𝑡) The radius of the sphere defining the particle horizon today is slightly smaller (by amount shown in blue in the equation above) if accounting for the dark energy-dominated late period. $/%

𝜂) = 3𝑡)

#/%

𝑡Qo$ −

3. (Numerical) Use your favorite plotting tool to reproduce curves in Figure 10.2.

Python code included in the distribution.

4. Assume the Universe today is flat with both matter (ΩP,) ) and a cosmological constant (Ωm,) ). a. (Numerical) Compute the conformal age or the particle horizon of the Universe and plot your result as a function of ΩP,) . Numerical integration may be necessary.

Plot it using your favorite electronic graphics tool.

T 𝑐𝑑𝑡̃ 𝑑w ≡ 𝑐𝜂 = Ë ̃ ) 𝑎(𝑡 ) ;T

;Z

But 𝑑𝑡 = ;Z 𝑑𝑎 = Ż .

Friedmann’s first equation, massaged as in Eq. (9.18) in the Lecture Notes: 𝑎̇ $ a b = 𝐻)$ ÍΩP,) 𝑎N% + Ωm,) Î 𝑎 This is for the flat Universe, so ΩP,) + Ωm,) = 1 → Ωm,) = 1 − ΩP,) → 𝑎̇ $ a b = 𝐻)$ ÍΩP,) 𝑎N% + (1 − ΩP,) )Î 𝑎 Solving for 𝑎̇ 𝑎̇ = 𝐻) :ΩP,) 𝑎N# + I1 − ΩP,) J𝑎$ So the integral above becomes from the beginning of the Universe 𝑎 = 0 to today 𝑎) = 1

Z& 𝑐𝑑𝑡̃ 𝑑w = 𝑐𝜂 = Ë =Ë ̃ ) 𝑎(𝑡 ) ) T

𝑑w =

𝑐 # Ë 𝐻) )

𝑐𝑑𝑎Ó 𝑎Ó𝐻) :ΩP,)

𝑎ÓN# + I1 − Ω

= P,)

J𝑎Ó$

𝑑𝑎Ó :ΩP,) 𝑎Ó + I1 − ΩP,) J𝑎ÓG

+

The prefactor h just gives the distance in units of 13.7 billion light years because #

&

𝑡) = h ≈ 13.7 billion years. &

If you put this into Wolfram Alpha, you will get a closed-form solution in terms of a hypergeometric function, but it’s just as easy to integrate this numerically. I do that in Python, which I also use to plot the dependence. The code is attached below. #!/usr/bin/python #------------------------------------------------------------------# Code to compute the conformal age or the particle horizon of the # Universe and plot the result as a function of Omega_(m,0). #------------------------------------------------------------------import matplotlib.pyplot as plt import numpy as np from mpmath import * from math import sqrt #------------------------------------------------------------------def f(a): # define the integrand return 1.0/sqrt(Omega_m0*a+(1.0-Omega_m0)*(a**4)) N = 100 # number of points in the plot x_min = 0.01 # minimum Omega_m,0 (avoid 0 -- singularity) x_max = 1.0 # maximum Omeega_m,0 = 1 x = np.mgrid[x_min:x_max:N*1j] const = 13.7 # prefactor c/H_0 y = const*np.zeros(N) for i in range(N): Omega_m0 = x[i] # set Omega_m,0 value for integration y[i] = const*quad(f, [0, 1]) # numerically integrate from 0 to 1 # print (x[i],y[i]) # print values on the screen fig, ax1 = plt.subplots() fig.subplots_adjust(left=0.20,bottom=0.14,right=0.9 ,top=0.9,wspace=0.5,hspace=0.35) plt.tick_params(axis='both', labelsize=14, pad=7) ax1.set_xlabel(r"$\Omega_{\rm m,0}$", size=16) ax1.set_ylabel(r"$d_p$ [billion light years]", size=16) ax1.set_title(r"Particle Horizon with Varying Mass Content", size=14) ax1.set_xlim([0,1]) #ax1.set_ylim([0,10]) ax1.plot(x,y,color='k',lw=2,label='') plt.savefig('fig_HW6.pdf',format='pdf') plt.show()

b. What is the current particle horizon size for a Universe with ΩP,) = 0.28? Use 𝐻) = 100ℎ km s N# Mpc N# , where ℎ ≈ 0.73.

We can get the value of the integral at ΩP,) = 0.28 from the code above, which # uses 𝑡) = h ≈ 13.7 billion years, so 𝐻) = 100 ℎ km s N# Mpc N# and ℎ = 0.73. &

Quick and dirty way to do that is increase the number of points in the code and uncomment the print line. Then read off the y-value for ΩP,) = 0.28. I find 𝑑w = 46.6 billion light years c. How would the previous estimate change if the value of ℎ ≈ 0.67 were used?

In part b. above, we used ℎ = 0.73. 𝑐 # 𝑑𝑎Ó 1 𝑑w = Ë → 𝑑w ∝ → 𝑑w ℎ = 𝑐𝑜𝑛𝑠𝑡. 𝐻) ) ℎ :ΩP,) 𝑎Ó + I1 − ΩP,) J𝑎ÓG 𝑑w,x0y ℎx0y = 𝑑w ℎ → 𝑑w,x0y = 𝑑w

ℎ

ℎx0y 0.67 𝑑w,x0y = 46.6 billion light years = 62.4 billion light years 0.5 Moral of the story here is that the size of the particle horizon and the age of the Universe depend on the value of the Hubble parameter, whose value is being updated regularly. This is why in a lot of places, including my textbook, the results are given in terms of ℎ, from the definition of the Hubble parameter 𝐻) = 100 ℎ km s N# Mpc N# .

d. What is the mass contained within the current particle horizon in solar masses for ΩP,) = 0.28? 𝜌P,) ΩP,) = 𝜌^_C`,) ## N% From 1.d: 𝜌^_C`,) = 1.37 × 10 𝑀⊙ Mpc , so 𝜌P,) = ΩP,) 𝜌^_C`,) ≈ 0.38 × 10## 𝑀⊙ Mpc N% The radius of the particle horizon for ΩP,) = 0.28 from the plot above is 𝑑w = 46.6 billion light years = 14.3 × 10d pc = 14.3 × 10% Mpc Therefore, the volume of the particle horizon is 4 4 𝑉 = 𝜋𝑑w% = 𝜋(2.92 × 10#$ ) Mpc % 3 3 And the total baryonic mass in terms of solar masses is 4 𝑀 = 𝜌P,) 𝑉 = I0.38 × 10## 𝑀⊙ Mpc N% J a 𝜋(2.92 × 10#$ ) Mpc % b 3 𝑀 ≈ 4.65 × 10$% 𝑀⊙

Chapter 11 1. Use your favorite plotting tool (do not do it by hand!) to reproduce the Figure 11.2.

As discussed in the lectures and notes, the Universe was first dominated by radiation, then by matter, and, finally, by dark energy. The scale of the Universe 𝑎(𝑡) scales with time as Radiation-dominated Universe, Eq. (8.17): 𝑎(𝑡) = 𝑎# 𝑡 #/$ Matter-dominated Universe, Eq. (8.13): 𝑎(𝑡) = 𝑎$ 𝑡 $/% Dark energy-dominated Universe: 𝑎(𝑡) ∝ 𝑒 qr(,& h& T .

From Part A, we found: 𝑎m (𝑡) = 𝑒 qr(,& h& (TNT& ) and that at the transition from matter- to dark energy-dominated Universe: 𝑡Qo$ = 0.63 𝑡) , 𝑎Qo$ = 0.73 Matching the expression for the matter-dominated Universe with the expression for the dark energy-dominated Universe (they must match at the transition): $/% N$/% 𝑎Qo$ = 𝑎$ 𝑡Qo$ = 0.73 → 𝑎$ = 0.73𝑡Qo$ →

N$/% 𝑎P (𝑡) = 0.73𝑡Qo$ 𝑡 $/% = 0.73 s

𝑡

$/%

t 𝑡Qo$ Matching the expression for the radiation-dominated Universe with the expression for the matter-dominated Universe (they must match at the transition): $/% %/$ 𝑡Qo# 1 1 #/$ 𝑎Qo# = 𝑎# 𝑡Qo# = 0.73 s t = → 𝑡Qo# = a b 𝑡Qo$ 𝑡Qo$ 3500 0.73 × 3500 %

$ 1 𝑡) → 𝑡Qo# = a b 0.63 𝑡) ≈ 0.73 × 3500 205,000 This is about the same as what we found in class (Lecture 8, slide 10): 𝑡Qo# = T&

$)),)))

Therefore, to summarize 𝑡 ∈ Í0, 𝑡Qo# Î: 𝑎(𝑡) = 𝑎# 𝑡 #/$ 𝑡 ∈ Í𝑡Qo# , 𝑡Qo$ Î: 𝑎(𝑡) = 𝑎$ 𝑡 $/% 𝑡 ∈ Í𝑡Qo$ , ∞Î: 𝑎(𝑡) = 𝑒 h& (TNT& ) ALTENRATIVELY:

𝑎̇ $ a b = 𝐻)$ (Ω_,) 𝑎NG + ΩP,) 𝑎N% + Ωm,) ) 𝑎 𝑎̇ $ = 𝑎$ 𝐻)$ (Ω_,) 𝑎NG + ΩP,) 𝑎N% + Ωm,) ) 𝑑𝑎 𝑎̇ = = 𝑎𝐻) :Ω_,) 𝑎NG + ΩP,) 𝑎N% + Ωm,) 𝑑𝑡

𝑑𝑎 𝐻) 𝑎9Ω_,) 𝑎NG + ΩP,) 𝑎N% + Ωm,) 𝑑𝑎

= 𝑑𝑡

= 𝑑𝑡 𝐻) 9Ω_,) 𝑎N$ + ΩP,) 𝑎N# + Ωm,) 𝑎$ 1 𝑑𝑎 Ë 𝑑𝑡 = Ë 𝐻) 9Ω_,) 𝑎N$ + ΩP,) 𝑎N# + Ωm,) 𝑎$ Integrate in terms of conformal time 𝑑𝜂 = 𝑑𝑡/𝑎 1 𝑑𝑎 Ë 𝑑𝜂 = Ë 𝐻) 9Ω_,) + ΩP,) 𝑎 + Ωm,) 𝑎G I 1 Z 𝑑𝑎" 𝜂 = Ë 𝑑𝜂′ = Ë 𝐻) ) ) :Ω_,) + ΩP,) 𝑎" + Ωm,) 𝑎" G This will give us 𝜂(𝑎). Invert it to get 𝑎(𝜂). Then plug 𝑎(𝜂) into T

I

)

)

𝑡(𝜂) = Ë 𝑑𝑡 = Ë 𝑎(𝜂" )𝑑𝜂" This will give us a parametric plot of 𝑎(𝜂) vs. 𝑡(𝜂), as shown below.

Chapter 12 1. Derive Eq. (12.14).

𝑑% 𝒑 𝑝$ 𝑃& (𝒙) = 𝑔& Ë 𝑓 (𝒙, 𝒑) (2𝜋ℏ)% & 3𝐸(𝑝)

2. Derive Eq. (12.16).

Chain rule: 𝜌̇ 𝑐 $ = −3(𝜌𝑐 $ + 𝑃)

𝑎̇ 𝑎

𝜕𝑃 𝜕𝑃 𝜕𝑇 = 𝜕𝑡 𝜕𝑇 𝜕𝑡 𝜕 𝑃 1 𝜕𝑃 𝜕𝑃 𝑃̇ 𝑃 𝑎N% sa𝜌 + $ b 𝑎% t − $ =0 → = 𝑐 $ 𝑎N% Xs𝜌̇ + $ t 𝑎% + 3𝑎$ 𝑎̇ a𝜌 + $ bY 𝜕𝑡 𝑐 𝑐 𝜕𝑡 𝜕𝑡 𝑐 𝑐 𝜕𝑃 3𝑎̇ (𝜌𝑐 $ + 𝑃) = I𝜌̇ 𝑐 $ + 𝑃̇J + 𝜕𝑡 𝑎 𝜕𝑃 𝜕𝑃 𝜕𝑇 3𝑎̇ (𝜌𝑐 $ + 𝑃) = = I𝜌̇ 𝑐 $ + 𝑃̇ J + 𝜕𝑡 𝜕𝑇 𝜕𝑡 𝑎 𝜕𝑃 𝜌𝑐 $ + 𝑃 = 𝜕𝑇 𝑇 3. Assuming blackbody radiation, convert 𝑇.,) = 2.73 K into a. eV From the notes, or elsewhere: 1 eV = 11604 K → 𝑇.,) = 2.73 K → 0.235 meV This means that the average energy of a CMB photon is 𝐸Ü = 0.235 meV b. 𝑛.,) in photons cmN% Using Wien’s law: ℎ𝜈zQ{| = 2.8 𝑘𝑇, where 𝑘 = 1.38 × 10N$% J K N# and 𝑎9 = 7.57 × 10N#f J mN% K NG 𝜌. 𝑎9 𝑇.G 𝑎9 𝑇.G 𝑎9 𝑇.% 𝑛. ≈ = = = 𝐸# ℎ𝜈 2.8 𝑘𝑇. 2.8 𝑘 % 𝑎9 𝑇.,) 𝑛.,) ≈ ≈ 399 cmN% 2.8 𝑘 c. 𝜌.,) in eV cmN% 𝜌.,) ≈ 𝐸Ü 𝑛.,) = (0.235 meV)(399 cmN% ) = 9.3 × 10N$ eV cmN% d. 𝜌.,) in g cmN% 1 g → 𝐸 = 9 × 10#% (6.24 × 10#b ) eV = 5.6 × 10%$ eV 9.3 × 10N$ eV cmN% 𝜌.,) ≈ = 1.66 × 10N%G g cmN% eV 5.6 × 10%$ g e. Ω.,)

𝜌.,) 9.3 × 10N$ eV cmN% Ω.,) ≡ = ≈ 1.88 × 10Ne 𝜌^_C`,) 5 × 10% eV cmN% Use 𝐻) = 100ℎ km s N# Mpc N# , where ℎ ≈ 0.73.

Chapter 13 1. Derive the Planck energy given in Eq. (13.9). m$ } ~ • 𝑐 ℎ 𝐺 = g $ = g cm$ s N$ s ~ • cm } g cm$ cm% U V s t s $ t = g cm$ s N$ s s gs 𝐴 + 2𝐵 + 3𝐷 = 2 𝐵−𝐷 =1 −𝐴 − 𝐵 − 2𝐷 = −2 e

#

€+ ,

#

Solution: 𝐴 = $ , 𝐵 = $ , 𝐶 = − $, so 𝐸M = 𝑐 e/$ ℎ#/$ 𝐺 N#/$ = : k , just as in Eq. (13.9).

2. How big was the Universe at the time of recombination at 𝑧 = 1100?

1 1 1 𝑎≡ = ≈ 1 + 𝑧 1 + 1100 1100 At recombination, the Universe was about 1/1100 of what it is today in term of its #

%

linear size, or U##))V ≈ 7.5 × 10N#) of today’s Universe’s volume.

Chapter 14 1. Show that the blackbody distribution of the CMB at the time recombination retains its blackbody form as the Universe expands.

1 𝑓I𝜈; 𝑇. J = $jℏ8 𝑒 ]‚- − 1 𝑇. 𝑇.,) 𝑇. 𝑇. ∝ (1 + 𝑧) → = 𝑐𝑜𝑛𝑠𝑡. → = → 𝑇. = 𝑇.,) (1 + 𝑧) 1+𝑧 1 + 𝑧) 1 + 𝑧 1 𝜈 𝑓I𝜈; 𝑇. J = = 𝑓U ; 𝑇.,) V $jℏ8 1 + 𝑧 𝑒 ]‚- ,) (#v4) − 1 â )ƒ./0 l) = 〈Θ(𝒏 â )〉 = 0.

2. Show that Θ(𝒏 3. Derive Eq. (14.29). Hint: Use the addition theorem ƒ

4𝜋 ∗ â⋅𝒏 = â ) 𝑌ƒ„ â ′) 𝑃ƒ (𝒏 å 𝑌ƒ„ (𝒏 (𝒏 2𝑙 + 1 â")

„lNƒ

4. Show that a half of the CMB photons observed now have energies below 3.5 𝑘 𝑇.,) .

2 4𝜋ℏ % 4𝜋𝐼8 (𝜈) 2𝐼8 (𝜈) 𝑐𝜈 𝑐 $ 𝜈 8𝜋ℏ 𝜈$ 𝑛8 (𝜈) = = = $jℏ8 = % $jℏ8 𝑐2𝜋𝜈 𝑐𝜈 𝑐 𝑒 ]‚- ,) − 1 𝑒 ]‚- ,) − 1

8𝜋ℏ 𝑛8 (𝜈) = % 𝑐

𝜈$ $jℏ8

8𝜋ℏ → 𝑛(𝐸) = % 𝑐

𝜈$ $j†

𝑒 ]‚- ,) − 1 𝑒 ]‚- ,) − 1 % $ %.e 8𝜋 𝑘𝑇 𝑥 %.e]‚- ,) 𝑁‡%.e ]‚- ,) ∫) 𝑛(𝐸 )𝑑𝐸 𝑐 % U2𝜋ℏV ∫) exp(𝑥) − 1 𝑑𝑥 1.75487 = = = = 0.73 ‰ % 𝑁`t`{s 2.40411 8𝜋 𝑘𝑇 ∫) 𝑛(𝐸)𝑑𝐸 U V (2𝜁(3)) 𝑐 % 2𝜋ℏ 5. Show that the speed of sound in plasma is 𝑣< = 𝑐/√3. 𝑑𝑃 𝑐< = k 𝑑𝜌 1 𝑑𝑃 1 % 𝑐 𝑃 = 𝜌𝑐 $ → = 𝑐 → 𝑐< = 3 𝑑𝜌 3 √3 6. Derive Eq. (14.3) from Eq. (14.2). 𝑑𝐼 4𝜋ℏ 𝜈% 𝐼8 (𝜈) ≡ = $ $jℏ8 𝑑𝜈 𝑐 𝑒 ]‚ − 1 𝑐 𝑐 𝑐 𝑑𝜈 𝜈$ 𝑐 𝜈= → 𝜆= → 𝑑𝜆 = − $ 𝑑𝜈 → =− =− $ 𝜆 𝜈 𝜈 𝑑𝜆 𝑐 𝜆 𝑐 % 𝑑𝐼 𝑑𝐼 𝑑𝜈 4𝜋ℏ U𝜆V 𝑐 4𝜋ℏ𝑐 $ 1 𝐼1 (𝜆) ≡ = = $ $jℏ+ U− $ V = e $jℏ+ 𝑑𝜆 𝑑𝜈 𝑑𝜆 𝑐 𝜆 𝜆 𝑒 ]‚1 − 1 𝑒 ]‚1 − 1