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Problem 1 Cosmic Microwave Background Ryden 2. Throughout this problem it is important to be aware that you are esti- mating these quantities. This gives you leeway to make a number of sim- plifications, but it also becomes important that you do not overstate your accuracy. If you give me an answer to this question with three or more digits of precision, I will deduct points. In general, it is important to be aware of how precisely you measure or compute a quantity, and to quote the answer accordingly. The Cosmic Microwave Background (CMB) is a blackbody spectrum with temperature T = 2.725 K. According to the Stefan-Boltzmann Law, an object with temperature T radiates energy at a rate of A × σT 4 , where A is the object’s surface area and σ is the Stefan-Boltzmann constant. By turning this picture inside out, you can see that this is also the energy absorbed by a perfect absorber sitting in a medium of temperature T. So, to get an estimate of how much energy you absorb per second, we assume you have about 2 m^2 of surface area and use σ = 5. 67 × 10 −^8 J m−^2 K−^4 s−^1 :
A × σT 4 ≈ 6 × 10 −^6 J s−^1 To use this to estimate a number of photons absorbed per second, we can find a “typical energy” of a photon in a blackbody spectrum, and divide the energy absorbed per second by this amount. A glance at figure 2.7 is helpful here: it shows amount of energy in a blackbody spectrum as a function of photon energy, and peaks around hν = 3kB T. We will use this in the answer below, although using kB T would be equally acceptable; this problem asks only for an order-of-magnitude calculation.
3 kB T = 1. 13 × 10 −^22 ⇒ 6. 3 × 10 −^6 J s−^1 / 1. 1 × 10 −^22 J = 6 × 1016 photons s−^1
Finally, how long would it take to raise your temperature by 10−^9 K? Well, ∆E = Cm∆T , where C is a specific heat, m is mass, ∆T is change in temperature, and ∆E is the corresponding change in energy. We can estimate our mass as 100 kg, we are given our specific heat, we know the desired change in temperature and the rate at which the CMB is pouring energy into us, so we can write:
6 × 10 −^6 J s−^1 × ∆t = Cm∆T ⇒ ∆t ≈ 70 seconds Problem 2 Neutrino Mass Limits Ryden 2. We are asked to minimize the sum me + mμ + mτ given the constraints (m^2 μ − m^2 e)c^4 = 5 × 10 −^5 eV^2 and (m^2 τ − m^2 μ)c^4 = 3 × 10 −^3 eV^2. The best way to do this is to use the constraints to express me + mμ + mτ entirely in terms of one of the m’s (I will be using me), set the derivative equal to zero (as usually done when minimizing) and solve. The calculation will be somewhat simpler if we express the constraints as follows:
m^2 μ − m^2 e = 5 × 10 −^5 eV^2 /c^4 = ∆μ,e
m^2 τ − m^2 μ = 3 × 10 −^3 eV^2 /c^4 = ∆τ,μ ⇒
mμ =
√ m^2 e + ∆μ,e
mτ =
√ m^2 μ + ∆τ,μ =
√ m^2 e + ∆μe + ∆τ μ We substitute, take the derivative:
d dmμ
( me +
√ m^2 e + ∆μ,e +
√ m^2 e + ∆μe + ∆τ μ
)
me √ m^2 e + ∆μ,e
me √ m^2 e + ∆μe + ∆τ,μ However, there is a problem. You can see that for positive real me the value of this derivative is always positive, i.e., the derivative of the sum of the masses can never be zero. So instead of minimizing the usual way, the minimum is found by setting the smallest mass equal to its minimum possible value: zero. By examining the constraints we were originally given, you can see that me has to be the smallest of the three. So,
me = 0
mμ =
5 × 10 −^5 eV/c^2 = 7. 1 × 10 −^3 eV/c^2
Comparing this to the Robertson-Walker metric as written in equation 3.19,
ds^2 =
dx^2 1 − κx^2 /R^20
we see that this equation would match up well if κ > 0. Since the only al- lowed values of κ are +1, 0, and -1, which correspond to closed, flat, and open universes respectively, this spherical metric corresponds to a flat geometry. Problem 4 Non-Euclidean Geometry Ryden 3. Before trying to calculate an answer for this problem, it is a good idea to think about what behavior you expect to see. Since you are solving a problem in a new kind of world (two-dimensional with positive curvature) where you have little intuition built up, you will need a way to check the answer you will get. To set this up, imagine you are standing at the north pole of your new world. Any place would do, but the math (and in some ways the visualiza- tion) is easier if you place yourself here. The trick is that if you stand at the north pole, all straight lines leading away from you are lines of longitude. Because of this, the angle an object occupies in your vision is the same as the angle of longitude it sweeps out. Now take some small object (the ds in the problem) and hold it up to your nose. No matter how small the object is, it should blank out most of your field of view if you hold it close enough to your eyes. In fact, held close enough to your face, it should appear to block out everything in front of you; in a 2D universe, this equates to an angle of π. Now send the object away. It should get smaller, but don’t assume it will get smaller the same way things do in a flat universe. Remember that since you are at the pole, the angle an object spans in your vision is the same as the angle it spans in longitude (in spherical coordinates, the angle dφ). So the minimum angle an object can subtend in your vision is ds/R, when it is sitting at the equator. Once the object has passed the equator, its behavior gets even more counterintuitive. Remember, again, that the angle an object takes up in your vision is the same as the angle φ it takes up. After it passes the equator, the object appears to grow, and by the time it has reached the south pole, it actually appears to cover half of the horizon once again!
Now that we have an idea of what to expect, we can begin calculating without being in serious danger of misstepping. I will be using Ω to represent the angle from the north pole, since the usual choice (θ) was taken by the wording of the problem. Now I warn you that there is a more precise (and complicated) way to calculate dθ than I will be using, and that the way I will show you is inaccurate at the poles (when Ω approaches 0 or π). We will calculate this by assuming the ds lies along a line of latitude, which is a good assumption near the equator, but a bad one when ds is close to the poles and the length ds is comparable to the circumference of the line of latitude. (This is picture will ultimately mean that ds wraps itself around the mathematical point that is your point of view - not what we want to happen, but that’s okay.) The angle of latitude swept out by ds is dφ = ds/C(Ω), where C(Ω) is the circumference of a line of longitude. We know that C(Ω) = 2πR sin(Ω), and that the distance to a line of latitude at Ω along the sphere is r = ΩR, therefore:
r R
C(Ω) = 2πR sin(Ω) = 2πR sin
( r R
)
dφ = 2πds/C(Ω) = 2πds/ 2 πR sin
( (^) r
R
)
dθ ≈ dφ ⇒
dθ ≈ ds/R sin
( r R
)
Checking this against our earlier guesses at how the dθ should behave, we see that this dθ 1) becomes large at Ω = 0 and Ω = π (or r = 0 and r = πR), 2) has a minimum at the equator (Ω = π 2 or r = π 2 R), and 3) has a minimum value ds/R. So this answer is, in fact, very close to what we originally imagined, and we know we got something that is, at least, pretty close to the real thing. The only way this answer is “off” is that at the poles, its value goes to positive infinity, rather than to π as we expected. This is because, as earlier stated, the picture we constructed here to calculate the value of dθ has the ds wrap around the mathematical point of the poles when it gets too close.
