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Solutions to Homework Assignment #9 – PHY
8-2. A) OK, for each value of r, we want to consider the corresponding circle about the center and to compute its radius R and its circumference C. Actually, computing the radius is easy. As we found in class, when the coefficient (grr) in front of the dr^2 term is 1, the radius is just R = r.
i) So, now we want to compute the circumference of the circle at r for the metric ds^2 = dr^2 + a^2 sin^2 (r/a)dθ^2. Since r will remain constant around the circle, the change in r (dr) is zero: dr = 0. Thus, along the circle we have
ds^2 = a^2 sin^2 (r/a)dθ^2 , or,
ds = a sin(r/a)dθ. The circumference is the length of this circle:
C =
∫ ds =
∫ (^2) π
0
a sin(2πr/a)dθ = (2π)a sin(r/a).
So, C/R = (2π)(a/r) sin(r/a). Let’s define x = a/r so that C/R = (2π)sinx^ x. Now, sin x < x, so C/R < 2 π. Note that C/R does become nicely 2 π in the limit as x → 0 (since sin x ≈ x for small x); that is, if the circle is small enough to be ‘local.’
ii) Once again, dr = 0 for our circle. Much as before, we find ds = a sinh(a/r)dθ, so the circumference is
C =
∫ ds =
∫ (^2) π
0
a sinh(2πr/a)dθ = (2π)a sinh(r/a).
Now, C/R = (2π)(a/r) sinh(r/a). Again, let’s define x = a/r. Then, C/R = (2π)sinhx x. Now, sinh x > x, so C/R > 2 π. Note that C/R once again nicely becomes 2π in the limit as x → 0 (since sinh x ≈ x for small x).
B) For this part, the coefficient (gθθ) of the dθ^2 term is just ρ^2. For any circle around the origin, dρ = 0 so C =
∫ ds =
∫ (^2) π 0 ρ dθ^ = 2πρ.^ Thus, the interesting question is: How long is the radius?
Let’s compute the actual length of the radius for each case. Along a radial line, dθ = 0.
a) So, for the first metric, the length R of the radial line from 0 to ρ is
R =
∫ ds =
∫ (^) ρ
0
a √ a^2 − ρ^2
dρ = a sin−^1 (ρ/a).
If you’re not familiar with this integral, you can work it out using a substi- tution like ρ = a sin θ. In other words, ρ = a sin(R/a). Thus, C = 2πa sin(R/a). Does this look familiar? It is the same relationship we found for metric (i). In fact, metric (i) and metric (a) represent the same geometry, they just use different coordinates. Can you see what the relationship is between r and ρ? As you might be able to tell, these are both metrics for a sphere of radius a.
b) For the second metric, the length R of the radial line from 0 to ρ is
R =
∫ ds =
∫ (^) ρ
0
a √ a^2 + ρ^2
dρ = a sinh−^1 (ρ/a).
If you’re not familiar with this integral, you can work it out using a substi- tution like ρ = a sinh θ. In other words, ρ = a sinh(R/a). Thus, C = 2πa sinh(R/a). This is exactly the relationship we found for metric (ii). It turns out that metrics (ii) and (b) once again represent the same geometry using different coordinates. Can you see what the relationship is between r and ρ in this case? As you might be able to tell, these are both metrics for a Lobachevskian space. c) For metric (i), we found ρ = a sin(R/a), so dρ = a cos(R/a)dR. Now, we just substitute:
ds^2 =
dρ^2 a^2 − ρ^2
a^2 cos^2 (R/a)dR^2 a^2
( 1 − sin^2 (R/a)
) (^) + a^2 sin^2 (R/a)dθ^2
a^2 cos^2 (R/a)dR^2 a^2 cos^2 (R/a)
= dR^2 + a^2 sin^2 (R/a)dθ^2. (1)
So, this is just the same as metric (a), for which we had R = r! Similary, for metric (ii), we found ρ = a sinh(R/a), so dρ = a cosh(R/a)dR. Now, we just substitute:
ds^2 =
dρ^2 a^2 + ρ^2
