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Homework 10
I’m going to try and remember to cite where I pull equations from in the lectures from now on when possible. I will refer to lecture equations by their lecture and slide number, but I won’t bother labeling order on a particular slide because there isn’t much risk for confusion. For example if I am citing an equation from the 6th slide of lecture 9 I will label it: L.9.6, with L so it is not confused with equations within the solutions.
Problem 1 (a) Because it is in a circular orbit r is constant and we have
L m = rv = r
d(rφ) dτ = r^2
dφ dτ = r^2
dφ dtshell
dtshell dτ = rvshell γshell. (1)
(b) By definition we have
vshell =
d(rφ) dtshell = r
d(φ) dtshell
so that
dtshell =
rdφ vshell
and integrating
∫ (^) tf
ti
dtshell =
∫ (^) φf
φi
rdφ vshell
r vshell
∫ (^2) π
0
dφ (4)
so that we have
∆tshell = 2 πr vshell
(c) Again we have
L = rv = mr d(rφ) dτ
rearranging we have
dτ =
mr^2 L dφ (7)
integrating we obtain
∆τ =
2 πmr^2 L
(d) By the definition of the shell coordinates from L.16.
dt =
rS r
dtshell. (9)
Integrating we obtain
∆t =
rS r
∆tshell. (10)
Problem 2 (a) Let’s just look at the potential for l = 4 first of all to get an idea of what is going on:
0 10 20 30 40
There is unstable orbit around r′^ = 3 and a (subtle) stable orbit around r′^ =
- (If you keep playing around with different l’s you will see that this is a general feature of this potential albeit the stable orbit is not always so clear from the plots). Thus if we want the smallest stable orbit we want the smallest r corresponding to the smallest possible angular momentum or equivalently the smallest l. From L.19.6 we have
r′^2 − l^2 r′^ + 3l^2 = 0 (11)
the solution to which is
r′^ =
l^2 ±
l^4 − 12 l^2 2
l^2 ± l
l^2 − 12 2
Thus the smallest possible l is l =
- Plugging in this value for l for r we get
r′^ = 6. (13)
Rewriting in ’unnatural’ units defined on L.19.3 we have
r = rS r′ 2
= 3rS. (14)
Problem 3 From L.19.7 we have
L m
= rvshell γshell =
8 GM
c^2
3 c 4
3 c/ 4 c
24 GM
7 c
12 crS √ 7
= 2 · 10 −^18 M.
And again from L.19.7 we have
ǫ ≡
E
mc^2
= A^1 /^2 γshell =
rS r
1 − v 2 c^2
4
) 2 = 1.^3.^ (16)
black and white). If you are confused about whats going on here just remember that all this funky stuff with GR and spherical coordinates doesn’t matter what- so-ever in this question- by looking at the effective radial potential we have reduced the problem to a one-dimensional problem that can simply be looked at as whether or not a ball has enough energy to roll out of a valley.
Problem 4 The effective potential for a photon is
Vγ,ef f =
r′^2
r′
To find the allowed circular orbits we set the first derivative equal to zero
∂Vγ,ef f ∂r′^
r′^3
r′
r′^2
r′^2
r′^3
r′^4
which yields r′^ = 3 (25)
thus
r =
2 r′ rS
3 GM
c^2
= 3rγ (26)
where rγ ≡ GM c 2 is the photon sphere radius. In order to find out if this orbit is stable we evaluate the second derivative of the effective potential at this radius:
∂^2 Vγ,ef f ∂r′^2
r′^3
r^4
Since − 812 < 0 the potential is concave down at this point and hence unstable. Alternatively we can look at the plot:
2 4 6 8 10
which makes it clear that there are no stable circular orbits and only one unstable circular orbit. Now b = 6 GM c 2 so b′^ = (^) r^2 Sb = 6. At the peak we have
Vγ,ef f |r′ (^) =3 =
Now since 1 b′^2
the photon will not be captured.
