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You are allowed to work together on the exercises, but everyone has to handle
inn their own solution. The assignment must be handed in in the special box
at the floor marked “7” in Niels Henrik Abels hus (math building) before the
deadline. Remember to fill in and attach a front page – front pages are found
nearby the box, or on the Internet.
Each exercise a) – k) counts 0-10 points, all together a maximum of 110 points.
Your Oblig 2 is approved if you score 20 points or more on Part 1 (50%), and
40 points or more on Part 2 (a bit more than 50%).
Introduction
According to Einstein’s general theory of relativity, the path traversed by a ray
of light (a photon) passing by a star (in our case, the Sun), is given by the
differential equation:
dr 2
dt
1−
− E2
RS
r
+
L2
=0
r2
where r = r(t) is the photon’s distance to the center of the star (as a function
of time, t), RS is the so-called Schwarzschild radius1 , E is the energy of the
photon and L is called the photon’s angular momentum.
Throughout the exerscise the magnitudes RS , E and L will be considered as
constants. We shall accept the given differential equation without any further
argueing about how it is deduced.
We will solve this equation and find out how the Sun’s gravity affects the path
of the light ray when it passes close to the Sun (the fact that light is also affected
by gravity is an important result in the theory of relativity).
NOTE: You can solve the exercises in any order and simply use the results from
exercises you haven’t solved yet. This is part of the reason why the exercises
focus on you showing how one derives the answer, and not the answer itself.
This has the added benefit that you will always to be able to check that your
answer is correct. But the main point is how you use mathematics to arrive at
the desired result.
Good luck!
1 If we compress the star so this becomes its new radius (3 km for a Sun-sized star), it will
collapse and become a black hole!
1
Part 1
In the first part of the exercise, problems a)-d), the challenge is to use Leibniz
notation for the derivative to manipulate magnitudes and variables to simplify
the differential equation. Remember that the chain rule tells us that for a
composed function r(φ(t)), the derivative of the function r with respect to t is
given by
dr dφ
dr
=
·
dt
dφ dt
a) In the equation on the previous page, L is defined as follows:
L := r2
dφ
dt
The angle φ is explained in figures 1 and 2. It varies with the distance, and is
a function of t. Use this (and the chain rule) to show that
dr
dr L
=
·
dt
dφ r2
b) In the following, we will make a substitution to simplify the calculations:
u :=
1
r
i.e. the variable u expresses the inverse distance. Far away it is close to 0, and
it reaches its maximum value when the photon reaches its minimum distance to
the Sun. Thus the magnitude u can be considered as a function of the distance
r, and therefore also as a function of the angle φ, through r = r(φ).
Apply this substitution to the result from a), and show that
dr
du
= −L
dt
dφ
c) Insert the result from b) into the differential equation
dr 2
dt
1−
− E2
RS
r
+
L2
=0
r2
and show that it can then be written in the simpler form
du
dφ
2
3
2
= RS u − u +
2
E
L
2
Figure 1: For exercise e). The homogenous equation describes a hypothetical
situation where light passes by the Sun in a straight line without changing
direction (i.e. a situation where gravity has been “turned off”).
d) Differentiate the equation with respect to φ (use the chain rule) and show
that it can be written as a second-order differential equation:
d2 u
+ u = ku2
dφ2
To simplify things, we have introduced the constant k := 32 RS .
Also notice that, according to the laws of physics,
to φ, and its derivative vanishes.
E
L
is a constant with respect
Part 2
The purpose of this part is to solve the differential equation
d2 u
+ u = ku2
dφ2
Our strategy will be to first find a solution to the homogenous equation
d2 u
+u=0
dφ2
3
and the find a particular solution to the inhomogenous equation. The final
solution will be the sum of the two.
e) First we will look at a case where the path of the light ray does not change
at all. If b is the minimum distance from the photon to the center of the star
and r = r(φ) is the distance when the angle is φ (see figure 1), show that
u=
cos φ
b
and use this to show that u then solves the homogenous equation:
d2 u
+u=0
dφ2
(Hint: b is a constant and does not depend on φ).
f ) From here on, we will call the solution we found in e) uh . In order to find a
general solution to the (inhomogenous) differential equation, we will also need
a particular solution us , such that
u = uh + us
Insert this into the equation
d2 u
+ u = ku2
dφ2
and then apply the following approximations :
u2s ≈ 0
uh + 2us ≈ uh
(Here we presume that the particular solution gives the photon a very small
deflection from the straight line represented by uh , so that the value of us is
very small and much smaller than uh ):
Show that the inhomogenous equation then can then be written as
d2 us
k
+ us ≈ 2 cos2 φ
dφ2
b
(Hint: To show this, you will also need u from exercise e), only here renamed
to uh .)
g) To find a solution of the inhomogenous equation we have to make a qualified
guess. Looking at the right side of the equation we are led to guess that the
solution looks like
4
Figure 2: How the angles φ and θ are connected. The reason for the choice of θ2
is, as the figure suggests, that the light is deflected both on its way towards the
Sun and away from it. Thus, the total angle of deflection becomes θ2 + θ2 = θ,
and calculating this is our final goal
us = A cos2 φ + B
where A and B are constants. Compute the constants A and B and show that
the particular solution is given by
us =
k
2 − cos2 φ
3b²
(Hint: We do not want to have any sinus terms in the solution, but by using a
well-known trigomometric rule you should be able to eliminate those terms.)
h) We define the angle of deflection θ (shown in the figure above2 ) so that
φ=
π θ
+
2
2
Use this and the addition formula for cosine to show that
cos φ ≈ −
θ
2
2 Strictly speaking, the relation only applies when the photon is sufficiently far from the
Sun that the solid and dotted lines are parallel. This can only happen as the distance between
the photon and the Sun approaches infinity (which will indeed happen when we introduce this
limit in exercise j). If you don’t see a problem with this, there is no need to worry about it understanding the figure is not important for solving the exercise.
5
(Hint: The reason we write ≈ above is that we have used yet another approximation3 : We assume that θ is a very small angle, and then
sin θ ≈ θ
You will need this small angle formula to arrive at the correct answer.)
i) If θ is a very small angle, we can also approximate θ2 ≈ 0. Use this approximation, along with the results from exercises e), g) and h) to show that
u≈
θ
2k
−
2
3b
2b
(Hint: Remember that u = uh + us , and that you know both solutions from
previous exercises. This is a good place to start.)
j) In order to find the angle θ, we now look at the case where the photon is very
far away from the Sun, so it will experience no more deflection. In this case
r → ∞, which means that u = 1r ≈ 0. Insert this value for u into the result
from i) and show that we get
θ≈
2 · RS
b
after back-substituting the value of k from exercise d).
k) Finally, an exercise with actual numbers! For the Sun, the Schwarzschild
radius is RS = 3 km. We send a light ray past the Sun so that it only momentarity touches the surface, which means that b is equal to the actual radius of
the Sun:
b = R = 696 000 km
Use this (and the result from exercise j) to calculate the angle of deflection θ
in this case. The answer you get will be in units of radians – calculate what
this corresponds to in degrees. Were we correct in making the assumption that
the light ray is only slightly deflected (in other words, is θ really a very small
angle)?
3 You can see this by looking at the unit circle for a very small angle and noting that the
arc length (in radians) will be measured along an almost vertical line whose length must then
be sin θ.
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