Selected Topics in Physics
a lecture course for 1st year students
by W.B. von Schlippe
Spring Semester 2007
Lecture 3
1. Particle in a Central Field
2. Kepler Problem
1
1. Particle in a Central Field
In 3D space, a force
where
G
G
r = r , rˆ = r r
is called a central force.
A central force can be represented as the gradient
of a function of the modulus of the vector r.
y
r
G G
F ( r ) = f (r )rˆ
F(r)
Consider the function
x
Its derivative w.r.t.
x
V (r )
is
∂V ( r ) dV ( r ) ∂r x dV ( r )
=
=
∂x
dr ∂x r dr
and similarly for the derivatives w.r.t
y
and z,
2
hence
⎛ ∂V ∂V ∂V ⎞ dV ( r ) ( x, y, z ) dV ( r )
∇V ( r ) ≡ ⎜
,
,
=
rˆ
⎟=
dr
r
dr
⎝ ∂x ∂y ∂z ⎠
(3.1)
Thus we can represent the central force in the following form:
G G
F ( r ) = −∇V ( r )
where the minus sign is conventional.
Newton’s equation of motion for a particle of mass
force field is therefore of the following form:
G
d 2r G G
m 2 = F ( r ) = −∇V ( r )
dt
m
in a central
(3.2)
V(r) is the potential energy function. The meaning of this terminology
will become clear in the following derivation.
3
To derive energy conservation, take the scalar product (“dot product”) with
G
G
r = dr dt
On the left-hand side we get
G G d ⎛ 1 G 2 ⎞
r ⋅r = ⎜ r ⎟
dt ⎝ 2 ⎠
Indeed:
(3.3)
1 dx 2 1 dy 2 1 dz 2
G G
+ yy
+ zz
=
+
+
r ⋅ r = xx
2 dt 2 dt 2 dt
1 d 2
d ⎛ 1 G 2 ⎞
2
2
=
x + y + z ) = ⎜ r ⎟
(
dt ⎝ 2 ⎠
2 dt
On the right-hand side we get
dV
G
r ⋅∇V ( r ) =
dt
(3.4)
Indeed: from Eq. (3.1) we have
G G
r ⋅ r dV
G
r ⋅∇V ( r ) =
r dr
4
also
dV ( r ) dV
=
dt
dr
dV
=
dr
dr
dt
⎛ ∂r
∂r
∂r ⎞
⎜x + y +z ⎟=
∂y
∂z ⎠
⎝ ∂x
G G
r ⋅ r dV
r dr
and comparing the r.h.s. with (3.5) we get the result (3.4).
Summarising our results we can write
d ⎛1 G
G
G
⎞
r ⋅ mr + ∇V ( r ) = ⎜ mr 2 + V ( r ) ⎟ = 0
dt ⎝ 2
⎠
(
hence
T≡
)
1 G 2
mr + V ( r ) = E = constant.
2
1 G 2
mr : Kinetic Energy ( K .E .) V ( r ) : Potential Energy ( P.E .)
2
(3.5)
5
We get a second conservation law if we take the vector product of (3.2)
(“cross product”) with vector r:
on the l.h.s. we get
G G d G G
r × r=
r ×r ;
dt
(
and on the r.h.s.:
)
G G G
r ×F = M
this is called the moment of force or torque. But since we are considering
central forces, the torque is equal to zero:
G
r × f ( r ) rˆ = 0
Now define the angular momentum:
G
G G
L ≡ mr × r
Thus
G
G
dL
= 0 and hence L = constant
dt
(3.6)
6
Summary: Newton’s equation of motion for a particle of mass m
a central force field
is of the form
in
G G
F ( r ) = f (r )rˆ
G G
d r
G
m 2 = F ( r ) = −∇V ( r )
dt
2
where
(3.2)
V(r) is the P.E. fn, and we could derive two conservation laws:
(i) conservation of energy:
1 G 2
mr + V ( r ) = E = constant.
2
(3.5)
and (ii) conservation of angular momentum:
G
G G
L ≡ mr × r = constant
(3.6)
7
2. Kepler Problem
The Kepler problem is the problem of the motion of a planet in the
gravitational field of the sun or, more generally, the motion of two masses
in gravitational interaction.
Thus consider two point masses, m1 and m2,
located at r1 and r2, respectively.
m1
y
The force exerted by
F12
F21=-F12
r1
m2
m2 on m1 is
G
mm
F12 = G G 1 G2 2
r1 − r2
G G
r2 − r1
G G
r1 − r2
r2
x
and the force exerted by m1 on m2 is
by Newton’s third law
G
G
F21 = − F12
8
Thus we have the following two equations of motion for the two masses:
m1m2 G G
G
m1r1 = G G G 3 ( r2 − r1 )
r1 − r2
(3.7)
m1m2 G G
G
m2 r2 = −G G G 3 ( r2 − r1 )
r1 − r2
(3.8)
If we add Eqs. (3.7) and (3.8), we get
G
G
p1 = m1r1 ;
or with
G G G
P = p1 + p2
G
G
m1
r1 + m2 r2 = 0
G
G
p2 = m2 r2
d G G
( p1 + p2 ) = 0
dt
(3.9)
(3.9a)
is the total momentum of the two-body system
therefore our result (3.9) can be expressed by
G
P = constant
conservation of total momentum
9
The radius vector of the centre of mass of
m1 and m2 is defined by
G m1rG1 + m2 rG2
R=
m1 + m2
(3.10)
hence taking its derivative we find with Eq. (3.9):
G G
R = P ( m1 + m2 ) = constant
and hence
G
MR = 0
(3.11)
(3.12)
This equation is identical with Eq. (3.9) (or (3.9a)), but now we have
a new interpretation: the entire two-particle system of mass
is moving without acceleration.
M=m1+m2
10
If we integrate Eq. (3.11), then we get
G G G
G
R = R0 + vt ; v = constant
G
G G
where R0 is an integration constant and v = R
Carry out a GT into a frame in which the c.o.m. is at rest and at the
origin (Exercise!): this is called the centre of mass frame or
centre of mass system (CMS).
This is illustrated on the next slide.
11
y
the centre of mass frame of
p1
particles
m1 and m2
it is defined by
m1
G
G
p2 = − p1
x
m2
but the distance from the origin
p2=-p1
t is different:
m1 G
G
r2 ( t ) = −
r1 ( t )
m2
at any instant
in the CMS, Eq. (3.11) takes on the following form:
G
MR = 0
12
Let us go back to the equations of motion of the system of two
interacting particles, but write it down for an arbitrary force.
Taking account of Newton’s third law we have:
G G
m1r1 = F
G
G
m2 r2 = − F
Divide (3.13) by
m1
and (3.14) by
G G G
r = r1 − r2
and
1
μ
=
1
1
+
m1 m2
(3.14)
m2, then subtract, hence
1 ⎞G
G G ⎛ 1
r1 − r2 = ⎜ +
⎟F
⎝ m1 m2 ⎠
Now define
(3.13)
(3.15)
relative coordinate
reduced mass μ
13
then (3.15) becomes
G G
μr = F
(3.16)
and if we add (3.13) and (3.14), then as before we get Eq. (3.12):
G
MR = 0
(3.12)
Thus our original equations (3.13) and (3.14) have been transformed
into two simpler equations (3.12) and (3.16).
Equations (3.13) and (3.14) both depended on two vectors, r1 and r2.
They are therefore coupled equations; in (3.12) and (3.16) we have
decoupled them.
In the particular case, when the force F is the gravitational interaction,
G
m1m2
F12 = G G G 2
r1 − r2
G G
r2 − r1
G G
r1 − r2
14
we get
m1m2
G
μ r = −G 2 rˆ
r
and if we note from the definition of the reduced mass that
m1m2 = μ ( m1 + m2 ) = μ M
then we get the equation of motion in the following form:
G
μ r = −G
μM
r
2
rˆ
(3.17)
which is the equation of motion of a particle of mass μ in the
gravitational field of a mass
coordinate system.
M
that is at rest at the origin of the
15
Our task is now to solve Eq. (3.17), and then go back to the coordinates
of the particles
m1 and m2
Recall our results found for central forces:
• Conservation of energy, and
• conservation of angular momentum
1 G 2
μ r + V ( r ) = E = constant.
2
G
G
dL
= 0 and hence L = constant
dt
where
V(r)
(3.18)
(3.19)
is defined by
G G
μM
∇V ( r ) = − F ( r ) = G 2 rˆ
r
16
hence
V (r ) = −
γ
r
with
γ = Gμ M
and we remember that
G
G G
L = μ r × r
Now we use conservation of angular momentum, Eq. (3.19), to simplify
the expression (3.18) for the energy.
First we note that the conservation of angular momentum implies
that the angular momentum vector has a constant direction.
But the cross product of two vectors is perpendicular to the plane
spanned by these vectors.
Therefore the vectors
G
G
r and r
stay in one plane. Without …
17
… loss of generality we can make this the
Therefore
xy
plane.
G
r = r ( cos ϕ ,sin ϕ , 0 )
and its derivative is
G
r = r ( cos ϕ ,sin ϕ , 0 ) + rϕ ( − sin ϕ , cos ϕ , 0 )
thus
G 2
r = r 2 + r 2ϕ 2
Recall:
and
G G
r × r = r 2ϕ
(3.20)
1 G 2
T = μ r : Kinetic Energy ( K .E .)
2
L =μ
2
2
(
G G
r ×r
)
2
= μ 2 r 4ϕ 2
(3.21)
hence with (3.20):
1 2
L2
T = μ r +
2
2μ r 2
18
hence
1 2
L2
γ
μr +
− =E
2
2
2μ r
r
(3.22)
We have separated the K.E. into two terms. The first term depends on
the radial velocity and the second term depends only on r. Thus the
second term is, mathematically, similar to a potential energy. This
leads us to the notion of an effective P.E.:
L2
γ
Veff ( r ) =
−
2μ r 2 r
10
5
0.2
-5
-10
0.4
0.6
0.8
1
This fn has a minimum at
r0 = L2 μγ
Veff ( r0 ) = −
μγ 2
2
2L
(3.23)
(3.24)
19
The total energy cannot be less than the minimum of the effective P.E.
since otherwise we would get a negative square of the radial velocity.
Thus
E≥−
μγ 2
2 L2
The equal sign corresponds to zero radial velocity, i.e. to circular motion.
The radius of the circular orbit is by Eq. (3.23) related to the energy E:
r0 = − γ 2 E
This is a remarkable result that we will remember when we come to
discuss the Bohr theory of the hydrogen atom!
For Emin<E<0 the motion lies between a minimum and a maximum
value of
r, which we get from the condition
E = Veff ( r )
Solving for
r
we get …
i.e.
L2
γ
E=
−
2μ r 2 r
20
rmax,min
γ
⎛ γ ⎞
2
L2
=−
± ⎜
⎟ +
2E
⎝ 2 E ⎠ 2μ E
where the upper (lower) sign corresponds to rmax ( rmin ).
For E >0 only the upper sign gives an acceptable value; the second
root must be rejected.
In astronomical terms E > 0 corresponds to the motion of a
non-recurring comet.
Proceed now to solving Eq. (3.22):
1 2
L2
γ
μ r +
−
=E
2
r
2
2μ r
(3.22)
To do this, recall Eq. (3.21) which we rewrite in the form of
L = μ r 2ϕ
21
hence
ϕ =
L
μr 2
and then we eliminate the time by noting:
dr dr dϕ dr L
=
=
dt dϕ dt dϕ μ r 2
and substituting into (3.22) we get
2
L ⎛ 1 dr ⎞
L2
γ
+
−
=E
⎜ 2
⎟
2
r
2 μ ⎝ r dϕ ⎠ 2 μ r
2
or
2
⎛ 1 dr ⎞
1 2μγ 1 2μ E
+
− 2
= 2
⎜ 2
⎟
2
L r
L
⎝ r dϕ ⎠ r
22
Now define
hence
1
u = , hence
r
du
1 dr
=− 2
dϕ
r dϕ
2
⎛ du ⎞
2 μγ
2μ E
2
+
−
=
u
u
⎜
⎟
L2
L2
⎝ dϕ ⎠
Differentiate w.r.t. φ
2u′u ′′ + 2uu′ −
2 μγ
u′ = 0
2
L
Here u’ is a common factor; therefore we have
either
( i ) u′ = 0, or ( ii ) u′′ + u =
μγ
L2
Option (i) is the circular motion; consider option (ii): this DEq has the solution
1 μγ
u = = 2 + A cos (ϕ + α )
r L
23
and finally the equation of the trajectory:
L2 μγ
r=
=
2
μγ
+
μγ ) cos (ϕ + α )
1
A
L
(
cos
+
ϕ
+
α
A
(
)
L2
1
r
θ
24