5. Systems of Particles
So far, we’ve only considered the motion of a single particle. If our goal is to understand
everything in the Universe, this is a little limiting. In this section, we take a small step
forwards: we will describe the dynamics of N , interacting particles.
The first thing that we do is put a label i = 1, . . . , N on everything. The ith particle
has mass mi , position xi and momentum pi = mi ẋi . (A word of warning: do not
confuse the label i on the vectors with index notation for vectors!) Newton’s second
law should now be written for each particle,
ṗi = Fi
where Fi is the force acting on the ith particle. The novelty is that the force Fi can be
split into two parts: an external force Fext
i (for example, if the whole system sits in a
gravitational field) and a force due to the presence of the other particles. We write
X
Fi = Fext
i + Fij
j6=i
where Fij is the force on particle i due to particle j. At this stage, we get to provide a
more precise definition of Newton’s third law. Recall the slogan: every action has an
equal and opposite reaction. In equations this means,
• N3 Revisited: Fij = Fji
In particular, this form of the third law holds for both gravitational and Coulomb forces.
However, we will soon find a need to present an even stronger version of Newton’s third
law.
5.1 Centre of Mass Motion
The total mass of the system is
N
X
M= mi
i=1
We define the centre of mass to be
N
1 X
R= mi x i
M i=1
– 67 –
,The total momentum of the system, P, can then be written entirely in terms of the
centre of mass motion,
N
X
P= pi = M Ṙ
i=1
We can now look at how the centre of mass moves. We have
!
X X X X X
Ṗ = ṗi = Fext
i + Fij = Fext
i + (Fij + Fji )
i i j6=i i i<j
But Newton’s third law tells us that Fij = Fji and the last term vanishes, leaving
X
Ṗ = Fext
i (5.1)
i
This is an important formula. It tells us if you just want to know the motion of the
centre of mass of a system of particles, then only the external forces count. If you
throw a wriggling, squealing cat then its internal forces Fij can change its orientation,
but they can do nothing to change the path of its centre of mass. That is dictated by
gravity alone. (Actually, this statement is only true for conservative forces. The shape
of the cat could change friction coefficients which would, in turn, change the external
forces).
It’s hard to overstate the importance of (5.1). Without it, the whole Newtonian
framework for mechanics would come crashing down. After all, nothing that we really
describe is truly a point particle. Certainly not a planet or a cat, but even something
as simple as an electron has an internal spin. Yet none of these details matter because
everything, regardless of the details, any object acts as a point particle if we just focus
on the position of its centre of mass.
5.1.1 Conservation of Momentum
There is a trivial consequence to (5.1). If there is no net external force on the system,
P
so i Fext
i = 0, then the total momentum of the system is conserved: Ṗ = 0.
5.1.2 Angular Momentum
The total angular momentum of the system about the origin is defined as
X
L= x i ⇥ pi
i
– 68 –
, Recall that when we take the time derivative of angular momentum, we get d/dt(xi ⇥
pi ) = ẋi ⇥ pi + xi ⇥ ṗi = xi ⇥ ṗi because pi is parallel to ẋi . Using this, the change in
the total angular momentum is
!
dL X X
ext
X XX
= xi ⇥ ṗi = x i ⇥ Fi + Fij = ⌧ + xi ⇥ Fij
dt i i j6=i i j6=i
P
where ⌧ ⌘ i xi ⇥Fext i is the total external torque. The second term above still involves
the internal forces. What are we going to do about it? Since Fij = Fji , we can write
it as
XX X
xi ⇥ Fij = (xi xj ) ⇥ Fij
i i6=j i<j
This would vanish if the force between the ith and j th particle is parallel to the line
(xi xj ) joining the two particles. This is indeed true for both gravitational and
Coulomb forces and this requirement is sometimes elevated to a strong form of Newton’s
third law:
• N3 Revisited Again: Fij = Fji and is parallel to (xi xj ).
In situations where this strong form of Newton’s third law holds, the change in total
angular momentum is again due only to external forces,
dL
=⌧ (5.2)
dt
5.1.3 Energy
The total kinetic energy of the system of particles is
1X
T = mi ẋi · ẋi
2 i
We can decompose the position of each particle as
xi = R + yi
where yi is the position of the particle i relative to the centre of mass. In particular,
P P
since i mi xi = M R, the yi must obey the constraint i mi yi = 0. The kinetic
energy can then be written as
1X ⇣ ⌘2
T = mi Ṙ + ẏi
2 i
1X X 1X
= mi Ṙ2 + Ṙ · mi ẏi + mi ẏi 2
2 i i
2 i
1 1 X
= M Ṙ2 + mi ẏi 2 (5.3)
2 2 i
– 69 –
So far, we’ve only considered the motion of a single particle. If our goal is to understand
everything in the Universe, this is a little limiting. In this section, we take a small step
forwards: we will describe the dynamics of N , interacting particles.
The first thing that we do is put a label i = 1, . . . , N on everything. The ith particle
has mass mi , position xi and momentum pi = mi ẋi . (A word of warning: do not
confuse the label i on the vectors with index notation for vectors!) Newton’s second
law should now be written for each particle,
ṗi = Fi
where Fi is the force acting on the ith particle. The novelty is that the force Fi can be
split into two parts: an external force Fext
i (for example, if the whole system sits in a
gravitational field) and a force due to the presence of the other particles. We write
X
Fi = Fext
i + Fij
j6=i
where Fij is the force on particle i due to particle j. At this stage, we get to provide a
more precise definition of Newton’s third law. Recall the slogan: every action has an
equal and opposite reaction. In equations this means,
• N3 Revisited: Fij = Fji
In particular, this form of the third law holds for both gravitational and Coulomb forces.
However, we will soon find a need to present an even stronger version of Newton’s third
law.
5.1 Centre of Mass Motion
The total mass of the system is
N
X
M= mi
i=1
We define the centre of mass to be
N
1 X
R= mi x i
M i=1
– 67 –
,The total momentum of the system, P, can then be written entirely in terms of the
centre of mass motion,
N
X
P= pi = M Ṙ
i=1
We can now look at how the centre of mass moves. We have
!
X X X X X
Ṗ = ṗi = Fext
i + Fij = Fext
i + (Fij + Fji )
i i j6=i i i<j
But Newton’s third law tells us that Fij = Fji and the last term vanishes, leaving
X
Ṗ = Fext
i (5.1)
i
This is an important formula. It tells us if you just want to know the motion of the
centre of mass of a system of particles, then only the external forces count. If you
throw a wriggling, squealing cat then its internal forces Fij can change its orientation,
but they can do nothing to change the path of its centre of mass. That is dictated by
gravity alone. (Actually, this statement is only true for conservative forces. The shape
of the cat could change friction coefficients which would, in turn, change the external
forces).
It’s hard to overstate the importance of (5.1). Without it, the whole Newtonian
framework for mechanics would come crashing down. After all, nothing that we really
describe is truly a point particle. Certainly not a planet or a cat, but even something
as simple as an electron has an internal spin. Yet none of these details matter because
everything, regardless of the details, any object acts as a point particle if we just focus
on the position of its centre of mass.
5.1.1 Conservation of Momentum
There is a trivial consequence to (5.1). If there is no net external force on the system,
P
so i Fext
i = 0, then the total momentum of the system is conserved: Ṗ = 0.
5.1.2 Angular Momentum
The total angular momentum of the system about the origin is defined as
X
L= x i ⇥ pi
i
– 68 –
, Recall that when we take the time derivative of angular momentum, we get d/dt(xi ⇥
pi ) = ẋi ⇥ pi + xi ⇥ ṗi = xi ⇥ ṗi because pi is parallel to ẋi . Using this, the change in
the total angular momentum is
!
dL X X
ext
X XX
= xi ⇥ ṗi = x i ⇥ Fi + Fij = ⌧ + xi ⇥ Fij
dt i i j6=i i j6=i
P
where ⌧ ⌘ i xi ⇥Fext i is the total external torque. The second term above still involves
the internal forces. What are we going to do about it? Since Fij = Fji , we can write
it as
XX X
xi ⇥ Fij = (xi xj ) ⇥ Fij
i i6=j i<j
This would vanish if the force between the ith and j th particle is parallel to the line
(xi xj ) joining the two particles. This is indeed true for both gravitational and
Coulomb forces and this requirement is sometimes elevated to a strong form of Newton’s
third law:
• N3 Revisited Again: Fij = Fji and is parallel to (xi xj ).
In situations where this strong form of Newton’s third law holds, the change in total
angular momentum is again due only to external forces,
dL
=⌧ (5.2)
dt
5.1.3 Energy
The total kinetic energy of the system of particles is
1X
T = mi ẋi · ẋi
2 i
We can decompose the position of each particle as
xi = R + yi
where yi is the position of the particle i relative to the centre of mass. In particular,
P P
since i mi xi = M R, the yi must obey the constraint i mi yi = 0. The kinetic
energy can then be written as
1X ⇣ ⌘2
T = mi Ṙ + ẏi
2 i
1X X 1X
= mi Ṙ2 + Ṙ · mi ẏi + mi ẏi 2
2 i i
2 i
1 1 X
= M Ṙ2 + mi ẏi 2 (5.3)
2 2 i
– 69 –
