Download 4.3 Newtonian Mechanics: Many Particles It`s easy to

4.3 Newtonian Mechanics: Many Particles
It’s easy to generalise the above discussion to many particles, we simply add an index
to everything in sight! Let particle i have mass mi and position ri where i = 1, . . . , N
is the number of particles. Newton’s law now reads
Fi = ṗi
(4.11)
where Fi is the force on the ith particle. The subtlety is that forces can now be
working between particles. In general, we can decompose the force in the following
way:
X
Fi =
Fij + Fext
(4.12)
i
j6=i
th
where Fij is the force acting on the i particle due to the j th particle, while Fext
is
i
th
the external force on the i particle. We now sum over all N particles
X
X
X
Fi =
Fij +
Fext
i
i
i,j with j6=i
=
X
i
(Fij + Fji ) +
i<j
X
Fext
i
(4.13)
i
where, in the second line, we’ve re-written the sum to be over all pairs i < j. At
this stage we make use of Newton’s third law of motion: every action has an equal
an opposite reaction. Or, in other words, Fij = −Fji . We see that the first term
vanishes and we are left simply with
X
Fi = Fext
(4.14)
i
P
where we’ve defined the total external force to be Fext = i Fext
i . We now define
P
the total mass of the system M = i mi as well as the centre of mass R
P
mi ri
R= i
(4.15)
M
Then using (4.11), and summing over all particles, we arrive at the simple formula,
Fext = M R̈
(4.16)
which is identical to that of a single particle. This is an important formula. It tells
that the centre of mass of a system of particles acts just as if all the mass were
concentrated there. In other words, it doesn’t matter if you throw a tennis ball or a
very lively cat: the center of mass of each traces the same path.
4.3.1 Momentum Revisited
P
The total momentum is defined to be P = i pi and, from the formulae above, it is
simple to derive Ṗ = Fext . So we find the conservation law of total linear momentum
for a system of many particles: P is constant if Fext vanishes.
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Similarly, we define total angular momentum to be L =
P
i Li . Now let’s see what happens when we compute the
time derivative.
X
L̇ =
ri × ṗi
i
!
=
X
ri ×
i
=
X
Fij + Fext
i
(4.17)
j6=i
X
i,j with i6=j
ri × Fji +
X
ri × Fext
i
(4.18)
i
The last term in this expression is the definition of total exterP
nal torque: τ ext = i ri × Fext
Figure 3: The magi . But what are we going to do
with the first term on the right hand side? Ideally we would netic field for two parlike it to vanish! Let’s look at the circumstances under which ticles
this will happen. We can again rewrite it as a sum over pairs
i < j to get
X
(ri − rj ) × Fij
(4.19)
i<j
which will vanish if and only if the force Fij is parallel to the line joining to two
particles (ri − rj ). This is the strong form of Newton’s third law. If this is true, then
we have a statement about the conservation of total angular momentum, namely L
is constant if τ ext = 0.
Most forces do indeed obey both forms of Newton’s third law: Fij = −Fji and
Fij is parallel to (ri − rj ). For example, gravitational and electrostatic forces have
this property. And the total momentum and angular momentum are both conserved
in these systems. But some forces don’t have these properties! The most famous
example is the Lorentz force on two moving particles with electric charge Q. This is
given by,
Fij = Qvi × Bj
(4.20)
where vi is the velocity of the ith particle and Bj is the magnetic field generated by
the j th particle. Consider two particles crossing each other in a “T” as shown in the
diagram. The force on particle 1 from particle 2 vanishes. Meanwhile, the force on
particle 2 from particle 1 is non-zero, and in the direction
F21 ∼ ↑ ×⊗ ∼ ←
(4.21)
Does this mean that conservation of total linear and angular momentum is violated?
Thankfully, no! We need to realise that the electromagnetic field itself carries angular momentum which restores the conservation law. Once we realise this, it becomes
– 18 –
a rather cheap counterexample to Newton’s third law, little different from an underwater swimmer who can appear to violate Newton’s third law if we don’t take into
account the momentum of the water.
4.3.2 Energy Revisited
The total kinetic energy of a system of many particles is T =
decompose the position vector ri as
ri = R + r̃i
1
2
P
i
mi ṙ2i . Let us
(4.22)
where r̃i is the distance from the centre of mass to the particle i. Then we can write
the total kinetic energy as
T = 12 M Ṙ2 +
1
2
X
2
mi r̃˙ i
(4.23)
i
Which shows us that the kinetic energy splits up into the kinetic energy of the centre
of mass, together with an internal energy describing how the system is moving around
its centre of mass. As for a single particle, we may calculate the change in the total
kinetic energy,
XZ
XZ
ext
T (t2 ) − T (t1 ) =
Fi · dri +
Fij · dri
(4.24)
i
i6=j
Like before, we need to consider conservative forces to get energy conservation. But
now we need both
= −∇i Vi (r1 , . . . , rN )
• Conservative external forces: Fext
i
• Conservative internal forces: Fij = −∇i Vij (r1 , . . . , rN )
where ∇i ≡ ∂/∂ri . To get Newton’s third law Fij = −Fji together with the requirement that this is parallel to (ri − rj ), we should take the internal potentials to satisfy
Vij = Vji with
Vij (r1 , . . . rN ) = Vij (|ri − rj |)
(4.25)
so that Vij depends only on the distance between the ith and j th particles. We also
insist on a restriction for the external forces, Vi (r1 , . . . rN ) = Vi (ri ), so that the force
on particle i does not depend on the positions of the other particles. Then, following
the steps we took in the single particle case, we can define the total potential energy
P
P
V = i Vi + i<j Vij and we can show that H = T + V is conserved.
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4.3.3 An example - gravity revisited
Let us return to the case of gravitational attraction between two bodies but, unlike
in Section 1.2.4, now including both particles. We have T = 12 m1 ṙ21 + 21 m2 ṙ22 . The
potential is V = −Gm1 m2 /|r1 − r2 |. This system has total linear momentum and
total angular momentum conserved, as well as the total energy H = T + V .
Determinism, combined with uniqueness of solutions, means that precisely the
same initial conditions yield exactly the same trajectories. The basis of the universal
law of gravity is the regularity of the planetary orbits. The Galilean and Keplerian regularities could not hold if the geometry of space, at least locally, were not
Euclidean. Without this determinism, science would not have progressed.
What does this imply for the past and future of the universe? Given the positions and velocities of all the particles in the universe at any one time, we could
theoretically calculate the entire future and history of the universe. Think about
that.
4.4 Virial Theorem
”If only theorists knew what goes into an experimental data point, and
if only observers would know what goes into a theoretical calculations,
they would both take each other much less seriously”
The famous Swiss astrophysicist Fritz Zwicky was educated in Theoretical Physics
at ETH Zürich, finishing his diploma in 1922. He was notoriously bad tempered and
once called his colleagues at Caltech ”spherical bastards - spherical because the are
still bastards from any angle you look at them”! He was the first person to apply
the virial theorem in astrophysics and found the first evidence for dark matter in
1933. The fact that most of the matter in the universe is non-baryonic is one of the
greatest unsolved problems in modern physics.
4.4.1 Global properties of multi-particle systems
Can we derive some global relations between observable properties of a system of
interacting particles? i.e. linking together mass, velocities, size etc.
Define a global quantity called ”the virial”:
X
G=
pi · ri
(4.26)
i
which does not have an obvious interpretation, but is related to the inter-particle
forces. The summation is over all particles in the system.
The total time derivative is
X
dG X
=
ṙi · pi +
ṗi · ri
(4.27)
dt
i
i
– 20 –
where the terms can be expanded as
ṙi · pi =
X
mi ṙi · ṙi = 2T
(4.28)
i
and
X
ṗi · ri =
i
X
Fi · ri .
(4.29)
i
The time average of this derivative over a time τ is
Z
1 τ dG
G(τ ) − G(0)
dG
=
dt =
.
dt
τ 0 dt
τ
(4.30)
whilst for the right hand side we have the corresponding time averaged quantities.
For periodic motions or equilibrium systems this term = 0, which gives the virial
theorem:
*
+
1 X
hT i = −
(4.31)
Fi · r i
2
i
where the angle brackets denote time averaged quantities.
4.4.2 Example 1. Ideal gas law
See board notes
4.4.3 Example 2. Self gravitating systems
If the force can be derived from potentials then we can write:
hT i =
1X
ri · ∇Vi .
2 i
(4.32)
If the particles interact via a central power law force F ∝ rn we can write V = krn+1 .
Thus
ri · ∇V = k(n + 1)rn+1
= (n + 1)V
i
(4.33)
from which we summarise:
hT i =
n+1
hV i
2
(4.34)
for which gravity n = −2 gives the scalar virial theorem: hT i = − 21 hV i.
We can calculate the total gravitational potential of a system by summing over
the energy required to assemble shells of matter from infinity to the final system. For
2
a uniform density sphere we find V = − 3GM
(See board notes on the application to
5R
a star).
– 21 –
Now we can rewrite the scalar virial theorem in a form that can be applied to
star clusters or galaxy clusters:
M=
hRg i hvg i
G
(4.35)
where the Rg and vg are the mean gravitational radius of the system and the mean
velocity dispersion of the components. For the Coma galaxy clusters, Rg ≈ 106 pc
and vg ≈ 1500km/s giving a virial mass of about 1015 M which is approximately 100
times the mass in stars L = 1013 L ≡ 1013 M for stars like the sun. The missing
mass problem (dark matter), is one of the great unsolved problems in physics research
today - 70 years after Zwicky made this discovery we still do not know what most of
the matter in the universe is made of.
– 22 –