How Much Momentum do the Plates Have???
Randall M. Richardson
Department of Geosciences, Bldg. 77
University of Arizona
Tucson AZ 85721
(520) 621-4950
[email protected]
Momentum, defined as the product of mass times velocity, is an important quantity that
can help us understand the dynamics of plate motion. It is closely related to inertia, for
which most of us have a good intuitive feel (a massive object has a lot of inertia, and it is
hard to change its velocity). Because inertia and momentum are related through mass,
most objects with a lot of momentum have a lot of inertia, and our intuition tells us that
it is hard to change the velocity of an object with a lot of momentum. Clearly, for
example, a speeding train has a lot of momentum, and few of us would volunteer to try
slowing it down by standing in front of it!
.
How, then, is momentum important for understanding the dynamics of plate motion?
If plates have a lot of momentum, then it makes sense to conclude that it takes a lot of
effort (i.e., large forces) to change the speed or direction of plate motion.
To explore this further, consider the hypothetical plate below, whose surface
dimensions are 5000 x 5000 km and which has a vertical thickness of 100 km.
z
5000 km
0k
500
m
y
v
100 km
x
About 20 such hypothetical plates could fit on the surface of the Earth, so this is a
reasonable size for the plate.
To get the momentum p for this plate, we need the mass m and the velocity v of our
plate. Let’s assume the plate has a velocity v of 100 mm/yr in the x direction, as defined
on the figure above. This is a reasonable plate velocity. The mass m is given by the
plate volume times its average density, which we will assume for simplicity is 3000
kg/m3. After all, we’re doing a geophysical calculation here, and we wouldn’t want to
use too many significant digits!
The mass m of the plate is then:
m =
=
=
=
5000 km x 5000 km x 100 km x 3000 kg/m3
(5000 x 103m) x (5000 x 103m) x (100 x103m) x 3000 kg/m3
(5 x 106) x (5 x 106) x (1 x 105) x 3000 kg
7.5 x 10 21 kg.
The speed (or velocity v, if we specify motion in the x direction) is:
v
= 100 mm/yr = 10-1 m/yr = 10-1 m/yr x (1 yr/3.15 x 107 s)
= 3.2 x 10-9 m/s.
Combining these, we have that the momentum p is given by
p
=
=
=
=
mxv
7.5 x 10 21 kg x 3.2 x 10-9 m/s
2.4 x 1013 kg m/s = 2.4 x 1013 kg m/s2 x s
2.4 x 1013 N s, the SI units for momentum.
What can we say about this, other than it looks like a big number? Let’s begin by
comparing it to something we can better visualize.
Consider the momentum of a speeding train. We’ll assume that the mass of the train is
about 1,000,000 kg (after all, we’re geophysicists, and we can assume anything!
Actually, a typical automobile has a mass of about 1000 kg, so it is reasonable to assume
that a (fully loaded?) train has the mass of about 1000 automobiles). We’ll also assume
it is traveling at 100 km/hr, which is also about 30 m/s (trust me, and if you don't, do
the calculation yourself). Then the momentum p is about 3 x 107 N s.
Thus it takes about 800,000 speeding trains to equal the momentum of our plate, and it
still looks like plates have a lot of momentum.
Let’s look at it another way, however, that has more relevance for understanding the
forces that drive plate motion. Consider the following thought experiment. We start
with the plate initially at rest and ask ourselves how long it takes to accelerate the plate
to typical plate speeds using some simple force. This is also equivalent to starting with
the plate moving at the typical plate speed and applying the same force in the opposite
direction and asking how long it will take to stop the plate.
What shall we choose for a force? Let’s consider blowing on one end with a pressure
equivalent to one atmosphere. This isn’t a very large force in the scheme of things, but
it will give us some feeling for how hard it is to change plate speeds. Since the plate is
initially at rest and we want it to move in the +x direction, let’s have the force act on the
yz face of the plate at x = 0 starting at time t = 0.
We need to know the total force acting on the yz face of the plate. Since pressure is the
force over an area, force equals the pressure times the area over which the pressure is
acting. The area A of the yz face is
A = 5000 km x 100 km = 5000 x (103 m) x 100 x (103 m) = 5 x 1011 m2.
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Now, a pressure of one atmosphere is 1 bar or about 0.1 MPa, the SI unit of stress,
where MPa stands for megapascals, or 106 pascals, and where 1 pascal equals 1 newton
per meter squared (i.e., 1 Pa = 1 N/m2). Thus 0.1 MPa = 105 N/m2. Therefore, the total
force F is given by the pressure times area, or
105 N/m2 x 5 x 1011 m2 = 5 x 1016 N.
F =
We wish to relate this force to a change in plate velocity. From first principles, we start
with Newton’s Second Law:
F = m x a,
where a is acceleration ∂ 2 u/∂t 2 , where u is displacement. Momentum p is:
p
= m x v.
The partial derivative of velocity v with respect to time is, however, just the acceleration
a, and thus,
∂p/∂t = m∂v/∂t = ma = F.
Therefore,
F = ∂p/∂t
Integrating both sides with respect to time t, gives
∂p
dt =
∂t
∫ F dt = ∫
∫ dp
where we have to be a little careful about the limits of integration. When we begin our
problem, we are at time t = 0 and at rest (i.e., p = 0). At the end of the experiment, we
will be at time t = t1, the time it takes to accelerate the plate to 100 mm/yr, and
momentum p1 will equal the final momentum, which we have already calculated as 2.4
x 1013 N s. Thus our problem is given by
∫
t1
F dt =
0
∫
t 1 ∂p
0
∂t
∫
p1
dt = dp
0
and we wish to solve to t1. F is a constant 5 x 1016 N throughout the experiment, so the
left-hand side becomes
∫
t1
0
(5 × 1016 N) dt = 5 × 101 6 N (t1 – 0) = 5 × 101 6 N t1 .
The right-hand side becomes
p1
∫ dp = p – 0 = 2.4 x 10
0
1
13
N s.
Equating the left- and right-hand sides gives
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5 x 101 6 N t1 = 2.4 x 101 3 N s,
and finally solving for t1 gives
t1 = 4.8 x 10– 4 s,
or about half a millisecond! That is, we can accelerate our plate to typical plate velocities
in less than a millisecond by blowing on one end of it! Equivalently, if no other forces
are acting on it, we can bring it to a stop from typical plate velocities by blowing on the
other end of it in less than a millisecond.
This is such an amazing result that it is appropriate to review our assumptions to see
where we could have been off by many orders of magnitude. We’ve already
established that our plate has typical dimensions and velocity. Besides, doubling the
speed or any one of the dimensions at most doubles the momentum, which is far too
little to change the conclusion that it takes about one millisecond.
Where else could we be wrong? Perhaps our analogy of blowing on one end of it is
drastically in error. Without going into the details, one force that is often mentioned as
a component of the driving mechanism for plate tectonics is the so-called ridge push
force (associated with the gravitational sliding of the plate off the elevated topography
of the mid-ocean ridge). This force is usually given as about 3 x 1012 newtons per meter
(N/m) of ridge length. If this force acts along the yz face, the total force is
F = 5000 km x 3 x 1012 N/m = 5000 x (103 m) x 3 x 1012 N/m
= 1.5 x 1019 N,
or 300 times bigger than the force in our test case. There are also people who say that
ridge push forces are too small to play a major role in the driving mechanism, especially
given that slab pull (due to the excess mass of the cold, subducted slab pulling on the
surface plate) is an order of magnitude larger than the ridge push force. This large slab
pull, however, is likely balanced by a comparably large resistance localized between the
trench and the end of the slab, so that the surface plates see little effect due to the slab.
In any case, the bottom line is that we have chosen a conservative (i.e., small) force for
our test case.
We therefore have to conclude that the plates have negligible momentum. In fact, the
entire mantle convection system, of which the plates are a part (and a part which may
well determine mantle convective patterns rather than the other way around), has
negligible momentum. Plate momentum is not negligible in the sense that it is a small
number, or even small compared to a speeding train, but small in the sense that it takes
very little time for reasonable forces to significantly change the velocity. How can this
be true? There are two points worth mentioning. First, although plates are indeed
very massive, they have speeds that are very, very small. Comparing the speed of a
plate (about 10-9 m/s) to that of, say, our speeding locomotive (about 10 m/s), we see a
difference of ten orders of magnitude. If we wanted to accelerate our plate to the speed
of a locomotive with the same force that we have assumed acts on one end, it would
take ten orders of magnitude longer, or about 4.8 x 106 s, which is about two months.
Geologically speaking, this is still very fast. Second, and more importantly, reasonable
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forces acting to drive and resist plate motion are actually fairly large because the
dimensions of plates are so large. Thus it is misleading to look at the momentum of the
plates just in terms of the size of the number. Instead, just as we have done, the most
useful exercise is to think of the momentum in terms of how long it takes to change the
velocity of a body with “reasonable” forces. In this sense, the plates have negligible
momentum.
What are the implications of negligible plate momentum on the driving mechanism for
plate tectonics? Well, first and foremost, plates respond essentially instantaneously to
changes in the forces acting on them. We observe, however, that plate velocities tend
to be approximately constant on the order of millions to tens of millions of years and
then can change on very short notice geologically (the Hawaiian-Emperor bend comes
to mind immediately). This must imply a delicate balance of forces that keeps the net
force on the plate essentially zero (i.e., the sum of forces driving and resisting plate
motion must cancel); otherwise the plates would accelerate. This means that the driving
mechanism must be very effectively buffered against changes in the net force acting on
the plates. Most likely the buffering comes from some kind of a velocity dependence
on the forces, so that, for example, resistive forces would increase greatly if the plate
were driven faster, and in this manner a balance is maintained. It also means that
relatively small forces, like the ridge force, may play an important role in driving plate
motion.
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