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sphere. And suppose that we shoot an object along the ground, with no forces on
it. Where will it go? It will appear to go ïn a straight line, but it has to remain
on the surface of a sphere, where the shortest distance between two poinfs 1s
along a great circle; so it goes along a great circle. If we shoot another object
similarly, but in another direction, it goes along another great circle. Because
we think we are on a plane, we expect that these two bodies will continue to
diverge linearly with time, but careful observation will show that if they go far
enough they move closer together again, as though they were attracting each
other. But they are nø£ attracting each other—there is just something “weird”
about this geometry. This particular ïllustration does not describe correctly the
way in which Einstein's geometry is “weird,” but ït illustrates that if we distort
the geometry sufficiently it is possible that all gravitation is related in some way
to pseudo forces; that is the general idea of the Einsteinian theory of gravitation.
12-6 Nuclear forces
W©e conclude this chapter with a brief discussion of the only other known
forces, which are called mœ%wclear ƒorces. These forces are within the nuclei of
atoms, and although they are much discussed, no one has ever calculated the
force between two nuelei, and indeed at present there is no known law for nuclear
forces. These forces have a very tiny range which is just about the same as
the size of the nucleus, perhaps 10~†13 centimeter. With particles so small and
at such a tiny distance, only the quantum-mechanical laws are valid, not the
Newtonian laws. In nuclear analysis we no longer think in terms of forces, and in
fact we can replace the force concept with a concept of the energy of interaction
of two particles, a subject that will be discussed later. Any formula that can
be written for nuclear forces is a rather crude approximation which omits many
complications; one might be somewhat as follows: forces within a nucleus do
not vary inversely as the square of the distance, but die off exponentially over a
cortain distance r, as expressed by #' = (1/z?) exp(—z/ro), where the distance 7o
is of the order of 10—13 centimeter. In other words, the forces disappear as soon
as the particles are any great distance apart, although they are very strong
within the 10~13 centimeter range. So far as they are understood today, the laws
of nuclear force are very complex; we do not understand them in any simple
way, and the whole problem of analyzing the fundamental machinery behind
--- Trang 250 ---
nuclear forces is unsolved. Attempts at a solution have led to the discovery of
numerous strange particles, the x-mesons, for example, but the origin of these
forces remains obscure.
--- Trang 251 ---
I2
MVor'Ek (ra ốổl IPoforeffteal FErrorggg/ (Ì)
13-1 Energy of a falling body
In Chapter 4 we discussed the conservation of energy. In that discussion, we
địd not use Newton's laws, but i§ is, oÝ course, of great interest to see how 1
comes about that energy is in fact conserved in accordance with these laws. For
clarity we shall start with the simplest possible example, and then develop harder
and harder examples.
The simplest example of the conservation of energy is a vertically falling
object, one that moves only in a vertical direction. An object which changes its
height under the inÑuence of gravity alone has a kinetic energy 7 (or K.E.) due
to its motion during the fall, and a potential energy ?møh, abbreviated (or
P.E.), whose sum is constant:
simu7 + mụgh = const,
K.E. P.E.
1'+UU = const. (13.1)
Now we would like to show that this statement is true. What do we mean, show ït
is true? Hrom Newton's Second Law we can easily tell how the objecE moves, and
1E is easy to fnd out how the velocity varies with time, namely, that it increases
proportionally with the time, and that the height varies as the square of the time.
So 1Ý we measure the height from a zero point where the object 1s stationary, 1W
1s no miracle that the height turns out to be equal to the square of the velocity
times a number of constants. However, let us look at it a little more closely.
Let us ñnd out đứrecfiu from Newtons Second Law how the kinetic energy
should change, by taking the derivative of the kinetic energy with respect to time
and then using Newton's laws. When we diferentiate smu2 with respect to time,
we obtain đT d đo đo
Trm (Sm02) = 3m20 Px... (13.2)
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since 7n is assumed constant. But from Newton”s Second Law, m(do/đf) = F}, so
đT/dt = Fo. (13.3)
In general, it will come out to be #'-ø, but in our one-dimensional case let us
leave 1 as the force times the velocity.
Now in our simple example the force is constant, equal to —?mng, a vertical
force (the minus sign means that it acts downward), and the velocity, oÝ course,
1s the rate of change of the vertical position, or heipht h, with time. Thus the
rate of change of the kinetic energy is —rng(dh/đf), which quantity, miracle of
miracles, is minus the rate of change of something elsel It is minus the time rate
of change of mmghl 'Therefore, as time goes on, the changes in kinetic energy and
in the quantity rmgh are equal and opposite, so that the sum of the two quantities
remains constant. Q.E.D.
W©e have shown, om Newton's second law of motion, that energy is con-
served for constant forces when we add the potential energy ?mgh to the kinetic
©n©rgy sinu2. Now let us look into this further and see whether it can be gener-
alized, and thus advance our understanding. Does it work only for a freely falling
body, or is it more general? We expect from our discussion of the conservation
of energy that it would work for an object moving from one point to another
in some kind of frictionless curve, under the inÑuence of gravity (Fig. 13-1). If
the obJect reaches a certain height h from the original height HỨ, then the same
formula should again be right, even though the velocity is now in some direction
other than the vertical. We would like to understand :ø0h# the law is still correct.
Let us follow the same analysis, ñnding the time rate of change of the kinetic
energy. This will again be rmø(du/đf), but rm(du/đf) is the rate of change of
the magnitude of the momentum, 1.e., the ƒorce ?n the đirection oƒ motion—the
Fig. 13-1. An object moving on a frictionless curve under the influence
Of gravity.
--- Trang 253 ---
tangential force ;¿. Thus
—= — = F0.
dc “ng
Now the speed is the rate of change of distance along the curve, đs/đf, and
the tangential force #‡ 1s not —rng but is weaker by the ratio of the vertical