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As another example, when friction is present it is not true that kinetic energy
is lost, even though a sliding object stops and the kinetic energy seems to be lost.
The kinetic energy is not lost because, of course, the atoms inside are jiggling
with a greater amount of kinetic energy than before, and although we cannot
see that, we can measure it by determining the temperature. Of course IÝ we
disregard the heat energy, then the conservation of energy theorem will appear
to be false.
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Another situation in which energy conservation appears to be false is when
we study only part of a system. Naturally, the conservation of energy theorem
will appear not to be true iŸ something is interacting with something else on the
outside and we neglect to take that interaction into account.
In classical physics potential energy involved only gravitation and electricity,
but now we have nuclear energy and other energies also. Light, for example,
would involve a new form oŸ energy in the classical theory, bu we can aÌso, iÍ we
want to, imagine that the energy of light is the kinetic energy of a photon, and
then our formula (14.2) would still be right.
14-5 Potentials and ñelds
W© shall now discuss a few of the ideas associated with potential energy and
with the idea of a fieid. Suppose we have two large objects A4 and Ö and a
third very small one which is attracted gravitationally by the bwo, with some
resultant force #'. We have already noted in Chapter 12 that the gravitational
force on a particle can be written as its mass, mm, times another vector, C, which
1s debendent only upon the øoszfzon of the particle:
F' —nC.
W© can analyze gravitation, then, by imagining that there is a certain vector Ơ
at every position in space which “acts” upon a mass which we may place there,
but which is there itself whether we actually supply a mass for it to “act” on
or not. has three components, and each of those components is a function
6Ÿ (z,, z), a funetion oŸ position in space. Such a thing we call a #eld, and we
say that the objects A and Ö generafe the field, ï.e., they “make” the vector Ơ.
When an obJect is put in a field, the force on it is equal to 10s mass times the
value of the fñeld vector at the point where the object is put.
W©e can also do the same with the potential energy. Since the potential
energy, the integral of (—force) - (ds) can be written as ?m times the integral of
(—ñeld) - (4s), a mere change of scale, we see that the potential energy (+, , 2)
of an object located at a poïnt (z, , z) in space can be written as ?m tỉmes another
function which we may call the pofenfial W. The integral ƒ C - ds = —Ù, just
as [ F'- dø = —U; there is only a scale factor between the two:
U== | E+ds— =m | C -ds — mộ, (14.7)
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By having this function (z,,z) at every point in space, we can immedi-
ately calculate the potential energy of an object at any point in space, namely,
U(z, , z2) = mmW(z, 0, z)—rather a trivial business, it seems. But it is not really
trivial, because it is sometimes muụch nicer to describe the field by giving the
value of W everywhere in space instead of having to give Ơ. Instead of having to
write three complicated components of a vector function, we can give instead the
scalar function . Furthermore, it is much easier to calculate than any given
component of Ý when the field is produced by a number of masses, Íor since
the potential is a scalar we merely add, without worrying about direction. Also,
the fñeld Œ can be recovered easily from , as we shall shortly see. Suppose we
have point masses ?n, ma, ... at the points 1, 2,... and we wish to know the
potential W at some arbitrary point p. 'This is simply the sum of the potentials
at ø due to the individual masses taken one by one:
ữ() » ANG. 1,2,... (14.8)
In the last chapter we used this formula, that the potential is the sum of
the potentials from all the diferent objects, to calculate the potential due to a
spherical shell of matter by adding the contributions to the potential at a poïnt
trom all parts of the shell. "The result of this calculation is shown graphically
in EFig. 14-4. It is negative, having the value zero at r = oo and varying as l/r
down to the radius ø, and then is constant inside the shell. Outside the shell
the potential is —Œmm/r, where rm is the mass of the shell, which is exactly the
same as it would have been ïif all the mass were located at the center. But it is
not eueruhere exactly the same, for inside the shell the potential turns out to
be —Œm/a, and is a constantl WZhen the potential is constant, there is no Jield,
or when the potential energy is constant there is no force, because iŸ we move an
$(r) = —Gm/r
ở(r) = CONSTANT = —Gm/a
Fig. 14-4. Potential due to a spherical shell of radius a.
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object from one place to another anywhere Iinside the sphere the work done by
the force is exactly zero. Why? Because the work done in moving the object from
one place to the other is equal to minus the change in the potential energy (or,
the corresponding field integral is the change of the potential). But the potential
energy is the sœme at any two points inside, so there is zero change in potential
energy, and therefore no work is done in goïng between any ÿwo points inside the
shell. The only way the work can be zero for all directions of displacement is
that there is no force at all.
This gives us a clue as to how we can obtain the force or the field, given the
potential energy. Let us suppose that the potential energy of an object is known
at the position (z,,z) and we want to know what the force on the object is.
lt will not do to know the potential at only this one point, as we shall see; 1%
requires knowledge of the potential at neighboring points as well. Why? How
can we calculate the #-cormponent of the force? (If we can do this, oŸ course, we
can also find the - and z-components, and we will then know the whole force.)
Now, if we were to move the object a small distance Az, the work done by the
force on the object would be the z-component of the force times Az, if Az is
sufficiently small, and this should equal the change in potential energy in going
from one point to the other:
AW =_—AU = lạ Az. (14.9)
We have merely used the formula ƒ F'- ds = —AU, but for a 0erw short path.
NÑow we divide by Az and so fnd that the force is
t„ =—AAU/Az. (14.10)
Of course this is not exact. What we really want is the limit of (14.10)
as Az gets smaller and smaller, because it is only ezacfửu right in the limit of
infinitesimal Az. “This we recognize as the derivative of U with respect to #,
and we would be inclined, therefore, to write —đŨ/dz+. But U depends on z, ở,
and z, and the mathematicians have invented a different symbol to remind us to
be very careful when we are diferentiating such a function, so as to remember
that we are considering that onh + 0aries, and and z do not vary. Instead