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same way. Why? Because one machine, when analyzed by Moe, has exactly the
same equations as the other one, analyzed by Joe. Since the eguations are the
same, the phenornena appear the same. So the proof that an apparafus in a new
position behaves the same as it did in the old position is the same as the proof
that the equations when displaced in space reproduce themselves. 'Pherefore
we say that the laus oƒ phụsics are sụmmetrical [or translatlional đisplacemenis,
symmetrical in the sense that the laws do not change when we make a translation
of our coordinates. OÝ course it is quite obvious intuitively that this is true, but
1E is interesting and entertaining to discuss the mathematics of it.
11-3 Rotations
The above is the first of a series of ever more complicated propositions
concerning the symmetry of a physical law. The next proposition is that it should
make no diference in which đirecfion we choose the axes. In other words, If we
--- Trang 216 ---
build a piece of equipment in some place and watch it operate, and nearby we
buïld the same kind of apparatus but put it up on an angle, will it operate in the
same way? Obviously ¡it will not if it is a Grandfather clock, for examplel If a
pendulum clock stands upright, ¡it works fine, but ïf ¡it is tilted the pendulum falls
against the side of the case and nothing happens. The theorem is then false in
the case of the pendulum clock, unless we include the earth, which ¡is pulling on
the pendulum. Therefore we can make a prediction about pendulum clocks 1Ÿ we
believe in the symmetry of physical law for rotation: something else is involved in
the operation oŸ a pendulum clock besides the machinery of the clock, something
outside it that we should look for. We may also predict that pendulum clocks will
not work the same way when located in diÑferent places relative to this mysterious
Source of asymmetry, perhaps the earth. Indeed, we know that a pendulum clock
up ïn an artificial satellite, for example, would not tick either, because there is no
effective force, and on Mars it would go at a diferent rate. Pendulum clocks đo
involve something more than just the machinery inside, they involve something
on the outside. Onece we recognize this factor, we see that we must turn the earth
along with the apparatus. Of course we do not have to worry about that, it is easy
to do; one simply waits a moment or ©wo and the earth turns; then the pendulum
clock ticks again in the new position the same as it did before. While we are
rotating in space our angles are always changing, absolutely; this change does not
seem to bother us very much, for in the new position we seem to be in the same
condition as in the old. 'This has a certain tendency to confuse one, because 1
1s true that in the new turned position the laws are the same as in the unturned
position, but it is nof true that as 0e turn a thíng ï€ follows the same laws as it
does when we are not turning it. IÝ we perform sufficiently delicate experiments,
we can tell that the earth ¡s rofa#ng, but not that it had rotated. In other words,
we cannot locate its angular position, but we can tell that it is changing.
Now we may discuss the efects of angular orientation upon physical laws.
Let us ñnd out whether the same game with Joe and Moe works again. 'This
time, to avoid needless complication, we shall suppose that Joe and Moe use the
same origin (we have already shown that the axes can be moved by translation
to another place). Assume that Moe”s axes have rotated relative to Joe's by an
angle Ø. The two coordinate systems are shown in Fig. l1I-2, which is restricted
to two dimensions. Consider any point P having coordinates (z,) in jJoe's
system and (z”,') in Moe's system. We shall begin, as in the previous case, by
expressing the coordinates zø“ and 3“ in terms of z, #, and Ø. To do so, we first
drop perpendiculars from ? to all four axes and draw 4Ö perpendicular to PQ.
--- Trang 217 ---
— (& ý)
ca N ysin8 (MOE)
S0 NI x
xcos0 ~Zr
PB (JOE)
Fig. 11-2. IWwo coordinate systems having different angular orienta-
tions.
Inspection of the fgure shows that #“ can be written as the sum of two lengths
along the ø-axis, and ø as the difference of two lengths along 4Ø. All these
lengths are expressed in terms of z, , and Ø in equations (11.5), to which we
have added an equation for the third dimension.
+“ = #øcos 8 + 1 sin 6,
ˆ = cos0 — zsin0, (11.5)
The next step is to analyze the relationship of forces as seen by the two observers,
following the same general method as before. Let us assume that a force #', which
has already been analyzed as having components „ and #2 (as seen by Joe), is
acting on a particle of mass rn, located at point Pín Fig. 11-2. For simplicity,
let us move both sets of axes so that the origin is at , as shown in Eig. l1-3.
Moe sees the components of #" along his axes as F and È;¿. F„ has components
along both the z/- and '-axes, and #„ likewise has components along both these
axes. To express #¿ in terms of F; and #„, we sum these components along the
z/-axis, and in a like manner we can express #2 in terms of #+ and F„. The
results are
tạ = Fạ cos Ø + Fý, sin Ø,
Tàu = Fy cosØ — F„ sìn Ú, (11.6)
đà, =F,.
Tt is interesting to note an accident of sorts, which is of extreme importance: the
formulas (11.5) and (11.6), for coordinates oŸ P and components of #", respectively,
are 0ƒ identical form.
--- Trang 218 ---
FyE-------=z F
_~“4 ụ
FT BÀ x:
Fig. 11-3. Components of a force in the two systems.
As before, Newton”s laws are assumed to be true in Joe°s system, and are
expressed by equations (11.1). The question, again, is whether Moe can apply
Newton”s laws—will the results be correct for his system of rotated axes? In
other words, if we assurne that Eqs. (11.5) and (11.6) give the relationship of the
mmeasurements, is it true or not true that
m(d°z! (dt?) = F:,
m(d2W' dt?) = Fụ, (11.7)
m(d°z! (dt?) = F..?
To test these equations, we calculate the left and right sides independently, and
compare the results. To calculate the left sides, we multiply equations (11.5)
by n, and diferentiate twice with respect to time, assuming the angle Ø to be
constant. This gives
m(d°+! (dt?) = m(dÊ+/df?) cos 9 + m(d®u /dt2) sin 0,
m(dSV /dt?) = m(d /đt2) cos 9 — rn(dŠ+/dt?) sìn 6, (11.8)