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to have to write cs everywhere, we shall use the same trick concerning units of
the energy, the mass, and the momentum, that we used in Eq. (17.4). Energy
and mass, for example, difer only by a factor c2 which is merely a question of
units, so we can say energy is the mass. Instead of having to write the c2, we
put # = mn, and then, of course, 1Ÿ there were any trouble we would put in the
right amounts of e so that the units would straighten out in the last equation,
but not in the intermediate ones.
Thus our equations for energy and momentum are
=m =mo/Wl1— 02, (176)
Ð =0 =1mo0/V1— 02.
Also in these units, we have
E3 — pˆ = mạ. (17.7)
For example, 1Ÿ we measure energy in electron volts, what does a mass of 1 electron
volt mean? It means the mass whose rest energy is 1 electron volt, that is, mọc?
is one electron volt. For example, the rest mass of an electron is 0.511 x 108 eV.
Now what would the momentum and energy look like in a new coordinate
system? To find out, we shall have to transform 4q. (17.6), which we can do
because we know how the velocity transforms. Suppose that, as we measure it,
an object has a velocity 0, bu we look upon the same object rom the point of
view Of a space ship which itself is moving with a velocity u, and in that system
we use a prime to designate the corresponding thing. In order to simplify things
at first, we shall take the case that the velocity ø is in the direction of u. (Later,
we can do the more general case.) What is 0”, the velocity as seen from the space
ship? It is the composite velocity, the “diference” between 0 and u. By the law
which we worked out before,
g= —, (17.8)
1—0U0
Now let us calculate the new energy F”, the energy as the fellow in the space
ship would see it. He would use the same rest mass, of course, but he would
--- Trang 328 ---
use ? for the velocity. What we have to do is square œ, subtract 1% from one,
take the square root, and take the reciprocal:
g2 — 02 — 2u + uŸ
— 1—9uu+u2u2)
— L— 20 +u202— 02+ 2u — u2
1 — t) = —
1— 2u + u2u2 Í
— l—02—u2+u2u?
—— I—2u+u2u2 `
_— (—?)(1—u?)
— (1-uo} `
'Therefore
1 c— 1— 0 (17.9)
V1i—2 v1—u2V1—w2.
The energy #7 is then simply rmọ times the above expression. But we want
to express the energy in terms of the unprimed energy and momentum, and we
note that
Pmm..... (mo/V1— 02) — (mou/V1— 0?)u
V1—02vli—u2 V1—u2 l
E=——, 17.10
TC (17.10)
which we recognize as being exactly of the same form as
; ‡— uz
Ÿ =———m.
Next we must fñnd the new mmomentum 7Ø. This is just the energy #7 times 0,
and is also simply expressed in terms of / and ø:
b— pyy rmo(1 — ưo) 0u— TU — Tnow
= %2 — — — ———— - —————— ————
P„ VI-—ø»2V1-u2 (1—-u°) VTI—w2V1—u2
/ Đ„ — tUE
= ——ễ,Ụ 17.11
--- Trang 329 ---
which we recognize as being of precisely the same form as
; % — UuÈ
# =———n.
vV1—-u2
Thus the transformations for the new energy and momentum in terms of the
old energy and momentum are exactly the same as the transformations for # in
terms oŸ £ and z, and zø“ in terms of #z and ứ: all we have to do is, every tỉme we
see £ in (17.4) substitute #, and every time we see # substitute ø„, and then the
cquations (17.4) will become the same as Eqs. (17.10) and (17.11). This would
imply, iƒ everything works right, an additional rule that ø, = ø„ and that ø = Ø¿.
To prove this would require our goïng back and studying the case of motion up
and down. Actually, we did study the case of motion up and down in the last
chapter. We analyzed a complicated collision and we noticed that, in fact, the
transverse momentum is øø‡ changed when viewed from a moving system; so we
have already verified that „ = ø„ and 7, = ø;. The complete transformation,
then, 1s
g — Đ„ — tu
% 1 — „2 hà
Ủy = Đụ;
gi (17.12)
Đy — Dz:;
E/ — b -~ uDx
v1—-u2
In these transformations, therefore, we have discovered four quantities which
transform like z, , z, and , and which we call the ƒour-uector mmomentum. Since
the momentum is a four-vector, it can be represented on a space-time diapgram
of a moving particle as an “arrow” tangent to the path, as shown in Fig. 17-4.
'This arrow has a time component equal to the energy, and its space components
Fig. 17-4. The four-vector momentum of a particle.
--- Trang 330 ---
represent its three-vector momentum; this arrow is more “real” than either the
energy or the momentum, because those just depend on how we look at the
diagram.
17-5 Four-vector algebra
The notation for four-vectors ¡is diferent than it is for three-vectors. In
the case of three-vectors, if we were to talk about the ordinary three-vector
mmomentum we would write it ø. If we wanted to be more specifc, we could say
it has three components which are, for the axes in question, ø„, ø„, and Øø;, Or