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* 'Phe electrons would actually win the race versus 0¿s2ble light because of the index of
refraction of air. A gamma ray would make out better.
--- Trang 300 ---
paperl When the electrons are going that fast their masses are enormous, but
their speed cannot exceed the speed oŸ light.
Now let us look at some further consequences of relativistic change of mass.
Consider the motion of the molecules in a small tank of gas. When the gas
is heated, the speed of the molecules is increased, and therefore the mass 1s
also increased and the gas is heavier. An approximate formula to express the
Increase of mass, for the case when the velocity is small, can be found by
expanding mo/1 — 02/c2 = mọ(1 — 02/c2)~1⁄2 in a power series, using the
binomial theorem. We get
mạ(1L— t2/cÈ) 8 = mạ(1 + }08/cÊ + 304/et +),
W© see clearly from the formula that the series converges rapidly when 0 is small,
and the terms after the first two or three are negligible. So we can write
~ 1 s( 1
m nọ + srnoU 2 (15.11)
in which the second term on the ripght expresses the increase of mass due to
molecular velocity. When the temperature inereases the ø2 increases proportion-
ately, so we can say that the increase in mass is proportional to the increase
in temperature. But since singu? 1s the kinetic energy in the old-fashioned
Newtonian sense, we can also say that the increase In mass of all this body of gas
is equal to the inerease in kinetic energy divided by e2, or Am = A(K.B.)/€2.
15-9 Equivalence of mass and energy
The above observation led Einstein to the suggestion that the mass of a body
can be expressed more simply than by the formula (15.1), if we say that the mass
is equal to the total energy content divided by c2. If Eq. (15.11) is multiplied
by c? the result is
mcŸ = mạc” + 3m0” + - -- (15.12)
Here, the term on the left expresses the total energy oŸ a body, and we recognize
the last term as the ordinary kinetic energy. Einstein interpreted the large
constant term, ?moe2, to be part of the total energy of the body, an intrinsic
energy known as the “rest energy.”
Let us follow out the consequences of assuming, with Einstein, that ứhe
energu oƒ a bod aluas eguals mc2. As an interesting result, we shall find the
--- Trang 301 ---
formula (15.1) for the variation of mass with speed, which we have merely assumed
up to now. W© start with the body at rest, when its energy is mọc”. Then we
apply a force to the body, which starts it moving and gives it kinetic energy;
therefore, since the energy has increased, the mass has increased——this is Implieit
in the original assumption. So long as the force continues, the energy and the
mass both continue to increase. We have already seen (Chapter 13) that the rate
of change of energy with time equals the force times the velocity, or
—— —= È'`-0. 15.13
di ° 5.13)
W© also have (Chapter 9, Eq. 9.1) that ' = d(mo)/dt. When these relations are
put together with the delnition of , Eq. (15.13) becomes
d(mc2) d(mb)
————=U'—.. 15.14
dc ””” đ 5.14)
We wish to solve this equation for ?mm. To do this we first use the mathematical
trick of multiplying both sides by 2mm, which changes the equation to
đm d(ựm?®)
2m) —— = 2m0 - ———. 15.15
cm) d‡ TT ' )
W© need to get rid of the derivatives, which can be accomplished by integrating
both sides. The quantity (2m) đưm/đf can be recognized as the từne derivative
of m2, and (2m0) - d(mo) /dt is the tỉme derivative of (mø)2. So, Eq. (15.15) is
the same as (m2) (m202)
d(m d(m“u
———=—.. 15.16
“— dị di (15.16)
Tf the derivatives of two quantities are equal, the quantities themselves differ at
most by a constant, say Œ. 'Phis permits us to write
m°c? = m?u° + Œ. (15.17)
W© necd to delne the constant Œ more explicitly. Since Eq. (15.17) must be true
for all velocities, we can choose a special case where ø = 0, and say that in this
case the mass is mo. Substituting these values into Eq. (15.17) gives
mặc? =0+ Œ.
--- Trang 302 ---
W© can now use thịs value of Ở in Eq. (15.17), which becomes
mỸc2 = mm 2u” + mặc”. (15.18)
Dividing by e? and rearranging terms gives
mÊ(1 — 0Ê/c2) = mã,
from which we get
m = mo/W1— 02/c2. (15.19)
This is the formula (15.1), and is exactly what is necessary for the agreement
between mass and energy in Eq. (15.12).
Ordinarily these energy changes represent extremely slight changes in mass,
because most of the time we cannot generate much energy from a given amount
of material; but in an atomie bomb of explosive energy equivalent to 20 kilotons
of NT, for example, it can be shown that the dirt after the explosion is lighter
by 1 gram than the initial mass of the reacting material, because of the energy
that was released, I.e., the released energy had a mass of l1 gram, according to
the relationship AE = A(mc2). Thịis theory of equivalence of mass and energy
has been beautifully verified by experiments in which matter is annihilated——
convcrted totally to energy: An electron and a positron come together at rest,
cach with a rest mass mmọ. When they come together they disintegrate and bwo
øamma rays emerge, each with the measured energy of mọc2. This experiment
furnishes a direct determination of the energy associated with the existence of
the rest mass of a particle.
--- Trang 303 ---
I6
Miolqfitisfic Freorgjgg «areeÏ WẤC@rtt©reftrrtt
16-1 Relativity and the phỉilosophers
In this chapter we shall continue to discuss the principle of relativity of
Binstein and Poincaré, as it afects our ideas of physics and other branches of
human thought.
Poincaré made the following statement of the principle of relativity: “Ac-
cording to the principle of relativity, the laws of physical phenomena must be