In classical mechanics, it is easy to see from the conservation laws =
applied in the center of mass that the kinetic energies of the daught=
er nuclei stand in the inverse proportion to their masses.
In relativistic mechanics the same approach leads to a calculation =
a little bit more involved but still pretty straightforward. In this =
case, I obtained for the ratio of the individual kinetic energies the=
expression
K1/K2 =3D (M-m1+m2)/(M-m2+m1), (1)=
=20
where M is the rest mass of the initial nucleus, and m1, m2 are the r=
espective rest masses of the daughter nuclei. It still is rather simp=
le. In particular, it shows that a nucleus with lesser rest mass (say=
, m2) gets greater kinetic energy, and vice versa.=20
The classical limit can be obtained from (1) by rearranging=20
K1/K2 =3D (M-m1-m2+2m2)/(M-m2-m1+2m1) (2)
=20
and then allowing m1+m2 to approach M. This immediately gives the cla=
ssical result
K1/K2 =3D m2/m1 (3=
)
Moses Fayngold,=20
NJIT
=20
-----Original Message-----
=46rom:=09Forum for Physics Educators on behalf of Ken Caviness
Sent:=09Fri 10/7/2005 3:20 PM
To:=09PHYS-L@LISTS.NAU.EDU
Cc:=09
Subject:=09Fraction of energy carried off
I have a physics question for you:
(modified from Young & Freedman, 11th ed., 8.88): A large nucleus at=
rest decays into 2 particles the ratio of whose masses is $B&V (J. =
What fraction of the total kinetic energy (after the decay) does eac=
h particle have? (This is a classical, non-relativistic treatment.)
First, feel free to do the problem before scrolling down to see my an=
swer below. I tried a couple different methods and show only the one=
that I like the most!
Next, notice what this means about how the fraction of total kinetic =
energy which each particle carries off depends on the fraction of the=
total mass which each particle got. Interesting, I think! Seems fu=
ndamentally significant to me.
Finally, and the reason I turn to the Forum for Physics Educators, co=
nsider doing the same problem relativistically. I don't get the same=
simple, neat answer, in fact, I get a complicated mess. Does this s=
till simplify somehow? Or is the peculiarly simple and appealing res=
ult only true in the classical limit? Maybe a different (and still e=
legant) result is always true? I've tried total energies instead of =
kinetic energies, still a mess. Besides, that wouldn't work in the c=
lassical limit anyway. Sigh! I would appreciate any thoughts on thi=
s.
A nifty result. The fraction of kinetic energy each got is the same =
as the fraction of the mass that the _other_ got. For instance, if a=
n object explodes into 2 pieces, getting 1/10 and 9/10 of the mass, r=
espectively, then they carry away 9/10 and 1/10 of the energy of the =
explosion, respectively. A useful corollary lets you treat any colli=
sion/rebound of two objects: shift into the center of mass coordinat=
e system, in which the incoming relative momenta are equal in magnitu=
de, and in that coordinate system the above fraction of energy relati=
onships will hold. Nice, don't you agree?
An earlier solution used everything in terms of speeds (p=3Dmv, K=
=3D(1/2)mv^2) to reach the same results, but took longer to get there=
. In keeping with that idea, in the relativistic case I'll use the r=
elativistic relationship between energy and momentum,
E^2 =3D p^2 c^2 + (m c^2)^2,
rather than the formulas relating each to velocity
p =3D $B&C (J m v, K =3D ( $B&C (J-1)m c^2,
where $B&C (J =3D 1/sqrt(1-v^2/c^2).
Here we go:
K_1 =3D E_1 - m c^2 =3D sqrt( p^2 c^2 + (m c^2)^2 ) - m c^2
K_2 =3D E_2 - M c^2 =3D sqrt( p^2 c^2 + (M c^2)^2 ) - M c^2.
Yes, it's the same momentum magnitude p in both cases, since momentum=
is conserved. This should result in some simplification, just as in=
the non-relativistic treatment, even though M and m are different. =
To proceed,
=3D [ sqrt( p^2 + m^2 c^2 ) - m c ]
------------------------------------------------------------
[ sqrt( p^2 + m^2 c^2 ) - m c + sqrt( p^2 + M^2 c^2) - M c ]
=3D m c [ sqrt((p/(mc))^2 + 1) - 1 ]
-------------------------------------------------------------------=
--
m c [ sqrt((p/(mc))^2 + 1 ) - 1 ] + M c [ sqrt((p/(Mc))^2 + 1 ) - 1=
]
=3D m [ sqrt((p/(mc))^2 + 1) - 1 ]
-----------------------------------------------------------------
m [ sqrt((p/(mc))^2 + 1 ) - 1 ] + M [ sqrt((p/(Mc))^2 + 1 ) - 1 ]
=3D m [ sqrt((p/(mc))^2 + 1) - 1 ]
-----------------------------------------------------------------
m [ sqrt((p/(mc))^2 + 1 ) - 1 ] + M [ sqrt((p/(Mc))^2 + 1 ) - 1 ]
Nothing like M/(M+m) showing up here, and I don't see how to proceed =
without using classical limit approximations, for example. Maybe the=
result is just a lucky coincidence? I would be disappointed if that=
were all there was to it.