2.5 Estimating Yield—Analytic
85
energy-momentum relationship of photons, but an ersatz justification can be argued
as follows. The non-relativistic result can be re-written as P = ρ < v
2 >/3 where
ρ is the mass density of the gas. Photons do not have mass, but for the purposes
of this quick argument we can use Einstein’s famous E = mc
2 equation to assign
an effective equivalent total mass to the total energy of all photons: m tot = E tot /c
2 .
Hence the density becomes ρ = m tot /V = E tot / (c
2 V ), and so the pressure becomes
P = E tot < v
2 >/ (3c
2 V ). Setting < v
2 > = c
2 , P = E tot / 3 V, or P = U/3 as claimed. In the
case of a “gas” of uranium nuclei of standard density of that metal, radiation pressure
dominates for per-particle energies greater than about 2 keV (see Exercise 2.14)
How does a gas of photons arise to give a radiation pressure in an exploding bomb
core? Fission fragments are bare nuclei and so are highly electrically charged. As they
decelerate, they naturally emit energy in the form of photons of wavelengths across
the electromagnetic spectrum. Much of the energy released in a nuclear explosion is
in the form of gamma-rays and X-rays which ionize the surrounding air.
For simplicity, we model the bomb core as an expanding sphere of radius r(t) with
every atom in it moving radially outwards at speed v. Do not confuse this velocity
with the average neutron speed v neut , which enters into τ. If the sphere is of density
ρ(t) and total mass M, its total kinetic energy will be
K core =
1
2
Mv
2
=
2π
3
ρ v
2 r
3
.
(2.88)
Now invoke the work-energy theorem in its thermodynamic formulation W =
P(t)dV, and equate the work done by the gas (or radiation) pressure in changing the
core volume by dV over time dt to the change in the core’s kinetic energy over that
time:
P(t)
dV
dt
=
d K core
dt
.
(2.89)
To formulate this explicitly, write, from (2.88), dK core /dt =(2π / 3)ρ r
3 (2vdv/dt),
put dV/dt =4π r
2 (dr/dt), and incorporate (2.87) to give
dv
dt
=
3P o
ρ r
e
(α/τ ) t
.
(2.90)
Note that in taking the derivative for dK core /dt, ρ r
3 can be treated as a constant
since it is (but for a factor of 4π /3), the mass of the core. To solve this for the radius
of the core as a function of time, we face the problem of what to do about the fact that
both ρ and r are functions of time. We deal with this by means of an approximation.
Review the discussion regarding core expansion following (2.83) above. As the
core expands, its density when it has any general radius r will be ρ(r) = Cρ o (R o /r)
3 ,
and criticality will hold until such time as ρ r = ρ o R o , or, on eliminating ρ, r =
C
1/2 R o . We can then define r, the range of radius over which criticality holds:
85
energy-momentum relationship of photons, but an ersatz justification can be argued
as follows. The non-relativistic result can be re-written as P = ρ < v
2 >/3 where
ρ is the mass density of the gas. Photons do not have mass, but for the purposes
of this quick argument we can use Einstein’s famous E = mc
2 equation to assign
an effective equivalent total mass to the total energy of all photons: m tot = E tot /c
2 .
Hence the density becomes ρ = m tot /V = E tot / (c
2 V ), and so the pressure becomes
P = E tot < v
2 >/ (3c
2 V ). Setting < v
2 > = c
2 , P = E tot / 3 V, or P = U/3 as claimed. In the
case of a “gas” of uranium nuclei of standard density of that metal, radiation pressure
dominates for per-particle energies greater than about 2 keV (see Exercise 2.14)
How does a gas of photons arise to give a radiation pressure in an exploding bomb
core? Fission fragments are bare nuclei and so are highly electrically charged. As they
decelerate, they naturally emit energy in the form of photons of wavelengths across
the electromagnetic spectrum. Much of the energy released in a nuclear explosion is
in the form of gamma-rays and X-rays which ionize the surrounding air.
For simplicity, we model the bomb core as an expanding sphere of radius r(t) with
every atom in it moving radially outwards at speed v. Do not confuse this velocity
with the average neutron speed v neut , which enters into τ. If the sphere is of density
ρ(t) and total mass M, its total kinetic energy will be
K core =
1
2
Mv
2
=
2π
3
ρ v
2 r
3
.
(2.88)
Now invoke the work-energy theorem in its thermodynamic formulation W =
P(t)dV, and equate the work done by the gas (or radiation) pressure in changing the
core volume by dV over time dt to the change in the core’s kinetic energy over that
time:
P(t)
dV
dt
=
d K core
dt
.
(2.89)
To formulate this explicitly, write, from (2.88), dK core /dt =(2π / 3)ρ r
3 (2vdv/dt),
put dV/dt =4π r
2 (dr/dt), and incorporate (2.87) to give
dv
dt
=
3P o
ρ r
e
(α/τ ) t
.
(2.90)
Note that in taking the derivative for dK core /dt, ρ r
3 can be treated as a constant
since it is (but for a factor of 4π /3), the mass of the core. To solve this for the radius
of the core as a function of time, we face the problem of what to do about the fact that
both ρ and r are functions of time. We deal with this by means of an approximation.
Review the discussion regarding core expansion following (2.83) above. As the
core expands, its density when it has any general radius r will be ρ(r) = Cρ o (R o /r)
3 ,
and criticality will hold until such time as ρ r = ρ o R o , or, on eliminating ρ, r =
C
1/2 R o . We can then define r, the range of radius over which criticality holds: