A particle of rest mass M disintegrates at rest into a particle of rest mass
m_{1}, a particle of rest mass m_{2}, and a high-energy photon
(a gamma ray). Find the maximum possible energy of the gamma ray in the rest
frame of the initial particle of rest mass M.

Solution:

- Concepts:

Relativistic "collisions", energy and momentum conservation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

The γ-ray will have its maximum possible energy if after the disintegration the two particles have no relative kinetic energy.

Energy conservation: Mc^{2}= (m^{2}c^{4}+ p^{2}c^{2})^{½}+ E_{γmax}.

Here m = m_{1}+ m_{2}.

Momentum conservation: pc = E_{γmax}.

Combine: E_{γmax}= ½Mc^{2}- ½m^{2}c^{2}/M = ½(M - (m_{1}+ m_{2})^{2}/M)c^{2}.

If m = M, then E_{γmax}= 0.

A π^{0} meson decays into two γ-rays while at rest or in flight:
π^{0 }--> γ + γ.

(a) If the decaying π^{0} has velocity **v** and rest mass m_{π}
, and the γ-ray is emitted at an angle θ with respect to the original direction
of the π^{0}, find the γ-ray energy as a function of m_{π}, **v**,
and θ.

(b) What is the maximum and minimum energy an emitted γ-ray can
have, and at what emission angles do these occur?

Solution:

- Concepts:

Relativistic "collisions", energy and momentum conservation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

(a) In the laboratory, we have from energy and momentum conservation:

γmc^{2}= hf_{1}+ hf_{2}, γmv = hf_{1}cosθ_{1}/c + hf_{2}cosθ_{2}/c, hf_{1}sinθ_{1}/c = hf_{2}sinθ_{2}/c. Here m is the mass of the π^{0}.

We want to find hf_{1}as a function of θ_{1}.

Eliminate cosθ_{2}: (γmv -hf_{1}cosθ_{1}/c)^{2}= (hf_{2}/c)^{2}(1-sin^{2}θ_{2}) = (hf_{2}/c)^{2}- (hf_{1}/c)^{2}sin^{2}θ_{1}.

Eliminate hf_{2}: (γmv -hf_{1}cosθ_{1}/c)^{2}+ (hf_{1}/c)^{2}sin^{2}θ_{1 }= (hf_{2}/c)^{2}= ( γmc- hf_{1}/c)^{2}.

Write out all the terms: γ^{2}m^{2}(v^{2}- c^{2}) - 2γmhf_{1}((v/c)cosθ_{1}- 1) = 0.

hf_{1}= mc^{2}/[2γ(1 - (v/c)cosθ_{1})].

(b) hf_{1max}= mc^{2}/[2γ(1 - (v/c))]. cosθ_{1}= 1, θ_{1}= 0,

E_{max}= (mc^{2}/2)[(1 + v/c)/(1 - v/c)]^{½}.

hf_{1min}= mc^{2}/[2γ(1 + (v/c))]. cosθ_{1}= -1, θ_{1}= 180,

E_{min}= (mc^{2}/2)[(1 - v/c)/(1 + v/c)]^{½}.

In the rest frame of the π^{0}the two γ's are emitted with energies (mc^{2}/2) in the forward and backward directions. In the lab frame they are Doppler shifted.

A beam of π^{+} particles (m_{π+} = 140
MeV/c^{2}) is accelerated, so it has momentum p_{π+} = 2 GeV/c
in the laboratory frame. A π^{+} particle decays into a positive
muon (m_{μ} = 105 MeV/c^{2}) and a muon neutrino (m_{νμ}
= 0 MeV/c^{2}) .

(a) What is the energy of the muon in
the rest frame of the π^{+} particles?

(b) What is the energy
of the muon in the laboratory frame if it is found traveling in the beam
direction?

(c) What is the energy of the muon in the laboratory frame
if it is found traveling at 0.25 rad with respect to the beam direction?

Solution:

- Concepts:

Relativistic "collisions", energy and momentum conservation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

(a) Energy conservation: m_{π+}c^{2}= (m_{μ}^{2}c^{4}+ p_{μ}^{2}c^{2})^{½}+ E_{ν}, E_{ν}= pc.

Momentum conservation: p_{μ}= p.

Combine: (m_{π+}c^{2}- E_{n})^{2}= m_{π+}^{2}c^{4}+ E_{ν}^{2}- 2m_{π+}c^{2}E_{ν}= m_{μ+}^{2}c^{4}+ E_{ν}^{2}.^{ }E_{ν}= ½m_{π+}c^{2}- ½m_{μ}^{2}c^{2}/m_{π+}.

E_{μ}= m_{π+}c^{2}- E_{ν}= ½m_{π+}c^{2}+ ½ m_{μ}^{2}c^{2}/ m_{π+}= ½(140 + 105^{2}/140) MeV = 109.375 MeV.

p_{μ}= ½(140 - 105^{2}/140) MeV/c = 30.625 MeV/c.

(b) The rest frame of the m_{π+}moves with speed v with respect to the lab frame. Let the direction be the z-direction.

p_{π+}= 2 GeV/c. γmv = 2 GeV/c. γv = c*2 GeV/mc^{2}= c*2000/140 = a*c.

v^{2}= c^{2}a^{2}/(1 + a^{2}), v = 0.99756 c. β = 0.99756. γ = (1 + a^{2})^{½}= 14.32.

We can now find the energy E_{μ}' of the muon in the laboratory frame.

p_{z}= p_{μ}, E_{μ}' = γE_{μ}+ γvp_{μ}= 2003.8 MeV.

(c) Now p_{z}= p_{μ}cos(0.25). E_{μ}' = γE_{μ}+ γvp_{μ}cos(0.25) = 1990.2 MeV

Cs_{55}^{137} is a common laboratory
radioactive source of electrons and gamma rays. Eight percent of the time Cs_{55}^{137} beta
decays to the ground state of Ba_{56}^{137}.
The net atomic mass difference between the two isotopes is 1.18 MeV/c^{2}. A 180 degree spectrometer is used to measure the beta decay
spectrum. The spectrometer has a
radius R = 3.8 cm.

(a)
Write down the reaction for the beta decay.

(b) Calculate the maximum momentum of the beta decay electron/positron.
Express the result in MeV/c.

(c) What is the vector direction of the spectrometer magnetic field relative
to the drawing?

(d) What is the magnetic field setting of the spectrometer for the maximum
energy of the electron/positron to arrive at the detector?
Provide a numerical answer with units.

Solution:

- Concepts:

Relativistic "collisions", energy and momentum conservation, motion of a charged particle in a magnetic field - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

(a) Cs_{55}^{137}--> Ba_{56}^{137}+ e^{-}+ ν (ν = anti neutrino)

(b) The source is at rest. If we assume the antineutrino is massless then the electron has the maximum energy if the energy of the antineutrino is zero. If the antineutrino has mass, then the electron has the maximum energy if the Ba nucleus and the antineutrino have no relative kinetic energy. Since the mass of the antineutrino is then extremely small, we can neglect it compared to the mass of the Ba nucleus.

Let M_{1}be the mass of the Cs nucleus and M_{2}be the mass of the Ba nucleus.

M_{1}c^{2}= (M_{2}^{2}c^{4}+ p_{2}^{2}c^{2})^{½}+ (m_{e}^{2}c^{4}+ p_{e}^{2}c^{2})^{½}(energy conservation)

p_{2}^{2}= p_{e}^{2}= p^{2}(momentum conservation)

The mass of the Ba nucleus is much greater than the mass of the electron. If we neglect the recoil of the Ba nucleus (M_{2}^{2}c^{4}>> p^{2}c^{2})^{ }then

M_{1}c^{2}- M_{2}c^{2}= (m_{e}^{2}c^{4}+ p^{2}c^{2})^{½}

The maximum momentum of the electron is then given by

p^{2}c^{2}= (M_{1}c^{2}- M_{2}c^{2})^{2}- m_{e}^{2}c^{4}= (1.18 MeV)^{2}– (0.511MeV)^{2}

pc = 1.064 MeV

(c) The field points into the page.

(d) R = p/(qB)

B = 1.064* 10^{6}eV * (1.6 *10^{-19}J/eV)/[ 1.6 *10^{-19}C * 0.038 m * 3*10^{8}m/s] = 9.33*10^{-2}T

A free atomic nucleus at rest with mass M makes a transition from excited to
ground state by emitting a γ-ray photon.

(a) Find the energy of the γ ray photon and the kinetic energy of the recoil
nucleus if the excitation energy was ΔE.

(b) Calculate the recoil energy of the ^{191}Ir nucleus if the
excitation energy ΔE = 129 keV.

Solution:

- Concepts:

Relativistic dynamics - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

Let M be mass of the excited nucleus and m the mass of the de-excited nucleus.

(a) energy conservation: Mc^{2}= γmc^{2}+ hν (1)

momentum conservation: hν/c = γmu (2)

ΔE = (M - m)c^{2}(3)

Rewriting (1): Mc^{2}- hν = γmc^{2}

Squaring: M^{2}c^{4}+ h^{2}ν^{2}– 2Mc^{2}hν = γ^{2}m^{2}c^{4}

Rewriting and substituting (2) and (3): -2Mhν = γ^{2}m^{2}c^{2}- M^{2}c^{2}- h^{2}ν^{2}/c^{2}

= γ^{2}m^{2}c^{2}- M^{2}c^{2}- γ^{2}m^{2}u^{2}.

= γ^{2}m^{2}(c^{2}- u^{2}) – M^{2}c^{2 }= m^{2}c^{2}– M^{2}c^{2}.

2Mhν = (m - M)(m + M)c^{2}= ΔE (m + M).

Rewriting:

Photon energy: hν = ΔE (m + M)/(2M) = ΔE (m_{ }- M_{ }+ M_{ }+ M)/(2M)

= ΔE(1 - ΔE /(2Mc^{2})).

Kinetic energy of the recoil nucleus: T = (ΔE)^{2}/(2Mc^{2}).

(b) The mass of the^{191}Ir nucleus is ~ 1.67*10^{-27}kg, Mc^{2}~ 1.79*10^{5}MeV.

T = (129 keV)^{2}/(3.59*10^{8}keV) = 4.64*10^{-5}keV = 46.4 meV.

The question of the proton stability is one of the center questions of Grand
Unification theories. The present experimental limit of the proton lifetime
from the Super Kamiokande experiment is τ_{p} > 1.6·10^{33} y.
More sensitive experiments have been proposed. One of the possible proton
decay modes is p --> e^{+ }+ π^{0}, followed by the decay of the
neutral pion into two photons: π^{0 }--> γ + γ

Assume that the protons are at rest.

(a) What is average flight distance of the neutral pion before its decay?

(b) What are the maximum and minimum energies the photons can have?

M_{p }= 938.27 MeV

M_{π }= 134.98 MeV

M_{e }= 0.511 MeV

τ_{π} = 8.4·10^{-17} s

Solution:

- Concepts:

Relativistic "collisions", energy and momentum conservation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

p --> e^{+ }+ π^{0}:

Momentum conservation: γ_{π}β_{π}M_{π}= γ_{e}β_{e}M_{e}. (γ_{π}^{2}– 1)M_{π}^{2}= (γ_{e}^{2}– 1)M_{e}^{2}.

γ_{e}^{2}= (γ_{π}^{2}– 1)M_{π}^{2}/M_{e}^{2}+ 1.

Energy conservation: M_{p}= γ_{π}M_{π}+ γ_{e}M_{e}.

Combining: M_{p}^{2}+ (γ_{π}M_{π})^{2}- 2M_{p}γ_{π}M_{π}= γ_{e}^{2}M_{e}^{2}= (γ_{π}^{2}– 1)M_{π}^{2}+ M_{e}^{2}.

γ_{π}= (M_{p}^{2}+ M_{π}^{2}- M_{e}^{2})/( 2M_{p}M_{π}) = 3.55, β_{π}= ((γ_{π}^{2}– 1)/ γ_{π}^{2})^{½}= 0.96.

Average flight distance: <x> = v_{π}γ_{π}τ_{π}= cγ_{π}β_{π}τ_{π}= 85.2 nm in the lab frame.

(b) Maximum and minimum energies the photons can have:

E_{max}= (M_{π}c^{2}/2)[(1 + v_{π}/c)/(1 - v_{π}/c)]^{½}= 468.1 MeV.

E_{min}= (M_{π}c^{2}/2)[(1 - v_{π}/c)/(1 + v_{π}/c)]^{½}= 9.7 MeV.

A pion (π) decays at rest into a muon (μ) and a neutrino (ν). In terms of
the masses m_{π} and m_{μ} (use the approximation m_{ν}
= 0), find:

(a) the momentum, energy, and velocity of the outgoing muon.

(b) In the rest frame of the outgoing muon, what is the energy of the neutrino?
What was the initial velocity of the pion in this frame?

Solution:

- Concepts:

Conservation of energy and momentum, the Lorentz transformation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

(a) In the rest frame of the decaying pion let the muon move in the positive x-direction and the neutrino move in the negative x-direction.

Energy conservation: M_{p}c^{2}= (M_{μ}^{2}c^{4}+ p^{2}c^{2})^{½}+ E_{ν}.

Here p is the momentum of the muon.

Momentum conservation: pc = E_{ν}.

Combine: (M_{p}c^{2}- E_{ν})^{2}= M_{p}^{2}c^{4}+ E_{ν}^{2}- 2M_{p}c^{2}E_{ν}= M_{μ}^{2}c^{4}+ E_{ν}^{2 }E_{ν}= ½M_{p}c^{2}- ½M_{μ}^{2}c^{2}/M_{p}= (c^{2}/(2M_{p}))(M_{p}^{2}- M_{μ}^{2})

p = E_{ν}/c = (c/(2M_{p}))(M_{p}^{2}- M_{μ}^{2}) = momentum of the outgoing muon.

E_{μ}= M_{p}c^{2}- E_{ν}= (c^{2}/(2M_{p}))(M_{p}^{2}+ M_{μ}^{2}) = energy of the outgoing muon.

v = pc^{2}/E_{μ}= c(M_{p}^{2}- M_{μ}^{2})/ (M_{p}^{2}+ M_{μ}^{2}) = speed of the outgoing muon.

(b) The muon moves with respect to the original rest frame of the pion with speed v.

The original speed of the pion in the muon's rest frame therefore was v. It was moving in the negative x-direction. The relative speed of the two frames is v.

β = v/c = (M_{p}^{2}- M_{μ}^{2})/ (M_{p}^{2}+ M_{μ}^{2})

γ = (1 - v^{2}/c^{2})^{-½ }= (M_{p}^{2}+ M_{μ}^{2})/(2M_{p}M_{μ})

Lorentz transformation of the (p_{0},**p**) = (E_{ν}/c, -**i**E_{ν}/c) 4-vector from the original rest frame of the pion to the rest frame of the muon:

E_{ν}'/c = γE_{ν}/c + βγ E_{ν}/c = (M_{p}/M_{μ})E_{ν}is the energy of the neutrino in the rest frame of the muon.

A pion (mass M_{p}) decays into a muon (mass M_{μ}) and a
neutrino (massless).

(a) Find the kinetic energy of the muon in the
rest frame of the pion.

(b) The muon is unstable, with a lifetime τ_{1}
in the pion's rest frame. Find the lifetime t_{0} of the muon in
its own rest frame.

Solution:

- Concepts:

Relativistic "collisions", energy and momentum conservation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

(a) Energy conservation: M_{p}c^{2}= (M_{μ}^{2}c^{4}+ p^{2}c^{2})^{1/2}+ E_{n}.

Here p is the momentum of the muon.

Momentum conservation: pc = E_{n}.

Combine: (M_{p}c^{2}- E_{n})^{2}= M_{p}^{2}c^{4}+ E_{n}^{2}- 2M_{p}c^{2}E_{n}= M_{μ}^{2}c^{4}+ E_{n}^{2}.^{}E_{n}= ½M_{p}c^{2}- ½M_{μ}^{2}c^{2}/M_{p}.

E_{μ}= M_{p}c^{2}- E_{n}= ½M_{p}c^{2}+ ½ M_{μ}^{2}c^{2}/ M_{p}.

T = E_{μ}- M_{μ}c^{2}= (M_{p}- M_{μ})^{2}c^{2}/(2M_{p}).

(b) τ_{0}= τ_{1}/γ. We need to find γ.

T = (M_{p}- M_{μ})^{2}c^{2}/(2M_{p}) = (γ - 1)M_{μ}c^{2}.

(γ - 1) = (M_{p}- M_{μ})^{2}/(2M_{μ}M_{p}), γ = ( M_{p}^{2}+ M_{μ}^{2})(/2M_{μ}M_{p}),

τ_{0}= τ_{1}2M_{μ}M_{p}/( M_{p}^{2}+ M_{μ}^{2}).

An excited nucleus of ^{57}Fe formed by the radioactive decay of ^{
57}Co emits a gamma ray of 1.44 *10^{4} eV. In the process,
there is conservation of energy and m_{0}c^{2} = γm_{a0}c^{2}
+ hf, where m_{0}c^{2} is the initial mass of the nucleus
and m_{a0}c^{2} is its mass after the emission of the gamma ray.
There is also conservation of momentum, hf/c = γm_{a0}u, where u is the
recoil velocity of the iron nucleus. The energy released by the reaction
is E_{r} = (m_{0} - m_{a0})c^{2}.

(a)
Show that hf = E_{r}(m_{0} + m_{a0})/(2m_{0})
= (1 - E_{r}/(2m_{0}c^{2}))E_{r}.

Thus hf < E_{r}:
part of E_{r} goes to the photon, and the other part supplies kinetic
energy to the recoiling nucleus.

(b) Set m_{0} = 57*1.7*10^{-27}
kg, and show that E_{r}/(2m_{0}c^{2})) ~ 1.3*10^{-7}.

Thus the fraction of the available energy E_{r} that appears as recoil
is small.

(c) Mössbauer discovered in 1958 that, with solid iron, a
significant fraction of the atoms recoil as if they were locked rigidly to the
rest of the solid. This is the Mössbauer effect. If the sample has a
mass of 1 gram, by what fraction is the gamma ray energy shifted in the recoil
process?

(d) A sample of normal ^{57}Fe absorbs gamma rays
of 14.4 keV by the inverse recoilless process much more strongly than it absorbs
gamma rays of any nearby energy. The excited nuclei thus formed reemit
14.4 keV radiation in random directions some time later. This is resonant
scattering. If a sample of activated ^{57}Fe moves in the
direction of a sample of normal ^{57}Fe, what must be the value of the
velocity v that will shift the frequency of the gamma rays, as seen by the
normal nuclei, by 3 parts in 10^{13}? This is one line width.

(e) A Doppler shift in the gamma ray results in a much lower
absorption by a nucleus if the shift is of the order of one line width or more.
What happens to the counting rate of a gamma-ray detector placed behind the
sample of normal ^{57}Fe when the source of activated Fe moves

(i) toward the normal ^{57}Fe,

(ii) away from it?

(f) If a 14.4 keV gamma ray travels 22.5 meters vertically upward, by
what fraction will its energy decrease?

[Gravitation redshift, a thought
experiment: Suppose a particle of rest mass m is dropped from the top of a
tower and falls freely with acceleration g. It reaches the ground with a
velocity v = (2gh)^{1/2}, so its total energy E, as measured by an
observer at the foot of the tower is E = mc^{2} + ½mv^{2} + O(v^{4})
= mc^{2} + mgh + O(v^{4}).

Suppose an observer has some
magical method of converting all this energy into a photon of the same energy.
Upon its arrival at the top of the tower with energy E the photon is again
magically changed into a particle of rest mass m' = E'/c. Energy conservation
requires that m' = m.

Therefore E'/E = mc^{2}/(mc^{2}
+ mgh + O(v^{4})) = (1 + gh/c^{2} + O(v^{4}))^{-1}
= 1 - gh/c^{2} + O(v^{4}).]

(g) A normal ^{57}Fe
absorber located at this height must move in what direction and at what speed in
order for resonant scattering to occur?

Solution:

- Concepts:

Relativistic decay, energy and momentum conservation, the Doppler shift - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved.

The Doppler shift can compensate for the gravitational redshift to produce optimum absorption. - Details of the calculation:

(a) energy conservation: m_{0}c^{2}= γm_{a0}c^{2}+ hf. (1)

momentum conservation: hf/c = γm_{a0}u. (2)

Define: E_{r}= (m_{0}- m_{a0})c^{2}. (3)

Show: hf = E_{r}(m_{0}- m_{a0})/(2m_{0}).

m_{0}c^{2}- hf = γm_{a0}c^{2}. (1)

m_{0}^{2}c^{4}+ h^{2}f^{2}- 2m_{0}c^{2}hf = γ^{2}m_{a0}^{2}c^{4}. square (1)

-2m_{0}hf = γ^{2}m_{a0}^{2}c^{2}- m_{0}^{2}c^{2}- h^{2}f^{2}/c^{2}= γ^{2}m_{a0}^{2}c^{2}- m_{0}^{2}c^{2}- γ^{2}m_{a0}^{2}u^{2}

= γ^{2}m_{a0}^{2}(c^{2}- u^{2}) - m_{0}^{2}c^{2 }= m_{a0}^{2}c^{2}- m_{0}^{2}c^{2}. insert (2)

2m_{0}hf = (m_{a0}- m_{0})(m_{a0}+ m_{0})c^{2}= E_{r}(m_{a0}+ m_{0}). rearrange

hf = E_{r}(m_{a0}+ m_{0})/(2m_{0}) = E_{r}(m_{a0}- m_{0 }+ m_{0 }+ m_{0})/(2m_{0}) = E_{r}(1 - E_{r}/(2m_{0}c^{2})).

(b) Set m_{0}= 57*1.7*10^{-27}kg, hf = 1.44*10^{4}eV.

E_{r}- E_{r}^{2}/(2m_{0}c^{2})) - hf = 0. E_{r}^{2}- 2m_{0}c^{2}E_{r}+ 2m_{0}c^{2}hf = 0.

E_{r}= m_{0}c^{2}- ((m_{0}c^{2})^{2}- 2m_{0}c^{2}hf)^{1/2}.

m_{0}c^{2}= 8.72*10^{-9}J, hf = 2.3*10^{-15}J, E_{r}= 2.3 *10^{-15}J.

E_{r}/(2m_{0}c^{2}) ~ 1.3*10^{-7}.

(c) If m_{0}= 10^{-3}kg, then m_{0}c^{2}= 9*10^{13}J, E_{r}/(2m_{0}c^{2}) ~ 1.27*10^{-29}.

The fraction that appears as recoil energy is now negligible.

The gamma ray energy is now hf = E_{r}. It has increased by E_{r}^{2}/(2m_{0}c^{2})) = 1.3*10^{-7 }E_{r}.

Δf/f = 1.3*10^{-7}.

(d) We want Δf/f = 3*10^{-13}.

Δf/f = (f' - f)/f = [(1 + v/c)/(1 - v/c)]^{1/2}- 1 if the samples approach each other.

We can solve for v/c ~ 3*10^{-13}. v ~ 10^{-4}m/s.

(e) The counting rate of a gamma-ray detector increases for both directions, because the sample^{57}Fe absorbs less.

(f) A 14.4 keV gamma emitted from^{57}Fe ray travels vertically upward in a uniform gravitational field.

We have Δf/f = -gΔh/c^{2}from E_{2}/E_{1}= 1 - g(h_{2 }- h_{1})/c^{2}. Δf/f = -2.45*10^{-15}.

Such a shift can also be produced if the absorber moves away from the source.

We would need Δf/f = [(1 - v/c)/(1 + v/c)]^{1/2}-1 = -2.45*10^{-15}, v/c = 2.45*10^{-15}away from source.

(g) The absorber has to move towards the source with v/c = 2.45*10^{-15}in order to increase, in its own frame, the frequency for optimum absorption, because the frequency was decreased by the gravitational redshift.

If the absorber is at rest we still have absorption, since Δf/f|_{g}<< 3*10^{-13}.