1. Energy Quantization within the Old Quantum Theory Formulation
of Gravity in the Schwarzschild case
by
Amino Adelmo Lavorini
aadelmo.lavorini@gmail.com
Abstract
The physics interpretation of the world is described in terms of fundamental constituents and their
interactions. Such description is given in terms of particles with different type of charge whose interaction
generates all known fields. Research tell us in Nature exists the hierarchy of fundamental interactions
for wich only gravity seems to give no appreciable effects on particle scattering. Simple kinematical
arguments will be used to give new perspective to the study of gravitational effects in particle physics
whose implications can be more importants. Following the historical development of particle physics,
the aim of this paper is to give an expression to the quantization of energy levels for the Schwarzschild
field. In what follows the old fashion quantum theory prescriptions will be adopted and a very simple
expression for the energy levels is given in general and for particular cases.
1 Introduction
The strength of interaction between particles is specified via a dimensionless quantity called coupling constant.
In Nature four fundamental interactions are known i.e. Electromagnetic (Em), Weak (W), Strong (S) and
Gravitational (G) interaction: their relative strenght is expressed by the following formulas [1]
αEm =
e2
4πǫ0 c
, αW =
GF (mc2
)2
( c)2
, αS(Q2
) =
1
β0 ln(Q2/Λ2
QCD)
, αG =
GN m2
4π c
, (1)
where:
GF
( c)3
= 1.1663787(6) × 10−5
GeV −2
,
GN = 6.70837(80)( c) × 10−38 GeV
c2
−2
,
β0 =
11CA − 4TF nf
12π
.
As illustrated in Ref. [2], GF is known as Fermi coupling constant and GN as gravitational one, β0 is the
1-loop beta-function coefficent, CA = Nc = 3 is the color-factor associated with gluon emission from a gluon
and TR = 1/2 is the color factor for a gluon to split to a q¯q pair; ΛQCD = Λ
(3)
MS
= 339(10) MeV is a
constant of integration, i.e. the non-perturbative scale of QCD: its value is extracted from the experiments
to evaluate the world-average value for αs(M2
Z) in the modified minimal substraction scheme.
A numerical estimation of the coupling constants is given: by using nf = 3, then β0 = 9
4π , and protons
as interacting particles one has
2. αS(m2
p) αEm
0.686(82) 7.2973525698(24)× 10−3
αW αG
1.026825125(46)×10−5
4.69963(56)×10−39
Table 1: The strength of the Em, W , S and G interaction by using Eq. (1) and Ref. [2]
From Tab. 1 the inequalitirs αs(m2
p) > αem > αW >> αG follows, consequently hierarchy of four
foundamental forces of Nature tell us how gravity does not play a keyrole in particle physics: as it is well
known the Gravitational coupling has a strenght much more weaker compared to the typical ones of the
other three; even using LHC energies, 14 T eV in the centre-of-mass system, no man would be able to see
gravitational effects in particle interactions.
In 1967 Prof. Bryce S. DeWitt published the trilogy of Quantum Theory of Gravity [3, 4, 5], in particular
He shown how calculate the differential cross-section of two identical scalar particles mediated by a single
graviton [5] exchange; in the centre-of-mass system and considering extreme relativistic limits was found
dσ
dΩ ER
= 4G2
N E2
cot2 θ
2
+ tan2 θ
2
+
sin2
θ
4
+
1
2
2
. (2)
Eq. (2) shows how hard is to see gravitational effects in a particle collision since dσ
dΩ ER
∝ G2
N E2
.
However the most important mechanical relativistic applications can be found in the the field of nuclear and
sub-nuclear physics: in this framework particles collides with typical velocities of order of c, the velocity of
light in vacuo.
To test the statement above, one can analize the kinematics of Bhabha scattering e+
e−
→ e+
e−
by using
the maximum beam energy, for electron, avaible in Ref. [2] as shown in Tab. (2): the kinematics of the
process, in the centre-of-mass system with colliding particles moving along the z-axis, gives the following
expressions for the energy and (3-)momentum [6]:
Ee+ = Ee− = E′
e+ = E′
e− =
√
s
2
, |pe+ | = |pe− | = |p′
e+ | = |p′
e− | =
s − 4m2
e
2
, (3)
the prime is referred to scattered particles. The ratio β = v/c of the outcoming particles, generally expressed
by computing ∂E/∂p, with E = p2c2 + m2c4, can be written by means of Eq. (3) as
β = 1 −
4m2
e
s
,
√
s = Ee+ + Ee− . (4)
VEPP-2000 VEPP-4M BEPC BEPC-II DAΦNE
Beam energy [GeV] 1.0 6.0 2.5 1.89÷2.3 0.510
CESR CESR-C LEP ILC CLIC
Beam energy [GeV] 6.0 6.0 100÷104.6 250÷500 250÷1500
KEKB PEP-II Super-B SuperKEKB
Beam energy [GeV]
e−
: 8.33
e+
: 3.64
e−
: 7÷12
e+
: 2.5÷4
e−
: 4.2
e+
: 6.7
e−
: 7.0
e+
: 4.0
Table 2: Maximum beam energy from Ref. [2]
2
3. By inserting data of Tab. (2) into Eq. (4), results of Tab. (3) below follows: Now, since results of Tab. (3) ∗
VEPP-2000 VEPP-4M BEPC
√
s [GeV] 2.0 12.0 5.0
β 0.999999869440039 0.999999996373334 0.999999979110407
BEPC-II DAΦNE CESR
√
s [GeV] 3.78÷4.6 1.02 12.0
β
0.999999963450084
0.999999975319479
0.99999949803927 0.999999996373334
CESR-C LEP ILC
√
s [GeV] 12.0 200÷209.2 500÷1000
β 0.999999996373334
0.999999999986944
0.999999999988067
0.999999999997911
0.999999999999478
CLIC KEKB PEP-II
√
s [GeV] 500÷3000 11.97 9.5÷16.0
β
0.9999999999997911
0.9999999999999942
0.999999996355133
0.999999994213409
0.999999997960001
SuperB SuperKEKB
√
s [GeV] 10.9 11.0
β 0.999999995604412 0.999999995683968
Table 3: Velocity values for scattered particles by using Eq. (4), and Tab. 2
and noting that a Gravitational field is as weak as the constraint v/c << 1 is satisfied [7], One would inquire
How can We disregard gravitational effects in the high-energy regime
if outcoming particles have β ∼ 1 after a scattering process?
In my opinion, the right way to proceed and overcome this difficoulty is to start from first principles:
we must use Einstein’s equations of gravitational field [7, 8] for a point-particle and then to quantize the
system †
: the quantization of the system in General Relativity start by resolving the gravidynamics equation,
the celebrated Wheeler-DeWitt equation [3]: unfortunately there are difficoulties to resolve it, so that how
can we proceed now?
In this pioneristic work I decided to adopts the prescriptions of the Old Quantum Theory.
2 Bohr-Sommerfeld quantization rule and Schwarzschild solution
of the Einstein equations
The Old Quantum theory is essentially a general method of calculating quantized quantities, based upon
the hypotheses of Bohr and the correspondences principle [9]. The determination of the quantization rules
constitues the central problem of the Old Quantum Theory: its solution is the Bohr-Sommerfeld quantization
rule
pdq = hn (n = 1, 2, · · ·, ∞) ,
it determines the allowed trajectories of phase space and the corresponding quantized energies.
∗As their consequence we can say that in the limit of high-energies β → 1
†It goes without say to satisfy the statement above one must resolve the scattering problem of a particle in a gravitational
field for β → 1
3
4. Wilson and Sommerfeld have generalized this rule to multiply-periodic system whose motion can be
represented by a sequence of functions, namely the conjugate momenta pr, as a function of the coordinate
qr only, then the quantization rule are the R relations
dqrpr = hnr (r = 1, 2, · · · , R) ; (5)
consequently nr defines the r-th quantized trajectory of the system and the quantized value of the corre-
sponding constant of the motion.
To use this method, one has to test the periodicity of a given physical system with finite motion condition
is understood fulfilled. In particular for systems with several degrees of freedom is necessary a complete
separation of variables in the Hamilton-Jacobi’s method [11] then, quantization rules [10] reduces to Eq. (5);
In classical mechanics the analogy to quantization rules in the old fashion quantum mechanics is placed of
conditionally periodic motion and action variables.
2.1 Motion in a gravitational field with central symmetry
In this paper the goal is to give the quantizated version of the energy level for the central static gravitational
field. As it is well known, the solution of such field was found by K. Schwarzschild in 1916. He obtains the
following expression of the line element
ds2
= 1 −
rg
r
c2
dt2
− r2
sin2
θdφ2
+ dθ2
−
dr2
1 −
rg
r
, rg =
2GN µ
c2
. (6)
With Eq. (6) one can construct the Lagrangian
L = −mc2
1 −
rg
r
1/2
1 −
r2
( ˙θ2
+ sin2
θ ˙φ2
)
c2 1 −
rg
r
−
˙r2
c2 1 −
rg
r
2
1/2
= −mc2
1 −
rg
r
1/2
1 − β2 , (7)
to calculate the three-momentum components, whose expression in the θ = π
2 plane reads
pr =
m ˙r
1 −
rg
r
3/2
1 − β2
, (8)
pφ =
mr2 ˙φ
1 −
rg
r
1/2
1 − β2
. (9)
The expressions of the angular momentum M and the energy E, i.e. the two conserved quantities of the
system, are
M = pφ =
mr2 ˙φ
1 −
rg
r
1/2
1 − β2
, E =
mc2
1 −
rg
r
1/2
1 − β2
and, since the line element is ds = 1 −
rg
r
1/2
cdt 1 − β2, the expressions above reads
M = mcr2 dφ
ds
, (10)
E = mc3
1 −
rg
r
dt
ds
. (11)
From Eq. (10), Eq. (11) and Eq. (6) one has the equation of motion for a particle of mass m
dr
ds
2
=
E2
m2c4
− 1 −
rg
r
1 +
M2
m2c2r2
. (12)
As it is well known, in a central field, the angular momentum can be written as an integral of area. To show
this, one can rewrite the expression of the angular momentum, Eq. (10), in the form
M = 2mc
df
ds
, (13)
4
5. where df/ds is the sectorial velocity, since M is a conserved quantity consequently df/ds is a conserved one
as well; this is known as constancy of sectorial velocity [11]. Writing
df =
M
2mc
ds ,
and using the expression of ds from Eq. (12) one has
df =
M
2mc
dr
E2
m2c4 − 1 −
rg
r 1 + M2
m2c2r2
1/2
. (14)
The area described by the orbit can be found integrating Eq. (14)
f =
M
2mc
rmax
rmin
dr
dr
E2
m2c4 − 1 −
rg
r 1 + M2
m2c2r2
1/2
; (15)
whose extremes of integration are the roots of the mass-shell relation in polar coordinates
E2
= p2
rc2
+
M2
c2
r2
+ m2
c4
1 −
rg
r
, (16)
i.e. by resolving Ar2
− Cr + B = 0 we have
rmin =
C
2A
−
√
C2 − 4AB
2A
(17)
rmax =
C
2A
+
√
C2 − 4AB
2A
(18)
where
A = p2
rc2
+ m2
c4
− E2
, B = M2
c2
, C = m2
c4
rg , .
The expression of the period T of revolution for a particle moving in such a field, can be found by using
Eq. (13), Eq. (11) and Eq. (14); by integration it from 0 to T one has
T =
rmax
rmin
dr
E
mc3 1 −
rg
r
dr
E2
m2c4 − 1 −
rg
r 1 + M2
m2c2r2
1/2
. (19)
3 Quantization procedure
At this stage one can apply the Bohr-Sommerfeld quantization rule, the Lagrangian allows separeted variables
since the classical system shows the same property, then our system is two fold periodic and its period of
revolution was calculated.
Since we have two constants of motion then there are two quantized quantities, the first one follows from
Eq. (9), then
pφ dφ = M dφ = hℓ → M = ℓ; (20)
as usual ℓ, with ℓ ∈ N, is the azimutal quantum number.
To calculate the second one we must have the expression for the radial momentum; it follows from Eq. (16)
pr =
E2
c2
−
M2
r2
− m2c2 1 −
rg
r
, (21)
then, by equating it to zero one has two conditions
r = 0 , (E2
− m2
c4
)r2
+ (m2
c4
rg)r − M2
c2
= 0 ,
thus
r± = −
m2
c4
rg
2(E2 − m2c4)
±
(m2c4rg)2 + 4(E2 − m2c4)M2c2
2(E2 − m2c4)
. (22)
5
6. The integral to be calculated reads
pr dr =
r+
r−
dr
E2
c2
−
M2
r2
− m2c2 1 −
rg
r
= hκ ; (23)
here κ is the radial quantum number, it is also an integer number. If we define x as x = 1/r, then Eq. (23)
becomes
− dx
E2
c2 − m2c2 − M2x2 + m2c2rgx
x2
, (24)
by introducing the notation
α =
E2
c2
− m2
c2
, β = m2
c2
rg , γ = −M2
, (25)
the integral of Eq. (24) can be written as
− dx
R(x)
x2
, R(x) = α + βx + γx2
;
its solution is [12]
− dx
R(x)
x2
= − −
R(x)
x
+
β
2
dx
x R(x)
+ γ
dx
R(x)
,
with
dx
x R(x)
= −
1
√
α
Artanh
2α + βx
2 αR(x)
,
and
dx
R(x)
= −
1
√
−γ
Arsin
2γx + β
√
−∆
∆ = 4αγ − β2
< 0.
Since notations of Eq. (25), Eq. (22) reads
r± = −
β
2α
±
√
−∆
2α
,
consequently solution of Eq. (23), before restoring r as variable, has the final form,
prdr = r R(r) +
β
2
√
α
Artanh
2α + β/r
2 αR(r)
+
γ
√
−γ
Arsin
2γ/r + β
√
−∆
r+
r−
= hκ . (26)
Noting that R(r) = pr, see Eq. (21), and
lim
x→∞
Artanh(x) =
−ıπ
2
,
the nonzero term in Eq. (26) is the third one only
−
√
−γ Arsin
2γ/r+ + β
√
−∆
− Arsin
2γ/r− + β
√
−∆
= hκ . (27)
Since β ∝ rg → 0 then 2γ/r± ≃ ∓
√
4αγ; defining
x =
2γ/r+ + β
√
−∆
> 0 , y =
2γ/r− + β
√
−∆
< 0
with x2
+ y2
∼ 2 > 1 as β → 0 we can use the expression [12]
Arsin(x) − Arsin(y) = −π − Arsin x 1 − y2 − y 1 − x2 .
6
7. Since
2γ
r±
= −
4αγ
β ∓
√
−∆
,
then Eq. (27) reads
1
√
−∆
β −
4αγ
β −
√
−∆
1 −
1
−∆
β −
4αγ
β +
√
−∆
2
+
− β −
4αγ
β +
√
−∆
1 −
1
−∆
β −
4αγ
β −
√
−∆
2
= − sin
hκ
√
−γ
; (28)
identity sin(α + β) = sin(α) cos(β) + cos(α) sin(β) was used to keep in account the π-term. At this stage we
can also simplify the expression above: writing
√
−∆ = β2 − 4αγ = β2 + A2 (γ = −M2
< 0)
then by using
√
1 + x ≃ 1 + x/2 with x = β2
/A2
we have
β −
4αγ
β ±
√
−∆
≃ β +
A2
β ± A + β2
2A
.
For its square and keeping only β2
terms, we have two cases for plus and minus sign respectively; these two
have the same solution, in fact since
β +
A2
β ± β2 + A2
2
≃ β +
A
±1 + β
A
2
,
and defining x = β/A, we have
β +
A
1 + β
A
2
∼ β + A 1 −
β
A
2
= A2
for the plus sign and,
β +
A
−1 + β
A
2
= β −
A
1 − β
A
2
∼ β − A 1 +
β
A
2
= A2
for the minus one; in the final result, 1/(1 ± x) ≃ 1 ∓ x was used for the ± case respectively. Then
1 −
1
−∆
β −
4αγ
β ±
√
−∆
2
≃ 1 −
A2
A2 1 + β2
A2
≃
β2
A2
=
β
A
,
since
1
β2 + A2
≃
1
A
1 −
β2
2A2
we can write Eq. (28) as
β
A
1 −
β4
4A4
= sin
hκ
√
−γ
. (29)
Since Eq. (25) and Eq. (20),
√
−γ = ℓ then
A2
= 4M2 E2
c2
− m2
c2
= 4 2
ℓ(ℓ + 1)
E2
c2
− m2
c2
.
It follows
A = 2M
E2
c2
− m2
c2
1/2
= 2 ℓ
E2
c2
− m2
c2
1/2
7
8. and
A4
= 16M4 E2
c2
− m2
c2
2
= 16 4
ℓ2
(ℓ + 1)2 E2
c2
− m2
c2
2
from ˆl2
|l = 2
ℓ(ℓ + 1)|l then (ˆl2
)2
|l = 4
ℓ2
(ℓ + 1)2
|l . Eq. (29) then reads
m2
c2
rg
2 ℓ E2
c2 − m2c2
1/2
1 −
(m2
c2
rg)4
64 4ℓ2(ℓ + 1)2 E2
c2 − m2c2
2 = sin
2πκ
ℓ
. (30)
3.1 Special cases
As it is well known sin(α) = sin(α+2πk), (k ∈ Z) then sin 2πκ
ℓ = sin 2πn
ℓ where n = κ+ℓ is the principal
quantum number; n/ℓ = 1 + κ/ℓ, then n/ℓ ≥ 1. If n/ℓ = ξ ∈ N then sin(2πξ) = 0, consequently from
Eq. (30) we have
1 −
(m2
c2
rg)4
64 4ℓ2(ℓ + 1)2 E2
c2 − m2c2
2 = 0
then
En,ℓ = ±mc2
1 +
ξm2
µ2
4n(ℓ + 1)m4
P
(31)
where µ is the mass of the central body and mP is the Planck mass. For n, ℓ → ∞ we have
lim
n,ℓ→∞
En,ℓ = ±mc2
(Energy of continuous spectrum).
For κ = 0 with n/ℓ = 1 for n, ℓ → 0 we have
lim
n,ℓ→0
En,ℓ = ±mc2
1 +
m2
µ2
4m4
P
(Energy of ground state).
If n/ℓ = α/2, an odd half-integer, with α ∈ N > 2 we have sin(πα) = 0, then
En,ℓ = ±mc2
1 +
αm2
µ2
8n(ℓ + 1)m4
P
. (32)
Disregarding the r4
g-term in Eq. (30) we have
En,ℓ = ±mc2
1 +
m2
µ2
2ℓ(ℓ + 1)m4
P sin2 2πn
ℓ
, (n/ℓ ∈ Q, ℓ = 1, 2). (33)
To draw the spectrum of Schwarzschild field we get a particle whose Schwarzschild radius is equal to the
Planck length, i.e. a particle with m = mP /2. For the sake of simplicity we also set µ = m; by using
Eq. (31), Eq. (32) and Eq. (33) we have
n ℓ En,l/mc2
n ℓ En,l/mc2
0 0 1.0625 4 3 1.0035
1 1 1.0078 5 3 1.0035
2 1 1.0078 5 4 1.0016
3 1 1.0078 6 3 1.0013
3 2 1.0026 6 4 1.0008
4 2 1.0026 6 5 1.0012
Table 4: Spectrum of Schwarzschild field with m = µ = mP from Eq. (31), Eq. (32) and Eq. (33).
8
10. 2
/mcnlE
-1.07
-1.06
-1.05
-1.04
-1.03
-1.02
-1.01
-1
-0.99
〉|0,0
〉|1,1 〉|2,1 〉|3,1
〉|3,2 〉|4,2
〉|4,3 〉|5,3
〉|5,4 〉|6,3 〉|6,4 〉|6,5
Negative Continuos Spectrum
Figure 3: Spectrum of Schwarzschild field from Tab. (4): Negative.
4 Conclusions and Outlooks
It is shown with simple kinematical arguments how the concept of weakness of gravity in particle physics
interactions must be revised. The naive old fashion quantum theory can be used to give an expression to
the quantization of energy levels for the Schwarzschild field. The framework used in this paper can be used
to quantize the most general expression of the line element ds for the central field with special focus on
solutions of Einstein equations for point-like particle will be my next goal.
I would thank Prof. H. Koppelaar and V. Ny´am´adi for their precious suggestions and warm encourage-
ment.
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11