Vol.3, No.2, 141-144 (2011) Natural Science
http://dx.doi.org/10.4236/ns.2011.32020
Using of the Generalized Special Relativity (GSR) in
estimating the proton (nucleon) mass to explain
the mass defect
Mahmoud Hamid Mahmoud Hilo1,2*, M. D. Abd Allah3, Khalid Mohammed Haroon1
1Department of Physics, Faculty of Education, Al-Zaiem Al-Azhari University, Om Durman, Sudan
2Department of Physics, Faculty of Science and Arts at Al-Rass, Qassim University, Al-Qassim, KSA
3Department of Physics, Faculty of Science, Sudan University of Science and Technology, Khartoum, Sudan;
*Corresponding Author:
Received 8 November 2010; revised 10 December 2010; accepted 15 December, 2010.
ABSTRACT
This paper presents theoretical investigation on
explanation of the mass defect estimating a new
value for the proton mass inside the nucleus in
the presence of the gravitational potential, the
work has been done by using a new theory
called the generalized special relativity (GSR).
Keywords: Generalized; Mass Defect; GSR;
Approximations
1. INTRODUCTION
The simplest nucleus, that of hydrogen, is a single
proton, an elementary particle of mass about 940 MeV,
carrying positive charge exactly opposite to the electron’s
charge, having a spin of one half and being a fermion (so
no two protons can be in the same quantum state).
The next simplest nucleus, called the deuteron, is a
bound state of a proton and a neutron. The neutron, like
the proton, is a spin one-half fermion, but it has no elec-
tric charge, and is slightly heavier (by 1.3 MeV) than the
proton. The binding energy of the deuteron (analogous
to the 13.6 MeV for the Hydrogen atom) is 2.2 MeV. A
photon of this energy could “ionize” the deuteron into a
separated proton and neutron. However, it is not neces-
sary to actually do this experiment to establish how
tightly the deuteron is bound. One need only weigh the
deuteron accurately. It has a mass of 1875.61 MeV. The
proton has a mass of 938.27 MeV, the neutron 939.56
MeV, so together (but some distance apart!) they have a
mass of 1877.93 MeV, 2.2 MeV more than the deuteron.
Thus, when a proton and a neutron come together to
form a deuteron, they must unload 2.2 MeV of energy,
which they do by emitting a photon (called a γ-ray at
these energies [1].
Both protons and neutrons, being fermions, obey the
exclusion principle, two protons with spin up cannot be
in the same state, although two with opposite spin direc-
tions could, and a proton and a neutron can occupy the
same spot at the same time [2].
Protons and neutrons are referred to as nucleons. The
total number of nucleons in a nucleus is usually denoted
by A, where A = Z + N, Z protons and N neutrons. The
chemical properties of an atom are determined by the
number of electrons, the same as the number of protons
Z. This is called the atomic number. Nuclei can have the
same atomic number, but different numbers of neutrons.
These nuclei are called isotopes, the Greek for “same
place”, since they are in the same place in the periodic
table [3].
These nucleons attract each other with a short range
but very strong force, called the nuclear force. The situa-
tion here is different from that for electrons in the atom,
where the strong central force tends to dominate. In the
nucleus the nucleons are attracted mainly by their im-
mediate neighbors. Nevertheless, it is a useful beginning
to think of this attractive force as being a potential well,
as seen by an individual nucleon, and think in terms of
the nuclei as filling the lowest available quantum states
in this well, just as we did for electrons in the atom. We
find, for example, that the Helium nucleus, 2p + 2n, is
tightly bound—the four nucleons can all occupy the
lowest state in the well. However, some larger nuclei,
like C, O, Fe are actually a little more tightly bound even
than He (about 8.5 MeV per nucleon as opposed to about
7.5 for He) because each nucleon is attracted to its close
partners, and there are more close partners in these larger
nuclei. It has also been argued that some of these higher
nuclei strongly resemble bound states of α - particles.
The total binding energy (usually expressed per nucleon)
of any nucleus is easy to find—just as for the deuteron

M. H. M. Hilo et al. / Natural Science 3 (2011) 141-144
142
above, the mass of the nucleus is found accurately, and
subtracted from the sum of the masses of the separate
nucleons [4].
2. GENERLIZED SPECIAL RELATIVITY
(GSR) THEORY
The Generalized Special Relativity theory is a new
form of the special relativity theory that adopts the
gravitational potential, and it gives the formula of rela-
tive mass to be as follows [5]:
00 0
2
00 2
v
gm
m
gc


(1)
where 00 2
2
1gc

 , and

denotes the gravita-
tional potential, or the field in which the mass is meas-
ured.
The derivation of the mass Eq.1 using the generalized
special relativity (GSR) can be find as follows:
In the special relativity (SR), the time, length, and
mass can be obtained in any moving frame by either
multiplying or dividing their values in the rest frame by
a factor

.
2
2
v
1c   (2)
To see how gravity affect these quantities it is a con-
venient to re-express

in terms of the proper time [2].
22
cdg dxdx



  (3)
Which is a common language to both special relativity
SR, and general relativity (GR). We know that in SR
Eq.3 reduces to [4].
22 220
,.
ii
cdcdt dxdxxct    (4)
2
22
1v
1..1
ii
ddxdx
dtdt dt
cc


  (5)
Thus we can easily generalize

to include the effect
of gravitation by using Eq.3 and by adopting the weak
field approximation where [4].
11 2233
00 2
1,
2
1
ggg
gc


 (6)
2
00 00
22
1v
..
ii
ddxdx
gg
dtdt dt cc


  (7)
When the effect of motion only is considered, the ex-
pression for time in SR take the form [2].
0
2
2
v
1
dt
dt
c


(8)
where the subscript 0 stands for the quantity measured in
the rest frame. While if gravity only affect time, its ex-
pression is given by [4].
0
00
dt
dt
g
 (9)
In view of Eqs.8-9 and 7 the expression
0
dt
dt

 (10)
can be generalized to recognize the effect of motion as
well as gravity on time, to get
0
2
00 2
v
dt
dt
gc


(11)
The same result can be obtained for the volume where
the effect of motion and gravity respectively reads [2].
2
02
v
1
VV c  (12)
0000
VgVgV (13)
The generalization can be done by utilizing (7) to find
that
2
000 0
2
v.
VVg V
c
  (14)
To generalize the concept of mass to include the effect
of gravitation we use the expression for the Hamiltonian
in general relativity, i.e. [6].
2
0
200
0000 0
22
00
00 00
22
0
dx
HcgTg d
cmc
gg
V





 


(15)
Using Eqs.14 and 15 yields
2 2
200 0
g
mc mc
cVV
  (16)
Therefore
00 0
2
00 2
v
gm
m
gc


(17)

M. H. M. Hilo et al. / Natural Science 3 (2011) 141-144
143
Which is the expression of mass in the presence of
gravitational potential and it name the (GSR) theory.
In view of Eq.1, and when we substitute the value of
00
g
, then the relative mass according to (GSR) is find to
be
02
2
22
2
1
2v
1
mc
m
cc








(18)
When the field is weak in the sense that
2
21
c

 (19)
And when the speed v is very low such that
2
2
v1
c
 (20)
Eq.8 reduces to
02
02
2
2
12
1
2
1
mc
mmm
c
c





 





(21)
Using the identity

11. 1
n
xnxforx 
one can also gets:
02
1mm c





(22)
And, when the field is so strong such that
2
21
c

 and
2
2
v1
c
 (23)
Then Eq.18 reduces to
02
2
mm c

 (24)
3. RESULTS AND DISCUSSION
The decrease in the nucleon mass inside the nucleus is
assumed to result from the fact that part of these masses
is consumed to bind nucleons together. The mechanism
by which these masses are converted to binding energy
is not clear. One tries here to explain the mass decrease
in terms of (GSR). To do this one can use the expression
of the mass in Eq.24 to get:
where
p

denotes the proton nuclear force, which can
be calculated using the nuclear potential V = 2.818 ×
10-14 joules.
2
2
14
27
13
sec
2.818 10,
1.67 10
1.67842515 10
p
p
m
p
V
m





 

(25)
where the subscript p stands for the proton. Substitute of
p

in Eq.24 yields:
27
1.66968853 10.
1.0056 .
p
pp
mkg
mamu


 (26)
This nucleon (proton) mass value is smaller than the
value of the free proton
1.0078 .
p
mamu

(27)
Also the binding energy (BE) per nucleon can be cal-
culated using the new value of the nucleon mass as fol-
low:
12
1.00781.005 0.0022
ppp
mmm amu

  (28)
where mp1 stands for the proton mass, and mp2 for the
proton mass value using the (GSR).
2931.5
0.0022 931.52.0493
p
BEEm cmMev
Mev
  
 (29)
The Deuterium consists of one proton and one neutron.
The mass of the Deuterium is:


2
12.0141 DD
M
MH u
(30)
Since the mass of the free neutron is:
1.0087
n
mamu

(31)
Thus the mass of the proton inside the 2
1
H
nucleus
is:
1.0054
pDn
mM m

 (32)
The proton mass defer it thus given by:
1.0078 1.0054
0.0024
p
m
amu


 (33)
4. CONCLUSIONS
The proton mass value found by Eq.27 using the
(GSR) theory, is approximately the same to that calcu-
lated in Eq.33 as a difference between the deuteron mass
MD (2
1
H
) and the neutron mass mp. The error percentage
is found to be not more than 0.2%, which shows the ac-
curacy in the calculations of the (GSR) theory. Adopting
of the gravitational potential in this work presents two
important results, these are, a new value for the proton
(nucleon) mass inside the nucleus and the decrease in the
proton mass value explains the increase in the calculated

M. H. M. Hilo et al. / Natural Science 3 (2011) 141-144
144
nucleus mass value (the mass defect).
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