Download Field theretical approach to gravity

Field theretical approach to gravity
A.I.Nikishov
Abstract
The metric of spherically symmetric ball of ideal liquid is considered in G2 - approximation with the help of theory of sources. Using the integral equations of this theory
gives the exterior metric depending upon the radius of the ball of matter in some terms
proportional to G2 .I argue that according to this metric from measurement outside the
ball one can obtain the radius b of the ball of matter
Interior solution The following notation is used
(1)
(2)
gik = ηik + hik + hik + · · · ,
1
(n)
(n)
h̄ik = hik − ηik h(n) ,
2
(n)
hik ∝ Gn ,
i, k = 0, 1, 2, 3,
ηik = diag(−1, 1, 1, 1),
(n)
h(n) = h(n)
,αα − h00 ,
∂h
= h,α ,
∂xα
α, β = 1, 2, 3.
(1)
(1)
In linear approximation in Hilbert gauge h̄αβ.α (m0 , r) = 0 we have
m0 G r2
( −3),
2b b2
4
m0 = πb3 µ.
3
(2)
We note that according to the theory of sources the Hilbert gauge h̄ik ,k = 0 means that
gauge degree of freedom are excluded. This is true for any h̄nik . In (2) µ is the density of
(2)
liquid. The function h̄ik (m0 , r) in this gauge satisfies the differential equation
(1)
hik (m0 , r) = −2φ(m0 , r)δik ,
(1)
h̄ik = −4φ(m0 , r)δ0i δ0k ,
(2)
φ(m0 , r) =
(1)
∇2 h̄ik (m0 , r) = −16πG(Tik + tik ),
(3)
This is the differential form of eq. (17.6) in Ch.3, §17 in J.Schwinger. Particles, sources. and
fields. Addison- Wesley, 1970.
(1)
Here Tik is the tensor of matter, tik -tensor due to 3-graviton interaction In this paper we
assume that tik is given by general relativity, see Ch. 7, §6 in S. Weinberg. Gravitation and
Cosmology. New York 1972.
(2)
(2)
In functions proportional to G2 , such as h̄ik (m0 , r), hik (m0 , r) and so on m0 indicate only
that these functions are obtained directly from equations of perturbation theory. In terms
proportional to G2 in the considered approximation it does not matter whether it stands
there m0 or the dressed mass m = m0 (1 + 3mG
) In the region inside the ball we have, see
b
A.I.Nikishov. arXiv: 1605.06305 v.I [physics gen -ph] 16 May 2016]
t00 = −
3
m2 G2
21r2
(∇φ)2 − 6µφ =
(54
−
),
8πG
8πGb4
b2
(1)
T00 + t00 =
(1)
T00 = 2µφ =
m2 G2
15r2
(36
−
),
8πGb4
b2
1
m2 G2 3r2
(
− 9),
4πGb4 b2
(4)
(1)
Tαβ
+ tαβ
4r2
2xα xβ
m2 G2
[δαβ −9 + 2 −
].
=
4
8πGb
b
b2
(5)
Now using the integral equation of theory of sources ( see Schwinger book)
Z
(2)
(1)
0
h̄ik (m , r) = 16πG d4 x0 D+ (x − x0 )[Tik (x0 ) + tik (x0 )],
(6)
we find
(2)
h̄αβ (m0 , r) =
m2 G2
18 r2 3 r4
9 x α xβ 2 r 2 xα xβ
(2)
[δ
(−5
+
−
)
−
+
] = h̄αβ (m, r),
αβ
2
2
4
2
4
b
5 b
7b
5 b
7 b
(7)
r2 3 r4
m2 G2 51
−
12
(
+
).
b2
2
b2 2 b4
(8)
(2)
h̄00 (m0 , r) =
(1)
The expression (7) remains valid when m0 is replaced by m because h̄αβ (m0 , r) = 0 in (2) for
α, β = 1, 2, 3, see the text below eq. (13). From (7) we have
0
(2)
h̄(2)
αα (m , r) = h̄αα (m, r) =
r2 r4
m2 G2
[−15
+
9
− 4 ].
b2
b2
b
(2)
(9)
(2)
Next from solutions h̄ik (m0 , r) we should obtain the solution hik (m0 , r):
1
(2)
(2)
hik (m0 , r) = h̄ik (m0 , r) − ηik h̄(m0 , r),
2
(2)
(2)
h̄(2) == h̄(2)
αα − h̄00 = −h .
(10)
Using the eqs. (9) and (8), we get from the second eq. in (10)
h̄(2) (m0 , r) =
r2 5 r4
m2 G2 81
(−
+
21
−
).
b2
2
b2 2 b4
(11)
From the first eq. in (10) (in which i, k are replaced by α, β) we obtain
(2)
hαβ (m0 , r)
m2 G2
61 69 r2 23 r4
9 xα xβ 2 r 2 xα x β
=
[δαβ ( −
+
)−
+
].
b2
4
10 b2 28 b4
5 b2
7 b4
(12)
Similarly from (10) when i = k = 0 we find
(2)
h00 (m0 , r) =
m2 G2 21 3 r2 1 r2
( − 2 + 2 ).
b2
4
2b
4b
(2)
(13)
(2)
Besides solutions hik (m0 , r) we need solutions hik (m, r). The latter solutions can be
(1)
obtained from the former ones by adding to them G2 -terms from hik (m0 , r) when in them
m0 is replaced by m(1 − 3 mG
); m is the dressed mass. Indeed we have
b
(1)
hik (m0 , r) =
m0 G
r2
m2 G2
r2
(1)
δik (3 − 2 ) = hik (m, r) +
δ
(3
− 9),
ik
b
b
b2
b2
m0 = m(1 − 3
mG
). (14)
b
Similarly from (2)
(1)
(2)
h̄00 (m0 , r) = −4φ(m0 , r) = h̄00 (m, r) +
2
m2 G2 r2
(6 2 − 18).
b2
b
(15)
Equations (10) remain valid also when m0 is replaced by m. So taking into account (9) and
equation (19) below from the second equation in (10) we find
h̄(2) (m, r) =
m2 G2 45
r2 5 r4
+
15
(−
−
),
b2
2
b2 2 b4
(16)
Similarly from the first equation in (10) with the help of (7) and (16) we get
(2)
hαβ (m, r)
25 39 r2 23 r4
9 xα x β 2 r 2 xα xβ
m2 G2
[δαβ ( −
+
)−
+
],
=
b2
4
10 b2 28 b4
5 b2
7 b4
(17)
In the same manner with the help of (19) and (15) we obtain
(2)
h00 (m, r) =
m2 G2 15 3 r2 1 r2
(− + 2 + 2 ).
b2
4
2b
4b
(18)
r2 3 r4
m2 G2 15
(
−
6
+
).
b2
2
b2 2 b4
(19)
Similarly from (8) and (15) we find
(2)
h̄00 (m, r) =
(2)
(2)har
(2)iso
We note that hαβ (m, r) differs from harmonic hαβ (m, r) and isotropichαβ (m, r) only by
gauge terms, in more detail see A.I.Nikishov. arXiv: 1605.06305 v.I [physics gen -ph] 16 May
2016. Equation (18) holds in all these systems.
(2)
Exterior solution. With the help of (5) from (6) we obtain the contribution to h̄αβ (m0 , r)
from r0 < b
Z
1
14 δαβ
4
xα xβ δαβ
d 3 x0
(1)
(Tαβ + tαβ ) = (mG)2 [−
−
b( 5 − 3 )].
(20)
16πG
0
x − ~x |
3 rb
5·7
r
3r
r0 <b<r 4π |~
Here were used the equations (4), (5) and (16) in A.I.Nikishov. arXiv: 1605.06305 v.I
[physics gen -ph] 16 May 2016). With the help of relation (see for example equation (70) in
A.I.Nikishov. Part. Nucl. v.32, No 1, p.5 (2001))
16πGtαβ = (mG)2 [
14δαβ 28xα xβ
−
]
r4
r6
(21)
(2)
we find the contribution to h̄αβ (m0 , r) from b < r0
Z
16πG
b<r0 ,r
d 3 x0
1
14 δαβ 7xα xβ 28 xα xβ δαβ
tαβ (x0 ) = (mG)2 [
−
+ b( 5 − 3 )].
0
4π |~x − ~x |
3 rb
r4
5
r
3r
(22)
Here were used the equations (A18) and (15) in arXiv: 1605.06305 v.I [physics gen -ph] 16
May 2016. The sum of these two contribution (20) and (22) gives
(2)
h̄αβ (m0 , r) = (mG)2 [−7
(2)
xα xβ 192 xα xβ δαβ
b( 5 − 3 )]..
+
r4
35
r
3r
(2)
(2)
(23)
We remind here that h̄αβ (m0 , r) = h̄αβ (m, r) because h̄αβ (m0 , r) = 0, see the second equation
in (2). It is usefull to note that the region b < r0 contributes to (23) essentially more then
the region r0 < b.
3
From the first equation in (10) and the relation
12 10
+ 2 ) = −h̄(m0 , r)
rb
r
(24)
5δαβ 7xα xβ
6δαβ
192b xα xβ δαβ
+( 2 −
( 5 − 3 )].
)+
4
rb
r
r
35
r
3r
(25)
6δαβ
rb
because it is
h(2) (m0 , r) = (mG)2 (
it follows
(2)
hαβ (m0 , r) = m2 G2 [
(2)
The expression for hαβ (m, r) is obtained from here by dropping the term
cancelled by the term in
(1)
(1)
hαβ (m0 , r):
(1)
hαβ (m0 , r) = hαβ (m, r) − m2 G2
6δαβ
,
rb
(1)
hαβ (m, r) = δαβ
2mG
.
r
(26)
Thus
5δαβ 7xα xβ
64b xα xβ δαβ
(3 5 − 3 )].
−
)+
(25a)
2
4
r
r
35
r
r
It is easy to check that the interior solution (18) continuously goes over to exterior one (25a):
2
mG
111
53 xα xβ
(2)
(2)
=
[
.
(27)
hαβ (m, r)
δαβ −
] = hαβ (m, r)
2
b
35
35 b
r→b
b←r
(2)
hαβ (m, r) = m2 G2 [(
We note that the linear approximation (2) holds in harmonic, isotropic and considered
here coordinates systems. Our system, given in G2 approximation by (17) and (25a) we
will call the preferred one, because the gauge degrees of freedom do not contribute to it.
Scattering of a particle by gravitational field
1. Scattering of scalar particle in Born approximation
dσ =
1 cos θ
1 1
1
π 2
π
rg (1 + 2 ) 3 2θ dθ = rg2 (1 + 2 ) 3 (1 − y 3 + · · · )dθ.
2
β sin 2
2
β y
15
(28)
Here β is velocity at infinity. For a massless particle β = 1.
2.Scattering of a classical zero mass particle
dσ =
π 21
r
(1 + a1 y + a3 y 3 + a4 y 4 + · · · )dθ,
2 g y3
y=
θ
2
(29)
Here ai , i = 1, 3, 4. are numbers of order unity. The quantum formula (28) does not contain
~. So the difference with (29 ) is surprising.
3.For graviton scattering in Born approximation we have
dσ =
π 2 cos8 2θ + sin8 2θ
r
dΩ.
4 g
sin4 2θ
(30)
The first (second) term in the numerator are contribution from nonspin flip (from spin flip)
scattering.
Modified 3-graviton vertex.
The 3-graviton vertex has the form
1
L = hik tik .
2
4
(31)
GR
In general relativity tik = t ik is not the gravitational energy-momentum tensor In fieldtheoretical approach we can calculate tik either by Belifante or Rosenfeld method. Both
method give the same result for space without matter. Where there is a matter we have to
add interaction terms. In approximation of point-like particles we can choose these interaction
terms in such a way that particles move along geodesics. Then for tik I pick up a specific
tensor which I call
MT W
t
T00 +
ik
For slowly moving particles we have
MT W
t
00 =
(0)
T00 + µφ +
8πG
,
(∇φ)2
(0)
T00 = T00 (G = 0).
(32)
This formula gives the positive gravitational field energy density as it should be. It agrees
with Newtonian interaction energy 21 µφ.It follows from here that in Newtonian GmR1 m2 masses
m1 , m2 are dressed ones (together with gravitational energy). This is impotent for interaction
of heavy compact object (neutron stars) as the essential part of their energy resides in the
gravitation field outsside the matter.
As Schwinger shown the Newtonian interaction 12 µφ leads to modification of the potential
φ → φ + 12 φ2 . In our case this means that phi → φ + φ2 . This is necessary for the correct
perihelion precession of a planet.
MT W
GR
For small angle graviton scattering t ik gives the same result as tik = t ik .
Conclusion.
We may expect the modification of GR results for the movement of a relativistic particle
in strong gravitational field.
5