Energy of 4-Dimensional Black Hole
Dmitriy Palatnik ∗
August 10, 2021
Abstract
In this letter I suggest possible redefinition of mass density or stress-energy for a particle. Introducing
timelike Killing vector and using Einstein identity E = mc
, I define naturally density of mass for general
case and calculate energy of black hole.
1 Introduction
Standard definition of density [1], [3], [4], µ, for mass,
γµ , (1)
where γ is determinant of metric for spacial line element, lacks manifest dependence on speed of motion, v,
or on gravitational interaction with other masses. Since this dependence is important in what follows below, I
should redefine the mass density. Consider, first, flat spacetime with Cartesian coordinates. Matter distribution
is specified by {µ;Ua}, field of density and field of 4-velocity, respectively. According to Einstein’s formula,
m = m∗
, (2)
for the mass field variation, δm, moving with speed, v, one may write,
δm = δm∗
; (3)
Consider smooth distribution of 3-velocity, v, and assume that measuring of density takes place in frame in
which matter locally at rest, i.e. v = 0. For differential of mass element measured experimentally by using
Newton formula f = ma, one obtains,
dm = dV
1− v2
; (4)
here µ∗ is density of mass, measured at rest. Formula (4) may be rewritten as
dm = µ∗dV
; (5)
where dΩ = cdtdV is 4-dimensional volume and ds2 = gabdx
adxb. In third line of (5) I’ve transfered to general
coordinate system with metric gab. There is an ambiguity in (5) because one might put there
−g in stead of√
γ as well. As it’s shown below,
−g seems to be more correct after accepting (with no proof) formula (15)
for stress-energy; in order to get standard formula (10) for stress-energy, I take (5) as is.
∗The Waterford, 7445 N Sheridan Rd, Chicago, IL 60626-1818 dmitriy.palatnik@yahoo.com
http://arxiv.org/abs/0704.1322v5
By definition (5) µ∗ is a (non-scalar) field independent of speed in difference with standard density of mass,
µ, (1), which depends on speed and also not a scalar. Suppose, now, that mass, dm, is at rest in gravitational
field g00 = 1 + 2
, where φ(x) is gravitational potential. From (5) it follows then,
dm = µ∗
γdV , (7)
where from (1), (7) follows µ∗ = µ
g00. Using standard technique, i.e. formula
δSm = (2c)
−g Tbcδgbc , (8)
where
Sm = −mc
ds ; (9)
is action of matter, one obtains stress-energy for distribution of masses [1],
. (10)
Connection between standard density of mass and density measured at rest is
µ = µ∗
. (11)
Timelike Killing vector field, ξa, satisfies equations,
∇bξa +∇aξb = 0 ; (12)
suppose, that spacetime is asymptotically flat and ξcξc → 1 on spacial infinity. If solution to (12) does exist,
then, due to stress-energy symmetry T bc = T (bc) and conservation ∇cT bc = 0, current Jc = T cbξb conserves
too: ∇c{T cbξb} = 0. Then, due to Gauss theorem conserving energy integral does exist,
−gT 0bξb . (13)
Accepting Einstein rule, E = mc2, as axiom, one might define mass element according to (13), as
−gT 0cξc . (14)
From (1), (10), (14) it follows formula for standard mass density:
µ = µ∗
· U cξc ,
which in general case depends on additional factor U cξc. One might attempt to find another action rather than
Sm = −mc
ds, because as one observes, stress-energy should be taken as
T ab = µ∗c
, (15)
in stead of (10), otherwise using (10) in (13) one would obtain infinite energy-mass for black hole. Besides,
formulae (8), (9), (14) lead to inconsistent algebraic equation for stress-energy:
T ab = T 0cξc
UaU b
It is most natural to change action (9), rather than formulae (8), (14). For case of Sz metric below, ξcU
Note, that in formula (15) µ∗ is genuine scalar density of mass; for the element of mass (14) one would obtain,
dm = µ∗
· U cξc . (16)
In order to obtain (15) one should consider actions of type Sm = −mc
Lm(α)ds, where Lm(α) is analog of
lagrangian, depending only on α = U cξc.
2 Energy of Sz Black Hole
Consider Schwarzschild solution,
ds2 =
1− 2kM
c2dt2 −
1− 2kM
dr2 − r2(dθ2 + sin2 θdφ2) . (17)
Solving (12) for metric (17), one obtains timelike Killing vector,
1− 2kM
, 0, 0, 0
. (18)
Substituting (14), (15), (17), (18) in (13), and using dV = drdθdφ, one gets,
E = c2
r2dr sin θdθdφµ∗(r, θ, φ) . (19)
Transfering to Cartesian coordinates, (r, θ, φ) → (x1, x2, x3), and introducing density of mass,
µ∗ = Mδ(x
1)δ(x2)δ(x3) , (20)
one obtains necessary expression for energy of black hole, E = Mc2. Note, that the same result could be derived
by using formula (16).
3 Modification of Killing Equation
Solutions to Killing eqs (12) do exist for restricted class of gravitational configurations; indeed, 4 components of
Killing vector should satisfy 10 equations. Idea of following (i.e. formula (22) bellow) belongs to Boris Tsirelson.
Consider, again, continuous matter distribution, specified by {µ∗ ; U c0}, scalar mass density field, µ∗(xc1), and
4-velocity field, U c0(xc1). Assume, that all other than matter fields are absent and only contribution of stress-
energy is (15). Due to equations of motion of matter, stress-energy is divergence-free, ∇c0T c0c1 = 0, and
symmetric. Then, current Jc0 = T c0c1ξc1 does have vanishing divergence, if
T c0c1 [∇c0ξc1 +∇c1ξc0 ] = 0 . (21)
From conservation of current, ∇c0Jc0 = 0, follows conservation of mass-energy (13). In stead of demanding
implementation of Killing eqs (12) in order to have (21) satisfied, consider expression (15) and demand that
Killing vector field satisfies following 4 equations:
U c1 [∇c0ξc1 +∇c1ξc0 ] = 0 . (22)
For Sz metric (17), solution to (22) with
(g00)
2 , 0, 0, 0
, (23)
is (18). Is it true that solving eqs (22) for empty space one may consider limit µ∗ → 0 with (23)? One could
use (16) for computing mass-energy of black hole; result is same as above. One might find solution to (22) for
Kerr black hole with metric [1],
ds2 =
1− rgr
dt2 +
2rgra
sin2 θdtdφ − ρ
dr2 − ρ2dθ2
r2 + a2 +
sin2 θ
sin2 θdφ2 ; (24)
gab∂a∂b =
r2 + a2 +
sin2 θ
2rgra
∂t∂φ −
2 − 1
∆ sin2 θ
1− rgr
2 , (25)
where
ρ2 = r2 + a2 cos2 θ ; (26)
∆ = r2 + a2 − rgr , (27)
and using expression for 4-velocity,
U c =
g00 + 2φ̇g03 + (φ̇)2g33
, 0, 0,
g00 + 2φ̇g03 + (φ̇)2g33
 ; (28)
here φ̇ ≡ dφ
. That is, if to demand ξ1 = ξ2 = 0, then eqs (22) are equivalent to
∂rξ0 − 2Γ001ξ0 − 2Γ301ξ3 + φ̇{∂rξ3 − 2Γ013ξ0 − 2Γ313ξ3} = 0 ; (29)
∂θξ0 − 2Γ002ξ0 − 2Γ302ξ3 + φ̇{∂θξ3 − 2Γ023ξ0 − 2Γ323ξ3} = 0 . (30)
Set of non-zero Christoffel symbols for metric (24) is
Γ001, Γ
02, Γ
13, Γ
23, Γ
00, Γ
03, Γ
11, Γ
12, Γ
22, Γ
Γ200, Γ
03, Γ
11, Γ
12, Γ
22, Γ
33, Γ
01, Γ
02, Γ
13, Γ
−g = ρ2 sin θ . (31)
The solutions of (29), (30) are,1
ξ(1)c0 = (g00, 0, 0, g03) ; ξ
= (1, 0, 0, 0) ; (32)
ξ(2)c0 = (g03, 0, 0, g33) ; ξ
= (0, 0, 0, 1) ; (33)
g00 = 1−
; g03 =
sin2 θ ; g33 = −
2 + a2 +
sin2 θ
sin2 θ
For energy of Kerr black hole use eqs (13), (15), (32); the result is
E = c2
dr sin θdθdφµ∗(r; θ)
g00 + φ̇g03
g00 + 2φ̇g03 + (φ̇)2g33
. (34)
Transferring to Cartesian coordinates in case φ̇ = 0, according to transformation of spheroidal coordinates to
Cartesian, specified in [5], (r, θ, φ) → (x, y, z), where
r2 + a2 sin θ cosφ ; y =
r2 + a2 sin θ sinφ ; z = r cos θ ; (35)
and using (20), one recovers anew E = Mc2. Using second Killing vector, (33), one obtains conserving integral,
Q = c2
dr sin θdθdφµ∗(r; θ)
g03 + φ̇g33
g00 + 2φ̇g03 + (φ̇)2g33
. (36)
4 A Simple Theorem About Killing Vectors
Here I should specify a theorem, claiming that if metric of spacetime doesn’t depend on coordinate xj , then
spacetime has respective Killing vector ξc = gjc; number of Killing vectors for spacetime is equal to number of
coordinates on which metric coefficients don’t depend. Proof. Rewrite Killing eqs (12) in form,
∂aξb + ∂bξa − 2Γcabξc = 0 . (37)
Substitute ξc = gjc for specific j. Then eq (37) reads
∂jgab = 0 . (38)
In case of eqs (22) one would have in stead of (38) equation Ua∂jgab = 0. One might even go step back and
use eqs (21) then one would obtain equation UaU b∂jgab = 0 in stead of (38). 4-velocity, U
a, should be taken
along the trajectory of a particle. It’s understandable that if metric doesn’t depend on coordinate(s) xj , then
does exist translational symmetry of the physical system in direction specified by that (or these) coordinate(s),
which means that respective charges (energy, momenta) do conserve.
1 They are also solutions of original Killing eqs (12).
5 Acknowledgement
I wish to thank Christian Network and Boris Tsirelson professor of Tel-Aviv University.2 I’m grateful for spirit
of support from Catholic Church and Lubavich synagogue F.R.E.E.
References
[1] L D Landau E M Lifshits The Classical Theory of Fields Pergamon Press 1975
[2] L A Khalfin B S Tsirelson Foundations of Physics vol 22 No 7 1992
[3] Robert M Wald General Relativity U of Chicago Press 1984
[4] S Weinberg Gravitation and Cosmology John Wiley and Sons Inc NY 1972
[5] A Anabalon et al. arXiv: gr-qc/1009.3030v1
2 See, e.g. [2]
	1 Introduction
	2 Energy of Sz Black Hole
	3 Modification of Killing Equation
	4 A Simple Theorem About Killing Vectors
	5 Acknowledgement

  In this letter I suggest possible redefinition of mass density, not depending
on speed of the mass element, which leads to a more simple stress-energy for an
object. I calculate energy of black hole.

Introduction
Standard definition of density [1], [3], [4], µ, for mass,
γµ , (1)
where γ is determinant of metric for spacial line element, lacks manifest dependence on speed of motion, v,
or on gravitational interaction with other masses. Since this dependence is important in what follows below, I
should redefine the mass density. Consider, first, flat spacetime with Cartesian coordinates. Matter distribution
is specified by {µ;Ua}, field of density and field of 4-velocity, respectively. According to Einstein’s formula,
m = m∗
, (2)
for the mass field variation, δm, moving with speed, v, one may write,
δm = δm∗
; (3)
Consider smooth distribution of 3-velocity, v, and assume that measuring of density takes place in frame in
which matter locally at rest, i.e. v = 0. For differential of mass element measured experimentally by using
Newton formula f = ma, one obtains,
dm = dV
1− v2
; (4)
here µ∗ is density of mass, measured at rest. Formula (4) may be rewritten as
dm = µ∗dV
; (5)
where dΩ = cdtdV is 4-dimensional volume and ds2 = gabdx
adxb. In third line of (5) I’ve transfered to general
coordinate system with metric gab. There is an ambiguity in (5) because one might put there
−g in stead of√
γ as well. As it’s shown below,
−g seems to be more correct after accepting (with no proof) formula (15)
for stress-energy; in order to get standard formula (10) for stress-energy, I take (5) as is.
∗The Waterford, 7445 N Sheridan Rd, Chicago, IL 60626-1818 dmitriy.palatnik@yahoo.com
http://arxiv.org/abs/0704.1322v5
By definition (5) µ∗ is a (non-scalar) field independent of speed in difference with standard density of mass,
µ, (1), which depends on speed and also not a scalar. Suppose, now, that mass, dm, is at rest in gravitational
field g00 = 1 + 2
, where φ(x) is gravitational potential. From (5) it follows then,
dm = µ∗
γdV , (7)
where from (1), (7) follows µ∗ = µ
g00. Using standard technique, i.e. formula
δSm = (2c)
−g Tbcδgbc , (8)
where
Sm = −mc
ds ; (9)
is action of matter, one obtains stress-energy for distribution of masses [1],
. (10)
Connection between standard density of mass and density measured at rest is
µ = µ∗
. (11)
Timelike Killing vector field, ξa, satisfies equations,
∇bξa +∇aξb = 0 ; (12)
suppose, that spacetime is asymptotically flat and ξcξc → 1 on spacial infinity. If solution to (12) does exist,
then, due to stress-energy symmetry T bc = T (bc) and conservation ∇cT bc = 0, current Jc = T cbξb conserves
too: ∇c{T cbξb} = 0. Then, due to Gauss theorem conserving energy integral does exist,
−gT 0bξb . (13)
Accepting Einstein rule, E = mc2, as axiom, one might define mass element according to (13), as
−gT 0cξc . (14)
From (1), (10), (14) it follows formula for standard mass density:
µ = µ∗
· U cξc ,
which in general case depends on additional factor U cξc. One might attempt to find another action rather than
Sm = −mc
ds, because as one observes, stress-energy should be taken as
T ab = µ∗c
, (15)
in stead of (10), otherwise using (10) in (13) one would obtain infinite energy-mass for black hole. Besides,
formulae (8), (9), (14) lead to inconsistent algebraic equation for stress-energy:
T ab = T 0cξc
UaU b
It is most natural to change action (9), rather than formulae (8), (14). For case of Sz metric below, ξcU
Note, that in formula (15) µ∗ is genuine scalar density of mass; for the element of mass (14) one would obtain,
dm = µ∗
· U cξc . (16)
In order to obtain (15) one should consider actions of type Sm = −mc
Lm(α)ds, where Lm(α) is analog of
lagrangian, depending only on α = U cξc.
2 Energy of Sz Black Hole
Consider Schwarzschild solution,
ds2 =
1− 2kM
c2dt2 −
1− 2kM
dr2 − r2(dθ2 + sin2 θdφ2) . (17)
Solving (12) for metric (17), one obtains timelike Killing vector,
1− 2kM
, 0, 0, 0
. (18)
Substituting (14), (15), (17), (18) in (13), and using dV = drdθdφ, one gets,
E = c2
r2dr sin θdθdφµ∗(r, θ, φ) . (19)
Transfering to Cartesian coordinates, (r, θ, φ) → (x1, x2, x3), and introducing density of mass,
µ∗ = Mδ(x
1)δ(x2)δ(x3) , (20)
one obtains necessary expression for energy of black hole, E = Mc2. Note, that the same result could be derived
by using formula (16).
3 Modification of Killing Equation
Solutions to Killing eqs (12) do exist for restricted class of gravitational configurations; indeed, 4 components of
Killing vector should satisfy 10 equations. Idea of following (i.e. formula (22) bellow) belongs to Boris Tsirelson.
Consider, again, continuous matter distribution, specified by {µ∗ ; U c0}, scalar mass density field, µ∗(xc1), and
4-velocity field, U c0(xc1). Assume, that all other than matter fields are absent and only contribution of stress-
energy is (15). Due to equations of motion of matter, stress-energy is divergence-free, ∇c0T c0c1 = 0, and
symmetric. Then, current Jc0 = T c0c1ξc1 does have vanishing divergence, if
T c0c1 [∇c0ξc1 +∇c1ξc0 ] = 0 . (21)
From conservation of current, ∇c0Jc0 = 0, follows conservation of mass-energy (13). In stead of demanding
implementation of Killing eqs (12) in order to have (21) satisfied, consider expression (15) and demand that
Killing vector field satisfies following 4 equations:
U c1 [∇c0ξc1 +∇c1ξc0 ] = 0 . (22)
For Sz metric (17), solution to (22) with
(g00)
2 , 0, 0, 0
, (23)
is (18). Is it true that solving eqs (22) for empty space one may consider limit µ∗ → 0 with (23)? One could
use (16) for computing mass-energy of black hole; result is same as above. One might find solution to (22) for
Kerr black hole with metric [1],
ds2 =
1− rgr
dt2 +
2rgra
sin2 θdtdφ − ρ
dr2 − ρ2dθ2
r2 + a2 +
sin2 θ
sin2 θdφ2 ; (24)
gab∂a∂b =
r2 + a2 +
sin2 θ
2rgra
∂t∂φ −
2 − 1
∆ sin2 θ
1− rgr
2 , (25)
where
ρ2 = r2 + a2 cos2 θ ; (26)
∆ = r2 + a2 − rgr , (27)
and using expression for 4-velocity,
U c =
g00 + 2φ̇g03 + (φ̇)2g33
, 0, 0,
g00 + 2φ̇g03 + (φ̇)2g33
 ; (28)
here φ̇ ≡ dφ
. That is, if to demand ξ1 = ξ2 = 0, then eqs (22) are equivalent to
∂rξ0 − 2Γ001ξ0 − 2Γ301ξ3 + φ̇{∂rξ3 − 2Γ013ξ0 − 2Γ313ξ3} = 0 ; (29)
∂θξ0 − 2Γ002ξ0 − 2Γ302ξ3 + φ̇{∂θξ3 − 2Γ023ξ0 − 2Γ323ξ3} = 0 . (30)
Set of non-zero Christoffel symbols for metric (24) is
Γ001, Γ
02, Γ
13, Γ
23, Γ
00, Γ
03, Γ
11, Γ
12, Γ
22, Γ
Γ200, Γ
03, Γ
11, Γ
12, Γ
22, Γ
33, Γ
01, Γ
02, Γ
13, Γ
−g = ρ2 sin θ . (31)
The solutions of (29), (30) are,1
ξ(1)c0 = (g00, 0, 0, g03) ; ξ
= (1, 0, 0, 0) ; (32)
ξ(2)c0 = (g03, 0, 0, g33) ; ξ
= (0, 0, 0, 1) ; (33)
g00 = 1−
; g03 =
sin2 θ ; g33 = −
2 + a2 +
sin2 θ
sin2 θ
For energy of Kerr black hole use eqs (13), (15), (32); the result is
E = c2
dr sin θdθdφµ∗(r; θ)
g00 + φ̇g03
g00 + 2φ̇g03 + (φ̇)2g33
. (34)
Transferring to Cartesian coordinates in case φ̇ = 0, according to transformation of spheroidal coordinates to
Cartesian, specified in [5], (r, θ, φ) → (x, y, z), where
r2 + a2 sin θ cosφ ; y =
r2 + a2 sin θ sinφ ; z = r cos θ ; (35)
and using (20), one recovers anew E = Mc2. Using second Killing vector, (33), one obtains conserving integral,
Q = c2
dr sin θdθdφµ∗(r; θ)
g03 + φ̇g33
g00 + 2φ̇g03 + (φ̇)2g33
. (36)
4 A Simple Theorem About Killing Vectors
Here I should specify a theorem, claiming that if metric of spacetime doesn’t depend on coordinate xj , then
spacetime has respective Killing vector ξc = gjc; number of Killing vectors for spacetime is equal to number of
coordinates on which metric coefficients don’t depend. Proof. Rewrite Killing eqs (12) in form,
∂aξb + ∂bξa − 2Γcabξc = 0 . (37)
Substitute ξc = gjc for specific j. Then eq (37) reads
∂jgab = 0 . (38)
In case of eqs (22) one would have in stead of (38) equation Ua∂jgab = 0. One might even go step back and
use eqs (21) then one would obtain equation UaU b∂jgab = 0 in stead of (38). 4-velocity, U
a, should be taken
along the trajectory of a particle. It’s understandable that if metric doesn’t depend on coordinate(s) xj , then
does exist translational symmetry of the physical system in direction specified by that (or these) coordinate(s),
which means that respective charges (energy, momenta) do conserve.
1 They are also solutions of original Killing eqs (12).
5 Acknowledgement
I wish to thank Christian Network and Boris Tsirelson professor of Tel-Aviv University.2 I’m grateful for spirit
of support from Catholic Church and Lubavich synagogue F.R.E.E.
References
[1] L D Landau E M Lifshits The Classical Theory of Fields Pergamon Press 1975
[2] L A Khalfin B S Tsirelson Foundations of Physics vol 22 No 7 1992
[3] Robert M Wald General Relativity U of Chicago Press 1984
[4] S Weinberg Gravitation and Cosmology John Wiley and Sons Inc NY 1972
[5] A Anabalon et al. arXiv: gr-qc/1009.3030v1
2 See, e.g. [2]
	1 Introduction
	2 Energy of Sz Black Hole
	3 Modification of Killing Equation
	4 A Simple Theorem About Killing Vectors
	5 Acknowledgement
