Download REST MASS DECREASE IN A ROTATIONAL FIELD, IS TO BE

REST MASS DECREASE IN A ROTATIONAL FIELD, IS TO BE ACHIEVED BY
EMISSION OF A CORRESPONDING RADIATION
Tolga Yarman, Metin Arik, Alexander L. Kholmetskii
ABSTRACT
When an object is bound, already at rest, to an attrational field, its rest mass must (owing to the law of energy
conservation, embodying the mass and energy equivalence of the Special Theory of Relativity), get decreased.
The mass deficiency coming into play, must have been discharged, somehow. A rocket model, in a free fall of the
object, say, in a gravitational field, evokes that the rest mass should be discharged, via emission of photons, from
the back of the falling object. The same is expected when the object is brought to a mere rotational motion
(although the object in rotation is supposed to be fixed, but the intensity of the centrifugal force of concern,
keeps on getting increased until a stationary state is attained). Thus, bringing an object to a rotational motion
means that the energy we furnish to it, serves to dump out of it, as much rest mass (were the speed of light taken
unity). This is, just like we furnish work to a refrigerator to dump out of it heat. Herein, it is shown that photons
of fundamental energy h /(2 π) are emitted, while the object is delivered to a rotational accelerational field, 
being the angular velocity of the rotating object. This fundamental energy, as seen, does not depend on anything
else (such as mass or charge of the object), and it is in well conformity with Bohr’s Principle of
Correspondence. This means at the same time that, emission will be achieved as long as the angular velocity
keeps on increasing, and will cease right after the object reaches a stationary rotational motion (thus, a constant
centrifugal acceleration). By the same token, one can affirm that even the rotation at a macroscopic level, is
quantized, and can only assume, given angular velocities. The rate of emission of photons of concern is, on the
other hand, proportional to the angular acceleration of the object. It is thus constant for a constant angular
acceleration, although the energy h /(2 π) of emitted photons will get increased with increasing  , until the
rotation reaches a stationary level, where after we expect no emission.
INTRODUCTION
According to Yarman’s gravitational approach,1 the rest mass of an object bound to, say, a
gravitational field, owing to the law of energy conservation, must get decreased as much as
the static binding energy coming into play. The static binding energy is the energy necessary
to furnish to the statically bound object in order to bring it, to infinity, free of any
gravitational field. Therefore, if the binding energy of concern at r, from the center of the host
body, is EB(r), then the object’s rest mass m0 measured at infinity - owing to law of energy
conservation embodying the mass and energy equivalence of the Special Theory of Relativity
(STR) - must get decreased, as much as EB(r)/c2, if brought from infinity, to a rest at r; c, here
is the speed of light in empty space.
The law of energy conservation, thus coupled with Newton’s law of gravitation, though here,
strictly restricted to masses at rest, with regards to each other, as derived by Yarman
(Reference 1), leads immediately to m(r), the rest mass of the object at r, as assessed by the
distant observer:
m(r)  m0eα(r) ,
(1)
where α(r) is given by
GM
α(r)  2
rc
,
(2)
G being the gravitational constant, and M is the mass of the host body.
1
Yarman, has further derived the de Broglie relationship, based on just the scheme briefly
sketched above, for a planet orbiting a star, just like he was able to derive it, in the case of an
electron rotating around a nucleus.2 The present authors furthermore have shown that the de
Broglie relationship can also be derived in the case of a rotating disc, meaning that, even in
this latter case, the “field” is quantized.3 In other words, the rotational velocity, even on a
macroscopic level, must be quantized, and can only be increased, bit by bit.
ROTATING DISC: ENERGY OF THE EMITTED PHOTONS IS PROPORTIONAL TO
THE ANGULAR VELOCITY
Let us consider an object of rest mass m 00 on a circular non-relativistic motion, of radius Rn,
carried with the velocity vn. For such a motion, for the nth stationary level, the present authors,
via their derivation of the de Broglie relationship were able to write (cf. Reference 3):
 Bn 
nh
nh

 2R n .
mv n m00v n
(3)
Note that for an object plugged at a given location at a distance Rn from the center, this latter
quantity remains constant, no matter how fast the rotation is. Let us then call it, straight R.
The above relationship naturally looks like the Bohr postulate. Yet in fact, we arrived at it, via
the derivation of the de Broglie relationship (cf. Reference 3). What is more important over
here, is that in Bohr’s hydrogen approach, one lands at quantized level, where greater n’s
mean states where the electron is less bound to the proton. Yet along with Eq.(3), it is clear
that greater n’s mean greater tangential velocities, at the same given location.
The above relationship numerically can serve to calculate n:
2Rm 00v n
.
n
h
(4)
Let us calculate n, for an object of 1 g, on the edge of our rotating disc, of R = 0.5 m, making
100 revolutions (i.e. approximately 6x100 radians) per second:
(5)
n  6x6x(0.5) 2 x103 x100 / (6.62x10 34 )  14x1032 .
Now, we propose to calculate how much the velocity should get changed to achieve a unit
quantum interval, i.e. n  1 . We can thus write
2Rm 00v n
(6)
n 
1 .
h
The static binding energy variation dB (as observed by an observer located at the center of
the rotating disc, bearing the same angular velocity as that of the disc), due to an infinitely
small change d in the angular velocity  , can be expressed as (cf. Reference 3)
R 2d
R 2d
dB  m00
 m0()
,
(7)
2R 2
2R 2
1 2
1 2
c
c
2
where m0() is the rest mass of the rotating object as assessed by an observer located at the
center of the rotating disc, thus rotating with the same angular velocity  , as that of the disc,
but then with a practically null tangential velocity. In effect*
m0()  m00 1 
2R 2
.
c2
(8)
Note that this equation can be obtained via integrating Eq.(7). [Or vice versa, Eq.(7) can be
obtained via differentiating Eq.(8).]
One can further calculate the overall relativistic energy m  () (were c taken unity), of the
object of concern, as assessed by an outside observer:
m  ()c2  m00c 2 1 
2R 2
x
c2
1
R
1 2
c
2
2
 m00c 2 .
(9)
All this, is in harmony with results originally reported by Yarman et al,4 and experimentally
checked out by Kholmetskii et al.5,6,7
Eq.(9) means, the overall energy remains the same and the original. How come?
The answer is that one needs to furnish energy in order to deliver a body to an attractional
force field, if the body itself, is unable to move toward deeper states all by itself (where, it will
be bound to the field, more and more intensely), through the gradient space created by the
field in consideration. In effect, consider a falling stone, above Earth. It gets accelerated
downward, and gains kinetic energy. When it hits the ground, it delivers to Earth its kinetic
energy, and then should weigh less, as delineated by Eq.(1). One other way, is to lower it,
quasitatically from one altitude to a deeper one, say, about half a meter, by your hand. Then
you have to spend as much energy as the kinetic energy the stone would acquire when it is left
to the free fall at the higher altitude, until it reaches the lower one, in question. So either, it
falls down all by itself, and acquires kinetic energy, which will be delivered to Earth, after it
hits the ground and comes to a rest, or we should furnish to it, as much energy, and lower it to
a deeper field.
*
Since the centrifugal force is
F()  m0()2R ,
(i)
one can write
dB  F()
Rd
Rd .
 F()
2 2

 R 
1  2 
c


(ii)
This is the infinitesimal energy expression, related to an object at the edge of a rotating disc, undergoing a
change in the angular velocity, where there is obviously no displacement. Therefore, unlike the classical
energy expression which is defined as [force] x [displacement, the object undergoes, under the influence of
force], here we have another form of energy, yet well yeld by the law of energy conservation.
3
Now, when we bring an object to a rotational motion, along with a fixed rotational radius, it
has nowhere to fall, though it will be getting bound to the centrifugal field more and more
intensely, until the rotation reaches a stationary state. This means we have to deliver to it, the
necessary energy, with regards to the given end.
So what we come to do, is really, to pump out of it, a given rest mass [just the way Eq.(1)
delineates]. Its overall relativistic energy, nonetheless, remains the same [cf. Eq.(9)].
It is, as if, we do work with regards to a refrigerator, in order to pump out of it heat. Thereby,
if we consider the cooled refrigerator, along with its heated up back side, the whole thing’s
energy remains the same.
This easily confirms our Eq.(9), thus as well as our root Eq.(8). In other terms, we can state
that Eq.(8) must indeed be written in the given way, so that Eq.(9) holds, and (as
controversial as it may seem at a first glance), the overall relativistic energy of the object
brought to a rotation remains the same.
Briefly speaking, what we are interested in, here really, is the rest mass pumped out of an
object, fixed at a given location of the disc, via rotation, until the disc attains a stationary
velocity, and this is expressed by Eq.(8). [What rotation effectively does, when it takes an
object from “zero velocity”, to a “stationary velocity”, according to the present approach,
again, is to dump out of it, a rest mass, as expressed in Eq.(8).]
Let us then calculate this energy around the velocity v  R , for just a unit quantum number
increase. For this, let us rewrite Eq.(7), for a finite increment B in the static binding energy:
R 2 
B  m00
.
(10)
2R 2
1 2
c
We will have to write   v / R , and will have to extract v from Eq.(6):
h
.
v 
2Rm 00
We insert this, in the previous equation:
h
h
R 2


2Rm 00
2
ΔBΔn1  m00

h
.
2
2R 2
2R 2
R 1 2
1 2
c
c
(11)
(12)
This is the radiation energy that will be emitted when the quantum number in the
accelerational field gets increased by unity, around the angular velocity  . It is of course
interesting that this fundamental energy turns out to be independent of the original mass of
the object, also independent of the radius at which the object is located. It just depends on  ,
which however naturally depends on the given quantities, within the framework of Eq.(4). In
any case, as will be elaborated on below, the same occurs, in the Bohr hydrogen atom, and the
corresponding energy, when the electron jumps from a relatively high level to the infinitely
close lower level, does not depend on anything else, but its angular velocity on the given
initial level.
4
Thereby, what we landed at here, is in full harmony with Bohr’s correspondence principle.†
Yet in contrast with the derivation of this latter principle, we made no approximation with
regards to the quantum state number, n, in deriving Eq.(12).
The frequency  of the radiation in question is

.

2
If   100 tours / s  2x100 radians / s , then
2x100

 100 Hz .
2
We can continue, from Eq.(10) on, via writing

ΔBΔn1  h n n ,
2
†
(13)
(14)
(15)
The magnitude of Bohr’s total energy En is as usual expressed as
2e 4 m
En  2 2 ;
(i)
n h
here m is the electron mass, and e is either proton’s or electron’s charge intensity. For a high n one can write,
for the magnitude of the energy of a photon that will be emitted from an upper level n+1, to the level n right
below,
E 
2 2 e 4 m  1
1  2 2 e 4 m  n 2  2n  n 2  4 2 e 4 m

 
 2 

.
2
h
(n  1) 2 
h 2 
n4
n 3h 2
n

Recall the following Bohr relationship between m and the given orbit radius r:
42e 4 mr  n 2 h 2 .
We can accordingly transform Eq.(ii):
4 2 e 4 m e 2 n 2 h 2 e 2 .
E  3 2  3 2 
rn
n h
rn h
Recall at the same time the Bohr velocity relationship, at the given level:
2r
2e2
vn 
 2rf 
,
T
nh
(ii)
(iii)
(iv)
(v)
where f is the inverse of the rotational period T, at the given level, and it can be written in terms of the
rotational angular velocity  , of the electron as
(vi)
  2f .
We can hence write
rf 
e2
.
nh
This together with Eq.(iv) leads to
e2
 .
E 
 hf  h
rn
2
(vii)
(viii)
This is the same relationship as Eq.(12) of the text (cqfd).
5
where now ΔBΔn1 represents the energy dumped out of the object for a delivery of it, in the
accelerational field, to a level such that Δn  1 , thus (accelerational-intensity-wise), deeper
than the level n+1, having started up from the level n, delineated by the angular velocity  n .
This latter quantity can be written from Eq.(4):
nh
.
n 
2m 00R 2
(16)
This leads to
ΔBΔn1 
nh
h
n .
2
2R m00 2
The total energy B dumped out, having started from ni=0, to land at nf is then,
nf
nh
h
n f2 h 2
B
dn

.
0 2R 2 m
8π 2 R 2 m00
00 2
(17)
(18)
Using Eq.(16), and denoting by  f the final angular velocity, we obtain
(f 2m00R 2) 2 1
2
B
 m00f R 2 ,
2 2
8π R m00
2
(19)
and this constitutes a cross check of our derivation for non-relativisitic rotation speeds [wich
we came to assume at the level of Eq.(12)].
TIME DEPENDENT ANALYSIS: RATE OF EMISSION AND SPECTRUM
So far, we have achieved a time-independent analysis, between the initial and final states of
the rotating object. And, our main finding is delineated by Eqs. (12), (13) and (14), i.e. when
the object is brought to a quantum level, deeper than its actual state, by just unity, it should
radiate as we assume, a photon of energy h(t) /(2 π) , where (t) is its angular velocity, at
time t.
In order to get advanced, we will have to assume, how (t) varies with respect time. The
easiest way is to assume a constant increase in the angular velocity, between the initial and
final states, i.e.
(20)
(t)  t ,
supposing that we start from (0)  0 ,  being a constant angular acceleration (which we
assumed).
Eq.(27), accordingly, becomes
2m 00tR 2
;
n(t) 
h
(21)
n(t) is the number of the quantum state the object is delivered to, at time t, at the given
location determined by the radius R.
6
This leads to the number of quantum states swept per unit period of time:
dn(t) 2m00R 2
.

dt
h
(22)
The energy dE(t)/dt, radiated through the given unit period of time, around time t, can be
derived from Eq.(17):
dE(t)
2π(t) m00R 2
h
 m00α 2 R 2 t (Joules/s) .
(23)
dt
2π
h
This is the result we land at, for the object of initial rest mass m00, located at R.
In fact the entire disc will radiate. Let  be the density of the disc material, and h the
thickness of the disc. Thereby for an infinitesimal initial rest mass dm of the disc, located
around the location r, the power dp(t) radiated at t, can, from the above equation, be written as
(24)
dp(t)  dmα2r 2 t  2rdrhα2r 2 t (Joules/s) .
The total power P(t) radiated by the disc extending to a radius R , at time t, is [using Eq.(20)],
then
R
1
(25a)
P(t)   2rdrhα 2 r 2 t  hαR 4(t) Joules/s .
0
2
The total energy K radiated in T seconds, through which the disc will attain a stationary
rotational velocity v(T), is therefore,
K
T
o
1
1
1
1
hα 2R 4 t  hα 2R 4T 2  Mα 2R 2T 2  M2(T)R2 (Joules) ,
2
4
4
4
(25b)
where we called M the total mass of the rotating disc.
What is interesting over here, is that the energy K happens to be, on the other hand, the kinetic
energy furnished to the disc in order to take it, to the given final velocity v(T). This means,
the energy furnished to the disc is indeed used to pump out, the given rest mass, from the disc.
It is like using work to pump out heat out of a refrigerator.
Now, we can determine the number of photons N(t) emitted by the rotating disc, per unit
period of time, around time t, via Eq.(12):
1
hαR 4(t)
2hαR 4
2
(t) 

 N photons/s ,
(26)
(t)
h
h
2
which comes to amount, interestingly, to a constant quantity, thus, for a given constant
angular acceleration.
This means, the energy of the emitted photons will get increased as the angular momentum
increases [cf. Eq.(12)], and it will go from 0 Hz up to, something like 300 Hz, but the rate at
which the photons will be emitted, will remain constant [and, is furnished by Eq.(26)].
7
Given that M was the total mass of the rotating disc, Eq.(26) can then be written as
R 2α
photons/s .
 M
h
(27)
Let us suppose the following data:   10 g / cm3 , h=2 mm, R=0.5 m,   50x 2 / s2 . Then
we can estimate N:
2hαR 4 10x103 x10x 2x103 x 6x50x 6x102
N

 6x1036 photons / s . (28)
34
h
6.62x10
Suppose we have a detector 1 m away this source. Then we will have a flux  of
  6x10 36 photons/(4πx10 4 cm 2)  2x10 31 photons / cm 2 .
(29)
This should well be detected. Recall that, sound-wise 300 Hz, corresponds to Note G, above
Note Middle C of Piano Forte. (Note that The Key Note A, has 440 Hz.)
Recall further that according to the present approach, we expect emission of radiation while
the rotor gets accelerated, and no radiation, when the rotor attains a stationary velocity.
DISCUSSION
The interesting findings we came to, in this article, are as follows:
o The energy delivered to bring an object to rotation, means we dump an equal amount of
rest mass from this object, in fact to deliver it, to the rotational acceleration field it is
bound to.
o The field is quantized, just like the atomistic electric field is. So is in effect, the
gravitational field created by a ponderable mass.
o The energy necessary to bring any object from a rotational field at the nth quantum state
determined by the angular velocity  n of it, to the next quantum state n+1, is h /(2 π) ; h is
the Planck Constant.
o This is well compatible with Bohr’s Correspondence Principle.
o As we start from rest, and bring, say a disc to rotation, along with a constant angular
acceleration  , it will keep on radiating photons of energy h ( t ) /(2 π) , where (t) is the
instantaneous angular velocity. The rate at which the disc will radiate, is   MR 2α / h
photons/s, for a constant angular acceleration. Here M is the mass of the disc, and R is its
radius.
o The emission will cease right after the disc attains a stationary rotation (with a constant
rotation speed).
o We have shown that the total energy radiated, amounts to the total kinetic energy delivered
to the disc.
o Thereby, while the rest mass of the object is decreased, its overall relativistic energy
remains the same as its original relativistic energy at rest.
o It is as if, we do work with regards to a refrigerator, in order to pump out of it, heat.
Consequently, if we consider the cooled refrigerator, along with its heated up back side,
the whole thing’s energy, naturally remains the same.
8
Now, we better say few words with regards to the process of stopping the rotating disc.
We can well stop the disc by bombarding it with photons (that had been previously emitted),
in the direction opposite to the direction of rotational motion; the absorption of these photons
by the disc material, will bring the disc (via momentum transfer to it), to a "cool stop",
meaning all energy furnished to the disc to stop it, goes to restore the original rest mass loss,
and there is no waste of energy besides, trough the stopping process.
This scheme well delineates a rigorous well-even-energy-balance, drawing energy-wise, no
problem whatsoever.
Evidently, we can stop the rotating disc, via friction, which involves a thermal process, and
we are inclined to call this process, a "hot stop". We can already guess that, because both the
disc and the break mechanism we will apply to it, to stop it, should necessarily get warmed
up, throughout; one can anticipate that the overall energy furnished this way, to stop the
rotating disc, is more than the energy necessary to bring it to a "cool stop".
Indeed some of the energy we furnish goes to heat up the disc, and the break mechanism.
Thus, only some of the energy we furnish, most likely, via a black body radiation, will be
devoted to restore the lost rest mass of the disc, originally brought to rotation.
We can guess then that, to stop the disc by such a process, effectively requires more energy
than the previous process of getting it, to a "cool stop".
On the other hand, the efficiency of the friction energy to stop the disc, seems to constitute a
fundamental thermodynamical problem, and we can guess that, only a portion of it could be
converted into negative work, to sop the disc, and restore its earlier lost rest mass.
The likelihood is that, we can tap at most a Carnot ratio of the heat energy delivered to the
disc, which is thereby converted into negative work to stop the rotating disc.
In short, we have no problem, in drawing a thorough energy balance, either when we bring
the disc into rotation, nor when we stop it.
Through the first process the energy we furnish to it, serves to pump rest mass out of it. And
through the second process, the energy we furnish, to bring it to a cool stop, as we just
defined, serves to restore the originally dumped rest mass.
**
Next we have to discuss the compatibility of our approach with Classical Electrodynamics
(CED). Classically we are taught that a neutral body would not radiate.
Yet, the fact is that, no one ever could accelerate in an appropriate way a neutral particle, say
a hydrogen atom. No one furthermore, in the first place seems to have thought about
accelerating a neutral particle, and testing a possible radiation, that may result out of it.
9
In effect a neutral particle can only be accelerated in a gravitational field, or in a centrifugal
field.
Thus one can estimate the wavelength of a photon coming out of a falling stone, right above
Earth, and this comes to be approximately 2x109 km, which amounts to a distance more than
ten times the distance from Earth to the Sun, and it is quite understandable that no one could
even come close to measure such a radiation.
But either the falling stone, or an object delivered to a centrifugal field, is to lose rest mass, as
we have shown throughout, and the rest mass deficiency coming into play should most likely
be discharged, via radiation.
How?
One way of conceiving such a process is the following: The nuclei and the electrons, in
atoms, thus electric charges in an object, getting bound more intensely, to a gravitational field,
are supposed to move, though just a little bit, away from each other, for the atoms in a falling
stone stretch, and this as much as the binding energy difference of the gravitational field,
coming into play.
Briefly, charges move from each other just a bit, while getting bound to an attractional field,
meaning in their own electric accelerational field, and accordingly radiate, which is then fully
understandable within the frame of the CED.
Recall that even Bohr atom model can tell us how much stretching, and accordingly radiation,
will come into play, if we consider straight a hydrogen atom: We have with the usual
notation, 42e 4 mr  n 2 h 2 [cf. Eq.(iii) in the foot note above, about the Bohr Correspondence
Principle ]. Here e is the charge intensity of either proton or electron, m is the electron mass,
and r is the Bohr atom radius.
Therefore when m decreases by , say, by exp(  α) [cf. Eqs. (1) and (2)] in a free fall, in a
gravitational field, then r will increase by the same factor. This will definitely cause a
radiation, responsible of the rest mass deficiency, via the displacement of the proton and
electron from each other, in fact corresponding to an energy discarge. (Note that their
atomistic quantum levels are not altered. The proton and the electron due to gravitational
binding, each though, must become less energetic, which makes that their fundamental level
orbit radius, is stretched.
As to a particle in rotation... Again, in the centrifugal field, there must be rest mass decrease,
as discussed throughout, and stretching of the sizes of atoms, in the same way (we just
explained). Thus radiation.
10
REFERENCES
1
T. Yarman, Foundations of Physics Letters, Volume 19, December, 2006.
2
T. Yarman, Superluminal Interaction Making the Basis of Quantum Mechanics. Academic
Publication, Lambert, 2011.
3
T. Yarman, M. Arik, A. L. Kholmetskii, A Critical Analysis Of Einstein's Non-Conform
Analogy Between Rotation And Gravitatiton - Part II: Harmony Between Non-Locality And
STR In Both Micro And Macro Worlds, Mathematics, Physics and Philosophy, in the
Interpretations of Relativity Theory II, 4 – 6 September 2009, Budapest.
4
T. Yarman T, V. B. Rozanov, M. Arik, 2007 Proc. Int. Meeting on Physical Interpretation of
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