A Jay Sep Emc 2

8/8/2019 A Jay Sep Emc 2

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Incomplete derivation of L=mc 2 (or speculation of E=mc 2) , its

critical analysis; and applications of Generalized form E =Ac 2m.

or

Some hidden aspects of Einsteins Sep. 1905 paper

Ajay Sharma

Fundamental Physical Society. His Mercy Enclave Post Box 107 GPO Shimla

171001 HP India

PACS 03.30.+p, 04.20.-q, 24.10.Jv, 98.54.Aj

Email [email protected] , www.AjayOnLine.us

Abstract

Einsteins Sep. 1905 paper [1] in which L=mc 2 (light energy mass

equation) is derived, is not completely studied; and is only valid under

handpicked or super--special conditions of involved parameters. The origin of

E=mc 2 from L=mc 2 is completely speculative in nature. This derivation

(under general conditions) contradicts the law of conservation of

matter/energy. In simple words it implies that when light energy is emitted

then its mass must increase. The decrease in mass can be equal to L/c 2 or

less or more than L/c 2.Thus self contradictions also exist in Einsteins

derivation. The factor c 2 has been arbitrarily brought in picture by Einstein.

Einstein applied binomial theorem in the end of the derivation then equation is

L=mc 2 is obtained. If the same derivation is applied in the beginning, then

result is m b =m a i.e. no equation is obtained which involves c 2 or L. Thus

conversion factor is arbitrarily brought in the picture. These are the limitations

of the derivation.

Apparently Einstein may had been aware of limitations and

complexities of his derivation; hence he only took super special values of

parameters . Einstein derived L=mc 2 under special conditions and

speculated from it E=mc 2 without mathematical derivation.

The same derivation also gives L mc 2 or L =Amc 2, where A is coefficient of proportionality. Thus Generalized Mass Energy

inter conversion equation is derived in other way as E =Ac 2m, where A is

1
mailto:[email protected]:[email protected]


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coefficient of proportionality. There are numerous values of coefficients of

proportionality in the existing physics.

According to E=mc 2 in burning of wood if

1/1000,000,000 kg (10 -9 kg) matter vanishes then energy emitted ( 910 7kg )

can derive a truck of mass 1,000kg to distance of 90 km. It is not confirmed,

yet.

It is believed that E=mc 2 is justified in Hiroshima and

Nagasaki atom bomb explosions. But according to valid and authentic reports

released in 1965 (published in book 1992) , in such cases 98 % energy is

unaccounted for. So in such cases E=mc 2 is not confirmed quantitatively.

How universe is formed? How big bang took

place? Why universe is expanding? These fundamental questions are not

explained by E=mc 2. An attempt has been made to explain such

unanswered questions with help of mathematical model based upon E

=Ac 2m. The universe has started its life (before Big Bang) with zeroans, this

perception is consistent with NASAs Wilkinson Microwave Anisotropy Probe

(WMAP) data, interpreted by Erickcek.

The conceptual and mathematical limitations of

Einsteins derivation are also reflected in school level text book. If it is use to

explain the binding energy, then it contradicts the universal equality of masses

of nucleons. Experimentally observed binding energy of deuteron is

2.2244MeV or 0.002388 u. Thus mass of proton and neutron inside nucleus

should decrease about 0.11854% . It is a clear contradiction, as masses of

nuclei are universally equal. However it can be explained with help of E

=Ac 2m as according to it the binding energy equal to 2.2244MeV can be

obtained from mass defect 2.38810 -14 u and in this case value of A is 10 10 .

Thus for small mass large amount of energy is emitted.

Similarly if we critically analyse Einsteins 1907 derivation of rest

mass energy then it implies that E rme =Mrest c 2 follows from the non-existent

equation. For example Einstein started with equation that body possesseskinetic energy, then condition of relativistic kinetic energy is used. But to

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finally obtain E rme =Mrest c 2. Einstein applied condition to penultimate equation

that body is at rest which means first equation is zero and other equations are

non-existent. Thus Einstein obtained E rme =Mrest c2 from non-existent equation,

which is not justified. It is just like situation that first floor of the building is

demolished and other ten floors are floating in air.

Expected Solar storm (100 million hydrogen bomb strength) of 2012 or 2013

can be associated with E =Ac 2m with high value of A. It implies

conversion of mass to energy on the surface of sun is not at constant rate,

hence variable value of A is justified. During such activities for smaller mass

more energy is being emitted.

Apart from this E =Ac 2m is useful in explaining many

other phenomena. Thus conceptual and mathematical basis of E=mc 2 is

extended. Hence E =Ac 2m is general equation and E=mc 2 is its special

case.

The basic mathematical and conceptual difference between

E =Ac 2m and E=mc 2 is that according to former the energy emitted can

be less or more than later. The theme of both the equations is that mass is

converted to energy or energy is converted to mass.

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1.0 Einsteins approach in Sep. 1905 paper [1] in whichL =mc 2 was derived and E =mc 2 was speculated?

In Einsteins derivation basic equation is

2

2

1

cos1

c

v

c

v

=

(1)

where is light energy emitted by body in frame (x,y,z) and* is light energy measuredin system (, , ), and v is velocity with which the frame or system (, , ) is moving.

Einsteins perception: Let a system of plane waves of light, referred to the system of coordinates (x, y, z), possess the energy l; let the direction of the ray (the wave-normal)

makes an angle with the axis of x of the system. If we introduce a new system of co-ordinates (, , ) moving in uniform paralleltranslation with respect to the system (x, y, z), and having its origin of coordinates in motionalong the axis of x with the velocity v.

Thus v is the relative velocity between system (x, y, z) and system (, , ) . The bodywhich emits energy is considered stationary in the system (x,y,z) and also remains stationaryafter emission of energy in the system (xy,z).

Let0 E

and0 H

are energies in coordinate system (x, y, z) and system (, , ) beforeemission of light energy, further 1 E and 1 H are the energies of body in the both systems

after it emits light energy.

Table I. Energies emitted before and after emission by body in EinsteinsSep. 1905 derivation.

Sr No System (x,y,z) System (, , )

1

2

Before Emission E 0

After Emission E 1

Before Emission H 0

After Emission H 1

Thus Einstein wrote various equations as Energy of body in system ( x,y,z )

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Energy before Emission = Energy after emission +0.5L + 0.5L

0 E = 1 E + L5.0 + L5.0 = 1 E + L (2)

Energy of body in system (, , )

0 H = 1 H + 0.5 L {(1 coscv

)+ (1+ ) coscv

} (3)

where2

2

1

1

c

v

= (4)

0 H = 1 H + L (5)

Or ( 0 H 0 E ) ( 1 H 1 E ) = L [ 1] (6)Einstein calculated as

( 0 H 0 E ) = 0 K +C =2

2vM b + C

( 1 H 1 E ) = K 1 +C =2

2vM a + C

Kinetic energy of body before emission of light energy,0 K (2

2vM b ) and kinetic energy

of body after emission of light energy, K 1 (2

2vM a ) .

0 K K 1= L {2

2

1

1

c

v 1} (7)

Einstein considered the velocity in classical region thus applying binomial theorem,

0 K K 1= L

+++ 1.....................

83

21

4

4

2

2

cv

cv (8)

Further Einstein quoted [1]

Neglecting magnitudes of fourth ( v4/c4) and higher( v6/c6, v8/c8 .) orders, we may place.

0 K K 1= L 22

2 cv (9)

2

2vM b

2

2vM a = L 2

2

2 c

v (10)

or L = ( ab M M ) 2c = 2mc (11)or Mass of body after emission (aM ) = Mass of body before emission (bM ) 2

c L . (12)

Then Einstein generalized the result for every energy and called mass of body is measure of

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0 K K = L

+++ 1.....................

83

21

4

4

2

2

cv

cv

(8) Now consider the same case when velocity is 10m/s or 36km/hr , then eq. (8) . Under thisconditions eq.(8) becomes

2

2vM b

2

2vM a = L [1+5.5510 -16 + 4.62910-31+ ..-1] (21)

The rest masses and relativistic masses are equal as in eq.(8) if the term5.5510-16 isneglected. Thus neglecting second and third terms in eq.(21) , we get

M b Ma = 0 ( light energy is continuously emitted)Mass of body before emission = Mass of body after emission (22)Thus above deduction is not justified as energy is emitted out of NOTHING, which isviolation of Law of Conservation of Energy.

By using eq.(1) in classical region and retaining terms up to second order of v2/c2 [this termis neglected when Mmotion= Mrest ] . Einstein has simply brought term c2 in the equations.

2.0Einstein took only SUPER SPECIAL values of variables and its effects.The following arguments can be given that Einsteins derivation is true under SUPER SPECIAL or HANDPICKED conditions1. Einstein [1] has put condition on state of the body: Let there be astationary body in thesystem ( x, y, z ), and let its energy--referred to the system ( x, y, z ) be E0. Let the energy of the body relative to the system (, , ) moving as above with the velocityv, be H0. The bodyalso remains stationary in system (x, y, z) after emission of energy.

Einstein also assumed that the body also remains stationary after emission of light

energy. But practically this condition (Light emitting body is stationary) is NOT obeyed in manycases.(i) The nuclear fission is caused by the thermal neutrons which has velocity 2,185m/s. Theuranium atom also moves as it is split up in barium and krypton, and EMIT energy.(ii) When a gamma ray photon of energy at least 1.02MeV, moves near the field of nucleus it

is split up in electron and positron pair [2]. The gamma ray photon is in motion

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and so is the state of electron and positron pair.(iii) Similarly the particle and antiparticle moves towards each other for annihilation. The particle and antiparticle collide then annihilation takes place. In nuclear fusion the atoms areset in motion.THESE PHENOMENA WERE NOT DISCOVERED IN EINSTEINS TIME.(iii) When a paper burns then it is also sets in motion and energy in various forms is emitted.Chemical reactions were discovered in Einsteins time.EINSTEIN NEVER DISCUSSED THIS PHENOMENON IN HIS WORKS.So Einsteins condition that body is STATIONARY, emits light energy and its massdecreases, is not justified.Other conditions on Einsteins derivation.

Einsteins Sep. 1905 derivation [1] of L = mc2

is true under SUPER SPECIALCONDITIONS OR HANDPICKED CONDITIONS only. It is justified below. In the

derivation of L = mc2 there are FOUR variables e.g.(a) Number of waves emitted,(b) l magnitude of light energy,(c) Angle at which light energy is emitted and(d) Uniform velocity (relative velocity), v

Nature of v

According to Einstein: v is the relative velocity between system (x, y, z) and system (, , ). If system (x,y,z) is at rest and system ( , , ) moves with velocity v, then v is relativevelocity. If the system (x,y,z) and system (, , ) both move with same velocity then relativevelocity v is zero. Further Einstein strictly took the value of velocity as UNIFORM.

The law of inter-conversion of mass and energy holds good if (i) Velocity v is in classical region(ii) Velocity v is in relativistic region(iii) Velocity v is zero(iv) Velocity v is variable or uniformThese variables have numerous values. The law of inter conversion of mass and energy holdsgood under all conditions, but Einstein has considered just one i.e. velocity is in classicalregion.

Einstein has taken SUPER SPECIAL or HANDPICKED VALUES of parameters. Thus for complete analysis the derivation can be repeated with all possible valuesof parameters. In all cases the law of conservation of momentum is obeyed (which isdiscussed in next sub-section).(A) The body can emit large number of light waves but Einstein has taken onlyTWO light

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waves emitted by luminous body.

Why one or n light energy waves are neglected ?(B) The energy of two emitted light waves may have different magnitudebut Einstein hastaken two light waves of EQUAL magnitudes ( 0.5L each ).

Why other magnitudes ( 0.5001L and 0.4999L) are neglected by Einstein ?(C) Body may emit large number of light waves of different magnitudes of energy makingDIFFERENT ANGLES(other than 0 and 180 as assumed by Einstein).Why other angles ( such as 0 and 180.05, 0.9999 and 180 etc.) are neglected by Einstein ?Thus body needs to be specially fabricated, only then it can emit energy under theseconditions. Further body should emit light energy only, not other forms of energy.(D) Einstein has taken velocity in classical region (v


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heat, sound, LIGHT ENERGIES and energy in form of invisible radiations simultaneously,along with invisible radiations. For proper description of heat energy-mass inter-conversionwe need equation equivalent to eq.(1). Similar is the case of other energies.(E) Einstein has only considered Light Energy and other energies are neglected to deriveL =

mc 2. Even while light energy is emitted heat energy is also emitted.

Further Einstein has considered that body sends plane light waves. But energycan also be emitted in the invisible region and Einstein did not mention at all about HEATand SOUND ENERGIES (emitted along with light energy). Thus energies other than lightenergy are also emitted but NEGLECTED by Einstein in the derivation. So energies are NOTtaken in account completely.

Table II The values of various parameters considered by Einstein andneglected by Einstein in the derivation of Light Energy Mass equation L

= mc 2.

Sr No

Parameters Einstein considered Einstein neglected(No reason was given by Einstein why parameters are neglected).

1 No. of lightwaves

Two Light Waves One, three, four or n waves

2Energy of light wave

Equal0.5L and 0.5L each

Energies of the order of 0.50001L and0.49999L are also possible. There are numeroussuch possibilities, which need to be probed.Bodies can emit more than two waves.

3 Angle 0 and 180 The angles can be 0 and 180.05 or 0.9999 and

180 are also possible. There are numerous such possibilities which need to be probed.

4 Velocity Classical regionThe velocity can be in relativistic region.The velocity v can also be zero i.e. v = 0

5Velocity Uniform

In classical regionThe law of inter-conversion of mass to energyalso holds good, when velocity is variable.

Deductions: Einstein has taken only super-special values of parameters, and neglectedmany realistic values.

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3.0 Effect of general values of variables

L mc2

or L =A mc2

and contradiction of Law of Conservation of Matter

3.1 Einsteins derivation predicts that L mc 2 or L =A mc 2.Thus energy emitted can be different from L = mc 2.

Einstein has considered a body emitting two light waves of energy 0.5L each just in oppositedirections. It is equally possible that body may emit two light waves on energy 0.50001L and0.49999L. Then equations equivalent to Einsteins equations can be written as [11-24]

Ho = H1 + 0.500001 L[( 1 v/c cos0)] + 0.499999 L [1 v/c cos 180)] (23)Ho = H1 + 0.500001 L[( 1 v/c)] + 0.499999 L [1+ v/c cos )] (24)Ho = H1 + 0.500001 L - 0.500001 L v/c + 0.499999 L + 0.499999 L v/cHo = H1 + L 0.000002 L v/c (25)Eo = E1 + L (2)(Ho Eo) = (H1 E1) + L 0.000002 L v/c L (26)

(Ho Eo) (H1 E1) = L 0.000002 L v/c L (27)

= L ( 1 0.000002 v/c) L (28)= L [ ( 1 0.000002 v/c) 1 ] (29)= L [ ( 1+ v2/2c2 )( 1 0.000002 v/c) 1 ]= L[ 1 0.000002 v/c + v2/2c2 1 ] (30)

K b K a = L [ 1 0.000002 v/c + v2/2c2 1 ] (31)K b K a = L [ 0.000002 v/c + v2/2c2 ] (32)M bv2/2 Mav2 /2 = L [ 0.000002 v/c + v2/2c2 ] (33)M b Ma = L [ 0.000004/cv + 1/c2] (34)mc2 = L[ 0.000004c/v + 1] (35)

L= mc2 / [ 0.000004c/v + 1] (36) 0.000004c/v = 0.0000043108 / 10 = 1200 (37)

L= mc2 / [1200 + 1] = mc2 / 1199 ( 38)L mc2 or L =A mc2 where A is coefficient of proportionality (many in existing physics). Thus when general

conditions are applied then Einsteins Sep. 1905 derivation itself PUTS FORTH theGENERALIZED FORM which is L =A mc2 [11-24]. Thus conversion factor between massand energy is NOT always c2, in Einsteins derivation. Hence the value of conversion factor other than c2 is clearly also supported from Einsteins derivation, if properly analysed.3.2 Einsteins Sep. 1905 derivation predicts

Increase in mass of luminous body when Light Energy is emitted.

We have eq.(34) as

M b Ma = L [ 0.000004/cv + 1/c2

] = 0.000004L/cv + L/c2

(34)Ma = M b +0.000004L/cv L/c2 (34)0.000004L/cv = 4L/106 310810 = 4L / 31015 = 133.3L/ 1017 (39)

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L/c2 = L/91016 = 1.111L/ 91016 = 1.1111017 (40)

0.000004L/cv L/c2 =133.3L/ 1017 1.111 x1017 IS POSITIVE QUANTITY

Ma = M b +0.000004L/cv L/c2 = M b + POSITIVE QUANTITY (41)

Mass increases when light energy is emitted.It is contradiction of Law of Conservation of energy. How does double increase happen?(i) How energy is emitted?(ii) How mass increases?HOW DOES DOUBLE INCREASE HAPPEN SIMULTANEOUSLY ? No experimentcan justify this.

Now %age difference (in energy) between Einsteins special derivation and Generalizedderivation can be calculated.%age increase = {L/c2 [ 0.000004/cv + L/c2]}100/ L/c2

= 1.2104 (42)

3.3 If the direction (angles) of waves is exchanged then results from Einsteins derivationare ENTIRELY DIFFERENT.

In previous case it is regarded as waves of energy 0.50001L makes angle 0 with x-axis , and wave of energy 0.49999L makes an angle 180 . In this case result iswhen body emits light energy its mass increases But if angles of the waves are interchanged then results are entirely different. It is justified below.

Ho = H1 + 0.499999 L[( 1 v/c cos0)] + 0.500001 L [1 v/c cos 180)] (43)Ho = H1 + 0.499999 L[( 1 v/c)] + 0.500001 L [1+ v/c cos )] (44)Ho = H1 + 0.499999 L 0.499999 L v/c + 0.500001 L + 0.500001 L v/cHo = H1 + L + 0.000002 L v/c (45)Eo = E1 + L (46)(Ho Eo) = (H1 E1) + L + 0.000002 L v/c L ( 47)

(Ho Eo) (H1 E1) = L + 0.000002 L v/c L (48)= L ( 1 + 0.000002 v/c) L= L [ ( 1 + 0.00002 v/c) 1 ]= L [ ( 1+ v2/2c2 )( 1 +0.000002 v/c) 1 ]= L[ 1 + 0.000002 v/c + v2/2c2 1 ] (49)

K b K a = L [ 1 + 0.000002 v/c + v2/2c2 1 ] (50)K b K a = L [ 0.000002 v/c + v2/2c2 ]M bv2/2 Mav2 /2 = L [ 0.000002 v/c + v2/2c2 ] (51)M b Ma = L [ 0.000004/cv + 1/c2 ] (52)mc2 = L[ 0.000004c/v + 1] (53)L = mc2 / [0.000004c/v + 1] (54)L= mc2 / [0.000004c/v + 1] = mc2 / [1200+1] = mc2 / [1201]

IN THIS CASE MASS DECREASES MORE THAN L/c2SO EINSTEINS DERIVATION DOES NOT GIVE FIXED VALUE OF ENERGYcorresponding to mass annihilated.

L mc2 or L =A mc2 (55)

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Where A is coefficient of proportionality.

Decrease in mass is more than L/c 2 M b Ma = L [0.00004/cv + 1/c2] (56)Ma = M b 0.000004L/cv L/c2 (57)Thus Einsteins derivation gives self contradictory results. L mc2 is re-justified. Henceit must be re-derived by new method [11-24].

3.4 The equation

2

2

1

cos1

c

v

c

v

=

(1)

is NOT meant for(i) sound energy,In Doppler effect change in frequency of sound is estimated, not variation in mass. Thuseq.(1) is not used. Likewise eq.(1) is not associated with any other energy. The speed of sound is 332m/s and 3108 m/s. Einsteins derivation is in classical region of velocity.(ii) heat energy,There is no equation like eq.(1) which relates variation of heat energy. The similar is thecase of other types of energies.(iii) chemical energy,(iv) nuclear energy, (v) magnetic energy, (vi) electrical energy,(vii) energy emitted in form of invisible radiations,(viii) attractive binding energy of nucleus(xi) energy emitted in cosmological and astrophysical phenomena(x) energy emitted in volcanic reactions(xi) energies co-existing in various forms etc. etc.

Then why results based upon eq.(1) are applied to above energies (i xi).The reason is that all energies have different type of nature, and the energies are not

confirmed to obey the same equation.Einstein initially derived light energymass inter-conversion equation L =mc2, thenspeculated every energy mass equation E =mc2 from L =mc2. As eq. ( 1) is onlymeant for light energy, not for other energies. Hence speculative transition to E =mc2

from L =mc2 is absolutely without any mathematical basis.

3.5 If the measuring system is at rest (v=0) and body emits two light waves as inEinsteins derivation then derivation is not applicable. v can also zero if system (x,y,z)and system (, , )move with same velocity.However in this case EXPERIMENTALLY when light energy is emitted mass decreases.It is serious limitation of Einsteins derivation.

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When the measuring system (, , ) is at rest v = 0 then * = (58)

Ho = H1 +L/2 +L/2 (59)Eo = E1 + L (2)(Ho Eo) (H1 E1) = 0

As body is at rest then (Ho Eo) or (H1 E1) cannot be interpreted as kinetic energy.Hence further derivation is not APPLICABLE.It is serious limitation of Einsteins derivation, which is regarded as general in nature.

3.6 If Binomial Theorem is applied in the beginning, then L =mc2 is not obtained. It is onlyobtained if Binomial Theorem is applied only in the end of derivation. Thus this derivation isnot general. If velocity of body is such that v 2/c2


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v=1m/s or 3.6 km/hr 0.0000 00015L/cv = 15/108 3108 = 5/1017L/c2 = L / 91016 =1.11L / 1017 (66)5/1017 >1.11L / 1017 (67)Mass after emission = Mass before emission +positive quantity (68)

It is contradiction of law of Conservation of Energy/Mass.(i) How energy is emitted?(ii) How mass increases?HOW DOES DOUBLE INCREASE HAPPEN? No experiment can justify this predictionfrom Einsteins Sep. 1905 derivation.It is the most serious limitation.

%age increase in energy = {L/c2 (0.0000 00015L/cv +L/c2}100/L/c2

153108100/109 = 450 (69)

If angles are 0 and 179.98

H0 =H1 + 0.5 L [ (1- v/c cos0 ) + (1-v/c cos 179.98 )] (62a) H0 =H1 + 0.5 L[1-v/c + 1+0.9999 99939v/c]

H0 =H1 + 0.5 L[2-0.0000 0006v/c] (63a)E0 =E1+L(Ho Eo) (H1 E1) = 0.5L[2-0.000 0006] L

K 0 K 1 = L[(1-0.0000 0003v/c) 1]=L[(1+v2/2c2)(1-0.0000 0003v/c)-1] (64a)

M bv2/2 Mav2 /2 =L[0.0000 0003v/c + v2/2c2]

M b Ma = 0.0000 0006L/cv +L/c2

Ma = M b + 0.0000 0006L/cv L/c2 (65a)0.0000 0006L/cv = 6/ 108 x3x108 = 20 /1017L/c2 = 1.11/1017Mass after emission = Mass before emission + positive quantity. It is contradiction of law of conservation of matter.

If angles considered are

Einsteins angle = 180 Now angle 180.001 H0 =H1 + 0.5 L [ (1- v/c cos0 ) + (1-v/c cos 180.001 )] (62b)

H0 =H1 + 0.5 L[1-v/c + 1+0.9999 99999v/c]H0 =H1 + 0.5 L[2-0.0000 00001v/c] (63b)E0 =E1+L(Ho Eo) (H1 E1) = L[2-0.0000 000005v/c] L

K 0 K 1 = L[(1-0.0000 000005v/c) 1]=L[(1+v2/2c2)(1-0.0000 00005v/c)-1] (64b)

M bv2/2 Mav2 /2 =L[0.0000 00005v/c + v2/2c2]M b Ma = 0.0000 00001L/cv +L/c2Ma = M b + 0.0000 00001L/cv L/c2 (65b)

0.0000 00001L/cv = L /cv = L / 109 x3x108 = L /3x1017 = 0.333/1017L/c2 = 1.11/1017

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Ma = M b + 0.0000 00001L/cv L/c2 Ma = M b + 0.333/1017 1.11/1017

Thus decrease in mass is less than L/c2.

%age difference = {L/c2 (0.0000 00001L/cv +L/c2}100/L/c2

= 0.0000 000013108 100 = 30

3.8 Relativistic Velocity: In the derivation Einstein has done calculations under classicalconditions of velocity (v


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H0 =H1 + L ( 1 0.000017453v/c) (70)E0 = E1 + L (2)(Ho Eo) (H1 E1) = L ( 1 0.000017453v/c) L

= L [( 1 0.000017453v/c)-1}= L [ (1+v2/2c2) ( 10.000017453v/c) 1] (71)

M bv2/2 Mav2 /2 = 0.000017453Lv/c +Lv2/2c2

M b Ma = 0.000034906L/cv + L/c2

Ma = M b + 0.000034906L/cv L/c2 (72)

0.000035L/vc = 35L/1063108 = 11.66L /1014 =1.166L / 1013

L/c2 = L / 91016 = 1.11L /1017

Mass after Emission = Mass before emission+0.000034906L/cv L/c2

Mass after Emission = Mass before emission+ positive quantityThus again when light energy is emitted mass INCREASES, which is contradiction of LawConservation of Matter.When two waves are emitted identically as in Einsteins case. Only difference is that in

this case one angle is regarded as 0.9999 instead of 0. The other angle remains 180.

Einsteins angle = 0 Angle considered = 0.9999 H0 =H1 + 0.5 L [ (1- v/c cos0.9999 ) + (1-v/c cos 180 )] (73) H0 =H1 + 0.5 L [ (1- v/c 0.99984773 ) + (1+v/c )]

H0 =H1 + 0.5 L 0.499923865 L v/c + 0.5 L + 0.5 L v/c (74)H0 =H1 + L +0.000076135 L v/cE0 = E1 + L(Ho Eo) (H1 E1) = L +0.000076135 L v/c L

= L [(1+0.000076135 v/c ) 1]= L [(1+v2/2c2)(1+0.000076135 v/c ) 1]= L [1+0.000076135 v/c + v2/2c2 1]= L [0.000076135 v/c + v2/2c2] (75)

K 0 K 1 = L [0.000076135 v/c + v2/2c2]M bv2/2 Mav2 /2 = L [0.000076135 v/c + v2/2c2]

M b Ma = [0.000076135 L/cv + L/c2] (76)Ma = M b 0.000076135 L/cv L/c2

Thus decrease in mass is more than predicted by Einsteins derivation.

Conclusion : Einsteins Sep. 1905 derivation NOT ONLY implies L mc2 or L =A

mc2 (E mc2 or E =A mc2), but also contradicts the Law of Conservation of Matter [11-24].Hence this observation has serious limitations.

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If Einsteins derivation is completely analysed then it also predicts(i) Mass of body decreases less than L/c2

(ii) Mass of body decreases more than L/c2

Then these are examples of self contradictions.

(iii) Light energy is emitted mass remains the same.HOW ENERGY IS EMITTED MASS REMAINS THE SAME ?

(iv) Light energy is emitted MASS INCREASES.How light and mass are emitted out of nothing.?

Complete mathematical interpretations

L mc2 or L =A mc2 Where A is coefficient of proportionality.

3.10 Binomial Theorem is applied in the beginning :Einstein has applied Binomial Theorem at the end as in eqs.(7-9) to obtain the equation (11)i.e. L = ( ab M M ) 2c = 2mc (11)

In Binomial Theorem is applied in the end the eq.(11) is not obtained.

Beginning (v/c


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Table III Comparison of RESULTS under Einsteins SUPER SPECIALCONDITIONS or HANDPICKED CONDITIONS and GENERAL CONDITION inSep. 1905 derivation. VELOCITY: Classical Region : Not applicable if v=0 and v is in relativistic region

Sr.

No

Energyof firstwave

Energy of secondwave

Angleof firstwave

Angleof secondwave

Mass after emissionMa

Mass after emission(ma)

%agedifference

1 0.5L 0.5L 0 180 Ma = M b L/c2 Decreases NA2 0.50001L 0.49999L 0 180 Ma = M b+ 0.00004L/cv

L/c2Increases 1.2106

3 0.5L 0.5L 0 180.01 Ma = M b +0.000000015L/cv L/c2

Increases 450

4 0.5L 0.5L 0 179.98 Ma = M b 0.00000006L/cv +L/c2

decreases< L/c2

1800

5 0.499999L

0.500001L 0 180 Ma = M b 0.000004L/cv L/c2

Decreases> L/c2

1.2106

8 0.500001L

0.499999L 0 180 Ma = M b+0.000004L/cv L/c2

Decreases< L/c2

1.2105

9 0.5L 0.5L 0 180.001 Ma = M b +0.000000001L/cv -L/c2

Decreases< L/c2

0.3

10 0.500001L

0.499999L 0 180 Ma = M b(v =1200m/s )

No increase No decrease

NA

11 L NA 90 NA Ma = M b L/c2 Decreases NA

12 L NA 89.99 NA Ma = M b +0.000034906L/cvL/c2

Increases 1.04106

Deductions: Law of conservation of energy is violated e.g. Einsteins derivation implies L mc2, energy may be emitted more than L/c2, when light energy is emitted mass increases.

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4.0 If body recoils when waves of different light energy are emitted.

The main equation in Einsteins derivation is

2

2

1

cos1

c

v

c

v

=

(1)

where v is relative velocity between light emitting body in system (x,y,z) and measuringsystemi.e. system (, , ).

(i) If light emitting body is at rest then relative velocity of measuring system is only v.

(ii) If light emitting body recoils away from the body VR , then relative velocity will bev + VR

(iii) If light emitting body is drifted towards the measuring system with velocity VR thenrelative velocity will be

v VR But in this case the recoil velocity VR is of the order of 510 32m/s and with velocity bodycan travel a distance of 1.5710-23 m in 10 years ( which is undetectable). Thus in this casebody is regarded as at rest.

Thus recoil velocity has no effect on the derivation.This equation is similar to

Mmotion=2

2

1c

v

M rest

(15)

as far velocity of measuring system is concerned. Let body is placed in reference frame(x,y,z) and its mass is measured in other frame (X,Y,Z) which is moving with velocity v.If the body moves away from the measuring frame , even then its mass changes. Thus in thisregard eq.(1) and eq.(15) has similar status.

How do mathematical equations vary when V R is taken in account ?Here main difference is that velocity is taken as ( v VR ) say (v + VR )

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2

2][1

cos1

c

V v

c

V v

R

R

+

+=

(77)

Now this equation can be applied to study variation in mass and light energy emitted as incase of eq.(1).

4.1 Conservation of momentum in general cases helps in calculations of recoil velocity.This case is similar to recoil of the gun but here two waves are emitted.The momentum is conserved irrespective of the fact that body remains at rest or recoils after emission of light energy [25].In case of Einsteins derivation momentum is confirmed in special and general cases.(i) When light emitting body remains at rest after emission. In this case recoil velocity iszero i.e. VR =0, it is calculated by applying law conservation of momentum.(ii)When light emitting body recoils due to emission of two waves in different directions .In this case velocity of recoil is NON-ZERO and calculated by applying law of conservationof momentum. Einstein did not consider that case.

The law of conservation of momentum can be used to calculate the velocity of recoilin this case also. Let the body of mass 1 kg emits in two waves in visible region of wavelength 5000A , it corresponds to 2hc/ or 7.951210-19 J , and the energy is divided intwo waves.Let body emits light energy (towards the observer, = 0) 0.50001L i.e.E1 = 3.9756810-19 J (78)and momentum p1 = E1/c = 1.3252210-27 kg m/s (79)Secondly, the body emits light wave of energy (away from the observer, = 180) 0.49999Li.e.E2 =3.9752210-19 J (80)

momentum p2= E2/c = 1.3251710-27

kg m/s. (81)Let us assume that when the body emits light waves of energy andrecoils (if it actually does) with velocity Vr .Initial momentum of waves + initial momentum of luminous body = 0 +0 (82)Final momentum of waves + final momentum of body = p1 +p2 +MV b (83)One wave (having momentum p2moves towards ) the direction in which body recoils andother wave moves in the opposite direction.Then according to law of conservation of momentum we get

0 = p1 +p2 +M bVr (84)

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VR = ( p1 +p2) /M b = (1.32522+1.32517) 10-27 =0.0000510-27

= 510-32 m/s (85)The velocity 510-32 m/s means body remains at rest.

A body is said to at rest if it does not change its position w.r.t. to surroundings.

It is analogous to observation that a CAR cannot move when head light is switched on.

Thus conservation of momentum requires that body should move with velocity 510-32 m/stowards the observer. With this velocity the body will recoil for distance equal toS(10 years) = 510-32 m/s 3.14107 10 = 1.5710-23 m (86)which is undetectable by all means hence conceptually body can be regarded as at rest.Thus body will tend to move with velocity 510-32 m/s( towards the observer) which isimmeasurably small by all means, hence the body remains at rest.This recoil velocity (VR ) i.e. 510-32 m/s is negligible compared to the velocity of themeasuring system.S (10 years) = vt =510-32 m/s 3.14107 x10 = 15.710-24 m =1.5710-23 m

Size of nucleus = 10-14 m (87)S(10 years) = 1.5710-9 size of nucleus (88)

Thus body moves a distance of 1.5710-23 m which is immeasurable. Hence body can beregarded as at rest, as in case of Einsteins derivation when two waves are emitted. Even if this velocity is taken in account then results are negligible.Even bigger numerical values are neglected in Physics e.g. in the relativistic variation of mass

Mmotion=2

2

1cv

M rest

(15)

Mmotion= Mrest [ 1+ v2/2c2 +3v4/8c4 + ..]

= Mrest [ 1+ 5.5510-18 +4.62910-31+ ..]Here velocity is regarded as 1m/s in classical region. Like wise velocity 510-32 m/s will beregarded as neglected.

Here even term 5.5510-18 is regarded as negligible, thus

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Mmotion= Mrest

Thus equations for recoil momentum and recoil kinetic energy KE recoil will bePrecoil = 510-32 kgm/s (89)KErecoil = 12.510-64 kgm2/s2 (90)

Due to this uniform relative velocity v of the system (, , ) will not changewithin measurable limits, however effect of Vr can be considered for completeness[11-24]. Itis discussed below.. 4.2 Einsteins Sep. 1905 derivation predicts

Increase in mass of luminous body when Light Energy is emitted.Recoil Velocity has NO effect on the results (as it changes the relative velocity v ).

If velocity of recoil is taken in account, then eq.(34) becomesM b Ma = L [ 0.00004/c (v+510-32 m/s) + 1/c2 ]= 0.00004L/c(v+510-32 m/s ) + L/c2 (91)

Ma = M b +0.00004L/c(v+ 510-32 m/s ) L/c2 (92)0.00004L/c(v+510-32 m/s ) = 4L/105x 3x108x[10+ 510-32 m/s ] = 4L / 31014 = 1.333L/ 1014L/c2 = L/91016 = 1.111L1017

0.00004L/c(v+VR ) L/c2 = 1.333L/ 1014 1.1111017 IS POSITIVE QUANTITY

Ma = M b +0.00004L/cv L/c2 = M b + POSITIVE QUANTITY (93)

Mass increases when light energy is emitted.

It is contradiction of law of Conservation of energy. How does double increase happen?(i) How energy is emitted?(ii) How mass increases?Thus Einsteins Sep. 1905 derivation leads to contradictory results frequently.

Recoil Velocity does not change the contradictory results.

Ho = H1 + 0.50001 L[( 1 v/c cos0)] + 0.49999 L [1 v/c cos 180)] (94)Ho = H1 + 0.50001 L[( 1 (v+VR )/c cos0)] + 0.49999 L [1( v+VR )/c cos 180 (95)

M b Ma = L [ 0.00004/c(v+VR ) + 1/c2 ] (96)mc2 = L[ 0.00004c/(v+ VR ) + 1] (97)

L= mc2 / [ 0.00004c /(v+5.310 31 ) + 1] (98) 0.00004c /(v+5.310 31 ) = 0.000043108 / (10+5.310 31 ) = 1200

L= mc2 / [ 1200 + 1] = mc2 / 1199 (99)

L mc2 or L =A mc2 (100)Where A is coefficient of proportionality.

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Mass increases even if the velocity of recoil is taken in account.

M b Ma = L [ 0.00004/c (v+VR ) + 1/c2 ] = 0.00004L/c(v+VR ) + L/c2 (101)

Ma = M b +0.00004L/c(v+VR ) L/c2 (102)0.00004L/c(v+VR ) = 4L/105 310810 = 4L / 31014 = 1.333L/ 1014L/c2 = L/9x1016 = 1.111L / 1017

0.00004L/c(10+510 32m/s ) L/c2 =1.333L/ 1014 1.111 1017 IS POSITIVEQUANTITY

Ma = M b +0.00004L/cv L/c2 = M b + 1.333L/ 1014 1.1111017 = M b + POSITIVE QUANTITY (103)

Hence mass increases in this case also, when light energy is emitted.

5.0 Derivation of E =Ac 2m by alternate method

(i) Newton [26] has stated that

Light and gross bodies are convertible to each other

in his book Optiks in 1704.Einstein tried to give mathematical derivation for Newtons perception in 1905. Under theSPECIAL or HANDPICKED conditions Einstein derived

or L = mc2 (11)Under the general conditionsL mc2 is also possible.From here Einstein SPECULATED whatever is true for light energy is true for everyenergy or energy content [ energy in Joules , m in kg. ] E i.e. E = mc2 or E mc2.Practically it is used to explain every energy as Einstein stated that

The mass of a body is a measure of its energy-content. Einstein himself used E (every energy) e.g. heat energy, attractive binding energy, energy inform of invisible radiations etc. In the derivation Einstein has only considered that body emitslight energy, but the luminous body can also emanate the heat energy or other forms of energy simultaneously.These energies are NOT considered by Einstein at all in mathematical calculations of L =mc2 .Thus higher the mass annihilated, higher the energy emitted

dE dm(ii) Einstein has derived under super special conditions that the conversion factor between

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mass and energy is c2. Also under general condition the conversion equation is in proportionality form i.e. L mc2 or E mc2 The value of conversion factor, other than c2 can also be justified in many other cases inwhich E = mc2 is not confirmed yet.

NON-CONFIRMATION OF E=mc 2 is in Chemical reactions.

The chemical reactions are the most abundant reactions in nature and were main sources of energy even in Einsteins days. But neither Einstein nor any other scientist tried to confirm E= mc2 in CHEMICAL reactions. Till date E=mc2 is not confirmed in Chemical reactions, but E = mc2 is regarded as true in such cases, which is not justified [2].If 10-9kg of matter is annihilated then energy equal to 9107J will be produced.

E = mc2

= 10-9

91016

kg m2

/s2

= 9107

J (104)This energy can derive a truck of mass 1,000 kg to distance of 90 km. Such or similar predictions are not experimentally confirmed inspecific experimentation . If we burn 1kg,hay/straw, some ashes and gases are produced. The mass of ashes and gases can be measuredin sensitive instruments. Then energy corresponding to mass annihilated can be calculated byusing E = mc2. Then experimentally energy emitted can be checked whether it isQUANTITATIVELY consistent with theoretical estimates or not.In terms of heat energy,

1 calorie = 4.18 J or 9107 J = 2.153107 calories (105)It will raise the temperature of mass of water 21,530 kg from 14.5C to 15.5C.

The energy emitted can be found less than predictions of E = mc2. It is preliminaryconclusion from observations, as even very little energy ( comparatively) is emitted whenhuge amount of wood , paper, straw etc. burn. Further reason for this conclusion is thatconceptually the chemical reactions (atoms are exchanged, making and breaking of bonds) areentirely different from nuclear reactions (new atoms are formed, nuclei are broken or fused).The conditions under which nuclear reactions proceed are entirely different from chemicalreactions. It would mean that the conversion factor between mass and energy is other than c2.

The reason given by scientists for non-confirmation of E=mc 2 in chemicalreactions is that in chemical reactions energy emitted is too less to be measured. Also in theexisting literature there are no evidences that ever serious attempts have been made for theconfirmation. Just possible in some experiments results may have not been inconsistent withE=mc 2, hence such experiments were discontinued due to lack of viable alternative. But nowsuch data can be accommodated in E =Ac2m. Now sophisticated instruments are being prepared to study various scientific phenomena, butno headway is made in confirmation of E=mc 2 in chemical reactions. Such experiments

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may definitely change the perception of E = mc2 in science.

(iii) In the laboratory[27-30] it is confirmed that using thermal neutron the Total KineticEnergy (TKE) of fission fragments that results from U235 and Pu239 is 20-60MeV is less than

predicted by E = mc2. This prediction is nearly 40 years old in the existing physics. Itfurther implies thatdE c2dm

(iv) Similarly mass of particle Ds(2317) has been found more than [31] current estimates

based upon E = mc2. Thus againdE c2dm

(v) Robert Serber (member of first American team entered Hiroshima and Nagasaki inSeptember 1945 to assess losses) had pointed out that efficiency of the Little Boy weapon[U235, 49kg ] that was used against Hiroshima was about 2% only [32]. The remaining 98%energy is not accounted for. It is assumed that all atoms dont undergo fission, thus materialis wasted. But the wasted material is not measured. How much energy is emitted in form of heat energy, light energy or energy in form of invisible radiations is not measured? Further Einstein has derived equation (under SUPER SPECIAL CONDITIONS) for light energy only,

but here energies in various forms are emitted.But it is not calculated that how much material is wasted and how

much energy is emitted? Einstein has simply considered emission of the LIGHT ENERGY by luminous body, but in nuclear explosion, heat energy, light energy, in form of invisibleradiations etc. is also emitted. Under ground nuclear explosions also justify dE c2dm [as

E = mc2 is not confirmed]. How E = mc2 is regarded as true for those energies for which it is not derived. Until such predictions are not precisely confirmed it is equally possible that conversion factor other than c2 is feasible i.e.

dE c2dm

Thus it is absolutely incorrect to state that E = mc2 is confirmed in uncontrolled nuclear explosions. The energy may co-exist in various forms. In nuclear fusion or fission explosions

various types of energies are emitted, and L = mc2 is derived for light energy under super special cases. So this derivation is not meant for other energies. So it is illogical to assume

that E = mc2 is correct for all energies even without experiments. Thus it should bementioned in such cases.

E = mc2 is not experimentally confirmed.

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(vi) The Big Bang Theory (the biggest energy releasing process in universe), is the mostsuccessful theory of understanding of the origin of universe.This theory assumes that whole mass of the universe (1055 kg, say) was in form of primeval

atom and then suddenly exploded [33]. According to E = mc2, this mass (1055 kg )would have been created from energy 91071 J.E= 1055 kg 91016 m2/s2= 91071 J (106)But from where this energy had come? How this energy changed to mass? This is hugeamount of energy is not is not regarded as to exist as such automatically. The answer to onequestion is other question.How whole the mass is condensed to a single point?From where energy was created for explosion?Which is the source of energy, causing universe to expand?If universe has started from the state of cipher, then will it change to state of cipher.But we reach at logical conclusions (justified in applications) if the conversion factor betweenmass and energy is considered other than c2, as conversion factor c2 does not explain those.It means thatdE c2dm

(vii) Webb [60] has reported results for time variability of the fine structure constant or Summerfield fine structure constant ( ) using absorption systems in the spectra of distant

quasars. It means the variations in the values of c are being discussed, if variation in value of c is confirmed then automatically

dE c2dm

[ In addition it can be justified that Einstein derivedL=mc 2 or speculatedE=mc 2 in1905 and nuclear reactions were discovered in 1920s. ThusE=mc 2 was applied after gap of 15 years in nuclear reactions. Then this equation is used as basis i.e. defined 1Atomic MassUnit (1amu) in terms of E=mc 2. Then all data was standardised in terms of 1amu (based

uponE=mc 2). Had Einstein would have discussed his derivation fully then E =Ac2mwould have been available for discussion in 1905 and afterwards. Any data which was foundinconsistent with E=mc 2 was neglected, as there was no alternative. Some limitations whenE=mc 2 is used are transparent, especially while explaining binding energy of nucleus anduniversal equality of masses of nucleons simultaneously. This aspect is separatelydiscussed. However these can be explained with help of E =Ac2m with value of A>>1.Also neither Einstein nor other scientists applied it for chemical reactions.

Further an indirect or unrelated or analogous example may be given here regarding

transformation of energy from one form to other.

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Energy in one form =k (energy in the other form) (107)

where k is conversion factor just like J (4.2 107 erg cal-1) in equation JH W = (108)But the same transformation factor does not exist if electrical energy is changed to soundenergy or light energy is changed to electrical energy etc. Thus in science in many cases,other phenomenon is a new phenomenon.] Thus from various deductions (including that of Einsteins Sep. 1905 paper) it is logicallyconcluded that

dE c2dmThe above proportionality, dE c2dm can be changed into equation by introducing a co-efficient of proportionality. The inception of coefficient of proportionality here is consistent

with centuries old perception of coefficient of proportionality in physics since days of Aristotle and Newton.According to AristotleSpeed is proportional to motive force, and inversely proportional to resistance.

v f/r or v = C f/r where is C is co-efficient of proportionality and determined experimentally. It is likeestablishing relation between quantities by method of dimensions and value of co-efficient of proportionality is experimentally determined.

According to Newtons First Law of motion as stated in The PrincipiaThe alteration of motion is ever proportional to the motive force impressed; and is made in

the direction of the right line in which that force is impressed.

Alternation in motion motive forceMotive Force Alternation in motion or F change in momentum or F maF=KmaThe value of K is regarded as unity to define unit of force in systems of units e.g. FPS, SI andCGS etc. However the values of coefficients of proportionality are experimentally

determined. In Newtons Law of Gravitation the coefficient of proportionality G isdetermined experimentally (as like other co-efficients), but in F=Kma, K is regarded as unityin systems of units just to simplify system of units, which is anomalous.

In caseconstant of proportionality varies from one situation to other then it isknown asco-efficient of proportionality e.g. co-efficient of thermal conductivity or viscosity etc.When more and more complex phenomena were studied or values of coefficients of proportionality were determined then it showed dependence on the inherent characteristics of

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the phenomena. So it varies from one situation to other.Thus removing the proportionality between dE and c2dm we get:

dE =Ac2dm (109)where A is a coefficient used to remove that sign of proportionality; it depends upon inherentcharacteristics of the processes in which conversion of mass to energy takes place and it isdimensionless. There are many existing derivations of this type.

The constant of proportionality varies from one situation to other depending uponinherent experimental conditions of the process. If it varies from one situation to other, thenconstant of proportionality is calledCOEFFICIENT OF PROPORTIONALITY . Thereare many examples about it. In E = Ac2m, A is a coefficient not a constant depending uponinherent characteristics of mass energy inter-conversion processes. The mass energy inter-conversion processes are numerous in nature, and each is independent to other. The conceptof co-efficients of proportionality is applicable to many equations in the existing literature.For example,

(a) Equation for Hubbles law

V=HS (110)The general range for value of H is 50 to 80 kilometers per second-Mega parsec (Mpc). Thus,there is difference of 60 percent between upper and lower limits of value of Hubblesconstant, hence it should be termed as Hubbles co-efficient.(b) Equation for Coulombs law

F = KqQ/r 2 = qQ/4 0r (111)value of 0= 8.854 10-12C2 N-1m-2 and is constant. Further the value of r varies from onesituation to other. For example values of r are for air 1.0006, for glass 4.9 to 7.5, for distilledwater 80.0, for barium-strontium-titanite 7500, hence values of K varies from one situation toother.

(c) Equation for decay constant radioactivity ,

The equation for T1/2 = 0.693/ Or = 0.693/T1/2 (112)

where is decay constant and its general trend is 1015 s-1 10-10 s-1.

(d) Equation for resistance of conductor,V I or V = I R

R = V/I (113)R = L /a (114)

where is resistivity of the conductor having area a and L is length of the conductor. Further the value of resistivity, for

(a) metals and alloys varies from 1.6 10-8 to 49 10-8 m,(b) for semi conductors it is 3.5 10-5 to 2300 m

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(c) for insulators 1010-1016 m.

(e) Equation for co-efficient of viscosity

F = Adv/dx (115)Higher the co-efficient of viscosity, more viscous is the fluid. In general the co-efficient of

viscosity varies from 1.05 10-3 poise to 19.2 10-6 poise. Similar is nature andcharacteristics of A.

(f) Co-efficient of thermal conductivity

The coefficient of thermal conductivity is expressed as the quantity of heat that passesthrough a unit cube of the substance in a given unit of time when the difference intemperature of the two faces is 1.

Q = KA [T1-T2]t/d (116)

The general range of variation of co-efficient of thermal conductivity K for variousconductors is

0.02 Wm-1K -1 to 400 Wm-1K -1

Table IV the various values of constant of proportionality in existing Physics

Sr. No 1 Equation Co-efficient of proportionality

Value

1 V=HS H, HubblesCo-efficient( popularlyknown asHubblesconstant )

50 to 80 kilometers per second-Mega parsec(Mpc).

2F = KqQ/r 2 = qQ/4 0r

r Relative permeability of medium

r are for air 1.0006, tofor barium-strontium-titanite 7500,

3 T1/2 = 0.693/ Decay constant

general trend is 1015 s-1 10 -10 s-1.

4V I or V = I R

R = V/I

R = L /a

Resitivity

(a) metals and alloysvaries from 1.6 10-8 to49 10-8 m,(b) for semi conductors itis 3.5 10-5 to 2300 m(c) for insulators 1010-1016

m.5 F = Adv/dx

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Co-efficient of viscosity

1.05 10-3 poise to19.2 10-6 poise.

6 Q = KA [T1-T2]t/d K Coefficient of

thermalconductivity

0.02 Wm-1K -1 to 400 Wm-1K -1

7 Aristotles speedv = C f/r C is coefficientof Proportionality

C to be determined for different media

8 Semi empirical binding energyformula

ais Different values to co-efficients are assigned

9 m Ac E = 2 A, conversionCoefficient

Numerous valuesin bizarre phenomena of mass energy inter-conversion.

Thus, the value of the co-efficient of thermal conductivity varies from one body toother. Similar is nature and characteristics of A in E = Ac2m.

equation for modulus of elasticity,equations of Faradays laws of electrolysisequation for critical velocity in fluid dynamics,equation for Newtons law of cooling,equation for induced dipole moment,equations for co-efficient for linear expansion, superficialexpansion, cubic expansion,equation for specific heat etc. etc.

Now consider the case that when mass is converted into energy. Let in some conversion

process mass decreases from iM (initial mass) to f M (final mass), correspondingly energy

increases from i E (initial energy) to f E (final energy). The eq. (109) gives infinitesimally

small amount of energydE created on annihilation of massdm . To get the net effect theeq. (109) can be integrated

= dmc AdE 2 (117)

Initial limit of mass = iM Initial limit of Energy = i E

Final limit of mass = f M Final limit of Energy = f E

Initially when mass of body isiM , then i E is the initial energy of the system. When mass

(initial mass, iM ) is converted into energy by any process under suitable circumstances the

final mass of system reduces to f M . Consequently, the energy of system increases to f E

the final energy. Thus f M and f E are the quantities after the conversion. Hence, eq. (109)

becomesor f E i E = 2 Ac [ f M iM ] (118)

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or m Ac E = 2 (119)

Energy evolved = 2 Ac (decrease in mass) (120)

Now it is natural to calculate the difference in mass by subtracting the final mass from initialmass as taken as in equation

f E i E = 2 Ac [ f M iM ] (118)

For example, a child is born and has weight 5 kg (in general sense). After few years, hisweight becomes 40 kg, and then we say his weight is increased by 35 kg i.e.

Increase in weight = Final weight Initial weight = 405 = 35 kg. (121)We never say that increase in weight of child is 35 kg i.e.

Increase in weight = Initial weight Final weight =540 = 35 kg. (122)Further energy is scalar quantity, it has only magnitude

Special Cases Now the following cases can be discussed for further understanding of the equation. f E i E = 2 Ac [ f M iM ] (118)(i) If the initial and final masses remain the same, then Mi = Mf then from eq. (118)

Ei = Ef (123)

i.e. under this condition when no mass is converted into energy, the energy remains the same.

(ii) If the characteristic conditions of the process permit then whole mass is converted intoenergy i.e. after the reaction no mass remains (Mf = 0)

f E i E = E = Ac2Mi (124)

Since Ei is the initial energy can be regarded as zero, then

Ef = Ac2 Mi (125)

If the characteristic conditions of the process permit then whole mass is converted into energy

i.e. after the reaction no mass remains (f M = 0)

iM Ac E 2= (126)

In this case energy evolved is negative implies that energy is created at the cost of annihilation of mass and the process is exo-energic nature (energy is emitted which may be inany form). Energy is scalar quantity having magnitude only, thus no direction is associatedwith it.Thus the generalized mass-energy equivalence may be stated asThe mass can be converted into energy or vice-versa under some characteristic conditions

of the process, but conversion factor may or may not always be 2c (9 1016 m2 /s2 ) or c -2.

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6.0 Applications of generalized mass energy interconversion equation m Ac E = 2

6.1 Chemical Reactions

Let wood or straw of mass 1kg is burnt under controlled conditions, consequently ashesand gases are emitted. The magnitude of ashes and masses are measured. Let the mass equalto 10-9 kg is annihilated , and equivalent amount of energy is emitted.If 10-9kg of matter is annihilated then theoretically energy equal to 9107J will be producedi.e.E = mc2 = 10-9 91016kg m2/s2 = 9107 J (104)This energy can derive a truck of mass 1,000 kg to distance of 90 km. Such or similar

predictions are not experimentally confirmed inspecific experimentation . In daily life itappears that this prediction may not be justified, as even huge amount of wood is burnt thenenergy appears to be emitted too less that above prediction to be justified [2]. It is not justifiedto regard E = mc2 true without experiments, rather it should be mentioned that E = mc2 isnot confirmed in this regard. THUS RESULTS ARE WIDE OPEN UNLESS SPECIFICEXPERIMENTS ARE NOT CONDUCTED.

Let experimentally energy observed is 4.5 107J corresponding to mass annihilated

10-9 kg, then value of A from m Ac E = 2 will be 0.5 i.e.

A = mc

E

2 = = 4.5 x107 / 9x1016x10-9 = 0.5 (127)

Thus in this case mass energy inter-conversion equation becomes

mc E =2

5.0 (128)

6.2 The total kinetic energy(TKE) of fission fragments of U 235 or Pu 239 is 20-60 MeV less

than predicted by E = mc 2.

The familiar fission reaction is 92U235 + 0n1 56Ba141 + 36Kr 92 + 30n1 + Q (202 MeV )

In laboratory [27-30] it has been experimentally confirmed that usingthermal neutrons the total kinetic energy of fission fragments that result from U235 or Pu239 is

20-60 MeV less than the Q value of the reaction predicted by E = mc2. These observationsare over 35 years old. In existing physics these inconsistent observations are explained in thefollowing ways.

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(i) It is typically assumed that energy is lost in unobservable effects [27-30]. If so then such

unobservable effects may also be applicable to those cases where E = mc2 is regarded as tohold good.(ii) Also attempts have been made to explain the total kinetic energy (or essentially totalenergy) of fission fragments by extending the successful liquid -drop model of Bohr and

Wheeler [27-30]. It implies the gravity of inadequacy of E = mc2 in this case.

The same can also be explained on the basis of E = Ac2 m. Let

according to E = mc2 the total kinetic energy (or essentially total energy) of fissionfragments of U235 or Pu239 is 200 MeV theoretically and experimentally observed energy is 25

MeV less i.e. 175MeV. Then according to E = Ac2 m the value of A is 0.875 i.e.

A = E / c2 m = 175 / 200 = 0.875 (129)

So in case of fission fragments of U235 or Pu239 the value of A is less than one i.e. 0.875 as ineq.(115), hence the energy emitted is less than predicted by E = mc2. In this case the valueof A other than unity is justified.6.3 Anomalous Observation The anomalous observation of excess mass of Ds(2317) can be understood with help of

m Ac E = 2 , as mass of the observed particle is found more [31] than predictions of 2

mc E = . In this case value of A will be less than one. For understanding consider

energy equal to 106J is converted into mass, then corresponding mass must be10 -10/9 kg . We

are considering the case that mass is found more than this. Let the mass be 10-10 10/ 9x9 kg.The value of A in this case is 0.9 as calculated from m Ac E = 2 i.e.

106 =A910161010-16 /81 = 9A 107 /81A= 0.9 (130)Thus corresponding to more energy less mass is observed.6.4 Annihilation of Antimatter

At the time of big bang equal amount of matter ( visible 1055 kg ) and antimatter were

produced. This antimatter was annihilated during hadron epoch (10-35 10-4 s) as nowantimatter is not found at all [35]. It would have released energy equal to 91071 J i.e.

E = 105591016 = 91071 J (131)Due to this energy temperature would have risen. However in this epoch temperature felldown from 1027K to 1012 K .Thus it is contradiction, as energy is released and temperaturefalls.

It can be explained with help of m Ac E = 2 . It can be explained if we assume thatnegligibly small amount of energy has been produced during hadron epoch, in equation

m Ac E = 2 . Let this energy be only 1 J. Now the value of A must be 1.1110-72 i.e.

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1= A 91071 Or A = 1.1110-72 (132)Thus,

mc E = 2721011.1 (133)

Hence antimatter is not observable as it is annihilated during hadron epoch without raising thetemperature. Thus this perception justifies big bang theory.6.5Creation of Mass Before Big Bang.

Basis of Big Bang Theory: The basis of Big Bang Theory is that time and space did notexist before Big Bang. The basic postulate of Big Bang Theory POSTULATES that initially(t=0) whole mass of universe 1055kg was in infinitely compact and singular state enclosing aspace even smaller than atomic particle , instantaneously exploded in gigantic detonation [33].How mass equal to 1055 kg was created? Why mass was not created more or less than 1055kg ?How it changed into singular state (state of infinitely large density) ?How the primeval atom exploded (from where energy came) ?These questions are unaddressed in Big Bang Theory. Here the measurement of time

started after 10-43 seconds of Big Bang, called the Plancks time.Findings based upon NASAs Wilkinson Microwave Anisotropy Probe (WMAP) .Erickcek and colleagues [36] deduced on the basis of NASAs Wilkinson MicrowaveAnisotropy Probe (WMAP) that existence of time is possible before Big Bang and universe

may be created from empty space. Normally microwave background radiations are mostlysmooth but Cobe satellite discovered some fluctuations. Erickcek and colleagues believed thatthese fluctuations contain hints that out universe bubbled off from previous one. Thus timeand space, hence universe existed before Big bang. Similar is conclusion of application of

m Ac E = 2 regarding origin of the universe.

Predictions based upon m Ac E = 2 are justified:

It describes HOW primordial atom was created in the pre-big bang world.

The explanation based upon m Ac E = 2 to understand the origin of primeval atom

i.e. state of pre-big bang universe leads to similar results as deductions based upon

WMAPs findings. According to Wilkinson Microwave Anisotropy Probes observations thetime and space existed before Big Bang, the primeval atom was created in due course of

time in existing space and then exploded. It is consistent with explanation of m Ac E = 2 and provided natural explanation for formation of universe (creation of mass, creation of primeval atom, its explosion and acceleration). As we are simply discussing the formation of primeval atom (which is not discussed yet), hence there is no contradiction of any existing

fact. The Primeval Theory based upon m Ac E = 2 is consistent with WMAPs

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observations:

The basic motivation behind Primordial Theory is HOW and WHY

primeval atom is created ?

In Primeval Theory it is assumed that initially space was created and whole the space wasfilled with infinitely large number of particles of zero mass or undetectable by any means,may be termed as Zeroans . The Zeroans are the most primitive constituents of the universe inspace, may be moving with infinitely large velocities. The Zeroans may have mass trillion-trillion times smaller than axioms, hypothetical particles proposed by Peccei [37], and arevirtually isolated. Such particles are currently regarded as constituents of dark matter.

The mass of photon is estimated or understood as less than 10-18 eV or 1.7710-54 kg .The numerous numbers of Zeroans were moving with infinitely large velocities acted as thesmallest possible (but just perceivable or imaginable) pulse of energy, may be coined asPrimordial Energy Pulse. It is the postulatory assumption in this perception, but now recentresearch based upon data from WMAP supports it, as it hints that universe bubbled of from previous universe. The zeroans are present in empty space even now.

In due course of time infinitely large number of Zeroans resulted or combinedtogether to form pulse of energy, 10100 J or less (smallest permissible units of energy) i.e.exceptionally -2 small. It may be called the Primordial Energy Pulse, which is subdivided innumerous parts or pulses of energy in empty space. As creation of universe is the intricate

process, thus such a low of amount of energy is perceived in its primeval origin. m Ac E = 2 is applicable to all peculiar cases where mass-energy inter-

conversion takes place, as in this case the conversion factor is not rigidly2c . Due to its

general nature m Ac E = 2 was applicable even in pre Big Bang era. Thus, the equation

m Ac E = 2 specifically predicts that in this primordial era, diminishingly small pulse of energy, say 10-100 J (or less), manifested or transformed itself in mass 1055kg, in due course of

time. Under this condition the value of uni A can be determined as:

uni A =mc

E

2 = 5516

100

1010910

= 1.111 10-172

(134)These values are too peculiar , as the situation is so and unaddressed. Now if the value of

energy is 10-100J and value of uni A is 1.11110-172 then mass can be calculated as

m =uni Ac

E 2

= 16172

100

10910111.1

10

= 1055 kg

(135)

Thus m Ac E = 2 , is the first equation which at least theoretically predicts that universe

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(1055kg) has been created ( and resulted to primordial or primeval atom ) from minuscule or immeasurably small amount of energy (10-100J or less). However actual process of creation of mass may be quite tedious and time consuming process. In free SPACE the motion of Zeroans hence process of creation of mass is continuous processes. The rate of creation is tooless to upset any existing calculations. As already mentioned this explanation is not possible

from 2mc E =

6.6 The Origin of Gravitation.

WHY DO BODIES ATTRACT EACH OTHER ?

The mass may be regarded is primary form of energy in nature, is converted to other forms of energy which may co-exist in various forms. In nuclear explosion mass is converted to heatenergy, light energy, energy in invisible form etc. If mass can be changed to other form of energy, then it can also be changed to gravitational energy. Also energy is transformedfrom one form to other. In electric bulb electrical energy changes to light energy, in heater electrical energy is converted to heat energy in radio electrical energy is converted into soundenergy, in cell chemical energy is changed to electrical energy, in photocell light energychanges to electrical energy, in a.c. dynamo mechanical energy is converted to electricalenergy there are many such examples. The inter-conversion of energy to other form may bewritten asEnergy in newly converted or transformed form =k (energy in the first form) (107)

where k is conversion factor just like J (4.2 107 erg cal-1) in JH W = .2

mc E = states that mass is converted into energy or vice versa, whereas according to eq. (107) energychanges from one form to other.

In uncontrolled nuclear fission or in nuclear reactors mass isconverted to light energy, heat energy, sound energy and energy in form of invisibleradiations is emitted etc or energy may co-exist in various forms. In nucleus the mass isconverted into the binding energy (attractive like gravitational energy). The attractive bindingenergy exists within nucleus and attractive gravitational energy exists on large scale, but both

arise from the annihilation of mass. Thus law of conservation of matter is obeyed. Electronsmove around the nucleus and heavenly bodies move around the sun.If mass changes to other forms of energy then it can also be changed to Gravitational Energy.In view of eq. (103) the analogous relation between mass annihilated and gravitational energy produced (measure of gravitational force or pull) can be written as

Gravitational energy ( g U ) = k Energy emitted in annihilation of mass ( m Ac 2 )

= k m Ac 2 (136)

where k is conversion factor which determines the extent of conversion of energy togravitational energy. Thus higher the value of A and higher the value of k more

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gravitational energy will be produced. The values of A and k depend upon inherentcharacteristics of the process.The inter conversion of energy to mass is continuous process. The fraction of mass (so produced) also changed into gravitational energy as described by eq. (136). This gravitationalenergy held together the created mass, if the gravitational energy simultaneously produced inone particular case is considerable then that matter remained in cohesive state. So, formation of mass of universe and origin of gravitation are both simultaneous processes. The gravitational energy is universally prevalent and is inherent property of bodies, it unitesthe bodies as these are produced.At the same time it is just possible that some particles which were created from PrimordialEnergy Pulse, may have not developed considerable amount of gravitational energy, hencenot condensed to bigger units and may be even as such now. These particles developed chargedue to friction. The different charges can be understood due to friction of particles in differentdirections. These may have mass trillion-trillion times smaller than axioms; hypothetical particles proposed by Peccei [37]. Such particles may be numerous in number and areconstituents of dark matter. This explanation account for dark matter as well.

Formation of Primordial Atom And Its Explosion Or Big Bang.

The lighter particles or bodies stuck together (under extreme conditions of temperature, pressure and gravitational energy) then their mass increased. Thus high temperature and highgravitational pull caused constituents of universe to contract to a single point. As the processof annihilation of mass to energy continued so the rise in temperature and increase ingravitational energy equally continued. Thus bodies were quickly attracted and condensed as being extremely hot they compressed to small size due to high gravitational energyconsequently radius of the universe decreased. This process was repeated again and again andwhole mass of the universe condensed to a single point in infinitely dense state.

The nature of gravitational force so developed, may be regarded assimilar to inter-atomic force. The inter-atomic force is attractive up to some extent (maximum

r = 5A) and when distance between the molecules decreases it turns strong repulsive force.As the gravitational energy increases (higher A , higher k ), so the size of constituents of universe decreases. As long as the size of compressed mass of universe is optimum, there isno considerable repulsion between constituents. Then the size of the universe is further decreased (due to extreme conditions of heat and gravitational energy) i.e. distance betweennuclei decreased beyond the optimum distance.

At this stage (size of universe is decreased beyond optimum size), theprimordial atom exploded as universe was in extremely unstable, repulsive and reactive

state. It is equivalent to repulsive inter-atomic force. At this stage, even small mass may have

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been converted to mammoth amount of energy as permissible by equation m Ac E = 2 , in

weird reaction. Due to this repulsive force primeval atom exploded. It is called Big Bangand ever since constituents of universe are receding away. Thus this perception not only justifies the origin of Big Bang of Universe but also explains its earliest origin.

6.7 Black Holes . The black holes are results of exploding stars. The black hole is regarded as end stage of starswhose mass is roughly 20-25 times (estimates may vary) the mass of the Sun. In such caseshuge amount of energy is produced. Stars explode when they are in super active state.

On the basis of m Ac E = 2 the reason for formation of black hole ( density

of the order of 1018kg/m3 ) is that due to small annihilation of mass, enormous amounts of heat and gravitational energies are produced. If the gravitational pull of a heavenly body is

exceptionally-2 higher even then visible light does not escape from it and remains invisible.According to this perception the pre-requisite for formation of the black hole is that it musthave unimaginably high value of gravitational energy. According to eq. (136) it is only possible if the value of A and k must be exceptionally high. Thus even smaller bodies can become back hole, the condition is that the gravitational pull must be such that even light cannot escape from the body.

This perception implies that for annihilation of small mass, hugeamount of gravitational energy is gained by body, consequently does not allow light to

escape. If this condition is satisfied then even bodies of smaller mass may become black holes. In view of this the lightest black holes are also possible, and some of them are observedto reside in the youngest galaxies. Even small heavenly bodies may form of black hole. Themicro or mini black holes have mass of the order of Planck Mass ( 1.22091019 GeV/c2 =2.17644108 kg). The value of A for black hole may be written as

bh A =mkc

U g 2

(137)

6.8 Dark Matter

About 80 % of the matter of the universe is regarded as dark matter (supposed to exist butinvisible). The dark matter can be understood in two in categories:

(i) MACHOs (Massive Astrophysical Compact Halo Objects), MACHOs are the big, strongdark matter objects ranging in size from small stars to super massive black holes. MACHOsare made of 'ordinary' matter, which is calledbaryonic matter. Astronomers search for MACHOs.

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(ii) WIMPs (Weakly Interacting Massive Particles). WIMPs, are the little weak subatomicdark matter candidates e.g. yet undetected gravitinos and photinos, undetected axions particleswith extremely small masses etc. which are thought to be made of stuff other than ordinarymatter, callednon-baryonic matter. Particle physicists look for WIMPs .An Explanation for Wimps on the Basis of E = Ac 2m

E = Ac2m favors that the dark matter in form of the lightest neutrinos, axioms etc.(a) The reason is that the heavier particles are annihilated to energy with value of Aextremely less than one say, A =10-10 or less

E = 10-10c2m or less (138)Thus the bulk of the mass of antecedents according to E = Ac2m (value of A less thanunity) is annihilated and small energy is emitted, thus they reduce to the lightest descendantse.g. neutrinos, axioms etc. which is consistent with existing perceptions. Or this theoryassumes that zeroans are the earliest constituents of the universe which may be regarded as predecessors of WIMPS.

(b) In the Primordial Theory of the Universe, it is assumed that the universe started its lifefrom the Zeroans and changed to Primordial pulse of energy. Thus particles like gravitons , photinos , neutrinos etc were created. This process is even now continuing in the emptyspace. Thus, lighter particles are possible which may be contributing to the Dark Matter. Thuslighter particles i.e. WIMPS are naturally preferred candidates for dark matter according toE = Ac2m.

Explanation for MACHOS on the Basis of E = Ac 2m

The MACHOS may have size from small stars to massive black holes. They may remaininvisible due to two reasons.(c) Firstly, they may have so strong gravitational field that even light energy may not beemitted. Thus, they remain invisible.

(d) Secondly, the visible energy they emit is instantly converted to some form of invisibleenergy or even mass (WIMPS). As the inter-conversion of energy from one form to other arefeasible processes i.e. visible radiations may change into infrared radiations.6.9 Gamma Ray Bursts. Gamma ray bursts (GRBs) are intense and short (approximately 0.1-100 seconds long) burstsof gamma-ray radiations and originate at very distant galaxies (several billion light yearsaway). GRBs are the most energetic events after the Big Bang in the universe and emit energyup to 1047 J in exceptionally short time. The origin of GRB can be understood on the basis

of this perception also. If for black hole (formation described above) the values of A and k are higher, then it contracts beyond a optimum limit due extreme conditions of gravitation and

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temperature. Thus it may further result in a detonation known as black bang emittingexceptionally high amount of energy in form of GRBs, due to high value of A . Thus theexplosion like big bang are continuing even now but at much smaller scale emitting energyin form of GRBs and corresponding to annihilation of small mass huge energy is emitted withhigh value of A . This discussion permits that GRBs may be emitted from black holes of smaller bodies comparatively.

According to 2mc E = if energy emitted is 1047 J, then mass annihilated will be

1.11 1030kg, that too in few seconds. This annihilated mass is comparable with mass of Sun

1.99 1030kg. [In fact in GRBs then energy emitted is experimentally estimated but the massannihilated is not, thus exact value of A cannot be precisely determined.] In GRB, the energyequal to 1047 J can be emitted from annihilation of mass of 10 kg. In this case the value of Agrb

is 1.11 1029

.

grb A =mc

E

2 = 1.11 1029 (138)

Likewise explanation for other cosmological phenomena and bodies can be given on the basisof the generalized mass energy inter-conversion equation, Like this energy emitted byQUASARS and other heavenly bodies can be explained.

7.0 E = mc 2 in case of deuteron contradicts UniversalEquality of Masses of Nucleons(Binding energy observed is 2.2244MeV or 3.965 10 -13 J )

Or

Conversion factor is not always c 2 in E = mc 2

There are two inherent observations;(i) firstly masses of nucleons ( neutrons and protons) are fundamental constants i.e. sameuniversally (inside and outside the nucleus in all cases)

(ii) and secondly nuclei possess BE ( mc2) due to mass defect.To explain these observations of deuteron (BE = 2.2244MeV), the difference in masses of nucleons must be 0.002388 u or about 0.11854% of masses of nucleons outside nucleus. Thustheoretically, according toE = mc 2 masses of nucleons must be less in nucleus, which is not justified as all data is based upon equality of masses of protons and neutrons in all cases. Themasses of nucleons are constants in category of atomic and nuclear constants. Conceptuallyaccording to E = mc2 the binding energy and mass defect cannot exist simultaneously.

If the applications of the generalized equation E = Ac2 m are invoked in

this regard, then it is capable of explaining both the observations simultaneously i.e. equality

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of masses of nucleons (assuming infinitesimally small mass defect) and binding energy. As

according to E = Ac2 m even due to infinitesimally small mass defect (2.38810-14 u ,say) the binding energy of deuteron can be 2.2244MeV due to presence of conversion factor

A. the conversion factor according toE = mc 2 is 2.38810-3 u. Thus E = Ac2 m,explains both the intrigues simultaneously i.e. equality of masses of nucleons and bindingenergy of deuteron. Theoretically it is clear that conversion factor is forced in equationE =

mc 2, in mathematical derivation.

7.1 Decrease in masses of proton and neutron in deuteron nucleus

In the experimental and theoretical nuclear physics the masses of nucleons (protons andneutrons) are fundamental physical constants in category of atomic and nuclear constants, and binding energy (energy required to break the nucleus) is an inherent property of all nuclei. Also it is equally true that all the mass energy inter conversionsare universally explained on the basis of E=mc2, where mc2 is mass defect.Mass defect =Mass of nucleons out side nucleus Mass of nucleons inside nucleus (139)Binding Energy =[Mass of nucleons out side nucleus Mass of nucleons inside nucleus ]c2 (140)This aspect is critically discussed in view of deuteron, which contains just one neutron

and proton.(a) The mass of proton is experimentally measured equal to 1.67262110-27kg,(1.007276u or 938.272029 MeV) and is same in all cases. Also the mass of neutron is1.67492710-27 kg (1.008664u or 939.565360 MeV) in all cases.(b) Experimentally binding energy (BE) of deuteron is measured by various methods[38-40] has been found to be 2.2244MeV (1amu = 931.494MeV, 1amu=1.660 538110-27 kg), which is equivalent to 0.002388u (3.98410-30 kg ) on the basisof E=mc2 i.e. masses of nucleons are converted into energy. The conversion factor

is c2 same for all masses.7.2 Decrease in mass of nucleons in nucleus is not justified .The phenomena of universal equality of masses of nucleons and origin of bindingenergy of nuclei have been studied critically and quantitatively [38-40]. The bindingenergy of the deuteron is experimentally observed as 2.2244MeV [14.9410-11J],according to E=mc2 it is equal to mass defect 0.002388amu. It means in the nucleusof deuterium, mass 0.002388u (of proton and neutron) is converted into bindingenergy.

The mass defect i.e. 0.002388u is comparable with sum of masses of the

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neutron and proton (2.01594 u), masses must decrease in nucleus considerably i.e.0.11845 % (compared to mass in free state). In deuteron there are only protons andneutrons, hence theoretically decrease in mass or mass defect 0.002388u is only at thecost of mass of proton (M p) and mass of neutron (Mn). There is no third entity whosemass may decrease. The mass of proton is 1.007276u and let decrease in mass of proton is half the mass defect (0.002388 u) i.e. 0.001194 u (which contributes towardsthe binding energy of deuterium). Then theoretically mass of proton in nucleus must be1.006082u (1.6706410-27 kg) and then decrease in mass of proton must be 0.1185 %.Also mass of neutron is 1.008664u and let decrease in mass of neutron is half the massdefect i.e. 0.001194u. Then mass of neutron in nucleus must be 1.00747 u(1.6729410-27 kg). Similarly is the decrease in mass of neutron in nucleus is 0.1185%.

Like this theoretically decrease in mass of nucleons in other nuclei can be estimated taking all relevant factors in account. More the value of bindingenergy more would be the decrease in mass of nucleons in the nucleus. The decrease inmasses of neutron and proton in deuteron, to explain the values of BE on the basis of

E = mc2, are shown in Table III.7.3 E = mc 2 , binding energy and mass of nucleons in deuteron .If mass of nucleons remains the same universally (fundamental fact used in existing

physics in all aspects of experimental and theoretical nuclear physics) i.e. mass defect

(m =0). Then according to E = mc2

, binding energy is also zero.Binding Energy (BE) = Mass Defect (m)c2 = 0 (141)Thus universal equality of masses of protons and neutrons ( means m =0) impliesthat mass defect must be zero theoretically, hence binding energy is also zero ( E=0 , .E =0.c2). If binding energy (energy required to break the nucleus) is zero then nucleuswill be unstable. Hence it is not experimentally justified.The masses of proton and neutron are same inside and outside the nucleus, it isexperimentally confirmed [2].

(i) When a gamma ray photon of energy 2.2244MeV hits the deuteron(binding energy of deuteron is 2.2244MeV), then deuteron is split up and proton andneutron are emitted. The mass of the released proton and neutron from nucleus is thesame as usual mass i.e. 1.007276 u and 1.008662 u.The masses of proton and neutron are never less in nucleus i.e. 1.006082u and

1.00747u as theoretically predicted by E = mc2 to explain binding energy. Noexternal energy is provided by the gamma ray photon to increase mass of proton andneutron inside the nucleus. If the masses of proton and neutron are less in nucleus, then

lighter protons and neutrons must be emitted. But masses of emitted proton and neutron

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are usual masses. Whatever is the energy of gamma ray photon that is provided todissociate the binding energy of deuteron as both are exactly equal i.e. 2.2244MeVeach. Gamma ray photon does not provide any energy to increase masses of proton andneutron.Consequences of equality of masses of nucleons: Thus mass defect and bindingenergy both must be zero,Mass defect =Mass of nucleons out side nucleus Mass of nucleons inside nucleus (139)Binding Energy =[Mass of nucleons out side nucleus Mass of nucleons inside nucleus ]c2 (140)The masses of protons and neutrons are equal both outside and inside the nucleus.

Effects: According to E = mc2 binding energy of nucleus can never originatewithout annihilation of adequate mass of nucleons. If mass defect m =0 (masses of nucleons are same inside and outside the nucleus) then binding energy is also zero;implying instability of nuclei which is never justified. Then what is the source for binding

energy on the basis of E = mc2 ? E = mc2 implies instability of deuteron, as binding energy (energy required to break the nucleus) is zero.

Importance of E = Ac 2 m : E = Ac2 m is applied even (on adhoc basis) in thisregard i.e. to simultaneously explain the binding energy and mass of nucleons in nucleus.

Consequently the generalized mass energy equation implies that for even annihilation of INCALCULABLE MASS defect (this assumption is for obeying universal equality of protons and neutrons), binding energy equal to 2.2244MeV (3.96510-13J ) is feasible dueto presence of conversion factor A. Let us assume that in this case mass defect is2.38810-14 u or 3.96510-41 kg and due to higher value of proportionality or conversioncoefficient the binding energy can be 2.2244MeV. Thus there is consistency betweenexperimentally observed facts (binding energy and equality of masses of nucleons) and

theoretical predictions based upon E = Ac2 m. Thus conversion factor may not be

regarded as c2 in all cases.The E = Ac2 m implies universal equality of masses of nucleons ( as mass

defect is negligible ) and binding energy is 2.2244MeV ( as conversion factor A is very

high). Thus E = Ac2 m explains both phenomena of equality of masses of

nucleons and binding energy simultaneously which is not so in case of L = mc2 .

7.4 Explanation on the basis of E = Ac 2 m

Einsteins Sep. 1905 derivation of mass energy inter conversion equation [1] has beencritically analyzed by the author [11-24], and found that this derivation is true under special

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conditions (only for certain values of parameters, not for all values of involved parameters).In this paper firstly Einstein [1] derived under SUPER-SPECIAL CONDITIONS the lightenergy-mass inter conversion equation i.e.

L

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