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 EquivalencePrinciple
 Ether andEquivalencePrinciple
 TECHNOLOGYFusion:The “ZEUS”machine
 ON THE EQUIVALENCE PRINCIPLEOF THE GENERAL THEORY OF RELATIVITY FREE FALL OF BODIES
 Let us consider a mass M of radius Ro (Fig. 1).    At a distance h from the center of this mass M, we place a spherical shell of mass m1 and radius R.   Moreover, at the center of the spherical shell we place another point mass m2, (m1 ≠ m2).   Now t= 0 (Phase I), we allow these three masses m1, m2, M to move freely under the influence of the force of universal attraction.   Let us also assume that after a time dt (Phase ΙΙ), υ1, υ2 and V are respectively the velocities of masses m1, m2 and Μ, relative to an inertial observer Ο.   Note: Masses m1, m2, and M are considered to be homogeneous and absolutely solid bodies.   In addition, velocities υ1, υ2 and V are considered to be positive numbers (that is, only the meters of their magnitudes are taken into account), while mass m2 is considered to be found always within the spherical shell m1.   The basic question that is being raised is the following:   At what velocities do masses m1 και m2 fall in the gravitational field of Mass M once simultaneously dropped in free fall from a height h?     The answer to this question is given below:
 PROOF   CASE Ι: When masses m1 and m2 are equal (m1 = m2).   In this case, and at time t = 0, in accordance with Newton’s first law on universal attraction, mass M attracts mass m1 by a force F.    Also, mass Μ attracts mass m2 by a force F΄.   Therefore,
 where,
 and
 However, given that masses m1 and m2 are equal, then forces F και F΄ whereby mass M attracts masses m1 and m2 will also be equal, that is: F = F΄         (4)Therefore, relations (1), (2), (3) και (4), yield:
 From relation (5), since masses m1 and m2 are equal (m1 = m2), it results that:                                 υ2 = υ1                           (6)   Thus, based on what precedes, we are led to the following conclusion:    Two equal masses m1 and m2, when simultaneously dropped in free fall from a height h in the gravitational field of a mass M, fall always at the same velocities υ1 and υ2, (υ1 = υ2).    CASE ΙΙ: When masses m1 and m2 are unequal, (m1 ≠ m2).   As demonstrated in Case Ι, when masses m1 and m2 are equal (m1=m2), they always fall at the same velocity (υ1 = υ2), in the gravitational field of a mass M.   The question that is being raised is the following:    If the two masses m1 and m2 are different (m1 ≠ m2), then which of the two falls at greater velocity in the gravitational field of a mass M, the smaller mass or the larger one?    The answer to this question is given below:   Let us assume that from the two masses m1 and m2, m1 < m2.    a) In this case and at time t = 0, in accordance with Newton’s first law on universal attraction, mass M attracts mass m1 by a force F and mass m1 also attracts mass Μ by an equal force F (based on Newton’s third law on action - reaction).   Therefore, in this case, the following relations will apply for both masses m1 and Μ:                                        F dt = m1 dυ              (7)                   and F dt = M dV1΄            (8)       where, dυ = (υ1 – 0)  and dV1΄ = (V1 – 0)      (9) From relations (7), (8) and (9) it results that:                             m1 υ1 = ΜV1         (10) b) Similarly, mass M attracts mass m2 by a force F΄ and mass m2 also attracts mass Μ by the same force F΄.    Thus, in this case, the following relations will apply for both masses m2 and Μ:     F΄ dt = m1 dυ΄             (11)and  F΄ dt = M dV2΄        (12)where, dυ΄ = (υ2 – 0)  and dV2΄ = (V2 – 0)      (13) Relations (11), (12) and (13) yield: m2 υ2 = ΜV2               (14)   Because, however, force F΄ is greater than force F, (F΄ > F), since attraction force F΄ between mass m2 and mass M is greater than attraction force F between mass m1 and mass M, given that m1 < m2, then from relations (8) and (12) it results that:                           F dt = M dV1΄            (15)                   and F΄ dt = M dV2΄            (16)   and given that F΄ > F as stated above, relations (15) and (16), yield:                       M dV2΄ > M dV1΄          (16.1)  And based on relations (9) and (13), from relation (16.1) it results that:                         V2 > V1                           (17)    Therefore, the larger mass m2 imparts to mass Μ greater velocity V2, while the smaller mass m1 imparts to mass Μ smaller velocity V1.   So, from relations (10) and (14) we obtain:
 and because from relation (17) V2 > V1, then relations (18) and (19) yield:
 Relation (21) will from now onwards be called the “basic relation” in our problem.   COMMENT: At this point, it should be stressed that by adding sides in relations (10) and (14), we obtain:
 where V is the resultant velocity which is simultaneously imparted to mass M by both masses m1 and m2.    Specifically:                             V = V1 + V2                       (23)   As it can be observed, relation (22) expresses (in absolute numbers) “the principle of conservation of momentum” in the system of the three masses m1, m2 and Μ.    Consequently, since in our problem m1 < m2, that is, m1 / m2 < 1, then relation (21) assumes only the following three forms:
 where m1/m2>0
 Let us now consider the above three cases (a),(b) and (c) :   1.  In the first place, relation (24) is rejected because:   According to the basic relation (21) in our problem:
 Now let us say:
 where, k = positive number (k < 1), that corresponds to two certain masses m1 and m2, (m1 < m2).    (e.g. if m1 = 5 kg and m2 = 100 kg, then m1 / m2 = k = 5 / 100 = 0,05).   Thus, from relation (26), we obtain:
 As it can be observed, inequality (28), is verified for all values of fraction υ2 / υ1 greater than k, Fig. 1 (a).
 Therefore, since fraction υ2 / υ1 is verified for all values which are greater than k, then it will apparently be verified, too, for value υ2 / υ1 = k+ε, where ε positive number ε ® 0. Evidently, is k+ε<1.    All the above apply for the basic relation (21) in our problem.   2. However, in relation (24) fraction υ2 / υ1 is verified only for values greater than 1, Fig. 1(a).   Therefore, relation (24) is never verified for value of fraction υ2 / υ1 =k+ε , (as it should occur according to the basic relation (21) in our problem).    Thus, for this reason relation (24) is rejected.   Similarly, for the same reason, relation (25.1) must also be rejected.    So from the three relations (24), (25)  and (25.1) , relation (25) will necessarily apply, that is:
 Relation (30) reveals that fraction υ2 / υ1 of velocities υ2 and υ1 of the two unequal masses m1 < m2, during their free fall in the gravitational field of mass M, is always between fraction k of masses m1 / m2 = k and number 1.   Relation (29) expresses the fundamental law of free fall of two unequal masses m1 and m2, (m1 < m2) in the gravitational field of mass M.    According to the fundamental law of free fall of bodies given by relation (29), we observe that relative to an inertial observer O, the smaller mass m1 falls at a greater velocity υ1 and that the larger mass m2 falls at a smaller velocity υ2 in the gravitational field of mass M.    NOTE: From three relations (24),(25) and (25.1) only relation (25), (namely the accepted relation) holds (simultaneously) for the three values m1/m2<1, υ2/υ1<1 and m1/m2>0, fig. (1.1) fig. (1.1) Relation (24) never holds (simultaneously) for the three values m1/m2<1, υ2/υ1>1 and m1/m2>0, fig. (1.2) fig. (1.2) Similarly, relation (25.1) never holds (simultaneously) for the three values m1/m2<1, υ2/υ1=1 and m1/m2>0, fig. (1.3) fig. (1.3)
 Yet, in the case where the two masses  m1 and m2 are equal (m1 = m2), (and therefore, velocities υ1 και υ2 will also be the same (υ2 = υ1), as proven in Case I), the fundamental law of free fall is formulated as follows:
 Finally, as it can be noted, the fundamental law of free fall, relation (29), mentioned above, clashes with the all-too-familiar Galileo’s law of free fall, according to which “all bodies, irrespective of their mass, fall at the same velocity in the gravitational field of a mass M”.    Consequently, on the basis of proof provided above, the law of free fall as formulated by Galileo is utterly erroneous.   Example: If, for instance, we simultaneously let one tank (armored fighting vehicle) and a cork (e.g. from a bottle) fall freely from a height h above the surface of the Earth (Moon, Asteroid, etc), then the cork will fall at a greater velocity than the tank, and these two bodies will never fall at the same velocity as Galileo erroneously asserts based on his Leaning Tower of Pisa Experiment.    COMMENT: In Fig. 1, another solution to the problem, that will reveal Galileo’s, as well as Einstein‘s great error (as regards the Equivalence Principle of the General Theory of Relativity), is to calculate quantitatively (numerically) relative to the inertial observer O velocities  υ1, υ2, V as functions of time t, that is:
 when masses m1, m2, M, (m1 ≠ m2), as well as height h are known.Apparently, velocities υ1 and υ2 that will result from relations (32) will be υ1 ≠ υ2.NOTE: Any physicist, mathematician, university who/which wishes to formulate and publish the above three functions (32) is welcome to do so.  2nd PROOF FOR THE PROBLEM (THE REVERSE PROOF)In Fig. 1 after a time dt > 0 (phase ΙΙ, with mass m2 being found at all times within the spherical shell ),m1 m2 , if we apply 1. The “principle of conservation of energy”, and 2. The “principle of conservation of momentum”, to the system of the three masses m1, m2, M, we obtain:
 where, G = theuniversalgravitationalconstant. Also, relations (R.1) can be written as follows: where: and Β = MV                                 (R.4)Also, Α > 0 and Β > 0..Note: In relations (R.1), υ1, υ2, V, h, h1, h2 are considered to be positive numbers.Also at time dt > 0, if the intervals covered by masses are m1, m2, M are dS1,dS2 ,dS respectively, then apparently the following relations are valid: By solving the system of equations (R.2), relative to υ1 and υ2, we obtain: where: In relations (R.5) number Κ is apparently,
 Therefore, we will have two cases: Ι. Κ = 0 or ΙΙ. Κ > 0. CASE Ι: When Κ = 0.In this case, because Κ = 0, relations (R.5) yield:
 Therefore, from relations (R.8) its results that:
 Thus, when Κ = 0 then υ1=υ2.Now we will prove that if υ1=υ2 , then m1=m2.Proof a. At first, we will prove that υ1=υ2 m1υ1=m2υ2 .Hypothesis: Let us assume that if υ1=υ2 then relation m1υ1=m2υ2 .is valid. According to Fig. 1 of our problem, the following relations are valid:
 Furthermore, from relation  m1υ1=m2υ2 of the hypothesis we obtain:
 On the basis of relation (R.11), relations (R.10) yield:
 or
 where F is the force of universal attraction between masses m1 and Μ at time t = 0, namely:
 and F’ is the force of universal attraction between masses m1 and Μ at time t = 0, namely:
 On the basis of relations (R.12.1) and (R.12.2), relation (R.12) yields:
 namely:
 However, (according to proof 1 of our problem), when masses m1 and m2 are equal (m1=m2), then the following will be valid:
 or namely:
 Yet, the fact that relation (R.12.4), to which we have concluded, is in accordance/consistent with relation  υ1=υ2 of our hypothesis, signifies that this hypothesis is valid.Therefore, it has been proven that υ1=υ2 m1υ1=m2υ2 b. We will then prove that m1υ1=m2υ2   m1=m2 In the above accepted relation m1υ1=m2υ2 by considering that υ1=υ2 (according to our hypothesis), we obtain:
 namely:
 Therefore, it has been proven that   m1υ1=m2υ2   m1=m2 On the basis of everything analyzed above, we come to the following conclusion:Conclusion ΙIn our problem (Fig. 1), when Κ = 0 (i.e. υ1=υ2), then m1υ1=m2υ2 and m1=m2.In other words, equal velocities correspond to equal masses. More analytically:
 CASE ΙΙ: Όταν είναι Κ > 0.Because υ1, υ2, m1 ,m2 , Β, Κ, > 0 in relations (R.5), in the second of relations (R.5), number , is at all times a positive number, namely >0. Apparently, number , is a appositive number also, that is,   >0Moreover, in relations (R.5), υ1> υ2,  namely υ1- υ2>0 , because:
 Following calculations, relation (R.13) yields:
 Given that the second side of relation (R.14) is a positive number, then: Therefore, from relation (R.15) it results that if in relations (R.5) Κ > 0, then υ1> υ2.Now we will prove that if υ1> υ2, then m1< m2.Proof a. At first, we will prove that υ1υ2 m1υ1m2υ2 . Hypothesis: Let us assume that if υ1υ2 , then relation m1υ1=m2υ2 will be valid.According to Fig. 1, the following relations will be valid: Also, from relation  m1υ1=m2υ2  of the hypothesis we obtain:
 On the basis of relation (R.15.2), relations (R.15.1) yield:
 or
 where F is the force of universal attraction between masses m1 and Μ at time t = 0, namely:
 and F’ is the force of universal attraction between masses  m2 and Μ at time t = 0, namely:
 Based on relations (R.15.4) and (R.15.5) relation (R.15.3) yields:
 namely:
 However, (according to proof 1 of our problem), when masses m1 = m2 are equal, the following will be valid:
 or namely:
 Nevertheless, the fact that relation (R.15.7), to which we have concluded, is not consistent with relation υ1υ2 in our hypothesis signifies that the hypothesis we have formulated is not valid. Consequently, when υ1υ2, then relation m1υ1=m2υ2 is not valid. Thus, relation m1υ1m2υ2will necessarily be valid.Therefore, it has been proven that , υ1υ2 m1υ1m2υ2.b. We will then prove that , m1υ1 0, then υ1>υ2, will apply, i.e. υ1υ2 (according to relation (R.15) proven earlier). This signifies thatυ1>υ2 m1υ1m2υ2 , based on the conclusion υ1υ2 m1υ1m2υ2 , referred to above.Therefore, in the second of relations (R.2), i.e. in relation
 (where B = MV), for the unequal positive numbers  m1υ1 and m2υ2, the following two cases will apply:
 Let us now examine the two cases (1) and (2) individually.b.1 In case (1):
 or
 If relation (R.15.11) is valid, then the following relation will be necessarily valid, as well:
 where ε is a positive number for which , .Yet, according to our problem  υ1>υ2, namely: which means that since relation (R.15.13) is valid, the following relation will also be valid:
 From relations (R.15.12) and (R.15.14) we obtain:
 namely:
 However, in relation (R.15.15), number 1 – 2ε is, 1 – 2ε < 1.Therefore, from relation (R.15.15) it results that:
 namely:
 Consequently, it has been proven that  m1υ1
 or
 So if relation (R.15.17) is valid, then the following relation will also be valid:
 However, according to our problem υ1>υ2, namely:
 This means that since relation (R.15.19) is valid, then the following relation will be valid, as well:
 Relations (R.15.18) and (R.15.20) yield:
 or namely:
 However, (as proven earlier – proof 1), when masses m1 and m2 are equal, (m1 = m2), the corresponding velocities υ1 and υ2 are equal, too (υ1=υ2); likewise for the corresponding momentums m1υ1 and m2υ2 are equal, too, namely:
 Yet, relation (R.15.22), clashes with relation (R.15.17) of case (b.2) in which
 Therefore, we come to the conclusion that case (b.2) is not valid.Thus, of the two cases (b.1) and (b.2), only case (b.1) is valid, namely, ,m1υ1 0 (i.e. υ1 > υ2 ), then m1υ1
 VERIFICATION Α. Verifying Conclusion Ι: When Κ = 0 (i.e. υ1=υ2), then m1=m2.ProofThe second of relations (R.2) yields:
 And based on what has been proven earlier, when υ1=υ2, then:
 Relations (R.16) and (R.16.1) yield
 or or: or
 Therefore, in our case (Κ = 0), since on the basis of relation (R.9) υ1=υ2, relations  (R.16.2) and (R.16.3) will yield
 Consequently, according to relation (R.16.4), Conclusion Ι is verified.Β. Verifying Conclusion ΙΙ: When Κ > 0 (i.e. υ1>υ2), then m1υ2, then:
 Relations (R.5) yield:
 Furthermore, relations (R.16.5) and (R.16.6) yield:
 or namely:
 In relation (R.16.7), given that Κ > 0 and Β > 0, in order for relation (R.16.7) to be valid, we must have:
 namely:
 Therefore, according to relation (R.16.8), Conclusion ΙΙ is verified.VARIOUS CONCLUSIONS We earlier proved that when υ1>υ2, then:
 Relation (R.17) successively yields:
 However, given that the kinetic energy Ε1 of mass m1 is:
 and the kinetic energy Ε2 of mass m2 is:
 relations  (R.17.1), (R.17.2) and (R.17.3) yield:
 In other words, the kinetic energy Ε1 of the smaller mass m1 is less than kinetic energy Ε2 of the greater mass m2, (m1υ2 2. J1υ2 (i.e. Κ>0), then:1. m10) from a state Α over to another state Β, ( ) then, the interval run by each body, its speed, acceleration, momentum, etc are at all times a function of this body’s mass.The above law is verified:1. In the two-body problem (Ν=2), such as in the following cases:   a. In the free-fall of a body within the gravitational field of another one.   b. In the movement of double stars.    c. In the movement of a planet around the Sun, etc. 2. In the three-body problem (Ν=3), such as in the following cases:   a. The problem of Fig.1, according to the two proofs referred to above.   b. The three-body (planet) system, e.g. Mercury-Venus-Earth, etc 3. In the N-body problem, such as in the following cases:   a. The solar system (Ν=10)    b. The movement of stars inside a galaxy, etc. As it can be observed, Galileo’s Pisa Experiment is a partial case of the fundamental N-body law stated above. Thus, assuming that Galileo let e.g. two bodies of different mass fall freely from the top of the Tower of Pisa, then Ν=3 in the above fundamental law (the third body is the Earth). If, for example, Galileo let four bodies of different mass fall freely in the same way, then Ν=5 in the above fundamental law.Also, if he let four bodies of different mass fall freely in the same way, then Ν=101 in the above fundamental law, and so forth…..Therefore, according to the fundamental N-body law, in all the above cases the bodies will fall towards the Earth at a different speed, acceleration, momentum, etc, and they will never fall at the same speed as Galileo wrongly asserts. Consequently, after everything stated above, it is quite clear that the Galileo’s “law of free fall” is wrong and apparently the “equivalence principle” (weak and strong equivalence principle) of the General Theory of Relativity is also a totally erroneous principle of physics. THE “NON-EQUIVALENCE PRINCIPLE” OF MASSES AND FIELDS As is well known, in the “Strong Equivalance Principle” of the General Relativity Theory, Einstein claims that: “All bodies, regardless of their mass, fall in the same way through the gravitational field of another body”.Essentially, the “Strong Equivalence Principle” is Galileo’s experiment (the leaning tower of Pisa) on an astronomical scale. Thus, according to the “Strong Equivalence Principle” if we let a small aluminium sphere (a), the Moon (b) and the Earth (c) fall freely through the gravitational field of a distant star M, then all three bodies fall in the same way. In other words, they fall at the same velocity or acceleration. (Forgive me, but I personally find this very funny!) (See Clifford M. Will, Was Einstein Right?, http://www.tsolkas.gr/forums/777.jpg http://www.tsolkas.gr/forums/888.jpg).After presenting the above, we shall now prove that everything claimed by Einstein in the “Strong Equivalence Principle” (and the weak equivalence principle) is completely false. The proof that we will employ is very simply and is as follows:Proof Let’s take, for example, the simple case of the problem of 2-bodies. Let’s assume (fig 1.4) that we have an inertial observer O’ and two masses m1 and Μ.Mass m1 is considered a point mass, while mass M is a uniform sphere with a radius of R. I addition, both masses m1 and M are considered totally solid bodies.We now place mass m1, at a distance h over the surface of mass M. fig. (1.4)At (t=0) we will allow mass m1 to fall freely from height h into the gravitational field of mass M. The two masses m1 and M, attracted to each other by the force of universal attraction, will collide after a time period (t > 0), with mass m1 moving towards mass M, and mass M towards mass m1 .Specifically, point mass m1 will hit the surface of mass M with a velocity of υ1 and mass M will hit the point mass m1 with a velocity of V. Obviously, the two velocities υ1 and V will have opposite directions.If we now apply to the system of masses m1-Μ the principle of conservation of energy and the principle of conservation of momentum when mass m1 is at height h, (t = 0) and then when the two masses collide (t > 0), we have: In relations (R.17.5), the numbers υ1 , V, h, and R are considered positive numbers and G is the constant of universal attraction. Solving the equation system (R.17.5) relative to υ1 and V, we have: Where υ1 and V are the velocities of masses m1 and Μ at the moment of collision, in terms of the inertial observer O’ .Obviously, the velocity υ1’ of mass m1 , in terms of an observer O who is on mass M, is:υ1’ = υ1  +  V PLEASE NOTE: In relations (A), if, for example, we allow for the following:a. Height h to be much smaller, in relation to radius R, i.e. approximately R+hR καιb. Mass m1 to be much smaller, in relation to mass M, i.e. approximately ,Then, from relations (A), we have:: relative to the inertial observer  O’, and relative to observer O, who is on mass M. .Where  is the intensity of the gravitational field of mass M on its surface. Attention! Based on relations (A.1), it appears that the free fall of bodies dropped from a height h is independent of their mass!That, however, is a grave error, because relations (A.1) express an approximate result (with minor or major deviations from the true values of υ1 and V) and in under no circumstances do they express the natural reality, as expressed in relations (A). For this reason, relations (A.1) must never be employed in strictly theoretical problems, such as the case of Galileo’s law on the “Free Fall of Bodies” and especially the formulation of the “Equivalence Principle” of the General Relativity Theory. Because if we do employ relations (A.1) in a purely theoretical problem, such as “The Equivalence Principle”, it is absolutely certain we will be led to incorrect theoretical conclusions (which was the case with Einstein in the formulation of “The Equivalence Principle”).We can obviously apply relations (A.1) for educational purposes only in Elemental Mechanics but never in Theoretical Mechanics in purely theoretical problems. In addition, in our problem fig (1.4) , we repeat our experiment with another mass m2, (m2>m1) without altering the values of M, R, and h.As a result, in this case relations (A) are: where υ2 and V’ are the velocities of masses m2 and Μ at the moment of collision, relative to the inertial observer O’. Relation (R.17.6) expresses the strict law on “the free fall of bodies”. In relation (R.17.6), where m2 > m1 it follows that  υ1>υ2 .In other words, in our problem (fig 1.4) we observe that larger masses fall with a lesser velocity (compared to smaller masses) within the gravitational field of mass M!Obviously, when masses m1 and m2 are equal, then from relation (R.17.6) it results that velocities υ1 and υ2 are equal as well (υ1=υ2).In addition, from the first relation (A) it emerges that velocity υ1 is always function of mass m1 , and from the first relation (B) that velocity υ2 is always function of mass m2. The same certainly applies to velocities V and V’ , , which are function of mass M.Consequently, from our observations above, the following two fundamental questions emerge:Question 1. In the law on “the free fall of bodies”, on the basis of which mathematical                     relation does Galileo prove that “all bodies, regardless of their mass,                      fall in the same way”?Question 2. In the “Equivalence Principle”, i.e. the “strong Equivalence Principle”,                    on the basis of which mathematical relation does Einstein prove that                    “all bodies, regardless of their mass, fall in the same way” within the gravitational                    field of another body, as in the example of the aluminium sphere (a), the Moon                    (b) and the Earth (c), mentioned above?Bear in mind that in the above example, the Earth (c) is 6.1024 times bigger (!!!) than the aluminium sphere (a), assuming that the aluminium sphere (a) weights, e.g., 1 kg. Meaning that, if we were to repeat the experiment (under the same conditions) in three separate stages, i.e.:Stage 1: With the Aluminium sphere (a) and distant star MStage 2: with the Moon (b) and distant star MStage 3: With the Earth (c) and distant star M,Then, in all three stages (first, second and third), the Aluminium sphere (a), the Moon (b) and the Earth (c) will fall with the same velocity υ1 onto the surface of distant star M, relative to an inertial observer Ο’;This is obviously a grave mistake on Einstein’s part, when he makes such claims, i.e. in the “Strong Equivalence Principle”. I am sorry, but everything Einstein claims in the “strong Equivalence Principle” (as per the above example) is not simply arbitrary and antiscientific but, above all, laughable…a. What answer do the relativists give to the above two questions?b. Can the relativists show us the mathematical relations employed in the above two questions (1) and (2)?After examining the above, we are led to the following basic conclusion:
 At this point, we need to stress that relations (A), relations (B) and relation (R.17.6) are complete, because they are strict expressions, in terms of quality and quantity, of the natural reality of the problem of the two bodies of fig. (1.4). In the above relations, if we make certain allowances, i.e. that mass m1 is much smaller than mass Μ,
 and mass m2 is much smaller than mass M,
 limiting, in this way, the general shape of our problem), then relation (R.17.6) yields:
 and, based on relations (R.17.7) and (R.17.8), the following results from relation (R.17.9): But even so, with the allowances (R.17.7) and (R.17.8) that we made, it emerges from relation (R.17.10) that υ1υ2 (with minor or major deviations) and never υ1=υ2, as Galileo claims in the law on “the free fall of bodies” and Einstein in the “Equivalence Principle”. (Their claim that that “all bodies, regardless of their mass, fall in the same way”). Based on the above, we can now formulate the following basic law:LAW: In the problem of 2 – bodies, as this is given in fig. (1.4), and based on relation (R.17.6):Velocities υ1 and υ2 and equal (υ1=υ2)only when masses m1 and m2 are equal (m1=m2).In any other case (with any assumptions or allowances), it is always υ1 υ2 and never υ1=υ2. EXAMPLEFrom the top of the Tower of Pisa, we release a mass m1=1kgr made from cotton wool..We then repeat the experiment with a mass m2=1018kgr , consisting of the material of a neutron star. If υ1 is the velocity of mass m1 as it falls onto the surface of the Earth and υ2 is the velocity of mass m2 (as it falls onto the surface of the Earth), we are seeking::How much greater the velocity with which the cotton wool mass falls onto the surface of the Earth is, compared to the velocity of the neutron star mass.The mass of the Earth is taken as M=6.1024kgr and masses m1 and m2 are considered point masses. SOLUTIONFrom relation (R.17.6) we have: Substituting the known values in the above equation, we have: Thus, the velocity υ1 of the cotton wool mass is 1,000000083 times greater than the velocity of  υ2 of the neutron star mass.As we can see, the cotton wool mass m1=1kgr falls to the surface of the Earth with a greater velocity than the neutron star mass  m2=1018kgr !!!That is a nice, tidy conclusion, which proves that Galileo was wrong!!!FIELDS OF GRAVITATIONAL AND INERTIAL FORCESLAW: All bodies, regardless of their mass, “fall” or move in exactly the same way (however great their mass may be) only through inertial force fields and never through gravitational force fields. In other words, inertial force fields and gravitational force fields have huge qualitative and quantitative differences and they are never equivalent to one another. Specifically, the basic differences between gravitational force fields and inertial force fields are as follows:1. Gravitational fields are objective, meaning they exist for all observers (inertial or non-inertial), without exception.2. Conversely, inertial force fields are perceptual, meaning they only exist for observers who are within the accelerating or decelerating reference system (S) and for no other observer. 3. Gravitational fields can never be uniform fields within a space (Ω), i.e. their intensity vectors cannot be parallel to one another, or have the same speed (>O) and direction.4. Conversely, inertial force fields may be uniform fields (e.g. a reference system (S) moving with linear, accelerating motion). 5. In a gravitational field (e.g. spherical symmetry), the intensity vectors of the gravitational field are always convergent, in relation to the centre of the mass that creates the gravitational field.6. Conversely, in any inertial force field, the field’s intensity vectors are always divergent or parallel and never convergent, in relation to the centre of the curvature of the reference system (S) of the inertial forces. In other words, the intensity vectors are always centrifugal, in relation to the observer, who is within the reference system (S).7. The causes that create gravitational fields (i.e. the masses) always influence one another dynamically (i.e. based on the laws of Newton).8. Conversely, the causes that create inertial force fields never influence one other dynamically (i.e. based on the laws of Newton).9. A gravitational field can never (not in a single point of its area, nor locally, nor macroscopically) be equivalent to an inertial or non-inertial reference system. This conclusion is known as the “Non-equivalence Principle of fields”. 10. The inertial mass mi is always equal to the gravitational mass mg and the two masses mi and mg are never equivalent to one another. This conclusion is known as the “Non-equivalence Principle of masses”.11. The “Non-equivalence Principle of fields” and the “Non-equivalence Principle of masses” always co-exist (they are irrevocably tied to one another) and are known collectively as the “Non-equivalence Principle”. 12. The “Non-equivalence Principle” (in terms of physics) expresses the exact opposite of what is expressed by the “Equivalence Principle” of the General Relativity Theory. INERTIAL AND GRAVITATIONAL MASSA COMMON MISTAKE MANY PHYSICISTS MAKE IN TERMS OF “VERIFYING” THE EQUIVALENCE PRINCIPLE OF THE GENERAL RELATIVITY THEORY.As is well known, the fact that a body’s inertial mass mi and its gravitational mass mg are equal, i.e.
 is a law formulated for the first time by Newton. This law (in terms of what we currently know as Physics) has been proven true through Nature, and verified by very accurate experiments, e.g. R. Eotvos, L. Sauthern, P. Zeeman, etc. Unfortunately, however, the equality mi = mg, is often used (incorrectly) as “proof” of the Equivalence Principle of the General Relativity Theory. Attention! This is where confusion arises, and the “sensitive” point of our problem, as, specifically:1. The equality mi = mg does not necessarily mean that masses mi and mg are equivalent. . 2. Conversely, however, the equivalence of masses mi and mg does mean that masses  mi and mg, are equal, i.e. mi = mg.Because, however (as we previously proved), the Equivalence Principle of the General Relativity Theory is false, it follows that the equivalence of masses mi and mg does not exist in Nature. Consequently, as case (2) above is not valid, case (1) is valid by necessity (and has been verified through experiments, as mentioned above), meaning that mi = mg is always valid, while masses mi and mg are never equivalent.Unfortunately, many physicists confuse and equate (mistakenly) the two concepts of equality and equivalence of masses mi and mg.That is obviously a grave mistake of the part of those physicists, because the equality and the equivalence of the masses mi and mg, are completely different “things” and should never be confused or taken as one and the same. Indeed, the equality mi = mg (which does, in fact, exist in Nature) must never be used to purportedly “prove” the “Equivalence Principle” of the General Relativity Theory, which is completely false, as we proved earlier.CONCLUSIONAfter the above discussion, we conclude that:1. Galileo’s law on the “free fall of bodies” is false.2. The “Equivalence Principle” (both strong and weak) of the General Relativity Theory is a false principle of Physics.3. A body’s inertial mass  mi is always equal to its gravitational mass mg, (mi = mg) and the two masses mi and mg are never equivalent  to one another. COMMENTOnce more, I would like inform all physicists (who believe that the Relativity Theory is correct) that: that are mistaken, and are doing nothing more than maintaining the “mummy” of Relativity Theory, a theory which has been dead since the moment of its birth, in the year 1905!I am certain that the Physicists of the future will laugh at this ridiculous theory of Physics, the “Relativity Theory”!
 CONCLUSIONS    Ι. GALILEO    As it is known, Galileo (based on his historical experiment conducted from the Leaning Tower of Pisa) asserts that all bodies, irrespective of their mass, fall at the same velocity in the gravitational field of a mass M (such as the Earth).    However, –and according to what has been demonstrated above– what Galileo asserts is utterly wrong, for instance when it comes to unequal masses m1 and m2 which are in free-fall in the gravitational field of mass M,either the gravitational field of mass M is non-uniform or uniform (Fig. 1).     ΙΙ. EINSTEIN    a) As it is well-known, in formulating the “Equivalence Principle” of the General Theory of Relativity, Einstein relied on the accuracy of the results of Galileo’s experiment.   Nevertheless, according to everything proven above, Galileo erred in his assertions, and consequently so did Einstein when he formulated the “Equivalence Principle”.    Specifically where the Equivalence Principle is concerned, Einstein by resting on the erroneous results of Galileo’s experiment asserts that the gravitational fields are equivalent to the fields of inertial forces, in the sense that gravitational fields cause all bodies, irrespective of their mass, to undergo thesame acceleration.   Yet, what Einstein asserts is utterly wrong, since only the fields of inertial forces have this property and not gravitational fields, as demonstrated earlier.    b) Moreover, Einstein’s Equivalence Principle asserts the following:     “A frame of reference, once in free fall in the gravitational field of a mass M, is locally equivalent to an inertial frame of reference”.   However, Einstein’s assertion is once again wrong for the following reason:    Let us assume that spherical shell m1(Fig. 2) is an “elevator” free-falling in the gravitational field of mass M.

Then, if an observer Ο΄ who is inside the “elevator” m1 places at a time t=0 a point mass m2 (m1≠m2) at the center of the spherical shell (elevator), mass m2 after a time t>0 will not remain at the center of the spherical shell (as Einstein holds) but will move from its center, as proven earlier (Fig. 1).

Conclusion

Therefore, the “Equivalence Principle” of the General Theory of Relativity fails to apply not merely “locally” but to any single point at the center of the spherical shell (elevator), as Einstein erroneously asserts based on his “elevator” thought experiments that led him to the formulation of the “EquivalencePrinciple” of the General Theory of Relativity.
Consequently (as proven above), because of the invalidity of the “Equivalence Principle” of the General Theory of Relativity, the inertial mass Μi of a body is never equivalent to gravitational mass Μg.
On the other hand, a body’s inertial mass Μi and its gravitational mass Μg are always equal, i.e. Μi=Mg, these two masses are never equivalent to each other as Einstein erroneously asserts in the “Equivalence Principle” of the General Theory of Relativity.
Thus, after everything analyzed above, we reach the conclusion that Galileo and Einstein (the latter relied on the experiment of the former) drew and formulated totally wrong conclusions in respect of free-fall, especially in the case of “Galileo’s experiment” on which the “Equivalence Principle” of the General Theory of Relativity was predicated.
Therefore (as demonstrated above), since the “Equivalence Principle” of the General Theory of Relativity is not valid, then the Theory of Relativity itself should be deemed as an utterly erroneous theory of Physics, failing to represent physical reality under all circumstances.

CRITIQUE

1. Critique of relation γ=g in Galileo’s experiment

As it is well-known, Einstein’s General Theory of Relativity postulates the following:
“Gravitational fields have the remarkable property of imparting the same acceleration to all bodies independent of their material composition and amount of mass”.

(See THE PRINCIPLE OF RELATIVITY, BY A. EINSTEIN – H.A LORENTZ – H. MINKOWSKI – H. WEYL, Page 114 DOVER PUBLICATIONS, INC).
Einstein based this conclusion on the all-too-familiar Galileo’s experiment (the experiment conducted from the Leaning Tower of Pisa).
All the above generate the following question:
Are Einstein’s assertions true or false?
The answer is that what Einstein asserts is false, for the following reasons:
With regard to the problem (Fig. 1) discussed earlier, it has been proven that:
a) In relation to an inert observer Ο, Fig.1 the two masses m1 and m2 (m1< m2), fall at different velocities υ1 and υ2, (υ1>υ2) in the gravitational field of mass Μ. (See  “The fundamental law of the free-fall”, relation (29)).
b) Conversely, Einstein by accepting in this case the conclusion of Galileo’s experiment as true, asserts that:
In relation to inert observer Ο, Fig. 1, the two masses m1 and m2 fall at the same velocity, i.e. υ1 2 at all times.
Yet, which of these two conclusions (a) and (b) is true?
Conclusion (a) or conclusion (b)?
The answer to this question is the following: Conclusion (a) is true for the reasons explained below:
Einstein’s major error lies in that he did not apply
1. The “principle of conservation of energy”, and
2. The“principle of conservation of momentum”
in order to find the accurate solution to the problem encountered in the system of the three masses m1,m2 and M; instead,  for masses m1and m2 he employed only Newton’s  first and second law , maintaining that:
For a body of any mass m which is in free fall inside the gravitational field of mass M, the following relation applies:

and Einstein continues......
Because force F which causes the body to move is its weight W, then:

and because:

relations (33) and (34) yield:

 Relation (37) (as it is well known), expresses the conclusion of Galileo’s experiment.   Einstein erred in his interpretation of the resulting relation (37), namely by asserting that:   “Because mass m of a body does not exist in relation (37), then acceleration γ is the same for all bodies that are free-falling in the gravitational field of a mass M, independent of these bodies’ mass”.    This is exactly why Einstein mentions the remarkable property of gravitational fields referred to above, which he bases on the conclusion of Galileo’s experiment expressed by relation (37).   However, here lies Einstein’s great error, because the accurate interpretation of relation (37) is the following:    Let us assume that there is a body of mass m which is in free fall in the gravitational field of a mass M.      Moreover, let us assume that during time t = t1>0, body m lies at a distance from mass M.    Then according to Newton’s first and second law, during time t1 its acceleration will be:
 As it can be observed in relation (39), mass m of the body does not exist.   However, this does not signify that all bodies, independent of their mass, fall freely in the gravitational field of mass M with the same acceleration.    The explanation to that is very simple, because:   As it can be observed in relation (39), acceleration γt is a function of distance  ht, that is: However, according to the “two-body problem”, distance ht relative to an inert observer Ο is always a function of masses m, M, of distance h (initially separating the two masses m and M at time t = 0, where their velocities equaled 0) and of time t, that is: Therefore, relations (39), (40) and (41) yield:
 Thus, the above lead to the following conclusion:The value of acceleration  γt of a body which is in free fall in the gravitational field of mass M at time   t = t1 is given by relation (39),  where is the distance between the two masses m and M at time t = t1.   This acceleration γt  is never independent of mass m of the body (as Einstein erroneously asserts on the basis of relation (37)), but is always a function of mass m of the falling body, of mass M, of the initial distance h and of time t, according to relation (42).     At this point it should be noted that during the free fall of body m in the gravitational field of mass M, distance ht between the two masses m and Μ constantly shortens, due to the universal force of attraction exerted by one mass over the other, a fact that was never taken into account during the performance of the Galileo experiment.    Consequently, the above conclusion opposes the one of Galileo’s experiment expressed by relation (37).    Therefore, gravitational fields do not have the remarkable property of imparting the same acceleration to all bodies independent of their mass, as Einstein falsely holds.     This property is observed exclusively in fields of inert  forces and absolutely not in any other field of forces in Nature.    Hence Einstein’s major error!!!2. Critique of Galileo’s experiment   As it is well known, the conclusion of Galileo’s experiment (at the Leaning Tower of Pisa) which is interpreted based on relation (37) and states that “all bodies fall at the same rate (velocity), irrespective of their mass,in the gravitational field of mass M” is utterly false for the reasons given below:    In the first place, Galileo’s experiment is a problem involving three bodies (special case); See problem in Fig. 1.   A direct consequence of  the resolution of this problem, is that, because mass M of the Earth is much greater than masses m1 and m2, (m1
 ii. Fraction υ2 / υ1 of velocities υ2 and υ1 always obeys the following relation:
 where,
 iii. When masses m1 and m2 are equal (m1 = m2), then velocities υ1 and υ2 are also the same, i.e. υ2 = υ1.
 Law 2: i. Momentum J1 of the smaller mass m1 is less than momentum J2 of the larger mass m2, hence                the relation:
 ii. When masses m1 and m2 are equal (m1 = m2), then the momentums are also the same, that is, J1 = J2 ή m1 υ1 = m2 υ2.
 Law 3: According to the “two body problem”:             i. The acceleration  of a body having mass m and falling freely from a height h in the              gravitational field of mass M, at time t1 is given by the following relation:
 where,   the distance between masses m and Μ at time t1.ii. Acceleration of the free-falling body is always a function of its mass m, of mass M, of height h and of time t, hence the relation: iii. This acceleration is never independent of the body’s mass m, as Galileo erroneously asserts These are therefore the three laws of free fall.
 DETECTOR OF GRAVITATIONAL FIELDS A spherical shell of mass m1, within which there is another mass m2 (m1 ≠ m2)  (Fig. 2), will be called “detector of gravitational fields”. Analytically: (a) If mass m2 is found at the center of the spherical shell m1, then the detector of      gravitational fields will be in its “constant” state and will be symbolized as G+, (Fig. 2a).(b) Conversely, if mass m2 is not found at the center of spherical shell m1, then the detector     of gravitational fields will be in its “inconstant” state and will be symbolized as G-,      (Fig. 2b).The great importance of the “detector of gravitational fields” lies in that it fully explains the concept of “Locally”, which the General Theory of Relativity uses in its various experiments (such as the well-known elevator experiment).
 Fig. 2 Note: If in Fig. 2 mass m1 consists of a light material, e.g. aluminum, wood, etc, and mass m2 consists of the material of a white dwarf, black hole, etc, then the above detector will be called “detector of gravitational fields of a thick mass”. CASE 1: Let us assume (Fig. 3) that there is an inertial frame of reference (S). An observer O who stands in this inertial frame of reference, places the detector of gravitational fields in its “constant” state G+ at a point P. Then (t = 0), he lets the detector of gravitational fields move freely .After a time t > 0, the detector of gravitational fields will be at the exact same point P in its “constant” state G+. Fig. 3 CASE 2: Lets us assume (Fig. 4), that an elevator (S) falls freely in the gravitational field of a celestial body of mass M0 (earth, moon, asteroid, etc.). An observer O who stands in the elevator places at a point P the detector of gravitational fields in its “constant” state G+. Then (t = 0), the observer lets the detector of gravitational fields move freely.According to the “three-body problem” (m1, m2, M0), (m1 ≠ m2) referred to above, after a time t > 0, the detector of gravitational fields will be in the elevator in its “inconstant” state G-. Fig. 4 CASE 3: Let us assume (fig.5) that an elevator cab (S) moves with acceleration γ away from the gravitational fields. An observer O who is found in the cab places at a point P the detector of gravitational fields in its “constant” state G+. Then (t = 0) the observer lets the detector of gravitational fields move freely.After a time t > 0, the detector of gravitational fields will be at another point P΄ again in its “constant” state G+. Fig. 5 CASE 4: Let us assume (Fig. 6) that an elevator cab (S) is at rest at the top of a tall building or a pillar of height h from the surface of a celestial body of mass M0 (earth, moon, asteroid, etc). An observer O who is found in the cab places at a point P the detector of gravitational fields in its “constant” state G+ (Fig. 6a).Then (t = 0) the observer lets the detector of gravitational fields move freely.According to the “three-body problem” (m1, m2, M0), (m1 ≠ m2) referred to above, after a time t > 0, the detector of gravitational fields will be at another point P΄ in its “inconstant” state G- (Fig. 6b). Fig. 6 RECAPITULATIONAccording to the three fundamental laws and the way of operation of the D.G.F. (DETECTOR OF GRAVITATIONAL FIELDS) that we mentioned above, we are now able to formulate the following law: THE FUNDAMENTAL LAW OF SEPARATION OF GRAVITITATIONAL FIELDS AND FIELDS OF INERTIAL FORCESLAW:An observer O which is within a frame of reference (S), O.XYZ that moves:a.    In any kind of movement on a plain field or in space (eg. rectilinear accelerating, circular, spiral etc), away from the gravitational field of a mass M, or b.    Falls freely within the gravitational field of the mass M, orc.    Is idle at a constant height h, above the mass M, ord.    Is an inertial frame of reference(see above cases) Then, the observer O, using a D.G.F, can easily know “locally” and particularly for each point Pi (Xi, Yi, Zi), i = 1, 2, 3… of the frame of reference O.XYZ whether the forces exercised on the point mass, which is located at the center Pi  of the D.G.F. are due to a gravitational  field (gravitational forces), or they are due to a field of inertial forces (inertial forces) or the  frame  of reference O.XYZ  is an inertial frame of reference.This foundational law that we mentioned above is fully contrary to the “Equivalence Principle” of the General Theory of Relativity, which (the Equivalence Principle), according to this law, is proven to be a wrong principle of Physics.Summarizing all the above, we come to the conclusion that:The “Equivalence Principle” of the General Theory of Relativity is wrong and consequently:1. The fields of the gravitational and inertial forces are never of the same nature (equivalent) as Einstein claims, but they are two completely different fields and these fields are different in nature (not equivalent). 2. The gravitational mass Mg of a body M is always equal to the inertial mass Mi, namely Mg = Mi, and these two masses Mi and Mg are neverequivalent to one another, as Einstein wrongly claims. TWO QUESTIONS PUT TO “EINSTEIN” Α. GRAVITATIONAL AND INERTIAL MASS   Let us assume that Ο.ΧYZ (Fig.1) is an inertial frame of reference and (Ο) is an observer standing in it.   Also, m0 and Μ0, (m0 < M0) are two masses which at time t=0 are at rest and lie at a distance h from each other.   We now let the two masses m0 and M0 move freely under the influence of the universal force of attraction.    After a time t > 0, the distance between the two masses m0 and M0 will be h΄, (h΄< h) and their velocities v and V respectively, (“two-body problem”). As it is well-known, according to the Theory of Relativity, when a mass moves relative to an inertial observer (Ο), then this mass increases in relation to the rest mass it had in the inertial frame of reference.    Therefore, in the case shown in Fig. 1, relative to the inertial observer (Ο) and at time t > 0, the two rest masses m0 and M0 will increase (mass = function of velocity) and the following relations will apply:                               m = m0 f (v)             (1)                                M = M0 f (V)             (2)where, m > m0 and M > M0.   Moreover, as it is well-known, according to the Theory of Relativity, when a mass moves under the influence of a gravitational field, then the inertial and gravitational masses are equivalent (according to the “equivalence principle”) and consequently since they are equivalent they will also be equal.   Thus, on the basis of the above, relative to the inertial observer (Ο) and at time t > 0, the following relations will apply for the two masses m and Μ in Relations (1) and (2):                               mi = mg                 (3)                                Mi = Mg                 (4)where, mi, Mi are the inertial masses of masses m and Μ respectively, and          mg, Mg are the gravitational masses of masses m and Μ respectively.    Therefore, after everything presented above, we are putting to “Einstein” the following question:   QUESTION I: In the example of Fig. 1 (“two-body problem”), let us consider e.g. the smaller mass m. Then:    Relative to the inertial observer (Ο) and at time t > 0, on the basis of the axiomatic foundation and the laws of the Theory of Relativity, and taking into account the fact that the mass increases with velocity (Relation (1)), in this case:a. By which mathematical formula A of the Theory of Relativity is inertial mass mi expressed, relative to inertial observer (Ο)?b. By which mathematical formula Β of the Theory of Relativity is gravitational mass mg expressed, relative to inertial observer (Ο)? c. How does relation mi = mg (as the Theory of Relativity maintains) relative to inertial observer (Ο) result from the above two mathematical formulas A and B?We would very much like “Einstein’s” answer to these questions.Β. GALILEO’S EXPERIMENT   As it is well-known, in formulating the “Equivalence Principle” of the General Theory of Relativity, Einstein accepts the results of Galileo’s experiment as utterly correct (the Tower of Pisa experiment).    Thus, in this case, we are putting the following question to “Einstein”:   QUESTION II: Based on the axiomatic foundation and the laws of the Theory of Relativity and taking into account the fact that the mass increases with velocity, can “Einstein” prove to us that the result of Galileo’s experiment is accurate?   That is, can “Einstein” prove (Fig. 2) that: Relative to an inertial frame of reference O.XYZ, two unequal masses (with rest masses) m0 and m0΄, (m0 < m0΄) which are let simultaneously to fall freely from this height h in the gravitational field of a third mass Μ0, (taking into account the fact that these masses increase with velocity during their free-fall), do these two masses m0 and m0΄ fall always at the same velocity?    Therefore, where is the full mathematical proof –according to the axiomatic foundation and the laws of the Theory of Relativity– of Galileo’s experiment accuracy?We would very much like “Einstein’s” answer to this question, also.COMMENT   Can “Einstein” provide an answer to the questions posed above?    From my point of view, NO!    Consequently, since Einstein cannot answer the above two questions, then the Theory of Relativity is an erroneous Theory of Physics due to its inconsistence and many internal contradictions, as stated above. NOTE: Any physicist, mathematician, university who/which objects to everything analyzed above, is welcome to publish his/her/its objections, so that a constructive dialogue can be established. Because undoubtedly dialogue provides the basis for scientific research. EPILOGUEEnough energy, time and money have been wasted on the Theory of Relativity so far.It is no use wasting any more, because:  The Theory of Relativity has been proven to be an utterly erroneous Theory of Physics!!!Copyright 2006: Christos A. Tsolkas                                              Christos Α. Τsolkas                                                                                                              October 2006 ©  Copyright 2001 Tsolkas Christos.  Web design by Wirenet Communications