LORENTZ TRANSFORMATIONS IMPLICITASSUMPTION AND ITS CONSEQUENCES

Louai Hassan Elzein Basheir∗(Dated: January 9, 2015)

This paper is prepared to show that the spacetime has wave properties and wave equation whileLorentz transformations represent the Doppler effect for that wave, and also shows the derivationof the correct relativistic factor and shows its consequences. It also shows the refutation of theLorentz-Fitzgerald length contraction and the refutation of the relativistic mass.

THEORETICAL BACKGROUND

1.1 The Lorentz Transformations

The Lorentz transformations are set of equationsin relativity physics that relate the space and timecoordinates of two systems moving at a constant velocityrelative to each other. Required to describe high-speedphenomena approaching the speed of light, Lorentztransformations formally express the relativity conceptsthat space and time are not absolute; that length ,time , and mass depend on the relative motion ofthe observer; and that the speed of light in a vacuumis constant and independent of the motion of theobserver or the source. The equations were developedby the Dutch physicist Hendrik Antoon Lorentz in 1904.1

For the important of the Lorentz Transformations, andbecause our findings in this paper are depending upon itspostulates, we are going at the beginning to transcribe itsderivation fully as it is mentioned in the reference "Conceptsof Modern Physics" 2:

1.2 Derivation of Lorentz Transformations

In an inertial frame of reference S the coordinatesof some event occur at the time t are x, y, z. Anobserver located in a different inertial frame S′ whichis moving with respect to S at the constant velocityv will find that the same event occurs at the time t′

and has the coordinates x′, y′, z′. (In order to simplifyour work, we shall assume that v is in the + direc-tion). The measurements x, y, z, t related to x′, y′, z′, t′:

First there is no differences between the corresponding

∗ Khartoum University, Physics College.P.O. Box: 7725 - Zip Code: 11123Email: louaielzein@yahoo.com

1 "Lorentz transformations." Chicago: EncyclopÃędia Britannica,2009.

2 Concepts of Modern Physics (6th ed. International Edition),Arthur Beiser, McGraw Hill, ISBN 007-244848-2.

coordinates y, y′ and z, z′ which are perpendicular tothe direction of the velocity, v, then:

z′ = z and y′ = y

a reasonable guess about the nature of the correct rela-tion between x and x′ is

x′ = k(x− vt) (1)

Here k is factor that does not depend upon either x or tbut may be a function of v.

To write the corresponding equation for x in term ofx′ and t′:

x = k(x′ + vt′) (2)

The time coordinates t and t′, however, are not equal.We can see this by substituting the value of x′ given byEq.(1) into Eq.(2). This gives

x = k2(x− vt) + kvt′

from which we find that

t′ = kt+

(1− k2

kv

)x (3)

Equations (1) and (2) to (3) constitute a coordinatetransformation that satisfies the first postulate of specialrelativity.

The second postulate of relativity gives us a way toevaluate k. At the instant t = 0, the origins of the twoframes of reference S and S′ are in the same place, ac-cording to our initial conditions, and t′ = 0 then also.Suppose that a flare is set off at the common origin of Sand S′ at t = t′ = 0, and the observers in each systemmeasure the speed with which the flares’ light spreadsout. Both observers must find the same speed c, whichmeans that in the S frame

x = ct (4)

and in the S′ frame

x′ = ct′ (5)

Substituting for x′ and t′ in Eq.(5) with help of Eqs.(1)and (3) gives

k(x− vt) = ckt+

(1− k2

kv

)cx

2

and solving for x,

x =ckt+ vkt

k −(

1− k2

kv

)c

= ct

k +v

ck

k −(

1− k2

kv

)c

= ct

1 +v

c

1−(

1

k2− 1

)c

v

This expression for x will be the same as that given byEq.(4), namly, x = ct, provided that the quantity in thebrackets equals 1. Therefore

1 +v

c

1−(

1

k2− 1

)c

v

= 1

and

k =1√

1− v2/c2(6)

Finally we put this value of k in Eqs.(1) and (3). Nowwe have the complete transformation of measurements ofan event made in S to the corresponding measurementsmade in S′:

x′ =x− vt√1− v2/c2

(7)

y = y′

z = z′

t′ =t− vx

c2√1− v2/c2

(8)

These equations comprise the Lorentz Transforma-tion.

Inverse Lorentz Transformation, which is convertmeasurements made in the moving frame S′ to theirequivalents in S. To obtain the inverse transformation,primed and unprimed quantities in Eqs.(7) to (8) areexchanged, and v is replaced by −v:

x =x′ + vt′√1− v2/c2

(9)

y = y′

z = z′

t =t′ +

vx′

c2√1− v2/c2

(10)

These equations, Eqs.(9) to (10), comprise the InverseLorentz Transformation.

1.3 Velocity Addition

FIG. 1. Velocity Addition

Special relativity postulates that the speed of light c infree space has the same value for all observers, regardlessof their relative motion. We must turn to the Lorentztransformation equations for the correct scheme of veloc-ity addition.

Suppose something is moving relative to both S and S′.An observer in S measures its three velocity componentsto be

Vx =dx

dtVy =

dy

dtVz =

dz

dt

while to an observer in S′ they are

V ′x =dx′

dt′V ′y =

dy′

dt′V ′z =

dz′

dt′

By differentiating the inverse Lorentz transformationequation for x, y, z, and t, we obtain

dx =dx′ + vdt′√

1− v2/c2

dy = dy′

dz = dz′

3

dt =dt′ +

vdx′

c2√1− v2/c2

and so

Vx =dx

dt=

dx′ + vdt′

dt′ +v

c2dx′

=

dx′

dt′+ v

1 +v

c2dx′

dt′

(11)

Relativistic velocity transformation

Vx =V′x + v

1 +v

c2V′x

(12)

1.4 Derivation of Time Dilation Formulafrom Lorentz Transformations

Let us consider a clock at the point x′ in the movingframe S′. When an observer in S′ finds that the timeis t′1, an observer in S will find it to be t1, where fromEq.(10),

t1 =t′1 +

v

c2x′√

1− v2/c2

after a time interval of to (to him), the observer in themoving system finds that the time is now t′2 according tohis clock. That is,

to = t′2 − t′1

The observer in S, however, measures the end of the sametime interval to be

t2 =t′2 +

v

c2x′√

1− v2/c2

so to her the duration of the interval t is

t = t2 − t1

t =t′2 − t′1√1− v2/c2

=to√

1− v2/c2

Then the time dilation is

t =to√

1− v2/c2(13)

where to is the proper time.

We define γ and β to be:

γ =1√

1− v2/c2and β =

v

c

Then the time dilation formula becomes

t = γto (14)

ANALYSIS

2.1.0 Derivation of Lorentz TransformationsDirectly from

Galilean Coordinate Transformations

The Galilean coordinate transformations are (seeFIG.1):

x′ = x− vt (15)x = x′ + vt′ (16)t = t′ (17)

y = y′

z = z′

We have equations (4) and (5) which represent experi-mental findings (also represent Einstein’s second postu-late), then substituting it into Eqs.(15) and (16), we find

x′ = x− vt= ct− vt= (c− v)t (18)

and

x = x′ + vt′

= ct′ + vt′

= (c+ v)t′ (19)

Dividing Eq.(19) into (18), we find

x′

x=

(c− v)t

(c+ v)t′

x′t′ = xt

(c− vc+ v

)

x′(x′

c

)= x

(xc

) c− vc+ v

x′2 = x2(c− vc+ v

)

4

Taking the square root, we get

x′ = x

√c− v

c + v(20)

= x

√(c− v)(c− v)

(c+ v)(c− v)

= xc− v√c2 − v2

= xc(

1− v

c

)√c2(

1− v2

c2

)

=x− v

(xc

)√

1− v2

c2

and

x′ =x− vt√1− v2/c2

(21)

Which is the Lorentz transformation for positioncoordinate.

Continuing with the derivation, we find

x′ = γ(x− vt)

x′ = γ(ct− vx

c

)

x′ = γc(t− v

c2x)

x′

c= γ

(t− v

c2x)

and

t′ =t− v

c2x√

1− v2/c2(22)

Which is the Lorentz transformation for time coordi-nate.

Now dividing Eq.(18) into (19), we obtain

x

x′=

(c+ v)t′

(c− v)t

xt = x′t′(c+ v

c− v

)

x(xc

)= x′

(x′

c

)c+ v

c− v

x2 = x′2(c+ v

c− v

)

Taking the square root, we get

x = x′√

c + v

c− v(23)

= x′

√(c+ v)(c+ v)

(c− v)(c+ v)

= x′c+ v√c2 − v2

= x′c(

1 +v

c

)√c2(

1− v2

c2

)

=

x′ + v

(x′

c

)√

1− v2

c2

and

x =x′ + vt′√1− v2/c2

(24)

Which is the inverse Lorentz transformation forposition coordinate.

5

Continuing with the derivation, we find

x = γ(x′ + vt′)

x = γ

(ct′ + v

x′

c

)

x = γc(t′ +

v

c2x′)

x

c= γ

(t′ +

v

c2x′)

and

t =t′ +

v

c2x′√

1− v2/c2(25)

Which is the inverse Lorentz transformation for timecoordinate.

Thus, from direct substitution of Eqs.(4) and (5),which represent experimental findings at the first place,into Galilean transformations, one will discover throughmathematical necessity the relativistic of the time andwithout postulated it.

2.2.0 The Necessity of The Lorentz Transformations

From Eqs.(18) and (19), then with the help of Eqs.(4)and (5), we obtain

x′ = (c− v)t

=(

1− v

c

)ct and

=(

1− v

c

)x (26)

And Eq.(19) yields

x = (c+ v)t′

=(

1 +v

c

)ct′ and

=(

1 +v

c

)x′ (27)

Solving Eqs.(26) and (27) for x′ and equating them, wefind

x

1 +v

c

=(

1− v

c

)x

x =

(1− v2

c2

)x

1− v2

c2= 1 , or γ = 1 consequently, (28)

v = 0 (29)

Substituting v = 0, the relativistic velocity, into Galileantransformations and Lorentz transformations yields

x = x′ and t = t′

which are valid consequences because v = 0 means thatall observers are aboard the same frame of reference.Because of the space been found to be related to thetime through the universal constant c (the speed of light-that is, x = ct), and due to substitution of x = ct intoGalilean transformations which yield Lorentz transfor-mations, the factor γ becomes a part of the Galileantransformations NOT a correction factor that applied toan incorrect type of transformations-that is, γ is a miss-ing part from Galilean transformations. Consequently,because v = 0 ⇒ γ = 1, and through investigation ofthe Galilean transformations (Eqs.(15) and (16)) we willdiscover that γ = 1 but v 6= 0 (notice that, we can write theGalilean transformations as: ( x′ = (γ = 1)(x − vt) and x =

(γ = 1)(x′ + vt′) ), we conclude that the Galilean trans-formations are not the complete derived form of trans-formations while the Lorentz transformations are.

2.3.0 The Correct Relativistic Factor

From Eqs.(20) and (23) we notice the factor√

c−vc+v and

its reciprocal√

c+vc−v therefore, let us denote this factor

with %. This factor represents the correct, or the completederived relativistic factor instead of the Lorentz factorγ. Therefore we have

6

FIG. 2. Ob′1, Ob′2 and the light source are aboard the same frame of reference, S′, thus the relative velocity between them iszero so both Ob′1, Ob′2 will agree about the wavelength that being observed. Although Ob1, Ob2 are aboard the same frame ofreference, S, they will disagree about the wavelength that being observed due to the asymmetry of the relativistic effect.

% = %(v) =

√c + v

c− v(30)

%−1 = %(−v) =

√c− v

c + v(31)

%(−v) + %(v)

= γ (32)

%(−v)%(v) = 1 (33)

%() = 1 (34)

%(−c) = 0 (35)

%(c) = ∞ (36)

and in any arbitrary direction in two dimension

%(β, θ) =

√1 + β cos θ

1− β cos θ(37)

From Eq.(20) and with the help of Eqs(5) and (4) weobtain

x′(v) =x

%(38)

ct′ =ct

%and

t′(v) =t

%(39)

Similarly, we obtain the same result from Eq.(23)-that is

t(v) = %t′ (40)

7

Equation (39) represents the time dilation-deflation for-mula while Eq.(20) or (38) represents space expansion-contraction formula where % is the relativistic factor.

2.4.0 Length Expansion

Using the relativistic factor γ we can also prove theexistence of the time dilation and space expansion butNOT the time deflation and the space contraction.

2.4.1 Proving Length Expansion from LorentzTransformations

Solving Lorentz transformations for v, the relative ve-locity (Eqs.(7) and (8)), yields

v =x

t− x′

γt(41)

v =c2t

x− c2t′

γx(42)

Equating Eqs.(41) and (42) gives

x

t− x′

γt=c2t

x− c2t′

γx

Multiplying through by xt, we get

x2 − xx′

γ= c2t2 − c2tt′

γ,

x2 − c2t2 = x

(x′

γ

)− c2t

(t′

γ

)(43)

Equation (43) implies that:

x =x′

γand t =

t′

γ, or

x′ = γx , t′ = γt

Similarly, when we use the inverse Lorentz transforma-tions, we obtain

x = γx′ , t = γt′

Thus, the general result is

x = γxo (44)t = γto (45)

where xo and to are the proper length and proper time.

Equation 44 proves the length expansion effectinstead of length contraction.

2.4.2 Lorentz-Fitzgerald Length Contractionvs

The Relativistic Factor %

A. Derivation of Lorentz-FitzgeraldLength Contraction3

FIG. 3. A horizontal pulse of light being reflected from avertical mirror in S, as viewed by observers in S′.

At first, we are going to mention the derivation of theLorentz-Fitzgerald length contraction then from it wewill prove the opposite of that effect, namely the lengthexpansion.

Consider the situation schematically illustrated inFigure 3, where observers in S have placed a mirrorM perpendicular to their X-axis, thus, observers in Sreason that

c =2∆x

∆t

where ∆t is the time interval required for the light pulseto travel A to M and back to A. Thus, the distance theymeasure between A and M is given by

∆x = ½c∆t, (46)

where ∆t should be interpreted as a length measurement.Observers in system S′ are operationally perform the

necessary measurements for the length as the following;consider the motion of the light pulse from A to M as

3 Marshall l. Burns, Modern Physics for Science and Engineering,p.44 to 48.* For the convenience of the reader we switch the symbols thathave been used in the reference to be consistence with that weuse in this paper-that is, S′, x′, y′, z′, t′ becomes S, x, y, z, t andvice versa.

8

FIG. 4. A horizontal pulse of light propagating towards avertical mirror in S, as viewed by observers in S′.

viewed by observers in the S′-frame and depicted in Fig-ure 4.

∆x′ = (c− v)∆t′A−B . (47)

Thus, the time interval ∆t′A−B is given by

∆t′A−B =∆x′

c− v.

FIG. 5. A horizontal pulse of light being reflected from avertical mirror in S, as viewed by observers in S′.

Now, consider the motion of the light pulse from Mback to the Y-axis, as viewed by observers in system S′.In this case, the distance ∆x′ is immediately obtainedfrom Figure 5 as

∆x′ = (c+ v)∆t′B−C . (48)

Solving the above equation for the time interval ∆t′B−Cgives

∆t′B−C =∆x′

c+ v.

which is not the same as ∆t′A−B as far as observers insystem S′ are concerned.

Let observers in S′ define4 the time interval ∆t′

as that time required for the pulse of light in S totravel from A to M and back to A, as viewed in theirreference frame. Clearly, then

∆t′ = ∆t′A−B + ∆t′B−C =∆x′

c− v+

∆x′

c+ v

which, when solved for ∆x′, yields

∆x′ = ½c∆t′(

1− v2

c2

). (49)

Substitution of Eq.(14) into Eq.(49) gives

∆x′ = ½cγ∆t (1− β2), (50)

Now, from Eq.(46) and (50) we have the famous Lorentz-Fitzgerald length contraction equation in the form

∆x′ =∆x

γ(51)

In view of this result, the length measurement obtainedby S′-observers on a moving object is always less thanthe corresponding proper length measured by S-observerson the object at rest (i.e., ∆x′ < ∆x), when v is notsmall and the length of the object is parallel to the axisof relative motion.

B. Derivation of The Relativistic Factor % fromLorentz-Fitzgerald Length Contraction

To prove the existence of length expansion effect in-stead of length contraction, we are going to use the samederivation of Lorentz-Fitzgerald length contraction. Thenwe have from equation 47

∆x′ = (c− v)∆t′A−B ,

But from Eq.(14), we get

∆t′A−B = γ∆tA−B = γ(½∆t),

Thus we have

∆x′ = (c− v) (½γ∆t),

= ½c∆tc− v√c2 − v2

= ½c∆t√c− vc+ v

4 Later we will find this definition is what led to the incorrectderivation of the length contraction relativistic effect, where therelativistic effect on length and time is not symmetrical in everydirection.

9

Finally, with substituting Eq.(46), we obtain

∆x′ = ∆x

√c− vc+ v

, or (52)

∆x′A−B =∆x

%

Similarly, equation 48 yields

∆x′ = ∆x

√c+ v

c− v, , or (53)

∆x′B−C = %∆x

Therefore,

∆x′A−B 6= ∆x′B−C (54)

Equations (52) and (53) confirm our previous resultsabout asymmetry of the relativistic effect on length dueto the direction (approaching or moving away), namelyEqs.(20) and (23).

Now when dividing Eq.(45) into (44) or Eq.(39)into (38), we find

x

t=xoto

= const. (55)

Let us denote this constant with ε. To find the value ofthis constant we divide Eq.(8) into (7) which gives

x′

t′= ε =

x− vtt− v

c2x

=

x

t− v

1− v

c2x

t

.t

t

But from Eq.(55) we find that xt = ε, then

ε =ε− v

1− ε vc2

Cross multiply

ε− ε2 vc2

= ε− v

ε2v

c2= v and ε = c (56)

Thus, Equation (55) becomes

x = ct and x′ = ct′

Which are the same equations (4) and (5).

3.1.1 Relativistic Energy and Momentum and thepresence

ofde Broglie Wave Velocity

The most famous relationship Einstein obtained fromthe postulates of special relativity concerns mass and en-ergy. The total energy E of the object is

E = γmc2 = mc2 + KE (57)

where Eo = mc2 is the rest energy.

We know we can express the total energy, E, interm of momentum, p, and rest energy Eo as

E2 = (mc2)2 + p2c2 (58)

Equating Eqs.(57) and (58), we find

γmc2 = ((mc2)2 + (pc)2)12

γ2(mc2)2 = (mc2)2 + (pc)2

(pc)2 = (γ2 − 1)(mc2)2

pc = mc2 (γ2 − 1)12

p = mc (γ2 − 1)12

p = β(γmc) (59)

multiplying by c, we obtain

pc = β(γmc2) = βE (60)

Substituting Eq.(60) into Eq.(58), we get

E2 = E2o + β2E2

(1− β2)E2 = E2o

E2 =E2o

1− β2

Thus, we have

E = γEo (61)

Substituting Eq.(61) into Eq.(60), we find

pc = βγEo

p = γv

(Eoc2

)

p = γmov

10

where mo is the rest mass so we can understand mov asthe initial momentum, po, then we have

p = γpo (62)

Dividing Eq.(59) into (61), we obtain

Ep

=γEoγpo

= constant =γmoc

2

γmov=

c2

v(63)

But we know that, the angular frequency of de Brogliewaves is

ω =γmoc

2

~(64)

and the wave number of de Broglie waves is

k =γmov

~(65)

Dividing Eq.(65) into (64) yields

ω

k=γmoc

2

γmov=

c2

v(66)

That is, the velocity c2/v which we found in Eq.(63)is the de Broglie wave velocity, vp. This exceeds thevelocity of the body v and the velocity of light c, sincev < c. The scientists reasoning the exceeding of thevelocity of light this way: "However, vp has no physicalsignificance because the motion of the wave group, not themotion of the individual waves that make up the group,corresponds to the motion of the body, and vg < c as it shouldbe. The fact that vp > c for de Broglie waves therefore doesnot violate special relativity."5

Now we have

ω

k=

Ep

=γEoγpo

=γmoc

2

γmov=

c2

v= constant (67)

Therefore,

c2

v= const. ⇒ v = const. (68)

The body velocity v has to be constant, which implies thecorresponding group velocity, vg has also to be constantbut that is not the case, where v = vg 6= constant. Thisincorrect result is mainly due to the use of the incorrectrelativistic factor, γ, and in the next sections we willderive the correct de Broglie wave velocity and solve thisconflict.

5 Arthur Beiser, Concepts of Modern Physics, p.103.

3.1.2 The Spacetime and The Spacetime WaveEquation

The concept of space and time are inextricably mixedin nature and a convenient way to express the result ofspecial relativity is to regard events as occurring in afour-dimensional spacetime in which the usual three co-ordinates x, y, z refer to space and a fourth coordinateict refers to time, where i =

√−1. Then we have

s2 = x2 + y2 + z2 − c2t2

where the quantity s is the length of the vector ~s. Thefour coordinates x, y, z, ict define a vector in spacetime,and we assume this four-vector remains fixed in space-time regardless of any rotation of the coordinate systembecause we assume the quantity s2 is invariant under aLorentz transformation6.

Now, we already proved that both time t and space xare subject to Doppler effect which leads to the conclu-sion that the spacetime may have wave properties-thatis, the quantity s may satisfy wave equation, where ~s isthe spacetime vector.

To answer this question let us differentiated twice thespacetime s with respect to time and with respect spaceand see if it is satisfy wave equation. Thus we have

s2 = r2 + (ict)2

where r is the magnitude of the spatial dimensions inany arbitrary direction (r2 = x2 + y2 + z2) and

r = ct (69)

Taking the first partial derivative with respect totime, t, we find

s2 = r2 − c2t2

s = (r2 − c2t2)12 then

s′t =1

2(r2 − c2t2)−

12 (−2c2t)

=−c2t

(r2 − c2t2)12

= −c2t

s(70)

Taking the first partial derivative with respect to posi-tion, x, we find

s′r =1

2(r2 − c2t2)−

12 (2r)

=r

(r2 − c2t2)12

=r

s(71)

6 Arthur Beiser, Concepts of Modern Physics, p.46.

11

Dividing Eq.(71) into (70), we obtain

∂s(r, t)/∂t

∂s(r, t)/∂r= −c

2t

r= −c and

(∂

∂t+ c

∂

∂r

)s = 0 (72)

Taking the second partial derivative with respect totime, t, we find

we have∂s(r, t)

∂t= −c2t(r2 − c2t2)−

12 then

∂2s(r, t)

∂t2= −c2

[t

(−1

2(r2 − c2t2)−

32 (−2c2t)

)

+ (r2 − c2t2)−12 (1)

]

= −c2[

c2t2

(r2 − c2t2)32

+1

(r2 − c2t2)12

]

= −c2(c2t2

s3+

1

s

)= −c2

(s2 + c2t2

s3

)

= −c2r2

s3(73)

Taking the second partial derivative with respect toposition, r, we find

we have∂sr(r, t)

∂r= r(r2 − c2t2)−

12 then

∂2sr(r, t)

∂r2= r

(−1

2(r2 − c2t2)−

32 (2r)

)

+ (r2 − c2t2)−12 (1)

=1

(r2 − c2t2)12

− x2

(r2 − c2t2)32

=1

s− r2

s3=

s2 − r2

s3

= −c2t2

s3(74)

Dividing Eq.(74) into (73), we obtain

∂2s(r, t)/∂t2

∂2s(r, t)/∂r2=(rt

)2= c2 and

∂2s(r, t)

∂r2=

1

c2∂2s(r, t)

∂t2(75)

The wave equation may be "factored":(∂

∂t+ c

∂

∂r

)(∂

∂t− c ∂

∂r

)s = 0 (76)

These last two equations are advection equations, oneleft travelling and one right, both with constant speed c.

We can also prove the satisfaction of wave equation for:

s2 = r2 − (ict)2 (77)

Taking the first partial derivative with respect to time,t, we find

s2 = r2 + c2t2

s = (r2 + c2t2)12 then

s′t =1

2(r2 + c2t2)−

12 (2c2t)

=c2t

(r2 + c2t2)12

=c2t

s(78)

Taking the first partial derivative with respect to posi-tion, r, we find

s′r =1

2(r2 + c2t2)−

12 (r)

=r

(r2 + c2t2)12

=r

s(79)

Dividing Eq.(79) into (78), we get

∂s(r, t)/∂t

∂s(r, t)/∂r=

c2t

r= c and

(∂

∂t− c ∂

∂r

)s = 0 (80)

Taking the second partial derivative with respect totime, t, we have

∂s(r, t)

∂t= c2t(r2 + c2t2)−

12

12

Then∂2s(r, t)

∂t2= c2

[t

(−1

2(r2 + c2t2)−

32 (2c2t)

)

+ (r2 + c2t2)−12 (1)

]

= c2[

1

(r2 + c2t2)12

− c2t2

(r2 + c2t2)32

]

= c2(

1

s− c2t2

s3

)= c2

(s2 − c2t2

s3

)

=c2r2

s3(81)

Taking the second partial derivative with respect toposition, r, we have

∂s(r, t)

∂r= r(r2 + c2t2)−

12

Then∂2s(r, t)

∂r2= r

(−1

2(r2 + c2t2)−

32 (2r)

)

+ (r2 + c2t2)−12 (1)

=1

(r2 + c2t2)12

− r2

(r2 + c2t2)32

=1

s− r2

s3=

s2 − r2

s3

=c2t2

s3(82)

Dividing Eq.(82) into (81), we obtain

∂2s(r, t)/∂t2

∂2s(r, t)/∂r2=(rt

)2= c2 and

∂2s(r, t)

∂r2=

1

c2∂2s(r, t)

∂t2(83)

Because, the quantity s satisfies wave equation, thatimplies the spacetime possess wave properties-that is, ithas frequency, wavelength, velocity, group velocity andundergoes Doppler effect, diffraction and interference.Therefore we expected any arbitrary smooth functionf that been represented using spacetime coordinates totake the form f(r − ct) [ or more generally f ( 1

%(r - ct)) ]

and satisfy wave equation.

4.1.0 Relativistic Kinematics and Dynamics

We can write Eqs.(38) and (39) as:

t(v) =to%

(84)

x(v) =xo

%(85)

ν(v) = %νo (86)

ν̄(v) = %ν̄o (87)

where ν̄ = 1λ is the spatial frequency of the spacetime

wave, while ν = 1t is the temporal frequency of it, and

to, xo, νo and ν̄o represent proper values.

We know that ω = 2πν , ν = cλ and λ = 2π

k ,then with the help of Eq.(86), we can obtain

λ(v) =λo%

(88)

k(v) = %ko (89)

ω(v) = %ωo (90)

Dividing Eq.(87) into (86) yields

ν = cν̄ (91)

and by multiplying Eq.(91) by 2π, we get

ω = ck (92)

Multiplying Eqs.(86) and (88), gives

λν = c (93)

Here ν, ν̄, ω, k and λ are variables of spacetime NOTthe light.

Multiplying Eq.(90) and (89) by ~, we obtain

E(v) = ~ (%ωo) = ~ω , or (94)

E(v) = %Eo (95)

and

p(v) = ~ (%ko) = ~k , or (96)

p(v) = %po (97)

Equation 95 and 97 are equivelant to equations 61 and62 where here we are using the correct relativistic factor% instead of γ factor.

13

Now dividing Eq.(96) into (94), we get

E

p=Eopo

=ωoko

= c = constant (98)

Equation 98 is equal to equation 67 and both comes fromdescription of particle-wave kinematics, thus we considerthe de Broglie waves and the spacetime waves (Eq.(75))are same thing. Therefore the correct velocity of the deBroglie wave is c, the speed of light.

4.2.0 Relativistic Momentumand

The Invariance of The Mass

A. Derivation of the Relativistic Momentum7

7 Arthur Beiser, Concepts of Modern Physics, p.22 to p.24.

FIG. 6. An elastic collision as observed in two different framesof reference. The balls are initially Y apart, which is the samedistance in both frames since S′ moves only in the x direction

Firstly, let us mention the derivation of relativisticmomentum:

Let us consider an elastic collision between twoparticles A and B, as witnessed by observers in thereference frames S and S′ which are in uniform relativemotion. The properties of A and B are identical whendetermined in reference frame in which they are at rest.The frames S and S′ are oriented as in figure 6, withS′ moving in the +x direction with respect to S at thevelocity v. Then at the same instant, A was thrown inthe +y direction at the speed VA while B was thrown inthe −y direction at the speed V ′B , where

VA = V ′B (99)

When the two particles collide, A rebound in the −ydirection at the speed VA, while B rebound in the +y′

direction at the speed V ′B . If the particles are thrownfrom position Y apart, an observer in S finds that thecollision occurs at y = ½Y and one in S′ finds that itoccurs at y′ = y = ½Y . The round trip time To for A asmeasured in frame S is therefore

To =Y

VA(100)

and it is the same for B in S′:

To =Y

V ′B

In S the speed VB is found from

VB =Y

T(101)

where T is the time required for B to make its round tripas measured in S. In S′, however, B′s trip requires thetime To, where

T = γTo (102)

according to our previous results. Although observers inboth frames see the same event, they disagree about thelength of time the particle thrown from the other framerequires to make the collision and return.

Replacing T in Eq.(101) with its equivalent in termsof To, we have

VB =Y

γTo(103)

From Eq.(100),

VA =Y

To

14

If we use the classical definition of momentum, p = mv,then in frame S

pA = mAVA = mA

(Y

To

)

pB = mBVB = mB

(Y

γTo

)(104)

This means that, in this frame, momentum will not beconserved ifmA = mB , wheremA andmB are the massesas measured in S. However, if

mB = γmA (105)

then momentum will be conserved.

In the collision of figure 6 both A and B are mov-ing in both frames, Suppose now that VA and V ′B arevery small compared with v, the relative velocity of thetwo frames. In this case an observer in S will see Bapproach A with the velocity v, make a glancing collision(since V ′B � v), and then continue on. In the limit ofVA = 0, if m is the mass in S of A when A is at rest,then mA = m. In the limit of V ′B = 0, if m(v) is themass in S of B, which is moving at the velocity v, thenmB = m(v). Hence Eq.(105) becomes

m(v) = γm (106)

We can see that if linear momentum is defined as

p = γmv (107)

then conservation of momentum is valid in specialrelativity.

We could alternatively regard the increase in anobject’s momentum over the classical value as being dueto an increase in the object’s mass. Then we would callmo = m the rest mass of the object and m = m(v) fromEq.(107) its relativistic mass, its mass when movingrelative to an observer, so that p = mv.

B. Refutation of The Relativistic Mass Principle

Examining the previous discussion about relativisticmomentum, which led to the concept of the relativisticmass (Eq.(106)) we notice the complete absence of thelength contraction or expansion effect from the deriva-tion of the relativistic momentum, where both observersare going to disagree about both, the length of time theparticle thrown from the other frame requires to makethe collision and return, and the length of distance theparticle thrown from the other frame travels to makethe collision and return.

If we use the Lorentz-Fitzgerald length contraction

to complete the derivation of the relativistic momentum,then Eq.(104) becomes

pB = mB

(YoTo

)(1− β2) and

pB = mBVA(1− β2) (108)

where Yo is the distance required for A to make its roundtrip as measured in S (the proper length). Therefore therelativistic mass becomes

m(v) = γ2m (109)

But we know that the Lorentz-Fitzgerald length contrac-tion is incorrect as we proved earlier, thus if we use thecorrect relativistic factor, %, we find

pB = mB

(% ½Yo + %−1 ½Yo% ½To + %−1 ½To

)= mB

(YoTo

),

pB = mBVA = pA (110)

and the relativistic mass becomes

m(v) = m (111)

That is, the momentum is conserved and the mass isinvariant .

Notice that although the ratio between time and distanceare the same in both reference frames i.e

(%Yo%To

and YoTo

),

the time length and space length are not equal.

For the sake of completion, let us consider the case ofthe length expansion as have been found in Eqs.(44).Thus we find

pB = mB

(γYoγTo

)

pB = mBVA = pA (112)

and the relativistic mass becomes

m(v) = m (113)

That is, the momentum is also conserved and the massis invariant.

Due to these results we conclude that the relativis-tic momentum relation (Eq.(107)) is invalid , and wealso conclude that the mass is invariant . Thus theclassical momentum expression remain valid in specialrelativity-that is,

p = mv (114)

15

We also will find the classical kinetic energy expressionremain as it is in special relativity

KE =

∫ s

0

Fds =

∫ s

0

d

dtmv ds,

=

∫ mv

0

v d(mv) = m

∫ v

0

v dv,

= ½mv2 (115)

where F is the force and s is the distance.

4.3.1 The Relativistic Velocity

The relativistic velocity formula is, as we found earlier(see Eq.(11)),

Vx =dx

dt=

dx′

dt′+ v

1 +v

c2dx′

dt′

But we already found that xt = x′

t′ = const. (see Eqs.(55)and (56)). The same results can be obtain by differenti-ating Eqs.(4) and (5), then we have

x = ct anddx

dt= c (116)

and

x′ = ct′ anddx′

dt′= c (117)

Substituting back into the formula we will find that therelativistic velocity is always equal to speed of light i.e(Vx = c ∀ x′, t′). Thus the relativistic velocity has nosignificance to calculate, only the relative velocity, v, isimportant, which by now will represents both the rela-tive velocity and the particle velocity, where the particleis going to be a frame of reference. The particles thathave same relative velocity also constitute one frame ofreference.

4.3.2 The Particle Velocity

From de Broglie wavelength of a particle postulate andEqs.(94),(96),(114) and (115) the wave number and an-gular frequency of the de Broglie waves associated witha body of mass m moving with velocity v are

k =m

~v (118)

ω =m

2~v2 (119)

Both ω and k are function of the body velocity v.The group velocity vg of the de Broglie waves associ-

ated with the body is

vg =dω

dk=dω/dv

dk/dv

Now

dω

dv=m

~v

dk

dv=m

~and so the spacetime group velocity turns out to be

vg = v (120)

The spacetime wave group or de Broglie wave group asso-ciated with a moving body travels with the same velocityas the body.

The phase velocity vp of spacetime waves (de Brogliewaves) is, as we found earlier,

vp =ω

k= c (121)

To find a formula for the velocity we can take one ofequations (84) to (90) and solve for the velocity v, thenwe have

t = to

√c− vc+ v

t2 = t2o(c− v)

(c+ v)

ct2 + vt2 = ct2o − vt2o

vt2 + vt2o = ct2o − ct2

v(t2 + t2o) = c(t2o − t2)

and

v = c

(t2o − t2

t2o + t2

)(122)

With the help of Eqs.(4) and (5), we find

v = c

(x2o − x2

x2o + x2

)(123)

Also we have

v = c

(ν2 − ν2oν2 + ν2o

)(124)

16

4.4.0 The Momentum and The KineticEnergy of a Particle

Substituting Eq.(123) into Eq.(114) and (115), we willfind the momentum to be

p = mc

(x2o − x2

x2o + x2

)(125)

and the kinetic energy to be

KE = ½mc2(

x2o − x2

x2o + x2

)2

(126)

and when the particle move with the speed of light, c,which is possible because the mass is invariant and isnot a function of the velocity. Then substituting v = cinto Eq.(85), we obtain

x = xo

√c− cc+ c

= 0

Thus the momentum and kinetic energy of a particle be-comes

p = mc (127)

KE = ½mc2 (128)

and the spacetime will shrink to zero in front of thatparticle and will expand to infinity (or the wave front tears)behind it,

tfront(c) =to%(c)

= (0)to = 0 (129)

xfront(c) =xo%(c)

= (0)xo = 0 (130)

where tfront and xfront are the time and space length onfront of the particle (spacetime wave that propagates along+ve x-axis).

tback(c) = %(c)to =to0

= ∞ (131)

xback(c) = %(c)xo =xo0

= ∞ (132)

where tback and xback are the time and space length behindof the particle (spacetime wave that propagates along −vex-axis). That is, the particle will penetrate the speed oflight potential barrier. This situation can be describedby Dirac delta function as in FIG.7 where the functionspikes when the relative velocity reach speed of light.

R%(v) =

½% ∀ |c− v| < 1

%,

0 ∀ |c− v| > 1

%.

FIG. 7.

∫ +∞

−∞R%(v)dv = ½%

∫ c+ 1%

c− 1%

dv = 1.

lim1%→0

R%(v) = δ(c− v) (133)

But

lim1%→0

R%(v) = limv→c

½%(v)

Then

½ limv→c

%(v) = δ(c− v) (134)

We have

γ = ½ ( %(v) + %(−v) ) (135)

Taking the limit as v → c, we find

limv→c

γ = ½ limv→c

%(v) + ½ limv→c

%(−v)

where

limv→c

%(−v) = %(−c) = 0

Therefore

limv→c

γ = ½ limv→c

%(v) (136)

17

From Eqs.(134) and (136), we have

limv→c

γ = δ(c− v) (137)

Now∫ +∞

−∞f(v)δ(c− v)dv = lim

%→0

∫ +∞

−∞f(v)R%(v)dv,

= ½ limv→c

∫ +∞

−∞f(v)%(v)dv,

= limv→c

½% f(c)

∫ c+ 1%

c− 1%

dv,

= limv→c

½% f(c)

(2

%

),

= f(c).

Here f(c) is any arbitrary smooth function. In our casethey are the kinetic energy function, KE(v), and the mo-mentum function, p(v) (see Eqs.(127) and (128)).

5.1.0 Particle in a Box

The wave nature of a moving particle leads to someremarkable consequences when the particle is restrictedto a certain region of space instead of being able to movefreely.

The simplest case is that of a particle that bouncesback and forth between the walls of a box.

From a wave point of view, a particle trapped in a boxis like a standing wave in a string stretching between thebox’s walls. The possible de Broglie wavelength of theparticle in the box therefore are determined by the widthL of the box. The general formula for the permittedwavelength is

λn =2L

nn = 1, 2, 3, ...

In the case of electron, the electron can circle a nucleusonly if its orbit contains an integral number of de Brogliewavelength. Hence the condition for orbit stability is

nλ = 2πrn n = 1, 2, 3, ... (138)

where rn designates the radius of the orbit that containn wavelengths, and the integer n is called the quantumnumber of the orbit. This is last equation can be writtenas

λn =2πaon

=cton

=c

nνoand

λn =λon

n = 1, 2, 3, ... (139)

where ao is the Bohr radius, the initial value or the lengthin the inertial frame of reference which compare otherframes of reference to.

Comparing equation 139 to equation 88, we find

% = n

That is, the relativistic factor is equal to the reciprocal ofthe principle quantum number, therefore the quantizedrelativistic factor, %m, is

%m = %(vm) = n , where m = 1,2,3,... (140)

where the quantized relative velocity or quantized par-ticle velocity, vm, is

vm = c

(m2 − 1

m2 + 1

), where m = 1,2,3,... (141)

and Eqs.(84),(85),(86) and (96) becomes

t

to=

1

n,

x

xo=

1

n,

λ

λo=

1

n,

ν

νo= n , where n = 1, 2, 3, ... (142)

Also we have

ppo

= n ,EEo

= n , where n = 1, 2, 3, ... (143)

When n = ∞ that implies the particle is moving withthe speed of light-that is,

n = %m(c) =

√c+ (vm = c)

c− (vm = c)= ∞

6.1.0 The Invariance of The Natural Constants

6.1.1 Einstein’s First Postulate - The Invariance ofPhysical Laws

We know the Einstein’s first postulate, which calledthe principle of relativity, that states: The lawsof physics are the same in every inertial frame ofreference. If the laws differed, that difference coulddistinguish one inertial frame from the others or makeone frame somehow more "correct" than another.

And the Einstein’s second postulate states: Thespeed of light in vacuum is the same in all inertialframes of reference and is independent of the motionof the source.8

8 University Physics with Modern Physics - Sears and Zemansky,P.1223 - electronic copy.

18

Now from Eqs.(116) and (117) we have

dx dt−1 = c (144)

dx′ dt′−1 = c (145)

Analysing the fundamental and the derived physical con-stants using the Planck’s units (see FIG.8) we will findthat, the constant c is a common factor between all of itand without any exception. If the speed of light is variantand a function of relative velocity, then all nature con-stants are going to change due to the change in the speedof light, which violates the Einstein’s first postulate andthe physical laws becomes variance.

The invariance of the natural constants and of themassmakes wave kinematics an inescapable kind of kinemat-ics nature has to take-that is, one variable increase, an-other decrease, but there product remain constant, butnot only the variable and its corresponding variable thatare change in inverse way but also there uncertaintiesare also wavy i.e (∆x∆t−1 > c), and that what form theuncertainty principle.

CONCLUSION

We conclude that, the spacetime has wave properties,and the spacetime waves and the de Broglie waves areone thing.

The spacetime wave travels with the speed of light invacuum and this velocity represents the correct de Brogliewaves velocity.

The relative velocity and the de Broglie group velocityor the spacetime group velocity are equal.

The Lorentz relativistic factor is incomplete and thecorrect relativistic factor has been derived.

The derivation of the Lorentz-Fitzgerald length con-traction is incorrect.

The relativistic effect due to the correct relativistic fac-tor is not symmetrical in every direction.

The mass is invariant and the relativistic mass is in-correct physical principle, and due to that the particlecan accelerate to the speed of light and penetrate thespacetime potential barrier, and the natural constantsalso have been found to be invariants.

The when the particle is restricted to a certain regionof space the new relativistic factor becomes equal to theprinciple quantum number.

19

FIG. 8. Planck Units.aa Planck units Source: http://en.wikipedia.org/w/index.php?oldid=566793775