(Dated: September 3, 2015)

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1 DOPPLER TRANSFORMATION AS EIGENVALUES OF LORENTZ TRANSFORMATION Louai Hassan Elzein Basheir 1 Physics College, Khartoum University, Sudan. P.O. Box: Zip Code: (Dated: September 3, 215) This paper has been prepared to show the derivation of the Doppler transformations as eigenvalues of the Lorentz transformations matrices and to conclusively proves that these eigenvalues are the correct relativistic factors which have to be use instead of the Lorentz factor. The paper also shows the forms of the Doppler matrices that can replace the Lorentz matrices, and discuss some of the consequences that follow the equivalence of Lorentz and Doppler transformations. 1

2 THEORETICAL BACKGROUND 1.1 Boost in the x-direction The Lorentz transformation for frames in standard configuration can be shown to be: ( t = γ t vx ) c 2 x = γ (x vt) (1) y = y z = z where: v is the relative velocity between frames in the x-direction, c is the speed of light, γ = 1 β 1 is the Lorentz factor, 2 β = v is the velocity coefficient, again for the x-direction. c Since the above is a linear system of equations, they can be written in matrix form: x γ iβγ x y z = 1 y 1 z (2) iβγ γ The above collection of equations apply only for a boost in the x-direction. The standard configuration works equally well in the y or z directions instead of x, and so the results are similar. For the y-direction: x 1 y z = γ iβγ 1 iβγ γ For the z-direction one obtains x 1 y z = 1 γ iβγ iβγ γ x y z x y z (3) (4) The Lorentz or boost matrix is usually denoted by Λ. Above the transformations have been applied to the four-position X, x x X = y z X = y z Page 2 of 14

3 The Lorentz transform for a boost in one of the above directions can be compactly written as a single matrix equation: X = Λ(v)X. (5) ANALYSIS 2.1 Eigenvalues of The Lorentz Matrix The Lorentz matrix has many properties, it is orthogonal where Λ 1 = Λ T, orthonormal where the magnitude of any of its unit column vector is equal to one (from this property one can obtain the value of the Lorentz factor γ), also it is symmetric where Λ = Λ T (for the Lorentz matrix with real elements) and it is hermitian (imaginary matrix). Since the Lorentz matrix has these properties, we expect to be able to derive eigenvalues from it, or simply to be able to write it in the following form: Λ(v)X = λx. (6) Equation (6) can be written as [Λ(v) λi] X = which implies that Λ(v) λi = and by solving this determinant one finds For the Lorentz matrix for x-direction: γ λ iβγ 1 λ 1 λ =, (7) iβγ γ λ and (γ λ) 1 λ 1 λ γ λ (iβγ) 1 λ 1 λ iβγ =, (γ λ) 2 (1 λ) 2 iβγ [ (1 λ) [ (1 λ)( iβγ)]] =, (γ λ) 2 (1 λ) 2 iβγ [ (1 λ) 2 ( iβγ) ] =, (γ λ) 2 (1 λ) 2 (1 λ) 2 (βγ) 2 =, (1 λ) [ 2 (γ λ) 2 (βγ) 2] =, which can factored: (1 λ) (1 λ) [(γ λ) + (βγ)] [(γ λ) (βγ)] =, (1 λ) (1 λ) [(γ + βγ) λ] [(γ βγ) λ] =, Page 3 of 14

4 which leads to the solutions: (γ βγ) λ = 1 λ = 1 λ = (γ + βγ) λ =, Thus the eigenvalues of the Lorentz matrix for x-direction: λ 1 = γ(1 β) λ 2 = 1 λ 3 = 1 λ 4 = γ(1 + β) (8) For y-direction and z-direction we obtain similar eigenvalues. Notice that λ 2 = λ 3 = 1 when the relative velocity between inertial frames is equal to zero, that is v = β =, γ = Eigenvectors of The Lorentz Matrix We know that the eigenvectors associated with eigenvalues have to be linearly independent and orthogonal, which implies its determinant has to be not equal to zero, so finding the eigenvectors matrix and exam its linear independency will check the validity of the derived eigenvalues (Eq.(8)). Substitute λ 1 into Eq.(7) yields the eigenvector associated with λ 1 for x-direction: γ γ(1 β) iβγ 1 γ(1 β) 1 γ(1 β) iβγ γ γ(1 β) βγ iβγ 1 γ(1 β) 1 γ(1 β) iβγ βγ (i(r 1 )) + R 4 R 4 βγ iβγ 1 γ(1 β) 1 γ(1 β) Page 4 of 14

5 R 1 βγ R 1, R 2 1 γ(1 β) R 2, 1 i 1 1 R 3 1 γ(1 β) R 3. Whereas x, y, z are lead variables, while is free variable. Thus we have x + i() = x = ct, (9) y =, z =. Hence the eigenvector associated with λ 1 for x-direction is v = ct. Substitute λ 4 into Eq.(7) yields the eigenvector associated with λ 4 for x-direction: γ γ(1 + β) iβγ 1 γ(1 + β) 1 γ(1 + β) iβγ γ γ(1 + β) βγ iβγ 1 γ(1 + β) 1 γ(1 + β) iβγ βγ ( i (R 1 )) + R 4 R 4 βγ iβγ 1 γ(1 + β) 1 γ(1 + β) R 1 βγ R 1, R 2 1 γ(1+β) R 2, 1 i 1 1 R 3 1 γ(1+β) R 3. Page 5 of 14

6 Whereas x, y, z are lead variables, while is free variable. Thus we have x i() = x = ct, (1) y =, z =. Hence the eigenvector associated with λ 4 for x-direction is w = ct. Substitute λ 2 or λ 3 into Eq.(7) yields the eigenvector associated with λ 2 and λ 3 for x- direction: γ 1 iβγ iβγ γ 1 R 1 γ 1 R 1, R 4 iβγ R 4. 1 (iβγ/γ 1) 1 (γ 1/iβγ) R 1 R 2 R 2, 1 (iβγ/γ 1) (iβγ/γ 1) + (γ 1/iβγ) R 4 (iβγ/γ 1)+(γ 1/iβγ) R 4. 1 (iβγ/γ 1) 1 The lead variables are x, and the free variable are y, z. Therefore we find =, x + (iβγ/γ 1)() = x =. (11) Page 6 of 14

7 Consequently, the eigenvectors associated with λ 2 and λ 3 are y y u = = +. z z Thus the eigenvectors matrix for the x-direction: M x = x ct ct y y. (12) z z Similarly, for the y-direction (Eq.(3)) we have M y = x x y ct ct. z z and for the z-direction (Eq.(4)): M z = x x y y. z ct ct Taking the determinants to eigenvectors matrices one finds det(v x ), det(v y ), det(v z ). Which proof the linear independency and the orthogonality of the eigenvectors. verifies that the Lorentz matrix is diagonalizable, that is Also it M 1 ΛM = D. Where M is the eigenvectors matrix, M 1 the inverse of eigenvector matrix, Λ the Lorentz matrix, D the diagonal matrix. Finding the inverse of M x : ct ct y z Page 7 of 14

8 R 1 ct R 1, R 2 y R 2, R 3 z R 3, R 4 R 4. 1/ct 1/y 1/z 1/ R 1 R 4 R /ct 1/y 1/z 1/ct 1/ R 4 2 R /ct 1/y 1/z 1/2ct 1/2 R 1 + R 4 R /2ct 1/2 1/y 1/z 1/2ct 1/2 Hence the inverse of M x is M 1 x = The inverse of M y : M 1 y = The inverse of M z : M 1 z = 1/2ct 1/2 1/y 1/z 1/2ct 1/2 1/x 1/2ct 1/2 1/z 1/2ct 1/2 1/x 1/y 1/2ct 1/2 1/2ct 1/2 Page 8 of 14

9 Again finding the inverse of the eigenvectors matrices assure its linear independency and its orthogonality. Hence the diagonal matrix for x-direction: D x = M 1 x Λ x M x, Then 1/2ct 1/2 D x = 1/y 1/z 1/2ct 1/2 1/2ct 1/2 = 1/y 1/z 1/2ct 1/2 = γ(1 β) 1 1 γ(1 + β) The diagonal matrix for y-direction: D y = 1 γ(1 β) 1 γ(1 + β) The diagonal matrix for z-direction: D z = 1 1 γ(1 β) γ(1 + β) = γ iβγ 1 1 iβγ γ ct ct y z γ(1 β)ct γ(1 + β)ct y z γ(1 β) γ(1 + β) λ 1 λ 2 λ 3 λ 4 Thus the diagonal matrix gives the same eigenvalues λ 1, λ 2, λ 3, λ 4 which farther verifies our findings. 2.3 The Doppler Transformations After we have verified the validity of deriving eigenvalues from Lorentz matrix, now becomes legal to write it in the form of equation (6), that is Λ(v)X = λ i X, i=1,2,3,4. (13) Page 9 of 14

10 Where λ 1 = γ(1 β), λ 2 = 1, λ 3 = 1, λ 4 = γ(1 + β). Multiply the R.H.S by the identity matrix yields λ i IX = DX, where I is the identity matrix and D is diagonal matrix. Therefore, equation (13) becomes Λ(v)X = DX. (14) From here on, the R.H.S of Eq.(14), and by mathematical necessity, is valid to replace the L.H.S. λ i x DX = λ i y λ i z i=1,2,3,4. λ i Let us define a function µ(+v) = γ(1 + β), then we have µ( v) = γ(1 β) and µ() = 1. The eigenvalue µ(+v) designates receding frame of reference, the eigenvalue µ( v) designates approaching frame of reference, while the eigenvalue µ() designates motionless frame of reference. Hence a general form of transformation matrix can be defined: D(v) = µ(v x ) µ(v y ) µ(v z ) µ(v) (15) where D is the equivalent transformation matrix, v the relative velocity, v x the relative velocity component in x-direction, v y the relative velocity component in y-direction, v z the relative velocity component in z-direction. Because the equivalent transformation matrix D transfers time and space in Dopplerian manner, we are going to name it the Doppler matrix and the factor µ the Doppler factor. Thus the Doppler transformation can be defined: x y z = and as matrix equation: µ(v x ) µ(v y ) µ(v z ) µ(v) x y z D(v)X = X. (16) Consequently, the Doppler transformations in x-direction: x γ(1 ± β) x y z = 1 y 1 z γ(1 ± β) Page 1 of 14

11 and as matrix equation: D x (v)x = X. where v y = v z = and v = v x. Finding the matrix multiplication one obtains the Doppler transformations equations which equivalent to the Lorentz transformations equations (Eqs.(1)): t = γ(1 ± β) t x = γ(1 ± β) x y = y z = z (17) The Doppler matrix works equally well in the y or z directions instead of x, and so the results are similar. For the y-direction: x y z = and as matrix equation: D y (v)x = X. 1 γ(1 ± β) 1 γ(1 ± β) where v x = v z = and v = v y. For the z-direction: x y z = and as matrix equation: D z (v)x = X. 1 1 γ(1 ± β) γ(1 ± β) where v x = v y = and v = v z. From Doppler matrices and its equations one recognizes the Doppler factor µ works as the relativistic factor instead of the Lorentz factor γ. x y z x y z Page 11 of 14

12 2.4 The Equivalence of The Lorentz Transformations and The Doppler transformations The Lorentz transformation is equivalent to the Doppler transformation but under the condition, that is the spatial coordinate (along the relative motion) is dependent upon the temporal coordinate. That is, if we arbitrarily choose the value of temporal coordinate then the spatial coordinate will not. The spatial and temporal coordinates are related through the constant c, the speed of light. To prove equivalence under this condition let us take the Lorentz transformations equations (Eq.(1)), the transformation for x-direction, therefore we have ( t = γ t vx ) c 2 But from Eqs.(9) and (1) we find the spatial variable x is related to the temporal variable t through the equation x = ct and x = ct, whereas x is lead variable while t is free variable. Thus substitute x = ct, we get ( t = γ t v(ct) ) c ( 2 = γ t vt ) c ( = γ 1 v c = γ (1 β) t ) t For x = ct, we obtain t = γ (1 + β) t For spatial transformation, if we choose x = ct, one obtains x = γ (x vt) ( ( x )) = γ x v c = γ (1 β) x For x = ct, we obtain x = γ (1 + β) x which definitely are the Doppler transformations for x-direction (see Eq.(17)). For x = (Eq.(11)), it is implies that t = /c =, therefore t = and x =, That is [ ] t v(x = ) = γ (t = ) =. c 2 x = γ [(x = ) v(t = )] =. Page 12 of 14

13 The absence of understanding this fact; the spatial and temporal coordinates are related through the constant c (also see Eq.(19)), led to great mistake, where in classical derivation of time dilation and length contraction formulae from the Lorentz transformations the substitution of zero into spatial or temporal coordinate has nothing to do with each other, that is [ ] t = γ t + v( x = ) = γ t, c 2 Time dilation formula. x = γ [ x v( t = )] = γ x and x = x γ, Length contraction formula. Noticing that; x = x 1 x 2 = x 2 = x 1 and because of the fact that x = ct then also t 1 = t 2 and t = t 2 t 1 =. 2.5 The Invariance of The Speed of Light and Its Consequences From equations (17), when we divide the temporal transformation equation into spatial transformation equation one discovers x t = γ(1 ± β) x γ(1 ± β) t x t = x t = const. (18) Let us denote this constant with ɛ. To find this constant we divide the Lorentz temporal transformation equation into the Lorentz spatial transformation equation (see Eqs.(1)). Remember the Lorentz and Doppler transformations equations are equivalent (see Eqs.(13) and (14)). Therefore we have ( x ) t x t = γ(x vt) ( γ t vx ) = c 2 v ( v ) ( x ) = ɛ, 1 c 2 t but from Eq.(18) we have x /t = x/t = ɛ, then ɛ = ɛ v 1 vɛ c 2, cross multiply ɛ vɛ2 = ɛ v which implies ɛ = c. (19) c2 Thus, equation (18) gives x = ct and x = ct, which implies (2a) dx/dt = c and dx /dt = c, (2b) d 2 x/dt 2 = and d 2 x /dt 2 =. (2c) These results (Eqs.(2a),(2b) and (2c)) invalidate the classical derivation of the relativistic velocity transformations and the relativistic acceleration transformations from the Lorentz transformations. Page 13 of 14

14 CONCLUSION We conclude that the spatial and temporal coordinates transfer from inertial frame of reference to another like wavelength and period consecutively, which gives clue that the spacetime is wavy space and transfers like electromagnetic waves, that is, it transfers through Doppler transformations. We also found that the relativistic factor is a function in velocity magnitude and direction (plus or minus), and we found the relativistic effect to be asymmetric along the direction of motion, that is, the approaching frame of reference and receding frame of reference are not equal relativistically; observers whom observe approaching frames of references measure (simultaneously) contraction in spatial length and temporal length, and observers whom observe receding frames of references measure (simultaneously) expansion in spatial and temporal length. We also found that the space and time transfer in manner that conserves the ratio between them, the speed of light. Therefore the spatial and temporal dimensions expand and contract simultaneously to keep the constant c invariant. We also found that due to the invariance of the speed of light (the division of time into space is equal to the speed of light) implies a fundamental relativistic kinematics formulae to be invalid, which are the relativistic velocity transformation and the relativistic acceleration transformation. Page 14 of 14