Jump to content

Anamitra Palit

Senior Members
  • Content Count

    51
  • Joined

  • Last visited

Community Reputation

-7 Poor

About Anamitra Palit

  • Rank
    Meson

Profile Information

  • Favorite Area of Science
    Relativity

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

  1. [One my have to refresh the page for viewing the formulas and the equations] We consider the transformation of the rank two contravariant tensor: \[\bar A^{\mu \nu}=\frac{\partial \bar x^{\mu}}{\partial x^\alpha}\frac{\partial \bar x^{\nu}}{\partial x^\beta}A^{\alpha\beta}\](1) Inverse transformation[for non singular transformations] \[A^{\alpha \beta}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\beta}}{\partial \bar x^\nu}\bar A^{\mu\nu}\](2) For the diagonal components[alpha=beta] \[ A^{\alpha \alpha}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\alpha}}{\partial \bar x^\nu}\bar A^{\mu\nu}\] (3) We consider the situation where the off diagonal components of A[alpha not equal to beta] are all zero The diagonal elements of A-bar are given by \[\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2 A^{\alpha\alpha}\] (4) For off diagonal elements of A-bar are given by ' \[\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}\] (4’) Subject to the situation that the off diagonal elements of A are all zero we consider the following three cases: Case 1. Assume A-bar^ mu nu=0for all μ≠ν for all barred reference frames[the general condition that off diagonal elements of A are zero continues to hold] we have \[\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}=0\] (4’’) We have (4'') irrespective of the transformation elements[all reference frames being considered]. From (4’’) A-bar^mu mu=0.Again from (4’) and (4’’) |A-bar ^mu nu becomes zero for μ≠ν. Then the tensor becomes null Case 2 Off diagonal elements are non zero for all A-bar[that the off diagonal elements of A are all zero continues to hold] \[\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}\] \[\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}\] \[\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}\] \[\Rightarrow\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2\delta^\alpha_\beta \delta^\alpha_\gamma=\frac{\partial \bar x^\mu} {\partial x^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}\] For β≠γ we have from (5) \[\frac{\partial \bar x^\mu}{\partial x ^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}=0\](6) \[\Rightarrow\frac{\partial \bar x^\mu}{\partial x^\beta}=0\]or \[\frac{\partial \bar x^\mu}{\partial x ^\beta}=0\] (7) \[\Rightarrow \bar A^{\mu\nu}=0\Rightarrow A^{\alpha\beta}=0\] (8) Equations represented by (7) and (8)are not true! We do have an enigma of a persistent nature. [In fact (8) implies that the metric tensor is the null tensor;the Riemann tensor and the Ricci tensor are null tensors; the Ricci scalar is zero valued.] Case 3 For A as per our initial postulation the off diagonal elements of A are all zero. For A-bar there we assume the existence of at least one mu ,nu pair in each barred frame such that A-bar^mu nu=0.These mu nu pairs may not be identical for all barred frames. Since we may have an infinitude of reference frames catering to our assumption it follows that for some specific (μ,ν)pair also A-bar mu mu is zero in an infinitely many frames of reference \[\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}\] Since A^m mu=0 for a specific pair in infinitely many frames,we have, \[\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}=0\](7) For the same mu nu pair (7) we have an infinitude of equations like (7) whee the transformation elements are distinct. We have included infinitely many barred frames in equation (7) \[\Rightarrow A^{\alpha\alpha}=0\Rightarrow A=0\] (8) Case 1 is an instance of case 3 Points to Observe: 1. Anti symmetric tensors transform to anti symmetric tensors in all frames of reference. Noe the null tensor is an antisymmetic tensor asides being a symmetric one.An arbitrary non zero[non null] tensor may be expressed as the sum of a symmetric an an antisymmetric tensor. Our analysis might be considered for an arbitrary tensor which is neither symmetric or antisymmetric.It reduces the null tensor making both the symmetric and the antisymmetric parts zero. 2.If a tensor has a component of identical value in all reference frames it has to be the null tensor. 3. Let us represent (1) in a standard matrix form \[\bar A=MAM^T\](9) where M is the transformation matrix For an arbitrary M even if M is non singular in nature, we will not necessarily have both A and A-bar as diagonal[both with the off diagonal elements as zero].If one of A or A-bar is diagonal with the other non diagonal then A ,A-bar both become null tensors as we have already seen! One may test the situation using a diagonal tensor with A and an arbitrary square matrix for M For an arbitrary transformation matrix ,a diagonal tensor A does not necessarily produce a diagonal A-bar. Only by specific valid choice of the space time transformation can we have both sides of (10) or (1) for that matter, as diagonal[components zero for alpha not equal to beta]. An arbitrary[non singular ] transformation will produce from a diagonal tensor a non diagonal. It is always possible to have infinitely many transformation for which the off diagonal elements are non zero[diagonal elements of the tensor in some reference frame are assumed to be non zero]
  2. Link to file on the Google Drive https://drive.google.com/file/d/1C2-ru6uuDIw9u_e4HQ1bdwa01oKysQ-B/view?usp=sharing Material in Latex[It might be necessary to refresh the page for viewing the formulas and the equations] The paper establishes mathematically that the Riemann Tensor is a zero tensor Riemann Curvature |Tensor \[R^{\mu\nu}_{\quad\gamma\delta}=g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}\] (1) Interchanging the dummy indices alpha and beta we have, \[R^{\mu\nu}_{\quad\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}\](2) Therefore, \[g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}\] (3) But \[R_{\beta\alpha\mu\nu}=-R_{\alpha\beta\mu\nu}\] (4) Thus we have \[g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=-g^{\beta\mu}g^{\alpha\nu}R_{\alpha\beta\gamma\delta}\] (5) \[R_{\alpha\beta\gamma\delta}\left[g^{\alpha\mu}g^{\beta\nu}+g^{\beta\mu}g^{\alpha\nu}\right]=0\] (6) With (6) alpha and beta are dummy indices;others are free indices.Treating the Riemann tensor components as variables[unknowns], there are sixteen of them for the sixteen alpha beta combinations any given gamma , delta and mu, nu combinations.Each equation has gamma,delta,mu and nu as constant quantities. In total we have 256 equations[four values for each gamma,delta,mu and nu].Since these equations are of homogeneous nature, \[R_{\alpha\beta\gamma\delta}=0\] (7) In the orthogonal system of coordinates we obtain from (6) for distince mu and nu \[R_{\mu\nu\gamma\delta}\left[g^{\mu\mu}g^{\nu\nu}+g^{\nu\mu}g^{\mu \nu}\right]=0\] (8) In (8) we have considered different values of mu and nu as well as different values for gamma and delta. One should also take note of the fact that there is no summation on mu and on nu. \[R_{\mu\nu\gamma\delta}g^{\mu\mu}g^{\nu\nu}=0\] (9) In (8) also there is no summation on mu and on nu. Since g_mu mu not equal to 0 and g_nu nu not equal to zero we have for unequal mu ,nu and unequal gamma delta, \[R_{\mu\nu\gamma\delta}=0\] (10) If the Riemann tensor is a zero tensor ten the Ricci tensor is also a zero tensor.The Ricci scala becomes a zero valued scalar. Riemann Curvature.pdf
  3. (t,x,y,z) could represent coordinates of four dimensional space in a general manner; they could be Cartesian,Spherical or any other system. Equation (1) of the initial post represents the most general type of a General Relativity metric in the orthogonal system: \[c^2d\tau^2= c^2g_{00} d t^2-g_{11} dx^2-g_{22} dy^2-g_{33} dz^2 \] [Inadvertently,with the string post of the discussion, after copy pasting I forgot to change the suffix with g_ii to appropriate values.]
  4. [One may have to refresh the page for viewing the formulas and the equations] 1. In our discussion we have considered time like separations at a point so that d tau^2>0 2. Towards the end of the last post..For arbitrary K,l and M we observe, point (2) has to be ignored 3. We had \[cdt=kdr=ld\theta=md \phi\].All three components in the last line have the dimensions of length. k is dimensionless while l and m have the dimensions of length. Dividing the Schwarzschild metric by (cdt)^2 we obtain \[\left(\frac {d\tau}{dt}\right)^2=\left(1-\frac{2Gm}{c^2 r}\right)-\frac{1}{k^2}\left(1-\frac{2Gm}{c^2 r}\right)^{-1}-\frac{1}{l^2} r^2-\frac{1}{m^2}r^2 \sin^2 \theta\] In the last equation we may set k=1 ,l=m=1 unit of length. Therefore, \[\left(1-\frac{2Gm}{c^2 r}\right)\ge \frac{1}{k^2}\left(1-\frac{2Gm}{c^2 r}\right)^{-1}+\frac{1}{l^2}r^2+\frac{1}{m^2}r^2 \sin\theta\] In the last equation has to hold for any k,l and m for which dtau^2 is positive.[as a particular instance we may set k=1 ,l=m=1 unit of length] and for arbitrary r beyond the event horizon:dt,dx,dy and dz are not present but their rlative sizes are present. For large values of 'r' the metric automatically becomes space like unless we make dtheta and dphi extraordinarily small with respect to dr and dt Plain text versions of the two stated formulas: [dtau/dt]^2=[1-2Gm/c^2r]-(1/k^2)[1-2Gm/c^2r]^(-)-(1/l^2)r^2-(1/m^2)r^2sin^2 theta Since time like separations are being considered on (+,-,-,-) [1-2Gm/c^2r]>(1/k^2)[1-2Gm/c^2r]^(-)+(1/l^2)r^2+(1/m^2)r^2sin^2 theta the above inequation will not hold for large r. For a general type of a metric given the space time location we have to be careful with the relative sizes of the coordinate intervals in order to have a time like or a space like interval.
  5. Metric: [Signature;(+,-,-,-)] \[c^2 d\tau^2=c^2 g_{00} dt^2-g_{11} dx^2-g_{00} dy^2-g_{00} dz^2\] (1) [In (1) g_00,g_11,g_22 and g_33 are the absolute values of the metric coefficients,the system of coordinates being orthogonal] By our choice let \[cdt=dx=dy=dz\] (2) Dividing both sides of (1) by (cdt)^2 We have \[ \left(\frac {d\tau}{dt}\right)^2=g_{00}-g_{11}-g_{22}-g_{33}\] (3) Since the left side of (3) is positive we have, \[g_{00}\ge g_{11}+g_{22}+g_{33}\](4) Taking Schwarzschild geometry \[1-\frac{2Gm}{c^2r}\ge \left[1-\frac{2Gm}{c^2r}\right]^{-1}+r^2+r^2{\sin}^2\theta\] (5) The above is not true for large r With \[c^2 d\tau^2=c^2 \left[1-\frac{2Gm}{c^2r}\right]dt^2-\left[1-\frac{2Gm}{c^2r}\right]^{-1} dr^2-r^2 d\theta^2-r^2 \sin^2\theta d\phi^2\] \[ cdt=kdr=ld\theta=md\phi \] Each term above has the dimension of lengthk is dimensionless while l and m have the dimensions of length. \[ \left(\frac {d\tau}{dt}\right)^2=\left[1-\frac{2Gm}{c^2r}\right]dt^2-\frac{1}{k^2}\left[1-\frac{2Gm}{c^2r}\right]^{-1}dr^2-\frac{1]{l^2}r^2d\theta^2-\frac{1}{m^2}r^2 \sin^2\theta d\phi^2\] \[\left[1-\frac{2Gm}{c^2r}\right]\ge \frac{1}{k^2}\left[1-\frac{2Gm}{c^2r}\right]^{-1}+\frac{1]{l^2}r^2-+\frac{1}{m^2}r^2 \sin^2\theta \] (5') As a special case we may set k=1 unit,l=1 unit,m=1 unit We now do have \[\left[1-\frac{2Gm}{c^2r}\right]\ge \left[1-\frac{2Gm}{c^2r}\right]^{-1}+r^2+r^2 sin^2\theta \] (5'') In the rest of the discussion we are considering the rectangular Cartesian system. At a given point on the manifold we take,[by our choice], \[cdt=kdx=ldy=mdz\] We divide both sides of (1) by (cdt)^2 \[ \left(\frac {d\tau}{dt}\right)^2=g_{00}-\frac{1}{k^2}g_{11}-\frac{1}{l^2}g_{22}-\frac{1}{m^2}g_{33}\](6) \[\left(\frac {d\tau}{dt}\right)^2=g_{00}-Kg_{11}-Lg_{22}-Mg_{33}\](7) For arbitrary K,L and M, we observe, (1) \[ g_{00}-Kg_{11}-Lg_{22}-Mg_{33}\ge 0\](2) \[ g_{00}-Kg_{11}-Lg_{22}-Mg_{33}=g_{00}-g_{11}-g_{22}-g_{33}\] These observations are not valid ones. The manifold is becoming threadbare. Two of the statements with out \[ \], have been given out in plaintext since the codes were not getting parsed
  6. You are requested to discard this thread
  7. Revised Post: [One may have to refresh the page to view the formulas/equations] In this writing we dive a highly restrictive relation for the metric coefficients in General Relativity We start with the the formula \[\sum g^{\alpha\beta}g _{\alpha\beta}=4\] (1) Summation extends over alpha and beta. In orthogonal systems (1) reduces to \[g_{00}g^{00}+g_{11}g^{11}+g_{22}g^{22}+g_{33}g^{33}=4\] (1' ) Applying the reversed Cauchy Schwarz inequality we have, \[\left[ g_{00}g^{00}- g_{11}\left(-g^{11}\right)- g_{22}\left(-g^{22}\right)- g_{33}\left(-g^{33}\right)\right]^2\\ \ge\left(g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right)\left(\left(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right)\] (2) Using (1) and (2) we obtain, \[ 16\ge \left|g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right|\\ \left|\left(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right|\] (3) Since for orthogonal coordinate systems we have the formula[1] \[g_{\alpha \alpha}g^{\alpha\alpha}=1\] with no summation on alpha in the above \[\Rightarrow g_{\alpha \alpha}=\frac{1}{g^{\alpha\alpha}}\] we now have, \[ \left[g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right]\\ \left[\left(\frac{1}{g_{00}}\right)^2-\left(\frac{1}{g_{11}}\right)^2-\left(\frac{1}{g_{22}}\right)^2-\left(\frac{1}{g_{33}}\right)^2\right]\le 16\] (4) Equation (4) is highly restrictive. On The Reversed Cauchy Schwarz Inequality: We consider 4 real numbers \[a_1,a_2\]and \[b_1,b_2\] We have \[\left(a_1b_1-a_2b_2\right)^2- \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_ab_2-a_2b_1\right)^2\ge 0\] \[\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\] (5) Next we take 2n real numbers a1,a2...an and b1,b2..bn By applying the Cauchy Schwarz inequality we have, \[\left(a_2b_2+a_3b_3....+a_nb_n\right)^2\le\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)\] \[\frac{\left(a_2b_2+a_3b_3....+a_nb_n\right)^2}{\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)}\le 1\] \[-1\le \frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}\le 1\] Therefore \[\frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}=\cos \theta\] \[\Rightarrow a_2b_2+a_3b_3....+a_nb_n=\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta \] \[a_1 b_1-a_2b_2-a_3b_3....-a_nb_n= \\ a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\] (6) \[\left(a_1 b_1-a_2b_2-a_3b_3....-a_nb_n\right)^2= \\ \left(a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\right)^2\] (7) Applying (5) on the right side of (7) we obtain, \[\left(a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\right)^2\ge \\ \left(a_1^2-a_2^2-a_3^2.....-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.....-b_n^2\right)\] (8) From (7) and (8) we have, \[ \left (a_1 b_1-a_2b_2-a_3b_3....-a_nb_n\right)^2 \ge\\ \left(a_1^2-a_2^2-a_3^2.....-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.....-b_n^2\right)\]References 1. Spiegel M,Theory and Problems of Vector Analysis with an Introduction to Tensor Analysis Schaum's series, McGraw-Hill Book Company,Singapore, 1974,Chapter8; Tensor Analysis, problem 48b,p194
  8. This post will be Revised.Pl wait. [it might be necessary to refresh the page or viewing the formulas/equations] In this writing we dive a highly restrictive relation for the metric coefficients in Genera Relativity We start with the the formula \[\sum g^{\alpha\beta}g _{\alpha\beta}=4\] (1) Summation extends over alpha and beta In the orthogonal systems (1) reduces to \[g_{00}g^{00}+ g_{11}g^{11}+ g_{22}g^{22}+ g_{33}g^{33}=4\] (1') Applying the reversed Cauchy Schwarz inequality to (1')we have, \[\left[ g_{00}g^{00}- g_{11}\left(-g^{11}\right)- g_{22}\left(-g^{22}\right)- g_{33}\left(-g^{33}\right)\right]^2\\ \ge\left(g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right)\left(\left|(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right)]\] (2) Using (1) and (2) we obtain, \[ 16\ge \left(g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right)\ \left[\left(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right]\] (3) Since for orthogonal coordinate systems we have the formula[1] \[g_{\alpha \alpha}g^{\alpha\alpha}=1\] with no summation on alpha in the above \[\Rightarrow g_{\alpha \alpha}=\frac{1}{g^{\alpha\alpha}}\] we now have, \[ \left [g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right]\\ \left[\left(\frac{1}{g_{00}}\right)^2-\left(\frac{1}{g_{11}}\right)^2-\left(\frac{1}{g_{22}}\right)^2-\left(\frac{1}{g_{33}}\right)^2\right]\le 16\] (4) Equation (4) is highly restrictive. The Reversed Cauchy Schwarz Inequality: We consider 4 real numbers \[a_1,a_2\]and \[b_1,b_2\] We have \[\left(a_1b_1-a_2b_2\right)^2- \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_ab_2-a_2b_1\right)^2\ge 0\] \[\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\] \[\left(a_1b_1-a_2b_2\right)^2\ge \left|a_1^2-a_2^2\right|\left|b_1^2-b_2^2\right|\] (5) Next we take 2n real numbers a1,a2...an and b1,b2..bn By applying the Cauchy Schwarz inequality we have, [\left(a_2b_2+a_3b_3....+a_nb_n\right)^2\le\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)\] \[\frac{\left(a_2b_2+a_3b_3....+a_nb_n\right)^2}{\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)}\le 1\] \[-1\le \frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}\le 1\] Therefore \[\frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}=\cos \theta\] \[\Rightarrow a_2b_2+a_3b_3....+a_nb_n=\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta \] \[a_1 b_1-a_2b_2-a_3b_3....-a_nb_n= \\ a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\] (6) Applying (5) on the right side of (6) we obtain, \[ a_1 b_1-a_2b_2-a_3b_3....-a_nb_n \ge\\ \sqrt{\left|a_1^2-a_2^2-a_3^2.....-a_n^2\right|}\sqrt{\left|b_1^2-b_2^2-b_3^2.....-b_n^2\right|}\] Or \[-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\\ \le a_1 b_1-a_2b_2-a_3b_3....-a_nb_n \] Therefore \[\left(a_1 b_1-a_2b_2-a_3b_3....-a_nb_n\right)^2 \ge \left|a_1^2-a_2^2-a_3^2.....-a_n^2\right| \\ \times \left|b_1^2-b_2^2-b_3^2.....-b_n^2\right|\] References 1. Spiegel M,THeory and Problems of Vector Analysis with an Introduction to Tensor Analysis Schaum's series, McGraw-Hill Book Company,Singapore, 1974,Chapter8; Tensor Analysis, problem 48b,p194
  9. In relation to the last posting one should take note of the following facts: We may consider two close events (t,x,y,z) and (t+dt,x+dx,y+dy,z+dz) on the particle's world line[space time path] and apply the Lorentz transformation on the infinitesimals dt, dx, dy and dz [asides t,x,y and z]. The reference frames involved have to be inertial obviously with uniform relative motion between themselves for the application of the Lorentz transformation. For that it is not necessary that the quantities dt/dtau,dx/dtau,dy/dtau and dz/tau have to be constants.In general they are variables. Flat space time, of course ,is supportive to relative acceleration between reference frames.If Special Relativity is identified with the Lorentz transformations it does not support acceleration between reference frames [for the transformations]
  10. In relation to the last comment made by Markus Hanke [One may have to refresh the page for viewing the formulas] We may have accelerating frames in Flat Space time.That is right.But in my discussion I used the relation(metric) \[c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2\] (1) The metric represented by (1) follows from the Lorentz Transformations. We can also derive the Lorentz transformations from the metric that is from (1)[Steven Weinberg, Gravitation and Cosmology, Chapter 2: Special Relativity]. Equation (1) characterizes Special Relativity When I talked of Special Relativity , I meant the effect of the metric represented by(1). The Lorentz transformations consider frames in relative uniform motion. The metric represented by (1):has it got anything to do with frames in relative acceleration which are possible in flat space time?What are the transformation laws there? Special Relativity and flat space time are not identical objects. Flat space time may have reference frames with relative acceleration between hem. This is not possible with Special Relativity because Lorentz Transformations consider only such frames of reference as have uniform relative motion between them[only inertial frames are considered in Special Relativity] \[\frac{dt}{d\tau}\] considered in the article is not the gamma between the reference frames. It is the gamma for the particle. \[\gamma=\frac{1}{\sqrt{1-\frac{v_p^2}{c^2}}}\] (2)where v_p is the particle velocity and not the relative velocity between reference frames. A particles motion in a specified reference frame may be considered independent of other reference frames and the relative velocity between them. Recalling (1) and dividing both sides by dtau^2 we obtain \[c^2=c^2v_t^2-v_x^2-v_y^2-v_z^2\] (3) The velocities are variable in respect of time.The term v_t relates to energy and energy of an accelerating particle is not constant. \[\left(v_t,v_x,v_y ,v_z\right)\] are variable quantities and not constants \[E=mc^2=m_0 \gamma c^2\](3) gamma in (3) is a variable gamma corresponding to the motion of the particle in a given frame of reference. The Lorentz transformations involve the relative velocity of separation between the two frames of reference and a different gamma[different from the particle gamma]. Both disappear when we formulate (1) from them: \[c^2dt^2-dx^2-dy^2-dz^2= c^2dt’^2-dx’^2-dy’^2-dz’^2\] (4) Each side of (2) is denoted by \[c^2d\tau^2\] Again from (2),by considering the invariance of proper time we may derive the Lorentz transformations uniquely : and gamma[between the two reference frames as distinct from the particle gamma] pops up . On Particle gamma: From (1) \[c^2 \left(\frac {dt}{d\tau}\right)^2=c^2-\left(\frac{dx}{d\tau} \right)^2-\left(\frac{dy}{d\tau} \right)^2-\left(\frac{dz}{d\tau} \right)^2\] \[c^2 \left(\frac {dt}{d\tau}\right)^2=c^2-v_x^2-v_y^2-v_z^2\] \[ \left(\frac {dt}{d\tau}\right)^2=c^2-v^2 \] \[ \frac {dt}{d\tau}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\](5) \[v^2=v_x^2+v_y^2+v_z^2\] is the particle velocity. The particle could at rest at the origin of one of the reference frames when is motion with respect to other reference frames in terms of dt,dx ,dy and dz when referred to (5) represents the gamma between the reference frames. Else it is the particle gamma.The invariance expressed by (4) does not sand in the way of the velocity components being variable.For example \[ dx’=\gamma\left(dx-vdt\right)\] \[ \Rightarrow \frac{dx’}{d\tau}=\gamma\left(\frac{dx}{d\tau}-v\frac{dt}{d\tau}\right)\] The constancy of v does not stand in the way of dx/dtau ,dt/dtau and dx’/dtau being variables.
  11. With Improved Latex as mentioned in the last posting[One may require to refresh the page for proper viewing] Four Acceleration \begin{equation}\left(c\frac{d^2 t}{d\tau^2},\frac{d^2 x}{d\tau^2},\frac{d^2 y}{d\tau^2}, \frac{d^2 z}{d\tau^2}\right)\end{equation} (1) Let \begin{equation}c^2N=c^2\left(\frac{d^2 t}{d\tau^2}\right)^2-\left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2\end{equation} (2) We consider the metric \begin{equation}c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2 \end{equation} (4) \begin{equation}\Rightarrow c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2- \left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2 \end{equation} (5) Differentiating (5) with respect to proper time we have, \[ c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0\] (6) By applying the Cauchy Schwarz inequality we have, \[\left(\frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}+\frac{d y}{d\tau}\frac{d^2y}{d \tau^2}+\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}\right)^2 \\ \ge \left(\left(\frac{d x}{d\tau}\right)^2+\left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau} \right)^2\right)\left(\left(\frac{d^2 x}{d \tau^2}\right)^2+\left(\frac{d^2 y}{d \tau^2}\right)^2+\left(\frac{d^2 z}{d \tau^2}\right)^2\right)\](7) or, \[\left(c^2\left(\frac{d t}{d \tau}\right)^2-c^2\right)\left(c^2\left( \frac {d^2 t}{d \tau^2}\right)^2-c^2N\right) \ge\left( c^2 \frac {d^2 t}{d\tau^2}\right)^2\left(c^2\frac{d t}{d\tau}\right)^2\] \[\left(\left(\frac{dt}{d \tau}\right)^2-1\right)\left(\left( \frac {d^2 t}{d \tau^2}\right)^2-N\right) \ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\] (8) \[ \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\] \[ \Rightarrow-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge 0\] (9) \[N\left(1-\left(\frac{dt}{d\tau}\right)^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2\] \[N\left(1-\gamma^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2\](10) The right side of (10) is always positive or zero. Therefore for N<=0 has to hold for the left side ,gamma[Lorentz factor] being always positive[> =unity]. N cannot be positive unless the particle is moving uniformly. If N is negative then from (1) we have \[c^2\left(\frac{d^2 t}{d\tau^2}\right)^2\le \left(\frac{d^2 x}{d\tau^2}\right)^2+\left(\frac{d^2 y}{d\tau^2}\right)^2+\left( \frac{d^2 z}{d\tau^2}\right)^2\] For a particle at rest (spatially) and N<0, \[\left(\frac{d^2 t}{d\tau^2}\right)^2\le 0\](11) Equation (11) will not hold, the left side being a [perfect square and hence positive or zero]unless \begin{equation}\frac{d^2 t}{d\tau^2}=0\end{equation} that is unless \begin{equation}\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant\end{equation} that is unless the particle is moving with a constant velocity. An accelerating particle will not cater to N<0. For N=0 we have from (10) \[\left(\frac{d^2x}{d\tau^2}\right)^2\le 0 \](12) Equation (12) is not a valid on unless the particle moves with a constant velocity. We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant velocity @ Markus Hanke: In special Relativity the frames of reference are inertial and hence they are in uniform relative motion. But the particles in a given inertial frame of reference[and hence in other frames of reference,inertial ]can be in a state of acceleration. The momentum of a particle is given by \[ p_i=m\frac{dx^i}{dt}=m_0\gamma\frac{dx^i}{dt}=m_0 \frac{dx}{d\tau}\] since \[\gamma=\frac {dt}{d \tau}\] The \[\gamma\] in the last line is the particle's gamma and it is expected to be a variable if the particle is in a state of acceleration. But in this article we see that the particle cannot accelerate.
  12. The initial posting may be understood with difficulty. I would repost the Latex form at my earliest convenience. In the mean time the viewer may consider the uploaded file. Norm 2.pdf
  13. Four Acceleration vector \begin{equation}\left(c\frac{d^2 t}{d\tau^2},\frac{d^2 x}{d\tau^2},\frac{d^2 y}{d\tau^2}, \frac{d^2 z}{d\tau^2}\right)\end{equation} (1) [[ct]=[x]=[y]=[z]] Let \begin{equation}c^2N=c^2\left(\frac{d^2 t}{d\tau^2}\right)^2-\left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2\end{equation} (2) We consider the metric \begin{equation}c^2d\tau^2=c^2dt^2-dx2-dy^2-dz^2 \end{equation} (4) \begin{equation}\Rightarrow c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\end{equation} (5) Differentiating (5) with respect to proper time we have, \begin{equation} c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}-\ frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\ frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\ frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0\end{equation} (6) By applying the Cauchy Schwarz inequality we have, \begin{array}{l}\left(\ frac{dx}{d\tau}\ frac{d^2 x}{d \tau^2}+\ frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}+\ frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}\right)^2 \\ \le \left(\left(\ frac{dx}{d\tau}\right)^2+\left(\ frac{dy}{d\tau}\right)^2+\left(\ frac{dz}{d\tau}\right)^2\right)\\ \times \left(\left(\ frac{d^2 x}{d\tau^2}\right)^2+\left(\ frac{d^2 y}{d\tau^2}\right)^2+\left(\ frac{d^2 z}{d\tau^2}\right)^2\right)\end{array} (7) \begin{equation}\left(c^2\left(\ frac{dt}{d \ tau}\right)^2-c^2\right)\left(c^2\left( \ frac {d^2 t}{d \tau^2}\right)^2-c^2N\right) \ge\left( c^4\ \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\end{equation} \begin{equation}\left(\left(\frac{dt}{d \tau}\right)^2-1\right)\left(\left( \frac {d^2 t}{d \tau^2}\right)^2-N\right) \ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\end{equation} (8) \begin{equation} \left( \frac {d^2 t}{d\tau^2}\right)^2 \left(\frac{dt}{d\tau}\right)^2-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge \left(\frac{ \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\end{equation} \begin{equation} \Rightarrow -{N}\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+{N}\ge \0\end{equation} (9) \begin{array}{N}\left({{1}-\left\frac{dt}{d\tau}\right)^2}\right)\ge \\ \left( \frac {d^2 t}{d \tau^2}\right)^2\end{array} \begin{equation}N\left(1-\gamma^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2\end{equation}(10) The right side of (10) is always positive or zero. Therefore N<=0 ,gamma[Lorentz factor] being always positive[> unity]. N cannot be positive unless the particle is moving uniformly. If N is negative then from (1) we have \begin{equation}c^2\left(\frac{d^2 t}{d\tau^2}\right)^2\le \left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2\end{equation} For a particle at rest (spatially) and N<0, \begin{equation}\left(\frac{d^2 t}{d\tau^2}\right)^2\le 0\end{equation}(11) Equation (11) will not hold, the left side being a [perfect square and hence positive or zero]unless \begin{equation}\frac{d^2 t}{d\tau^2}=0\end{equation} that is unless \begin{equation}\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant\end{equation} that is unless the particle is moving with a constant velocity. An accelerating particle will not cater toy N<0. For N=0 we have from (10) \begin{equation}\left(\frac{d^2x}{d\tau^2}\right)^2\le 0 \end{equation}(12) Equation (12) is not a valid on unless the particle moves with a constant velocity. We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant speed.
  14. Ghideon "Please explain how that addition is valid, the components you add are not arbitrary numbers." The squares of the components have been added. Its true that these numbers (squares of the components)are not independent. They are related by (3.1) and by(3.2). That does not matter.We can always add the two sides of a pair of equations. Next Issue[Independent of reply to Ghideon] For proper viewing one may have to refresh the page. We start with the norm of proper velocity[metric signature:(+,-,-,-)] \begin{equation}c^2=c^2 v_t^2-v_x^2-v_y^2-v_z^2\end{equation} \begin{equation}c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dt}{d\tau}\right)^2\end{equation} Differentiating with respect to propertime, \begin {equation}c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}-\frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0\end {equation} (1) \begin{equation}\Rightarrow v.a=0\end{equation}(2) We choose k such that [k] =T so that ka has the dimension of velocity We have, \begin{equation}\Rightarrow v.ka=0\end{equation}(3) We do have, \begin{equation}v.v=c^2\end{equation}(4) From (3) and (4) \begin{equation}\Rightarrow v. \left(v-ka\right)=c^2\end{equation} (5) By adjusting the value [but maintaining its dimension as that of time] we always do have equation (5) If \begin{equation}\left(v-ka\right)=v'\end{equation} is a proper velocity then we have v.v'=c^2 in opposition to v.v'>=c^2 If \begin{equation}\left(v-ka\right)=v'\end{equation}$ is a not a proper velocity then we have from the reversed Cauchy Schwarz inequality, \begin{array}{l}\left(c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\right)^2\ge\\ \left(c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\right)\left(c^2\left(\frac{dt'}{d\tau'}\right)^2-\left(\frac{dx'}{d\tau'}\right)^2-\left(\frac{dy'}{d\tau'}\right)^2-\left(\frac{dz'}{d\tau'}\right)^2\right)\end{array}(6) or,\begin{equation}\left(c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\right)^2\ge c^2c'^2\end{equation} (7) \begin{equation}c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\ge cc'\end{equation} or, \begin{equation}c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\le -cc'\end{equation} v.v'>=cc' or v.v'<=-cc' But v.v'=c^2. Therefore the solution is c'=c that is v' is a proper velocity.||But we assumed /postulated at the very outset that v' is not a proper velocity. Reversed Cauchy_Schwarz 5.pdf
  15. Important revisions/corrections have been made with the 'Extra Bit'.Relevant portion has been provided in Latex. The entire file[revised file] has been uploaded. Metric \begin{equation}c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2 \end{equation} (1) \begin{equation}c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\end{equation} \begin{equation}c^2=c^2{v_t}^2-{v_x}^2-{v_y}^2-{v_z}^2\end{equation}(2) We consider two proper velocities on the same manifold \begin{equation}c^2=c^2{v_{1t}}^2-{v_{1x}}^2-{v_{1y}}^2-{v_{1z}}^2\end{equation}(3.1) \begin{equation}c^2=c^2{v_{2t}}^2-{v_{2x}}^2-{v_{2y}}^2-{v_{2z}}^2\end{equation}(3.2) Adding (3.1) and (3.2) we obtain \begin{equation}2c^2=c^2\left({v_{1t}}^2+{v_{2t}}^2\right)-\left({v_{1x}}^2+{v_{2x}}^2\right)-\left({v_{1y}}^2+{v_{2y}}^2\right)-\left({v_{1z}}^2+{v_{2z}}^2\right)\end{equation} \begin{equation}2c^2=c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2x}}\right)^2-\left({v_{1y}}+{v_{2y}}\right)^2-\left({v_{1z}}+{v_{2z}}\right)^2-2v_1. v_2\end{equation} \begin{equation}2c^2+2v_1. v_2=c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2x}}\right)^2-\left({v_{1y}}+{v_{2y}}\right)^2-\left({v_{1z}}+{v_{2z}}\right)^2\end{equation} Since $$v1.v2\ge c^2 $$ we have \begin{equation}c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2x}}\right)^2-\left({v_{1y}}+{v_{2y}}\right)^2-\left({v_{1z}}+{v_{2z}}\right)^2\ge 4c^2\end{equation}(4) \begin{equation}\left(v_1+v2\right ).\left(v_1+v_2\right)\ge 4c^2\end{equation} If $v_1+v_2$ is a proper velocity then \begin{equation}c^2=c^2\left(v_{1t}+v_{2t})\right)^2-\left(v_{1x}+v_{2x})\right)^2-\left(v_{1y}+v_{2y})\right)^2-\left(v_{1z}+v_{2z})\right)^2\end{equation} \begin{equation}c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2+2v_1.v_2\end{equation} \begin{equation}c^2=c^2+c^2+2v_1.v_2\end{equation} Therefore \begin{equation}v_1.v_2\le -½ c^2\end{equation} which is not true since \begin{equation}v.v=c^2\end{equation} Therefore $v_1+v_2$ is not a four vector if $v_1$ and $v_2$ are four vectors Again if $v_1-v_2$ is a four vector then \begin{equation}c^2=c^2\left(v_{1t}-v_{2t})\right)^2-\left(v_{1x}-v_{2x})\right)^2-\left(v_{1y}-v_{2y})\right)^2-\left(v_{1z}-v_{2z})\right)^2\end{equation} \begin{equation}c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2-2v_1.v_2\end{equation} \begin{equation}c^2=c^2+c^2-2v_1.v_2\end{equation} \begin{equation} ½ c^2=v_1.v_2\end{equation} But the above formula is not a valid one. Given two infinitesimally close four velocities their difference is not a four velocity. Therefore the manifold has to be a perforated one. The manifold indeed is a mesh of worldlines and each world line is a train of proper velocity four vectors as tangents. A particle moves along a timelike path and therefore each point on it has a four velocity tangent representing the motion. The manifold is discrete and that presents difficulty with procedure like differentiation. Four352.pdf
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.