Jump to content

Anamitra Palit

Senior Members
  • Content Count

  • Joined

  • Last visited

Community Reputation

-11 Bad

About Anamitra Palit

  • Rank

Profile Information

  • Favorite Area of Science

Recent Profile Visitors

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

  1. We consider a linear vector space V of dimension n. W is a proper subspace of V.We take a vector 'e' belonging to V-W and N vectors y_i belonging to W;i=1,2,3…N;N>>n the dimension of V. All y_i cannot obviously be independent the number N being greater than the dimension of V the parent vector space; k of the y_i vectors are considered to be linearly independent where k is the dimension of W. The rest of the y_i are linear combinations of these k, basic y_i vectors of W. We consider sums αi=e+yi;i=1,2…N (1) Now each alpha_i=e+y_i belongs to V-W. We prove it as follo
  2. First we consider the fact that the norm-square of the four acceleration vector is negative or zero. Indeed \[c^2=c^2\left(\frac {dt}{d \tau}\right)^2-\left(\frac {dx}{d \tau}\right)^2-\left(\frac {dy}{d \tau}\right)^2-\left(\frac {dz}{d \tau}\right)^2\] (1) Differentiating both sides with respect to time we obtain \[c^2\frac{d^2 t}{dt^2}\frac{dt}{d\tau}-\frac{d^2 x}{dt^2}\frac{dx}{d\tau}-\frac{d^2 y}{dt^2}\frac{dy}{d\tau}-\frac{d^2 z}{dt^2}\frac{dz}{d\tau}=0\] (2) We transform to an inertial frame where the particle is momentarily at rest. \[c^2\frac{d^2 t}{d\tau^2}\frac{dt}{d \
  3. Forum expert Markus Hanke declined believing in the relation v1.v2>=c^2 that stems from the reversed Cauchy Schwarz inequality.In case he happens to believe in the stated relation my discussion has made a substantial contributed to the forum and it[my discussion] is far from being of a speculative nature. Kino seems to have accepted this relation v1.v2>=c^2 [The relation v.v=c^2[gamma^2-1+1/gamma^2] is not a valid one . Nevertheless v1.v2>=c^2 is a valid formula] I have corrected a loose statement made by the forum expert. |his indeed has a non speculative contribution.
  4. The acceleration four vector is indeed space like or null[for uniform motion] The norm of the acceleration four vector as suggested by Kino is negative[or zero , for uniform motion].There are different methods of proving the stated fact. One is presented here as a prelude to the point referred to finally. We start with the metric \[c^2d\tau ^2=c^2dt^2-dx^2-dy^2-dz^2\] (1.1) \[c^2=c^2 \left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\] (1.2) Differentiating both sides of (1.2) with respect to proper time
  5. The section of my second last post from after equation (17.2) and before the section " The Sum and Difference of Two Proper Velocities " has been revised We restate (13),(14),(15) ..(17.1) and then proceed.. \[v.v= c^2\](13) \[v.ka =0\](14) Therefore \[v.\left(v-ka\right) = c^2\] (15) We could expect from the last equation \[\left(v-ka\right)\] to be a proper velocity. Then norm^2=c^2. But we can vary k and make the norm different from c^2 so that v-ka is not a proper velocity. Let \[\left||v-ka\right||=\bar c^2 \ne c^2\] (16) We adjust the value of k so that
  6. Equation (7) of the last post has been rewritten[it could not be parsed in the last post and there was no time left for editing]: \[v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2}\sqrt{c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2\] For writing (3.3) in the last post we have the following considerations For X>=0, X>=X cos theta Therefore, a1 b1+X>=a1b1+X cos theta or, a1b1-X cos theta>=a1 b1-X X:product of two square roots, a positive number For X>=0 \[X\ge X
  7. The current form of the article in Latex [It might be necessary to refresh the page for viewing the ormulas and the equations] Mathematical Theorem If \[\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\ge 0\] then either \[\left(a_1b_1-a_2b_2\right)\ge \sqrt{a_1^2-a_2^2}\sqrt{b_1^2-b_2^2}\](1.1) or \[\left(a_1b_1-a_2b_2\right)\le -\sqrt{a_1^2-a_2^2}\sqrt{b_1^2-b_2^2}\](1.2) Proof \[\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_1b_2-a_2b_1\right)^2\] \[\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\
  8. \[X_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0\] The coefficient matrix is independent of gamma and delta. For the same alpha and beta various gamma and delta should produce the same value for the corresponding component ,that is, X_{alpha beta gamma delta}=X_{alpha beta gamma’ delta’} If the coefficient matrix is non zero(or zero) for some (gamma,delta ) pair it should remain non zero (or zero) for other (gamma, delta) pairs.
  9. 1)General perspective always pertain to the special cases.Whatever ensues from: \[X_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+g^{\alpha\nu}g^{\beta\mu}\right)=0\] will also apply to the Riemann tensor. For constant delta and gamma there are 16 variables of the type X_{alpha beta gamma delta} for four values of each alpha and beta..These are treated as the unknowns. There are 16 equations for the four values of each of mu and nu. We have sixteen linear homogeneous equations with sixteen variables[unknowns].If the determinant of the coefficient matrix is non zero then
  10. It is not a correct point.General perspectives cannot be violated in the special cases.Why don't you go through my last post in this discussion?
  11. [It might be necessary to refresh the page for proper viewing] Thanking Kino for his first comment.[posted on Saturday,3:43 pm] It is true that the product of a symmetric and antisymmetric tensor is always zero. Nevertheless we may consider the following: \[R_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0\](1) [A significant aspect of (1) is that it holds for any arbitrary(mu,nu) pair] Keeping gamma and delta constant[and we maintain so throughout this post] we vary alpha and beta for each equation in (1). In order to obtain new
  12. [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}\fr
  13. 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) Therefo
  14. (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.]
  15. [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
  • 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.