No, not at all..... Get it peer reviewed...take it to the masses... Why you do not???? Because you will reveal yourself as an Idiot to the world instead of just in a pseudoscience forum. tell you what chinglu...You will certainly get a prize for the champion question avoider....You have that in the bag!!!
This is not a peer review forum. So, obviously you cannot refute the argument I presented. Move along.
It's been refuted many times...The answers though label you as a fool. Of course this isn't peer review...That's what I've been saying here and in the other thread......So obviously any one with anything worthwhile would get it peer reviewed at a reputable outlet....Get it???
Minkowski Metric The Minkowski metric, also called the Minkowski tensor or pseudo-Riemannian metric, is a tensor eta_(alphabeta) whose elements are defined by the matrix (eta)_(alphabeta)=[-1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1], (1) where the convention c=1 is used, and the indices alpha,beta run over 0, 1, 2, and 3, with x^0=t the time coordinate and (x^1,x^2,x^3) the space coordinates. The Euclidean metric (g)_(alphabeta)=[1 0 0; 0 1 0; 0 0 1], (2) gives the line element ds^2 = g_(alphabeta)dx^alphadx^beta (3) = (dx^1)^2+(dx^2)^2+(dx^3)^2, (4) while the Minkowski metric gives its relativistic generalization, the proper time dtau^2 = eta_(alphabeta)dx^alphadx^beta (5) = -(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2. (6) The Minkowski metric is fundamental in relativity theory, and arises in the definition of the Lorentz transformation as Lambda^alpha_gammaLambda^beta_deltaeta_(alphabeta)=eta_(gammadelta), (7) where Lambda^alpha_beta is a Lorentz tensor. It also satisfies eta^(betadelta)Lambda^gamma_delta=Lambda^(betagamma) (8) eta_(alphagamma)Lambda^(betagamma)=Lambda_alpha^beta (9) Lambda_alpha^beta=eta_(alphagamma)Lambda^(betagamma)=eta_(alphagamma)eta^(betadelta)Lambda^gamma_delta. (10) The metric of Minkowski space is diagonal with eta_(alphaalpha)=1/(eta_(alphaalpha)), (11) and so satisfies eta^(betadelta)=eta_(betadelta). (12) The necessary and sufficient conditions for a metric g_(munu) to be equivalent to the Minkowski metric eta_(alphabeta) are that the Riemann tensor vanishes everywhere (R^lambda_(munukappa)=0) and that at some point g^(munu) has three positive and one negative eigenvalues. http://mathworld.wolfram.com/MinkowskiMetric.html
Can you please explain what that’s supposed to mean? What is “-1 0 0 0” supposed to mean? The Minkowski metric is given by n_ab = diag(1, -1, -1, -1) where diag(1, -1, -1, -1) is a diagonal metric with the numbers 1, -1, -1 and –1 as the diagonal components. That means that the components of the Minkowski metric are n_00 = 1, n_11 = n_22 = n_33 = -1. A bit of advice – Your notation is very sloppy making it hard to understand. E.g, it took a few seconds to realize that alphabeta is really “alpha beta”. Not separating them makes if confusing to read. That’s what spaces are actually for. The same thing holds here dtau^2 = eta_(alphabeta)dx^alphadx^beta First off since you can’t use Greek letters then you shouldn’t spell them out. Use the convention that Wald does in his GR text. In his text he uses letters from the beginning of the alphabet such as a, b, c, etc. Also try using T instead of “tau” so we have instead dT^2 = n_ab dx^a dx^b which is much easier on the eyes.
That is not my work, and I did give a link...here it is again.... http://mathworld.wolfram.com/MinkowskiMetric.html It appears that in copying and pasting, it didn't transpose as it should have.
