# Space time is reality Pseudo

Discussion in 'Pseudoscience' started by chinglu, Oct 19, 2013.

1. ### chingluValued Senior Member

Messages:
1,637
OK, you retreat.

Messages:
26,023
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!!!

5. ### chingluValued Senior Member

Messages:
1,637
This is not a peer review forum.
So, obviously you cannot refute the argument I presented.

Move along.

Messages:
26,023
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???

8. ### chingluValued Senior Member

Messages:
1,637
Let's see the refutation.

Messages:
26,023
It's been posted many times, and you ignore it....Let's see the peer review?

Messages:
26,023
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

(3)
= (dx^1)^2+(dx^2)^2+(dx^3)^2,
(4)
while the Minkowski metric gives its relativistic generalization, the proper time

(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

(7)
where Lambda^alpha_beta is a Lorentz tensor. It also satisfies

(8)
eta_(alphagamma)Lambda^(betagamma)=Lambda_alpha^beta
(9)
(10)
The metric of Minkowski space is diagonal with

eta_(alphaalpha)=1/(eta_(alphaalpha)),
(11)
and so satisfies

(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

11. ### pmbBannedBanned

Messages:
228
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

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.

Last edited: Nov 18, 2013