# Is This Spacetime Geometry Mathematically Conceivable?

Eugene Shubert
Is it possible to invent a non-Riemannian geometry to justify the existence of a "metric" of the form:

1/ds^2 = 1/dt^2 – 1/(dx^2 + dy^2 + dz^2)

Eugene Shubert
http://www.everythingimportant.org/relativity

isn't that just
ds2 = dt2- (dx2+dy2+dz2)
?

Umm, I would agree with schwarzchildradius here. Just multiply through and suddenly you get rid of the nasty fractions.

Yes I do remember elementary algebra, good for me. you can invert that equation.

plus
Would you like to expain in more detail how you think you can invert that fraction to get the required result?

Considering
I thought flipping a fraction such as 1/3^2 would result in 3^-2.

Doesn't it change the exponent?

C'mon guys...

1/ds2 = 1/dt2 – 1/(dx2 + dy2 + dz2)
1/ds2 = (dx2 + dy2 + dz2 - dt2)/[ (dx2 + dy2 + dz2)(dt2) ]

ds2 = [ (dx2 + dy2 + dz2)(dt2) ]/(dx2 + dy2 + dz2 - dt2)

Which just doesn't look any cleaner.

edit: changed to using integrated superscript.

Last edited by a moderator:
plus
Originally posted by suffian
C'mon guys...

1/ds2 = 1/dt2 – 1/(dx2 + dy2 + dz2)
1/ds2 = (dx2 + dy2 + dz2 - dt2)/[ (dx2 + dy2 + dz2)(dt2) ]

ds2 = [ (dx2 + dy2 + dz2)(dt2) ]/(dx2 + dy2 + dz2 - dt2)

Which just doesn't look any cleaner.

edit: changed to using integrated superscript.

That was my point.

Eugene Shubert
Let me suggest the physical meaning to the expression above.

I’m thinking of ds as an invariant that represents a differential increment of proper time. That would imply that the total amount of elapsed proper time t' would equal t/sqrt (1-1/V^2) where V^2 = (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2. I would interpret V^2 > 1 to be a superluminal velocity.