# Mathematica 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

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

#### Eugene Shubert

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?

#### 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.

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.

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