View Full Version : Is This Spacetime Geometry Mathematically Conceivable?
Eugene Shubert
May20-03, 12:30 AM
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
schwarzchildradius
May20-03, 07:57 PM
isn't that just
ds2 = dt2- (dx2+dy2+dz2)
?
Eugene Shubert
May20-03, 10:09 PM
No, of course not. Remember your elementary algebra.
Eugene Shubert
http://www.everythingimportant.org/relativity
Brad_Ad23
May21-03, 12:56 AM
Umm, I would agree with schwarzchildradius here. Just multiply through and suddenly you get rid of the nasty fractions.
schwarzchildradius
May21-03, 04:16 AM
Yes I do remember elementary algebra, good for me. you can invert that equation.
Would you like to expain in more detail how you think you can invert that fraction to get the required result?
Considering
May22-03, 10:01 AM
I thought flipping a fraction such as 1/3^2 would result in 3^-2.
Doesn't it change the exponent?
suffian
May22-03, 12:45 PM
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.
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
May22-03, 02:44 PM
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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