Is This Spacetime Geometry Mathematically Conceivable?
Is it possible to invent a nonRiemannian 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
ds^{2} = dt^{2} (dx^{2}+dy^{2}+dz^{2}) ? 
No, of course not. Remember your elementary algebra.
Eugene Shubert http://www.everythingimportant.org/relativity 
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.

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

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/ds^{2} = 1/dt^{2} – 1/(dx^{2} + dy^{2} + dz^{2}) 1/ds^{2} = (dx^{2} + dy^{2} + dz^{2}  dt^{2})/[ (dx^{2} + dy^{2} + dz^{2})(dt^{2}) ] ds^{2} = [ (dx^{2} + dy^{2} + dz^{2})(dt^{2}) ]/(dx^{2} + dy^{2} + dz^{2}  dt^{2}) Which just doesn't look any cleaner. edit: changed to using integrated superscript. 
Quote:

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 (11/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. 
All times are GMT 5. The time now is 03:30 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums