How Does the Singularity Behave as t Approaches 0 in the Kasner Solution?

  • Thread starter Thread starter Logarythmic
  • Start date Start date
  • Tags Tags
    Singularities
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
45 replies · 6K views
I'll work on this, though I've been studying for 12 hours now. ;) If that's all I need to know then thanks for your help.
 
Physics news on Phys.org
Take a break and good luck with the rest.
 
I think I will need some help with the rest aswell. ;) No MatLab installed here...
 
Unless I can use [tex]p_1p_2+p_2p_3+p_3p_1=0[/tex] aswell?
 
Sure you can. It's a consequence of the other two relations between the p's.
 
I get

[tex]p_1 = -\frac{p_3}{2} \pm \frac{1}{2} \sqrt{-3p_3^2+2p_3+1} + \frac{1}{2}[/tex]

and

[tex]p_2 = \frac{1}{2} \left(-p_3 \pm \sqrt{-3p_3^2+3p_3+1} + 1 \right)[/tex]

but what does this tell me? They cannot have the same sign but then what?
 
Logarythmic said:
Unless I can use [tex]p_1p_2+p_2p_3+p_3p_1=0[/tex] aswell?

You are working way too hard. What does this say about the possibility that all of the p's are positive or all are negative?
 
They cannot all be positive nor negative, but one positive and two negative or two positive and one negative. Is that right?
 
Almost. Except you can't have two negative p's either.
[tex]p_1+p_2+p_3=1[/tex]. What would this tell you about p3 if p1 and p2 are negative?
 
Then [tex]p_3 > 1[/tex]?
 
You're catching on. But the sum of the squares should be one too!
 
There we are. So it's either [1,0,0] or [+,+,-] and the behavior is strange. ;)
 
Yes. If you want to see an attempt to use this strange behavior look up Mixmaster cosmologies sometime.
 
Can you briefly tell me something about it?
 
Sorry, better get to work here. Besides, other people tell the story better than I.
 
I'll look it up tomorrow, now I need some sleep. Thanks for all your help.