Singularities and Kasner solution

In summary: Kasner solution?In summary, the singularity as t \rightarrow 0 in the Kasner solution behaves like a perfect fluid.
  • #1

Homework Statement

Investigate the possible behaviour of the singularity as [tex]t \rightarrow 0[/tex] in the Kasner solution.

Homework Equations

The metric for the Kasner solution is given by

[tex]ds^2 = c^2dt^2 - X_1^2(t)dx_1^2 - X_2^2(t)dx_2^2 - X_3^2(t)dx_3^2[/tex]

The Attempt at a Solution

I have no clue...
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  • #2
Try writing down a more specific form of the Kasner soln. The spatial metric functions can be written as powers of t. What are the conditions on the exponents coming from Einsteins eqns?
  • #3
I don't get it.
  • #4
You can't do anything with the metric as written - it's too general. What ARE X_1,X_2 and X_3 in the Kasner soln. You may have to look it up...
  • #5
All I can find about this is that substituting the metric into the Einstein equations gives

[tex]\frac{\ddot{X}_i}{X_i} - \left(\frac{\dot{X}_i}{X_i}\right)^2 +3\left(\frac{\dot{X}_i}{X_i}\right)\left(\frac{\dot{a}}{a}\right) = \frac{4\pi G}{c^4}\left(\rho - \frac{p}{c^2}\right)[/tex]

in which [tex]a^3 = X_1X_2X_3[/tex].
  • #6
That's a start. You shouldn't have a loose index i floating around though. Assuming you can get the correct Einstein equations (and there should be two), Kasner is a vacuum solution, so put rho=p=0. Put X_i=t^p_i. Turn this into equations in the constants p_i. Are you supposed to actually derive Kasner or just describe it's properties? It might be a good idea to just look up the solution to see what you are aiming for.
  • #7
I think I'm just supposed to describe the behaviour of the singularity. I don't like this book, it has too many errors. ;) Why should I use [tex]X_i = t^{p_i}[/tex]? According to my book this is a perfect fluid model.
  • #8
Ok, I found that this is a particulary simple behaviour. Also

[tex]\frac{\dot{X}_1 \dot{X}_2}{X_1X_2} + \frac{\dot{X}_2 \dot{X}_3}{X_2X_3} + \frac{\dot{X}_3 \dot{X}_1}{X_3X_1} = \frac{8 \pi G}{c^4}\rho[/tex]

Putting in [tex]X_i = t^{p_i}[/tex] and using [tex]\rho=0[/tex] gives me

[tex]0 \propto \frac{1}{t^2}[/tex]

Is this true?
  • #9
It gives you an algebraic condition on the p's that must vanish. What is it? Again there is another Einstein equation. It will give you another algebraic condition.
  • #10
So [tex]p_1p_2 + p_2p_3 + p_3p_1 = 0[/tex]?
I also got that

[tex]\frac{\dot{a}}{a} = \frac{1}{3} \left( \frac{\dot{X}_1}{X_1} + \frac{\dot{X}_2}{X_2} + \frac{\dot{X}_3}{X_3} \right)[/tex]

but then I get some a aswell..?
  • #11
That is one alright. Since we are not actually trying to derive this you should find

[tex]p_1+p_2+p_3=1[/tex] and
  • #12
Yeah that I got..
  • #13
Good. So there are actually lots of Kasner solutions corresponding to different values of the p's. Find a specific example and describe it's behavior.
  • #14
Well, what I don't understand is HOW to describe it's behaviour.
  • #15
Ok, here's a set of p's.

[tex]p_1=1/3, p_2=(1+\sqrt{3})/3, p_3=(1-\sqrt{3})/3[/tex]

How would you describe this behavior as t->0?
  • #16
Well, I will always get

[tex]\frac{\dot{a}}{a} = \frac{1}{3t}[/tex]

so for [tex]t \rightarrow 0[/tex]

[tex]\dot{a} \rightarrow \infty[/tex]
  • #17
Sure. But describe qualitatively the behavior of the scale factors themselves (the X_i's).
  • #18
Sorry, I don't know what to say about them.
  • #19
That's ok. In a matter dominated universe (a=t^(2/3), I think) a->0 as t->0. Is that true here?
  • #20
Humm, if I solve

[tex]\frac{\dot{a}}{a} = \frac{1}{3t}[/tex]

I get

[tex]a = t^{1/3} \rightarrow 0[/tex]

for [tex]t \rightarrow 0[/tex].

But if I use [tex]a_0[/tex] instead, I get

[tex]a = a_0 \frac{1}{3} ln(t) \rightarrow - \infty[/tex].
  • #21
You don't have to solve a DE for the scale factors. You already did that. a_i=t^p_i. Where the p_i's are what I sent you a few posts back. Note there are three different scale factors - the universe is anisotropic. And, hint, one of the scale factors is not like the others.
  • #22
I don't get it. What about [tex]t \rightarrow 0[/tex]?
  • #23
And how do I get [tex]a_i = t^{p_i}[/tex]?
  • #24
Logarythmic said:
And how do I get [tex]a_i = t^{p_i}[/tex]?

Huh? It's a Kasner solution. The scale factors are t^p1, t^p2 and t^p3. I sent you a sample set of p's. What's the behavior of each as t->0?
  • #25
Well then every [tex]a_i \rightarrow 0[/tex] as for the matter dominated universe.
  • #26
Are all of the p's positive?
  • #27
No the third one is not. So this one behaves like [tex]\frac{1}{t}[/tex] and thus goes to infinity. Right?
  • #28
Absolutely right. Two dimensions contract as t->0 and one expands. But this is only one of the many Kasner solutions. Are there any Kasner solutions where all p's are positive? Or are they all like this? I think this is the actual question you want to answer.
  • #29
Well wouldn't [tex]\vec{p} = [1,0,0][/tex] be a solution?
  • #30
Right again. One dimension expanding, two static. But can they all be positive (not zero)? Can two be negative and the third positive? I think once you've answered these you can say what the limiting behavior of the Kasner's are.
  • #31
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.
  • #32
Take a break and good luck with the rest.
  • #33
I think I will need some help with the rest aswell. ;) No MatLab installed here...
  • #34
Unless I can use [tex]p_1p_2+p_2p_3+p_3p_1=0[/tex] aswell?
  • #35
Sure you can. It's a consequence of the other two relations between the p's.

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