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

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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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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?
 
I don't get it.
 
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].
 
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.
 
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.
 
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?
 
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.
 
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..?
 
Yeah that I got..
 
Well, what I don't understand is HOW to describe it's behaviour.
 
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]
 
Sorry, I don't know what to say about them.
 
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].
 
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.
 
I don't get it. What about [tex]t \rightarrow 0[/tex]?
 
And how do I get [tex]a_i = t^{p_i}[/tex]?
 
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?
 
Well then every [tex]a_i \rightarrow 0[/tex] as for the matter dominated universe.
 
No the third one is not. So this one behaves like [tex]\frac{1}{t}[/tex] and thus goes to infinity. Right?
 
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.
 
Well wouldn't [tex]\vec{p} = [1,0,0][/tex] be a solution?
 
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.