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

[Logarythmic]

Homework Statement

Investigate the possible behaviour of the singularity as t \rightarrow 0 in the Kasner solution.

Homework Equations

The metric for the Kasner solution is given by

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

The Attempt at a Solution

I have no clue...

 

[Dick]

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?

 

[Logarythmic]

I don't get it.

 

[Dick]

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...

 

[Logarythmic]

All I can find about this is that substituting the metric into the Einstein equations gives

\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)

in which a^3 = X_1X_2X_3.

 

[Dick]

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.

 

[Logarythmic]

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 X_i = t^{p_i}? According to my book this is a perfect fluid model.

 

[Logarythmic]

Ok, I found that this is a particulary simple behaviour. Also

\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

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

0 \propto \frac{1}{t^2}

Is this true?

 

[Dick]

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.

 

[Logarythmic]

So p_1p_2 + p_2p_3 + p_3p_1 = 0?
I also got that

\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)

but then I get some a aswell..?

 

[Dick]

That is one alright. Since we are not actually trying to derive this you should find

p_1+p_2+p_3=1 and
{p_1}^2+{p_2}^2+{p_3}^2=1.

 

[Logarythmic]

Yeah that I got..

 

[Dick]

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.

 

[Logarythmic]

Well, what I don't understand is HOW to describe it's behaviour.

 

[Dick]

Ok, here's a set of p's.

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

How would you describe this behavior as t->0?

 

[Logarythmic]

Well, I will always get

\frac{\dot{a}}{a} = \frac{1}{3t}

so for t \rightarrow 0

\dot{a} \rightarrow \infty

 

[Dick]

Sure. But describe qualitatively the behavior of the scale factors themselves (the X_i's).

 

[Logarythmic]

Sorry, I don't know what to say about them.

 

[Dick]

That's ok. In a matter dominated universe (a=t^(2/3), I think) a->0 as t->0. Is that true here?

 

[Logarythmic]

Humm, if I solve

\frac{\dot{a}}{a} = \frac{1}{3t}

I get

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

for t \rightarrow 0.

But if I use a_0 instead, I get

a = a_0 \frac{1}{3} ln(t) \rightarrow - \infty.

 

[Dick]

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.

 

[Logarythmic]

I don't get it. What about t \rightarrow 0?

 

[Logarythmic]

And how do I get a_i = t^{p_i}?

 

[Dick]

And how do I get a_i = t^{p_i}?

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?

 

[Logarythmic]

Well then every a_i \rightarrow 0 as for the matter dominated universe.

 

[Dick]

Are all of the p's positive?

 

[Logarythmic]

No the third one is not. So this one behaves like \frac{1}{t} and thus goes to infinity. Right?

 

[Dick]

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.

 

[Logarythmic]

Well wouldn't \vec{p} = [1,0,0] be a solution?

 

[Dick]

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.

 

[Logarythmic]

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.

 

[Dick]

Take a break and good luck with the rest.

 

[Logarythmic]

I think I will need some help with the rest aswell. ;) No MatLab installed here...

 

[Logarythmic]

Unless I can use p_1p_2+p_2p_3+p_3p_1=0 aswell?

 

[Dick]

Sure you can. It's a consequence of the other two relations between the p's.

 

[Logarythmic]

I get

p_1 = -\frac{p_3}{2} \pm \frac{1}{2} \sqrt{-3p_3^2+2p_3+1} + \frac{1}{2}

and

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

but what does this tell me? They cannot have the same sign but then what?

 

[Dick]

Unless I can use p_1p_2+p_2p_3+p_3p_1=0 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?

 

[Logarythmic]

They cannot all be positive nor negative, but one positive and two negative or two positive and one negative. Is that right?

 

[Dick]

Almost. Except you can't have two negative p's either.
p_1+p_2+p_3=1. What would this tell you about p3 if p1 and p2 are negative?

 

[Logarythmic]

Then p_3 > 1?

 

[Dick]

You're catching on. But the sum of the squares should be one too!

 

[Logarythmic]

There we are. So it's either [1,0,0] or [+,+,-] and the behavior is strange. ;)

 

[Dick]

Yes. If you want to see an attempt to use this strange behavior look up Mixmaster cosmologies sometime.

 

[Logarythmic]

Can you briefly tell me something about it?

 

[Dick]

Sorry, better get to work here. Besides, other people tell the story better than I.

 

[Logarythmic]

I'll look it up tomorrow, now I need some sleep. Thanks for all your help.