# Negative scale factor RW metric with scalar field

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1. Nov 10, 2015

### Salah93

1. The problem statement, all variables and given/known data

The aim is to find a solution for the scale factor in a Robertson Walker Metric with a scalar field and a Lagrange multiplier.

2. Relevant equations

I have this action

$$S=-\frac{1}{2}\int d^{4}x\sqrt{-g}[R(g_{\mu\nu})+\lambda(g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-1)+L_{m}]$$

In conformal for Robertson Walker metric $$ds^{2}=a^{2}(\eta)(d\eta^{2}-\delta_{ij}dx^{i}dx^{j})$$

where $$\phi$$ is a scalar field (that depends only on conformal time ), $$\lambda$$ a Lagrange multiplier, $$L_m$$ lagrangian matter that I set to 0 and R the Ricci Scalar.

So the Ricci Scalar is $$R=-6\frac{a''}{a^{3}}$$ ()' is derivative with respect to conformal time (I found the value from the value that one finds for R from universal proper time, that is $$R=-6(\frac{\ddot a}{a}+\frac{\dot a^2}{a^2})$$ )

So the action is

$$S=-\frac{1}{2}\int d^{4}xa^{4}[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\eta^{00}\partial_{0}\phi\partial_{0}\phi-1)]=-\frac{1}{2}lim_{V\rightarrow\infty}V\int d\eta[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\phi'^{2}-1)]a^{4}$$

3. The attempt at a solution

I have the various $$L=L_{V}=-\frac{1}{2}V[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\phi'^{2}-1)]a^{4}$$ so because there is only the multiplicative factor V they all lead to the same eq. of motion. So I can omit the limit and take V=1. So the action is now

$$S=-\frac{1}{2}\int d\eta[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\phi'^{2}-1)]a^{4}=-\frac{1}{2}\int d\eta[-6aa^{''}+\lambda a^{4}(a^{-2}\phi'^{2}-1)]$$

And without considering surface elements the action become
$$S=-\frac{1}{2}\int d\eta[6a'^{2}+\lambda a^{4}(a^{-2}\phi'^{2}-1)]$$

Now I want to find eq. of motion:

Variation w.r.t a gives:
$$6a''-\frac{1}{2}[\frac{\partial(\lambda a^{4})}{\partial a}(a^{-2}\phi'^{2}-1)-a\lambda\phi'^{2}2]=0$$

W.r.t $$\lambda$$:
$$\frac{\phi^{'2}}{a^{2}}-1=0\rightarrow\phi'^{2}=a^{2}$$

W.r.t $$\phi$$:
$$\frac{d}{d\eta}(2\phi'\lambda a^{2})=0$$

$$\phi'\lambda a^{2}=\lambda a^{3}=\eta+K\rightarrow\lambda=\frac{\eta+K}{a^{3}}$$
K is a real constant

And so in the end I have:

$$6a^{''}+\eta+K=0$$

Integrating

$$a'(\eta)=-\frac{\eta^{2}}{12}-\frac{K\eta}{6}+C$$

And again:

$$a(\eta)=-\frac{\eta^{3}}{36}-\frac{K\eta^{2}}{12}+C\eta+B$$

B and C real constants.

Now, if $$\eta$$ goes to $$+\infty$$ is a problem because the scale factor becomes negative. And so the only possibility for this case in a flat Robertson Walker metric is that I have(setting B=0) $$a(\eta)=0$$ and then after a finite interval again $$a(\eta)=0$$ becuase then it will be negative and I think there isn't a physical meaning for negative scale factor. So, where are the errors? Because I can't interpret well this result.
If I imagine it as a 1-D motion of a particle I can study without any problems the system. But here I am studying the evolution of a particular case of the universe. How can I interpret negative scale factor(if all above is correct)?

I hope that the question is written well. Thank you

2. Nov 11, 2015

### Salah93

Can at least someone check if the variation of the action w.r.t a is correct?

3. Nov 11, 2015

### Salah93

Ok, I think that in general what I did is right. But I made a big mistake!

$$\frac{d}{d\eta}(2a^3\lambda)=0$$ gives $$a^3\lambda=A$$ that is obvoius! A is a constant

And then I obtain $$a(\eta)=-\frac{A}{6}\eta^2+B\eta+C$$. B and C are real constants.

And how can I get the value of the constant A? If A is positive there is a good behaviour at $$\eta$$ that goes to +infinity. (and the universe expands for ever). But for A negative the scale factor is negative! So I can't give a physical interpretation for it!