# How do I solve this cosmological model FRW?

Tags:
1. Apr 19, 2015

### unscientific

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

(a)Find how $\rho$ varies with $a$.
(b) Show that $p = \frac{2}{\lambda^2}$. Find $B$ and $t_0$.
(c) Find $w$ and $q_0$. What values of $\lambda$ makes the particle horizon infinite? Find the event horizon and age of universe.
(d) Find luminosity distance $D_L$ in terms of redshift $z$. Find $q_0$ by expanding.

2. Relevant equations

3. The attempt at a solution

Part(a)
For $V_0 = 0$, we can see that $\rho = P = \frac{1}{2} \dot \phi^2$. Thus $w=1$. For dependency on $a$:
$$\ddot \phi + 3(\frac{\dot a}{a})\dot \phi = 0$$
$$\dot \rho + 12(\frac{\dot a}{a}) \rho = 0$$
$$a^{-12}\frac{d}{dt}(\rho a^{12}) = 0$$
$$\rho \propto a^{-12}$$

Part (b)
I'll replace the $p$ by $x$ to avoid confusion with pressure $P$. Given $a(t) = t^x$ and $\phi = BM ln(\frac{t}{t_0})$, we have $\frac{\dot a}{a} = \frac{x}{t}$ and $\dot \phi = \frac{BM}{t}$ and $\ddot \phi = -\frac{BM}{t^2}$.

Substituting into equation of motion:
$$\frac{-BM}{t^2} + 3\left(\frac{x}{t}\right)\left(\frac{BM}{t}\right) - \frac{\lambda V_0}{M} \left( \frac{t}{t_0} \right)^{-\lambda B}$$
$$BM(3x - 1) - \frac{\lambda}{M} V_0 t^2 \left( \frac{t}{t_0} \right)^{-\lambda B} = 0$$

Substituting into FRW equation:
$$\frac{x^2}{t^2} = \frac{8\pi G}{3} \left[ \frac{2}{2}\left(\frac{BM}{t}\right)^2 + V_0 \left( \frac{t}{t_0} \right)^{-\lambda B} \right]$$
$$x^2 = \frac{B^2}{6} + \frac{8 \pi G}{3}V_0 t^2 \left( \frac{t}{t_0} \right)^{-\lambda B}$$
Using our result from the equation of motion:
$$x^2 = \frac{B^2}{6}+ \frac{8 \pi G}{3} \left[ \frac{BM^2}{\lambda} (3x-1) \right]$$
$$x^2 - \left(\frac{B}{\lambda}\right)x + \left( \frac{B}{3\lambda} - \frac{B^2}{6} \right) = 0$$

Can't seem to get $x$ solely in terms of $\lambda$, am I doing something wrong?

Part(d)
The metric for a flat, isotopic and homogeneous universe is given by
$$ds^2 = -c^2 dt^2 + a(t)^2 \left[ d\chi^2 + S^2(\chi) \left( d\theta^2 + sin^2\theta d\phi^2 \right) \right]$$

Flux is given by $F = \frac{L}{4\pi D_L^2}$. From the metric, proper area is given by $A = 4\pi(a_0 \chi)^2 = 4\pi \chi^2$. But due to redshift, photons are delayed by $\nu_0 = \frac{\nu_e}{1+z}$. Thus we have
$$D_L = \chi(1+z)$$
where $\chi$ is the comoving distance.
This is only in first order, how do I expand it in 2nd order?!

Last edited: Apr 19, 2015
2. Apr 20, 2015

### unscientific

Would appreciate help on parts (b) and (c), but I made a slight bit of progress on part (d).

Part (d)
We know that $D_L = \chi (1+z)$. I now need to re-express $\chi$ in terms of $z$. For a light-like geodesic,
$$\chi = c \int \frac{1}{a(t)} dt$$
I read a useful trick is $dz = d(1+z) = - \frac{\dot a}{a^2} dt = -(1+z) H(z) dz$. Substituting in,
$$\chi = c \int_0^z \frac{1}{H(z)} dz$$
$$D_L = c(1+z) \int_0^z \frac{1}{H(z)} dz$$
Using $H(z) = H_0 \left[ 1 + (1+q_0)z + \cdots \right]$:
$$D_L \approx \frac{c(1+z)}{H_0} \int_0^z \frac{1}{1 + (1+q_0)z} dz$$
$$D_L \approx \frac{c(1+z)}{H_0} \int_0^z \frac{1}{1 + (1+q_0)z} dz$$
$$D_L = \frac{c(1+z)}{H_0(1+q_0)} ln \left[ 1 + (1+q_0)z \right]$$
$$D_L \approx \frac{c(1+z)}{H_0(1+q_0)} \left[ (1+q_0)z - \frac{\left[ (1+q_0)z \right]^2}{2} \right]$$
$$D_L \approx \frac{c}{H_0}z \left[1 + \frac{z}{2}(1-q_0) \right]$$

Last edited: Apr 20, 2015
3. Apr 22, 2015

### unscientific

bump on part (b)..

4. Apr 23, 2015

### unscientific

bump part (b)

5. Apr 24, 2015

### unscientific

Bump on part (b) - How do I get $x$ in terms of $\lambda$?

6. Apr 25, 2015

### unscientific

Would appreciate help with part (b) please

7. Apr 26, 2015

### unscientific

any luck with solving $x(\lambda)$?

8. Apr 30, 2015

### unscientific

anyone had a go with part (b)?

9. May 4, 2015

### unscientific

Still can't see how you can find $p(\lambda)$..

10. May 7, 2015

### unscientific

anyone else tried part (b) yet?

11. May 10, 2015

### unscientific

bumpp part (b)

12. May 14, 2015

### unscientific

bump on (b) and (c)

13. May 16, 2015

### unscientific

bump on part (b) first

14. May 18, 2015

### unscientific

bump on (b)

15. May 22, 2015

### unscientific

part (b) bumping

16. May 23, 2015

### unscientific

bump on part (b)

17. May 25, 2015

### unscientific

bumpp on part (b)

18. May 31, 2015

### unscientific

bumpp - I think there is a trick somewhere (dimensional analysis or something)

19. Jun 1, 2015

bump