Solving Cosmology Problem on Page 12 of Damtp.cam.ac.uk

[latentcorpse]

I'm looking through the solution to a problem and am stuck with some of the maths.
The solution can be found here:
http://www.damtp.cam.ac.uk/user/r.ribeiro/supervisions_files/Cosmo1sol_M2010.pdf

The pages are numbered in the bottom right corner. I am on p12.

We have an expression [itex]1+z \simeq \frac{1}{\dots}[/itex] half way down the page. I can get to here fine.

Then she solves for z. How has she done this? Is it binomial? If so how has that worked? We have two terms here not just the usual [itex](1+x)^n[/itex]

And there is a new term that appears when we solve for z: [itex]H_0^2(t-t_0)^2[/itex]. Where's that from?

Then under that we "invert" to find [itex]t-t_0[/itex] Iliterally have no idea what has happened here!

Thanks for any help!

 

[fzero]

The first part is just a Taylor expansion in powers of [tex]t-t_0[/tex]. You can use the binomial expansion to do this, but if that's giving you problems just compute the necessary derivatives.

The "inversion" is just solving a quadratic equation. Did you even try working this part out for yourself before posting?

 

[latentcorpse]

The first part is just a Taylor expansion in powers of [tex]t-t_0[/tex]. You can use the binomial expansion to do this, but if that's giving you problems just compute the necessary derivatives.

The "inversion" is just solving a quadratic equation. Did you even try working this part out for yourself before posting?

Hmm. well say you have [itex](1+x)^{-1}=1-x[/itex] where for us [itex]x=H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2[/itex]

then [itex](1+x)^{-1}= 1-H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2[/itex] but I'm missing a term?And then for the next bit i think i was confused by the word "inverting". Nonetheless, it's a quadratic so we should be able to use [itex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]

In the form [itex](1+\frac{1}{2}q_0 ) H_0^2 x^2 - H_0 x - z = 0 [/itex] with [itex]x=t-t_0[/itex]

we get [itex]x = \frac{H_0 \pm \sqrt{H_0^2 + 4z(1+\frac{1}{2}q_0)}}{(2+q_0)H_0^2}[/itex]
which doesn't look very promising

 

[fzero]

Hmm. well say you have [itex](1+x)^{-1}=1-x[/itex] where for us [itex]x=H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2[/itex]

then [itex](1+x)^{-1}= 1-H_0(t-t_0) - \frac{1}{2} q_0 H_0^2(t-t_0)^2[/itex] but I'm missing a term?

You have to include terms of order [tex]x^2[/tex] to obtain all of the terms of order [tex](t-t_0)^2[/tex].

And then for the next bit i think i was confused by the word "inverting". Nonetheless, it's a quadratic so we should be able to use [itex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]

In the form [itex](1+\frac{1}{2}q_0 ) H_0^2 x^2 - H_0 x - z = 0 [/itex] with [itex]x=t-t_0[/itex]

we get [itex]x = \frac{H_0 \pm \sqrt{H_0^2 + 4z(1+\frac{1}{2}q_0)}}{(2+q_0)H_0^2}[/itex]
which doesn't look very promising

They've expanded the square root in powers of [tex]z[/tex].

 

[paranormal]

I understand that working through complex equations and mathematical expressions can be challenging. In order to solve the cosmology problem on page 12 of the Damtp.cam.ac.uk website, it is important to have a solid understanding of the underlying principles and concepts involved.

In this particular problem, the expression 1+z \simeq \frac{1}{\dots} is a result of applying the cosmological principle, which states that the universe is homogeneous and isotropic on large scales. This allows us to use the Friedmann equations, which relate the expansion rate of the universe (Hubble parameter, H) to the density and pressure of the universe.

The solution then proceeds to solve for z by using a binomial expansion, which is a mathematical technique used to approximate a complex expression by expanding it into simpler terms. In this case, the expression 1+z is expanded into two terms, 1 and z, which allows us to isolate z and solve for it.

The term H_0^2(t-t_0)^2 that appears in the solution is a result of the Friedmann equations, specifically the term for the expansion rate of the universe (H) squared. This term represents the energy density of the universe, which is a key factor in understanding the evolution and expansion of the universe.

Finally, the inversion process in the solution is a common technique in mathematics, where both sides of an equation are multiplied by the reciprocal of a term in order to isolate a variable. In this case, the inversion is used to solve for t-t_0, which represents the age of the universe.

I hope this explanation helps clarify some of the mathematical steps involved in solving the cosmology problem. I encourage you to continue studying and practicing mathematical techniques in order to deepen your understanding of cosmology and other scientific fields.