# Solving General Relativity ODE with Table on p98

• latentcorpse
In summary, the equations given on p97 can be solved for a and \tau by eliminating the energy density and using a substitution. The solutions are given by a=C \sin^2(\eta/2) and \tau=C\eta/2, where C is a constant. The conformal time, \eta, is a mathematical parameter that simplifies the equations, and can be related to the original time coordinate, \tau, through the integral \tau = \int_0^{\eta} a(\eta)...d\eta.
latentcorpse
im hoping someone has access to the book "General Relativity" by Wald as it'll be easier to see what's going on but notheless i'll try and put all the details in

on p97 we have the equations

$3 \frac{\dot{a}^2}{a^2}=8 \pi \rho - \frac{3}{a^2}$
$3 \frac{\ddot{a}}{a}=-4 \pi \rho$

where $\dot{a}=\frac{da}{d \tau}$

and i need to show the solutions are as shown in the table on p98:

$a=\frac{1}{2}C(1-\cos{\eta})$
$\tau=\frac{1}{2}C(\eta-\sin{\eta})$

i've spent ages on thsi and keep getting nowhere

Well you obviously need to know something about the density to say anything about the solutions. If I'm not mistaken, those solutions you mentioned are those for a matter-dominated closed universe. Does that help?

yes. that's correct, they are for a matter dominated universe. but i already took that into account and set P=0 where P is pressure in the equations i gave.

Combine the Friedmann equations into one by eliminating the energy density then write this equation in terms of conformal time. After that you can just plug your expression for a into the equation and see that it is a solution. As for t, you can get that from the definition of conformal time.

yeah so that gives

$\dot{a}^2-\frac{C}{a}+k=0$
but k=+1 for 3-sphere (which is the one I am trying to do)
so i need to solve
$\dot{a}^2-\frac{C}{a}+1$

i still don't know how to solve that
especially seeing as they get an answer in terms of $\eta$ not $\tau$
surely there's a way of solving ti rather than just checking the expression for a is in fact a solution?
also you mentioned getting $\tau$ from the definition of conformal time. is that in this same book?

$$\dot{a}^2-\frac{C}{a}+k=0$$

is a separable diff. equation:

$$\left( \frac{da}{d \tau} \right)^2 = \frac{C}{a} -k \Leftrightarrow \frac{da}{\sqrt{\frac{C}{a} -k}} = d\tau \Leftrightarrow d\tau = C^{-1/2} \frac{a^{1/2} da}{\sqrt{1-\frac{ka}{C}}}$$

which can be solved with a substitution

$$\frac{ka}{C} = \sin^2 \phi$$.

EDIT: Never mind, I see you've assumed that rho goes like a^-3 (matter domination). I was thinking of a more general case where rho is unknown. So just solve the equation using the method described above. Or simpler yet, rewrite that equation in terms of conformal time eta and solve directly for that using a similar technique. Conformal time:

$$d\eta = \frac{1}{a}dt$$

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so i was going to do it just for the 3-sphere geometry in which k=1 but if i leave it as k as you have above, will i get a more general answer that will save me ahving to do the integral multiple times.

anyway that $a^{1/2}$ is annoying
i get $da=\sin{2 \phi} d \phi$

and then $d \tau = C^{-1/2} \frac{a^{1/2} \sin{2 \phi}}{\cos^2{\phi}}$

$$a = C \sin^2 \phi \Rightarrow a^{1/2} = C^{1/2} \sin \phi$$

Where I have set k to 1. Note also the square root in the denominator, so there is only one power of cosine there which is canceled by the cosine inside sin(2phi). However this will give you a as a function of t. If you want a as a function of eta then rewrite the differential equation in terms of eta (chain rule), which incidentally leads to a simpler integral:

$$d\eta = C^{-1/2} \frac{a^{-1/2} da}{\sqrt{1-\frac{a}{C}}}$$

Again k=1.

so yeah that would actually give

$d \tau = 2 \sin^2{\phi} d \phi$

but yeas i do want both $a$ and $\tau$ in terms of $\eta$. you talk about $\eta$ as if it's some well known quantity, $\eta=\frac{t}{a}$, what does this represent and where does it come from? i haven't seen it before.

anyway $\dot{a}=\frac{da}{d \eta} \frac{d \eta}{d \tau}=\frac{da}{d \eta} \frac{1}{a}$

so $\frac{1}{a} \frac{da}{d \eta} -\frac{C}{a}+1=0$
which rearranges to the integral you had above
then i let $\frac{a}{C}=\sin^2{\phi},da=2 \sin{\phi} \cos{\phi} d \phi, a^{\frac{1}{2}}=C^{\frac{1}{2}} \sin{\phi}$
and get
$d \eta=\frac{C^{-1/2}C^{1/2}\sin{\phi}2\sin{\phi}\cos{\phi} d \phi}{\cos{\phi}}=2 \sin{\phi} \cos{\phi} d \phi$
then let $u=\sin{\phi},du=\cos{\phi} d \phi$
and get
$d \eta=2 u du \Rightarrow \eta=u^2=\sin^2{\phi}=\frac{a}{C}$
which isn't what I'm after...

$(\frac{1}{a} \frac{da}{d \eta})^2 -\frac{C}{a}+1=0$

gives

$$d\eta = C^{-1/2} \frac{a^{-1/2} da}{\sqrt{1-\frac{a}{C}}}$$

Which gives the solution you are afrer.

Note that $$a^{-1/2} = \frac{C^{-1/2}}{sin\phi}$$ so everything cancels out.

Also $\eta \neq \frac{t}{a}$. The definition of eta is $d\eta = \frac{a_0}{a}dt$ and a is a function of t (a0 can be normalized to 1).

Eta is the conformal time which is just a different time coordinate chosen such that it expands with the universe just like the space coodinate in Robertseon-Walker metric. It's used in cosmology because it simplifies some of the equations (as it does in this case). It's weird that you don't have it in your book, though maybe that's because it's a book on general relativity and not cosmology. In any case you can think of it purely as a mathematical parameter to arrive at the solutions you posted.

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hi again.

i integrated that and got
$\eta=\frac{2 \phi}{C}$ which i presumably now try and substitue back into my equations.

so i had $\phi=\sin^{-1}(\sqrt{a/C})$
which i can put back in but then i get

$a=C \sin^2(c \eta /2)$ which is close to what i want but not quite so I am obviouslly screwing up somewhere but i can't see it. and how would i get the $\tau$ expression from that?

Actually $\eta=2 \phi$

Remember that one C also come from da so all the C's cancel also.

So $a=C \sin^2(\eta /2)$

Now recall the trigonometric identity:

$\cos(2x) = 1 - 2\sin^2(x)$

Also since eta is defined as,

$d\eta = \frac{d\tau}{a} \Rightarrow \tau = \int_0^{\eta} a(\eta) d\eta$

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## 1. What is General Relativity and why is it important?

General Relativity is a theory of gravity developed by Albert Einstein that explains how objects with mass interact with one another in the universe. It is important because it has been proven to accurately describe the motion of celestial bodies and has led to significant advancements in our understanding of the universe.

## 2. What is an ODE and how is it related to General Relativity?

An ODE (Ordinary Differential Equation) is a mathematical equation that describes the relationship between a function and its derivatives. In the context of General Relativity, ODEs are used to solve for the curvature of space-time and the motion of objects in the presence of large masses.

## 3. Why is a table used to solve ODEs in General Relativity?

A table is used to solve ODEs in General Relativity because it allows for a systematic and organized approach to finding solutions. By breaking down the problem into smaller, more manageable steps, a table helps to simplify the complex calculations involved in solving ODEs.

## 4. What are the limitations of using a table to solve ODEs in General Relativity?

While tables can be helpful in solving ODEs, they are limited in their ability to handle more complex scenarios. For example, they may not be able to account for changing conditions or multiple variables, which can result in less accurate solutions. In these cases, more advanced mathematical methods may be needed.

## 5. How can solving ODEs in General Relativity contribute to our understanding of the universe?

Solving ODEs in General Relativity allows us to make predictions about the behavior of objects in space and time, which can then be tested and verified through observations and experiments. By studying these solutions, we can gain a deeper understanding of the fundamental laws of physics and the structure of the universe.

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