DIfferential equation for understanding cosmological expansion -

[andrewkirk]

I am trying to understand cosmological expansion and how it is possible to see objects that are receding from us faster than the speed of light. This is explained in words at http://www.mso.anu.edu.au/~charley/p...DavisSciAm.pdf" and I have tried to describe a simple mathematical model to understand how this works, as follows:

Assume the universe is expanding at an arithmetic (rather than a geometric) rate k relative to distance at time t0. By this we mean that, if we denote the distance between points x and x' at time t by d(t,x,x') then, for t1>t0:

​

d(t1,x,x') = (1+k(t1-t0)) d(t0,x,x').

Consider points x1 and x2 such that k*d(t0,x1,x2) = b*c where b>1. In other words, at time t0, x2 is receding from x1 at a speed faster than light.

I want to derive a formula for the distance y(t) from x1 of a photon emitted from x2 at time t0 in the direction of x1, as a function of time.

The velocity towards x1 of the photon at time t, when it is y(t) away from x1, is the speed of light c, less the speed of expansion at time t, which is:

​

-dy(t)/dt = c – k * d(t0,x(t),x1)

where x(t) is the location of the photon at time t. That location is currently y(t) away from x1 which, given our expansion rule must be equal to k * (t-t0) * d(t0,x(t),x1). Hence d(t0,x(t),x1) = y(t)/(1+k*(t-t0)), whence:

​

-dy(t)/dt = c – k * y(t) / (1+k*(t-t0))

We need to solve this ordinary differential equation to get a formula for y(t). The trouble is I never studied ODEs in sufficient depth to be able to solve this.

I could program a numeric solution given numeric values for k and d(t0,x2,x1) but an analytic solution would be much more illuminating.

Can anyone solve the ODE?

Thank you very much for any help!

 

[HallsofIvy]

-dy(t)/dt = c – k * y(t) / (1+k*(t-t_0))
That's a simple separable equation- it's just a matter of integrating.

-\frac{dy}{c- ky}= \frac{dt}{1- k(t- t_0)}
To integrate on the left, let u= c- ky so du= -kdy and dy= -du/k. The integral becomes
\frac{1}{k}\int \frac{du}{u}= \frac{1}{k}ln|u|+ C= ln|u^{1/k}|+ C
= ln|(c- ky)^{1/k}+ C

To integrate on the right, let v= 1- k(t- t_0)= 1- kt_0- kt so that dv= -kdt and dt= -(1/k)dv. The integral becomes
-\frac{1}{k}\int \frac{dv}{v}= \frac{1}{k}ln|v|+ D
ln|v|^{1/k}= ln|1- k(t-t_0)|^{1/k}+ D

Setting those two integrals equal,
ln|(c- ky|^{1/k}+ C= ln|1- k(t-t_0)|^{1/k}+ D
Combining the two constants into E= D- C,
ln|c- ky|^{1/k}= ln|1- k(t-t_0)|^{1/k}+ E
Taking the exponential of both sides,
(c- ky)^{1/k}= F(1- k(t-t_0))^{1/k}
where, now, e^{E} except that allowing it to be positive or negative let's us get rid of the absolute values.

Of course we can take the k power of both sides:
c- ky= G(1- k(t-t_0))[/itex]<br /> with G= F^k<br /> <br /> That&#039;s a <b>linear</b> equation!

 

[andrewkirk]

Thank you very much for your post HallsOfIvy. I'm afraid I can't see how you get from the original equation

-\frac{dy}{dt}= c - \frac{ky}{1+ k(t- t_0)}

to the separated form:

-\frac{dy}{c- ky}= \frac{dt}{1- k(t- t_0)}

It's possible that I put the question confusingly by not using Tex, which unfortunately doesn't always work as expected for me on this website - it seems to depend which computer I'm using. I've used Tex for the above formulae though and they are displaying very nicely.

 

[JJacquelin]

I don't intend discussing the topic of cosmological expansion, nor the over simplified model which leads to the very simple ODE solved in the attached page :

 

[andrewkirk]

Thank you very much JJacquelin for your suggested solution.

Unfortunately the ODE to be solved is

(1) \frac{dy}{dt} + \frac{ky}{1 + k(t- t_0)} = c

not

(2) \frac{dy}{dt} + \frac{ky}{1 - k(t- t_0)} = c

I tried using your method to solve equation (1) but the cancellations that made the equation (2) separable did not occur in (1) because the negative sign gave
Y=\frac{C}{t- t_0+1/k} rather than Y=C(t- t_0+1/k) and things got progressively messier from there. I tried a different substitution of

y(t)=\frac{F(t)}{t- t_0+1/k} which cleaned things up a bit but still didn't yield a separable equation.

Can you suggest a way to adapt the solution to (2) to give a solution to (1)?

Thanks again.

 

[JJacquelin]

Why not following the same method ? It is even more simpler.

 

[andrewkirk]

Sorry - I think I might have mixed up a sign somewhere. I'll work through it again before I take up any more of your time.

 

[JJacquelin]

Solving the other EDO :

 

[andrewkirk]

Thanks very much JJacquelin - you are a star!

 

[JJacquelin]

No merit at all. Just the use of the method that anybody learn in the first beginning of ODE studies. If this suffises to become a star, there would be so many stars in the world !