Equation to give the lookback time as a function of redshift

happyparticle
Messages
490
Reaction score
24
Homework Statement
Inverting ##z = H_0(t_0 - t_e) + (1 + \frac{q_0}{2}H_0^2(t_0 - t_e)^2)## to give the lookback time as a function of redshift ##t_0 - t_e = H_0^{-1}[z - (1 + \frac{q_0}{2})z^2]##
Relevant Equations
##H_0## is the Hubble's constant
##q_0## is the deceleration parameter
##z## is the redshift
Hi,
I'm currently reading the introduction to cosmology second edition by Barbara Ryden and at the page 105, the author says we get ##t_0 - t_e = H_0^{-1}[z - (1 + \frac{q_0}{2})z^2]## by inverting ##z = H_0(t_0 - t_e) + (1 + \frac{q_0}{2}H_0^2(t_0 - t_e)^2)##.

However, I can't figure out how she got this result.
Any help will be appreciate.

Thank you
 
Physics news on Phys.org
happyparticle said:
Homework Statement: Inverting ##z = H_0(t_0 - t_e) + (1 + \frac{q_0}{2}H_0^2(t_0 - t_e)^2)## to give the lookback time as a function of redshift ##t_0 - t_e = H_0^{-1}[z - (1 + \frac{q_0}{2})z^2]##
Relevant Equations: ##H_0## is the Hubble's constant
##q_0## is the deceleration parameter
##z## is the redshift

Hi,
I'm currently reading the introduction to cosmology second edition by Barbara Ryden and at the page 105, the author says we get ##t_0 - t_e = H_0^{-1}[z - (1 + \frac{q_0}{2})z^2]## by inverting ##z = H_0(t_0 - t_e) + (1 + \frac{q_0}{2}H_0^2(t_0 - t_e)^2)##.

However, I can't figure out how she got this result.
Any help will be appreciate.

Thank you
I believe you have some typographical errors. The formula for ##z## should read $$z \approx H_0(t_0-t_e) + \left(\frac{1+q_0}{2} \right)H_0^2(t_0-t_e)^2.$$
This is an approximate expression that assumes ##H_0(t_0-t_e)## is small. So, the equation expresses ##z## to second order in ##H_0(t_0-t_e)##.

For convenience, let ##x = H_0(t_0-t_e)## and ##b = \large \frac{1+q_0}{2}##. So, we may write the relation as $$z \approx x+ bx^2$$ where ##x## is a small first-order term. When inverting this, you only need to get an approximate expression for ##x## in terms of ##z## that is accurate to second order in ##z##.

Can you see a way to do that?
 
  • Like
Likes MatinSAR, happyparticle and Delta2
I can't see it. I tried a taylor's series but I don't get the same result.

I made a mistake. I think it works with a Taylor's series around z=0.

Thank you! I would not have seen it without your help.
 
Last edited:
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top