Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why are high redshift measurements of supernovae required... in order to measure the equation of state parameter of dark energy?

The luminosity distance can be written as \begin{equation} d_{L}(z) = \frac{c}{H_{0}}\left(z-\frac{1}{2}(1+q_{0})z^{2}\right), \end{equation} where \begin{equation} q_{0} = \frac{1}{2}\left(1+3w_{\small{DE}}\Omega_{\small{DE}}\right). \end{equation} Here, $c$ is the speed of light, $H_{0}$ is the Hubble constant, $z$ is the redshift of the galaxy hosting the observed supernova, $w_{\small{DE}}$ is the equation of state parameter of dark energy, and $\Omega_{\small{DE}}$ is its density parameter. I was taught that because $q_{0}$ doesn't enter until quadratic order, we need to measure high-$z$ supernovae. However, there are a lot of practical issues in measuring high-$z$ supernovae.

Now consider this. We can also write \begin{equation} d_{L}(z) = c(1+z)\int_{0}^{z}\frac{dz^{\prime}}{H_{0}\left[ \Omega_{\small{M}}(1+z')^{3}+\Omega_{\small{DE}}(1+z')^{3(1+w_{DE})}\right]}. \end{equation} Here, I cannot see any reason to prefer high-$z$ supernova of low-$z$ supernovae. It would seem, then, that we can avoid the practical difficulties by simply using a different formula.

Where am I going wrong?

share|cite|improve this question
Those $z$s should be $z'$s inside the integrand? – Chris White Apr 8 '13 at 20:27
@ChrisWhite Oops. Looks like it's been edited now though. – user12345 Apr 9 '13 at 6:56
up vote 6 down vote accepted

Be careful when trying to intuit how sensitive the integral formulation is to changes in $w$. The equation of state parameter only enters as part of the exponent of $1+z$, so for $z \approx 0$, $w$ has approximately no effect: $1^0 \approx 1^\epsilon$.

To illustrate with equations, suppose you already know $\Omega_\mathrm{M}$ and $\Omega_\mathrm{DE}$ exactly and only wish to measure $w$, assumed to be independent of $z$. If you say $w = 0 + \delta w$, then the luminosity distance becomes $$ d_L(z) = \frac{c(1+z)}{H_0} \int_0^z \frac{1}{\Omega_\mathrm{M}(1+z')^3 + \Omega_\mathrm{DE}} \left(1 - \frac{3\Omega_\mathrm{DE}\log(1+z')\delta w}{\Omega_\mathrm{M}(1+z')^3 + \Omega_\mathrm{DE}}\right) \mathrm{d}z' $$ to first order. You need to go to high enough $z$ such that the contribution of the second part of the integrand (weighted by the overall factor out in front) becomes significant enough to amplify the effect of nonzero $\delta w$. Even though the $\Omega_\mathrm{M} (1+z')$ terms make this contribution smaller with increasing $z'$, every little bit counts, and the larger $z$ is the larger the deviation will be.

To illustrate graphically, below is a plot I made using $\Omega_\mathrm{M} = 0.3$, $\Omega_\mathrm{DE} = 0.7$, and $H_0 = 70~(\mathrm{km}/\mathrm{s})/\mathrm{Mpc}$.

luminosity distance

The middle line is for $w = 0$, the one above that for $w = -0.05$, and the one below for $w = 0.05$. The lines really only start to separate out past about $z = 2$. Probably a more fair comparison assuming a constant fractional uncertainty in distance measurements is to plot the fractional change $$ \frac{d_L(z)\big\vert_w - d_L(z)\big\vert_0}{d_L(z)\big\vert_0}, $$ as shown below.

fractional change

Clearly if your supernova surveys are limited to $z < 0.5$ (as the early ones most certainly were), uncertainties in your luminosity distance measurement in the amount of $2\%$ will be comparable to the effect of varying $w$ by $0.05$ (which is a huge variation, by the way), requiring you to have many independent supernovae and very small systematic uncertainties.

Furthermore, the further in redshift you are willing to go, the larger the volume you survey, allowing you to get more datapoints.

share|cite|improve this answer
+1 for awesome. – DilithiumMatrix Apr 8 '13 at 23:45
+1 for awesome indeed. – user12345 Apr 9 '13 at 7:34

You can fit a value for Dark Energy (DE) using near or far supernovae (SNe), the benefit of distant SNe is purely from a statistics perspective: you want to minimize your errors on your best fit value of DE ($\Omega_{DE}$). To determine the uncertainty in your fit ($\sigma_{\Omega_{DE}}$) you have to do error propagation which will show a dependence on the overall range of $z$ values you're sampling.

So, you can determine $\Omega_{DE}$ by fitting a plot of $d_L(z)$, right? Imagine you have N data points scattered near z = 0 compared to having two groups of N/2 data points, one near the origin, and one near z~1. In the former case, your uncertainty on the best fit line will be dominated by the scatter in your data points, while in the second case the uncertainty is going to be the scatter, scaled to the slope between the centers of mass of each group.

If that's unclear, I can try to make a schematic

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.