Easing H 0 Tension by Invoking Local Inhomogeneity
5.2 Mapping D L back to Density Contrast
In the previous section we have seen that it is possible to derive a fully analytical set of radial null geodesics equations. Our goal now is to use these equations to obtain a new set of differential equations to map an observed DL(z) to a LTB model. In the coordinates we chose, a LTB solution is determined uniquely by the function k(r) , so we will have a total of three independent functions to solve for η(z), r(z), k(z). Since we have already two differential equation for the geodesics, we need an extra differential equation.
This can be obtained by differentiating with respect to the redshift the luminosity distance DL(z)
where DobsL (z) is the observed luminosity distance. In our case we will use the best fit function obtained using the method developed in ch.4. Now we have the set of equations we were looking for
Since we will solve our differential equations with respect to the the variable z, we need to transform the partial derivatives respect to η and r in eqs.(5.3,5.4) according to the chain rule:
where h(η, r) is a generic function in the coordinates (η, r). After this substitution the equations contain only functions of the redshift z, and derivatives respect to z. The differ-ential equations obtained in this form need to be further manipulated in order to re-write them in a canonical form in which the derivatives appear all on one side, since after the application of the chain rule to eqs.(5.3,5.4) derivative terms likedr(z)dz ,dη(z)dz ,dk(z)dz are also on the right-hand side. After a rather complicated algebraic manipulation done using
MATHEMATICA™ we get : In the above expressions we have expressed all the trigonometric functions in terms of the equivalent expressions in terms of tan(X) according to
S = p
We have also used the dimensionless version of the solution in terms of K(z), T (z) de-rived in the previous section.
As it can be seen the above three equations are not linear in the derivative terms, but the second one only involves {r0(z), K0(z)}, while the other two involve all the three functions {r0(z), K0(z), T0(z)}. This suggests that we can first solve for r0(z) in terms of and then substitute into other 2 eqs. to get:
K0(z) = t(2tK(z)3/2 9(1 + t2)r(z) + (3 + t2)ST (z)
These two equations now only involve K0(z), T0(z) in a linear form, so they can be solved directly, and then the result for K0(z) can be substituted in the equation for r0(z).
After some rather cumbersome algebraic manipulations we finally get:
dT (z) The density can be expressed as
ρ =H0(1 + t2)2k(z)3
Now we are ready to convert the luminosity distance into the density contrast.
5.3 Result
Here we show our preliminary results. Since we have not yet obtained the data from Keenan [7], we are not able to include their plots of observational data of density contrast.
As our goal is to compare our inverted density contrast with the one obtained in [7], we will follow their syntax and define fields 1, 2, 3 as what are shown in fig. 5.1. In the same figure we can also find that only field 1s and 3 contain enough data points, so we will analyze these two fields only. After removing 5 outliers, we successfully fit mobs− mF RW in field 3 with a reduced χ2 ∼ 0.77, and show that indeed SNe in field 3 are brighter than expected in fig.5.2. Statistically the fitting also passes the null hypothesis as the reference
Figure 5.1: This plot shows the sky map of all SNe and cepheids in our dataset. Three fields are specified in Keenan’s work [7] as the three regions with density contrast data.
Our targets of interest are field 1 and field 3 which contain enough data points to fit the luminosity distance curve. For the sake of clarity we will keep using the same color for field 1 and field 3 as [7] later on.
model [3] has a larger reduced χ2 ∼ 0.91 . In contrast as shown in fig.5.3, for field 1 where most higher redshift SNe lie in, the fit we get after removing 4 outliers is a simple shift in magnitude. The reduced χ2 ∼ 0.55 is again much lower than what vanilla FRW model could achieve [3]. Finally we invert each fitted curve within the 68% confidence band and get an envelop for the density contrast as shown in fig.5.4 and 5.5. According to sec.5.1 K0is not fixed, but actually the density contrast is almost independent of K0as shown in fig.5.6. So we decide to choose a specific K0 = −0.1 as an example since we believe that we are actually in a void. Finally we compare this inverted density profile to the one from [7]. Qualitatively our results for fields 1 and 3 is consistent with what was observed through luminous density, indicating that indeed local structures could alter the luminosity distance significantly.
Figure 5.2: This plot shows the 68% confidence band of the field 3 ∆m fit, along with the data points in this region. The deleted data points are in a darker color. The dashed curves are the 68% confidence band envelop and the vanilla curve is the best fit. The fitting model is chosen to be 5 functions of the form Φ(r) = r3 according to the dimensional argument of the polyharmonic spline interpolation method. The gray curve is the result from Riess 2016 [3].
Figure 5.3: This plot shows the 68% confidence band of the field 1 ∆m fit, along with the data points in this region. The deleted data points are in a darker color. The dashed curves are the 68% confidence band envelop and the vanilla curve is the best fit. The fitting model is chosen to be a simple constant shift. The gray curve is the result from Riess 2016 [3].
Figure 5.4: This plot shows the 68% confidence band of the inverted density contrast of the field 3, with K0 = −0.1. Clearly we can see a ∼68% significant 10% under-dense around z = 0.02 to 0.08 or 100 ∼ 400 Mpc. One can directly compare this plot to Keenan’s using conversion d(M pc) = H0−1z = 4400 z M pc. One important feature in Keenan’s result is the overdense region at around z = 0.1, and as we can see such feature is in the 68% confidence band of our result. The gray curve is the inverted density contrast of the FRW model with parameters from Riess 2016 [3].
Figure 5.5: This plot shows the 68% confidence band of the inverted density contrast of the field 1, with K0 = −0.1. Clearly we can see a ∼95% significant 10% under-dense everywhere. One can directly compare this plot to Keenan’s using conversion d(M pc) = H0−1z = 4400 z M pc, and find that the two agree with each other pretty well. The gray curve is the inverted density contrast of the FRW model with parameters from Riess 2016 [3].
Figure 5.6: This plot shows the inverted density contrast of the best fit in the field 3, under different K0. The blue, green, red curves correspond to K0 = −0.1, 0, and 0.1 respectively.