How can I plot the luminosity distance as function of redshift?

Questions
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1

Hi everyone, As the title said, I want to plot this function using mathematica:

distance_luminosity.png

where the only variable is the redshift z, how can I do that with the function Plot in mathematica ? The goal si to obtain the dL(z) as function of redshift z. Thanks.

2

you need value of parameters..

3

plot.png
plot.png1205×302 6.84 KB

4

Thank you so much, amazing!

5

Okay, I have another problem. I need to define this parametres :

de.png

de.png

where w is:

w.png

and then I’ll put the density rho into the previuos equation (the equation in the first post ). Now, I set w_0 and w_a equal to 1 and on mathematic I wrote :

output.png
output.png508×188 3.55 KB

Where I wrong?

6

Perhaps you meant to r[z_] instead of r[_z]. Instead of d(ln(u+1)) try writing D[Log[1+u],u] and integrating in the variable u, as in

Integrate[f[u] D[Log[1 + u], u], {u, 0, 1}]
7

Thak you for the answer. I try in this way :

7382final.png
7382final.png658×483 27.2 KB

but the command “Plot” run indefinitely (the program remain in “Running” state).

8

Use NIntegrate instead of symbolic integration:

c = Integrate[3 (4 + 3 u/(1 + u)) D[Log[u + 1], u], {u, 0, 1}];
r = E^c;
d[z_] := 300000 NIntegrate[(1 + 
       z)/(75 (.3 (1 + u)^3 + 
         r (1 + u)^(3 (1 + (1 + u/(1 + u)))))), {u, 0, z}];
Plot[d[z], {z, 0, 10}]
9

Yeeahh it works!!! Thank you so much! I don’t understand why it didn’t work with Integrate. Again, thank you!

10

Probably because for every value of z it wastes a long time (a couple of seconds on my system) trying (unsuccessfully) to compute the integral symbolically, over and over again:

c = Integrate[3 (4 + 3 u/(1 + u)) D[Log[u + 1], u], {u, 0, 1}];
r = E^c;
dSymbolical[z_] := 
  300000 Integrate[(1 + 
       z)/(75 (.3 (1 + u)^3 + 
         r (1 + u)^(3 (1 + (1 + u/(1 + u)))))), {u, 0, z}];
Timing[dSymbolical[.1]]
Timing[N[dSymbolical[.1]]]