How Is the Particle Horizon Distance Calculated in a Universe with Dark Energy?

In summary, the homework statement says that you need to numerically integrate and report the particle horizon distance today for the currently favored model \Omega_M=1-\Omega_{DE}=0.25,\omega=-1. Assume the scaled Hubble constant to be h = 0.72, and report the particle horizon in billions of lyr (Gyr).
  • #1
June_cosmo
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Homework Statement


Numerically integrate and report the particle horizon distance today for the currently fa-
vored model [itex]\Omega_M=1-\Omega_{DE}=0.25,\omega=-1[/itex]. Assume the scaled Hubble constant to be h = 0.72, and report the particle horizon in billions of lyr (Gyr).

Homework Equations

The Attempt at a Solution


Horizen distance [itex]d=\int_0^{t0}\frac{dt}{a(t)}[/itex], so in a flat universe [itex]a(t)=(t/t0)^{2/(3(1+3\omega))}[/itex],
so that we have $$d=\frac{2}{1+3\omega}H_0^{-1}$$,but this has nothing to do with [itex]\Omega_M=1-\Omega_{DE}=0.25[/itex]?
 
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  • #2
The [itex] a(t) \propto t^{\frac{2}{3}}[/itex] result is only true for a flat universe with [itex]\Omega_{DM} =0[/itex]. You need to review the Friedmann equations for the case with [itex]\Omega_{DM} \neq 0[/itex].
 
  • #3
phyzguy said:
The [itex] a(t) \propto t^{\frac{2}{3}}[/itex] result is only true for a flat universe with [itex]\Omega_{DM} =0[/itex]. You need to review the Friedmann equations for the case with [itex]\Omega_{DM} \neq 0[/itex].
Oh that's right! So how do I derive a(t) from Friedmann equations?
 
  • #4
Well, can you write down the first Friedmann equation for H in terms of the Ω parameters? Once you have done that, remember that [itex] H = \frac {\dot a}{a}[/itex]. Then you should be able to write a differential equation for a that you can numerically integrate.
 
  • #5
phyzguy said:
Well, can you write down the first Friedmann equation for H in terms of the Ω parameters? Once you have done that, remember that [itex] H = \frac {\dot a}{a}[/itex]. Then you should be able to write a differential equation for a that you can numerically integrate.
Thank you! So that would be [itex]\frac{H^2}{H_0^2}=\frac{0.25}{a^3}+0.75,[/itex]if we combine this with [itex]H=\frac{\dot{a}}{a}[/itex] and I solved this equation (online), it gave me http://www4f.wolframalpha.com/Calculate/MSP/MSP226920fg7hgi3c9658be000033830f2a7886e3e4?MSPStoreType=image/gif&s=20&w=550.&h=47. ? (a(x) means a(t) here). I think this isn't right?
 
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FAQ: How Is the Particle Horizon Distance Calculated in a Universe with Dark Energy?

What is particle horizon distance?

Particle horizon distance is the maximum observable distance in the universe. It is the distance that light has traveled since the beginning of the universe and the farthest point from which we can receive light.

How is particle horizon distance calculated?

The particle horizon distance is calculated using the age of the universe and the speed of light. It is estimated to be about 46.5 billion light years.

Why is particle horizon distance important?

Particle horizon distance is important because it gives us a limit to how far we can see in the universe. It also helps us understand the size and age of the universe.

Can particle horizon distance change over time?

Yes, particle horizon distance can change over time. As the universe expands, the particle horizon distance also increases. This means that we can see farther into the universe as time goes on.

Is particle horizon distance the same for all observers?

No, particle horizon distance is not the same for all observers. It depends on the location and motion of the observer in the universe. This is known as the cosmological principle, which states that the universe looks the same to all observers, but the distance to objects may vary.

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