Hubble's Constant with coincident galaxiesby astrofunk21 Tags: coincident, galaxies, hubble constant, hubble flow 

#1
Feb413, 08:41 PM

P: 22

The value of the Hubble constant is around 70 km s^{1} Mpc^{1}. Consider two galaxies moving with the Hubble flow of the Universe that are currently separated by 1 Mpc and assume that the rate of expansion of the Universe is constant with time.
A. At what time in the past were the galaxies coincident? This is presumably the origin of the Universe. B. In this nonaccelerating Universe, how does the Hubble constant vary as a function of time? What was the Hubble constant when the Universe was half its current age? This should be an easy question, but I am having some trouble with it. I feel that (A) should just be Hubble Time right? If you guys could lead me in the right direction or give some hints that would be great. 



#2
Feb413, 09:21 PM

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Welcome to PF astrofunk21,
EDIT: You are correct that for this nonaccelerating universe, the age of the universe (a.k.a. the amount of time since the galaxies were coincident) is equal to the Hubble time 1/H_{0}, but you need to show this. 



#3
Feb413, 09:34 PM

P: 22

v=H_{0}*D
Sorry about the post. Yea I see that H_{0} would change if v is constant. So now in terms of part (A), wouldn't it just be when "D" is 0? And for (B), is it just a matter of switching the velocity to a distance over time? 



#4
Feb413, 09:44 PM

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Hubble's Constant with coincident galaxiesSo, if you write down Hubble's Law, you're writing: v_{0} = H_{0}d_{0} where v_{0} is the recession velocity of two galaxies today and d_{0} is their separation today. But a Hubble's law applies at any time in the past. v = Hd where v is the recessional speed at some time in the past, and d is the separation of the two galaxies at that time. The tricky thing is that H, which is called the Hubble parameter, does not necessarily have the same value at this time in the past as it does today. (The value today, H_{0}, is called the Hubble constant). In general, H changes with time. So, if v = v_{0} = constant, then what you're saying is that Hd = constant. Since you already know d vs. t, you can get H vs. t from this relation. 



#5
Feb413, 09:59 PM

P: 22

Yes I know a function haha, my issue is interpreting the questions and understanding what to do. Now for part (B) my guess would be to get time into the function of course so I got this:
H(t) = vt/d Is there any truth to that? EDIT: Okay that was completely wrong. Scratch that 



#6
Feb413, 10:22 PM

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We already know that d(t) = v_{0}t, since v(t) = v_{0} = constant, right? v(t) = H(t)*d(t) H(t) = v_{0}/d(t) Can you take it from here? 



#7
Feb513, 12:45 AM

P: 22

Yea I understand all that. I guess I'm just missing the connection. Never really derived formulas ever.
It would be better for me to figure it out rather than you tell it to me though. Don't want you to give me the answer even though to most you probably already have. 



#8
Feb513, 12:46 AM

P: 22

I mean all I see is that it's almost the inverse of time.
H(t) = 1/t 



#9
Feb513, 01:03 AM

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So now it should be trivial to answer second part of that question: what was the value of the Hubble parameter at half the age of the universe? If H(now) = H_{0}, what does H(now/2) equal? 



#10
Feb513, 01:06 AM

P: 22

Oh wow okay, maybe I tried making this a little harder than it really should have been. Thanks for the help/patience, sorry if I was a little ill prepared.



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