# Calculate separation of galaxies as a function of time

1. May 9, 2005

### gerardo

I'm working on a problem that has me baffled mathematically. What equation can I use to calculate the future distance between two galaxies not gravitationally bound receding at a hubble constant of 70km/sec/Mpc as a function of time. Stated in another way, if two distant galaxies start off today at a separation of 1.46Mpc, how far apart will they be in 4 billion years, assuming a hubble constant of 70km/sec/Mpc?. I know as a function of distance how to calculate the veloicty V = Hd but how can I write an equation to calculate distance between galaxies as a function of time?. Thanks anyone have any sugggestions? hopefully this wont involved integrals or diff eq...

2. May 9, 2005

### ohwilleke

A differential equation or integral is the natural approach.

Velocity is the first derivative of distance with respect to time, and velocity is itself a function of distance in this case.

D(t)= Integral with respect to time of D(t)' where D(t)'=H*D(t)would be the basic formulation as a differential equation. Solve for D(t), plus in the current distant and current time as integration constants and you have your answer.

Last edited: May 9, 2005
3. May 10, 2005

### gerardo

yikes can you help me get started on this differential equation? I understand the basic concept of derivative and integrals ( only took calc 1 and 2 20 years ago ) but am still not sure how to solve for D(t) thanks again! i never took diff eq. would appreciate anyones help.

Last edited: May 10, 2005
4. May 10, 2005

### ohwilleke

Well $$D(t)=Be^H^t+C$$, where H is the Hubble's Constant, B is a constant used to make sure the units add up and C is a constant of integration set to fit your $$D(t)=1.46Mpc$$ and C=0 at t=0 should work. Plug it into the question and it seems to work.

BTW, I taught myself Calculus in 1987, so that is what, 18 years ago. I guess two years makes a difference.

Last edited: May 10, 2005
5. May 10, 2005

### gerardo

It's starting to make some sense in that it involves e, yes!, however e ^(ht) is unitless, so im not sure what value to assign for B to give it a distance unit. Let me give an example and perhaps you could show me how you would use Be^(ht) + C to solve it. Thanks in advance.
If two galaxies today are 1.46 mega parsecs apart ( 1 megaparsec = 3.086 X 10^22 meters ) and they expand apart at Hubble constant of 70km/s/megaparsec, how far apart will they be in 4 billion years?

H boils down to 2.2683 X 10^-18/sec and t = 1.26734 X 10^17sec (4 billion years) so Ht = 0.2863373 and e^.2863373 = 1.33154 but this is unitless. also how does this take into account the starting speed of 1.46Mpc * 70km/sec/Mpc.

I do recall the derivative of e^x is e^x though. pitifull , huh. I rechecked and I actually took calc 1 in 1976, 29 years ago. wow. If you have the time, what would you calculate the distance between these two galaxies starting at 1.46 Mpc apart for 4 billion years later? thanks again.

6. May 10, 2005

### gerardo

ahhhh if B= 1.46 Mpc ( the starting distance ), then at t = 0 e^0 = 1 and D(t) = 1.46
so at 4 billion years, d(t) = 1.46 * e(ht)? thanks

7. May 11, 2005

### ohwilleke

With regard to the units, I'm suffering from lawyer's disease. I didn't actually figure out one way or the other what they units were, and so I covered my bases.

Hubble's constant in SI units has the units 1000m/sec/3.086*10^22m, so the units of Ht if t is expressed in seconds and the convention value is used is a dimensionless 1/3.086^19, so indeed, the B=1.46 Mpc works. Incidentally, 1/H is known as the Hubble time, and is 13 billion years, give or take. Hubble time is commonly known as the "age of the universe".

8. May 11, 2005

### gerardo

it worked i have a basic grasp of it now thanks the correlary question is a bit harder however, how would the same calculation be done if the Hubble constant changes over time, for instance how would the differential equation look for the same problem but where the hubble constant increased smoothly from 70km/sec/Mpc to 80km/s/Mpc over the time period? Thanks again

9. May 12, 2005

### SpaceTiger

Staff Emeritus
Draw a graph. What would a "smooth" change imply for the dependence of H on t? These equations are completely separable, so there are no fancy diff eq. tricks needed, you just need to know how to do integrals like:

$$\int{\frac{dD}{D}}=ln(D)+C$$

10. May 12, 2005

### gerardo

hmmmm a smooth change would imply a linear relationship between H and time. I was able to approximate the answer for H increasing from 70 to 80 over 4 billion years by dividing the problem into 4 separate time sequences and using and average H for each billion year segment and adding up the distances, but this isnt very satisfying mathematically.

we know V ( velocity at any time = H * D so V(t) = h(t) * D(t) and H(t) is a linear function [y= mx+b] also V(t) = D'(t) so i guess h(t) = D'(t)/D(t) so that gets me to your integral of dD/D = lnD but how do i plug this equation into the D = Be^(ht) + C , if i substitue D'(t)/D for h in this equation i would arrive at

D = B e^((D'(t)/D)*t) + C and it loses me there , looks messy.

any other hints? thanks very much. i'm just a doctor, not an astrophysicist, Jim.....

11. May 12, 2005

### SpaceTiger

Staff Emeritus
Right, so you have:

$$H(t)=At+B$$

where A and B are constants you can calculate from the information given. This leaves an equation like:

$$\frac{dD}{dt}=(At+B)D$$

and is separable as

$$\int{\frac{dD}{D}}=\int{(At+B)dt}$$

You already know the solution to the left side, so now all you have to do is integrate the right side and solve for D.

12. May 14, 2005

### chronon

Of course you could say that the galaxy is moving away from us at 102.2 km/s and so the simplest thing to do would be to assume that it continues moving away at this speed and so multiply by 4 billion years to get 1.29E19 km further away. This does mean assuming that the Hubble constant is decreasing rather than being constant, but it is more realistic cosmologically - assuming the Hubble constant is constant over time as people have been doing here implies that the universe is the same at different times, which is steady state cosmology.

13. May 14, 2005

### SpaceTiger

Staff Emeritus
A truly constant Hubble constant will also lead to exponential expansion, the kind of thing we expect from inflation and future evolution with a cosmological constant.

I think the OP should stick with what's stated in the problem, though.

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