# Hubbles Law

1. Apr 9, 2006

### griff

Can someone please explain whats going on with Hubble's constant. I understand that to calculate the age of the universe you need to use the relationship; t=1/H. But I don't understand how you get the Hubble's constant into a format that you can use.

I've been told that you get a Mpc in Km and divide the Hubble's constant by that value (3.08 exp19). And then put this into the above relationship. But I don't see how this works.

Thanks for any help that is offered.

2. Apr 9, 2006

### Astronuc

Staff Emeritus
3. Apr 9, 2006

### lightgrav

Hubble's constant is similar to an acceleration ... but not quite the same.
H=Dv/Dx is the velocity difference divided by the difference in locations
(while a=dv/dt is the change in velocity divided by the change in time).

So the difference between our velocity and its (measured by red-shift)
Dv = H Dx . Presuming that the velocity of each OBJECT has not changed,
the relative locations increase by their relative velocities as time changes:
Dx = Dv dt + Dx(o) ... here, Dx(o) is the difference in locations at dt = 0.
. . . ^
To find the change in time that has occurred since they were together,
(as in the naive big-bang model) we set Dx(o) equal to 0.
Replacing Dx above gives : Dv = H Dv dt , so dt = 1/H is that time change.

If objects in the universe have been slowing down (or speeding up!),
an acceleration term must be included (above the ^ after line 7).

4. Apr 9, 2006

### nrqed

1/H is only valid in an idealized ''empty'' universe.

But going back to the units, notice that H is in kilometers per second per Megaparsecs, that is (k/s)/Mpc. This is a weird combination. its a speed per unit distance or a distance per unit time per unit distance!! You must realize that the distances cancel out so that this is a complicated way of saying that H has units of one over time, simply.

To be more specific, lets say you use 71 (km/s)/Mpc. Then you write 1 Mpc as 3.08 x10^19 km, as you said, and this gives (71/3.08x10^19) (km/s)/km = 2.31 x10^(-18) /s. Taking one over this gives a time (in second).

Patrick

5. Apr 10, 2006

### griff

Thanks for the help, incidentally do you need to change to units of the constant if your just using it to work out the distance of an object from its red shift using v=Hd or can you just use the units that it is already in for that.

6. Apr 10, 2006

### nrqed

No. Well, to be more precise...if the speed is in km/s, all you do is to use d= v/H as you wrote and the distance will come out in Megaparsecs, since km/s over (km/s per Megaparsecs) gives a result in Mpc.

Patrick