Density in the friedmann equation

[RedX]

I'm a little confused about the density \rho in the equation:

H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho - \frac{kc^2}{a^2}

Measuring \rho at a single instant in time seems easy. But \rho changes with time. The time dependence of \rho is given as \rho=\frac{M}{a(t)^3} where M is a constant. But to determine M from a measurment of \rho, doesn't one have to know a(t), which is what the equation is trying to find?

 

[bcrowell]

But to determine M from a measurment of \rho, doesn't one have to know a(t), which is what the equation is trying to find?

No, it's a second-order differential equation, so the general solution is going to have two adjustable constants in it that have to be adjusted to match the boundary conditions. Fo comparison, the differential equation \ddot{y}=-y has solutions of the form y=A\cos(t+b). You don't need to know A and b in order to determine that that's the general form of the solution.

 

[George Jones]

Suppose H_0 and \rho_0, the values of H and \rho at the present time, are measured. Evaluating the Friedmann equation at the present time gives a_0, the present value of the scale factor.

\rho = \frac{M}{a^3},

then gives

\rho = \rho_0 \frac{a_0^3}{a^3}.

 

[RedX]

Suppose H_0 and \rho_0, the values of H and \rho at the present time are measured. Evaluating the Friedmann equation at the present time gives a_0, the present value of the scale factor.

\rho = \frac{M}{a^3},

then gives

\rho = \rho_0 \frac{a_0^3}{a^3}.

Assume k=0. Then if you measure \rho_0, then don't you have H_0? Or vice versa: if you measure H_0 don't you have \rho_0?

That's what seems to be implied by the equation:<br /> H^2 = \frac{8 \pi G}{3}\rho <br />, so I'm still not sure how to get a_0

 

[George Jones]

That's what seems to be implied by the equation:<br /> H^2 = \frac{8 \pi G}{3}\rho <br />

This isn't the equation that you gave in your original post.

 

[Altabeh]

This isn't the equation that you gave in your original post.

I think he just said he assumes k to be zero.

That's what seems to be implied by the equation H^2 = \frac{8 \pi G}{3}\rho. So I'm still not sure how to get a_0

You have to be careful here: As bcrowell said, a_0 and \rho_0 are adjustable according to the conditions of material distribution. If you don't have a given a_0, then you cannot predict what probably the value of \rho_0 would be even when the universe is isotropically flat i.e. when k=0. This is simply because you still have those adjustable constants involved within the Friedmann equation so assuming here we must have a known H_0 due to reading it off the equation H^2 = \frac{8 \pi G}{3}\rho immediately is correct but yet again the equation H = \left(\frac{\dot{a}}{a}\right) is an ODE involving a constant of integration to be set by hand.

AB

 

[George Jones]

Yikes, I should be banned from making posts from home when my daughter is awake (like now). Sorry RedX, I neither read nor thought k = 0. I'll get back to this tomorrow.

 

[George Jones]

Okay, assume matter only and suppose measurements of H_0 and \rho_0 satisfy

H_0^2 = \frac{8 \pi G}{3} \rho_0,

which means that k = 0. This, in turn, means that the scale factor is not pinned down. In spite of this, one can solve for the time evolution of \rho and H. Daughter is asleep, but wife wants to watch a mystery movie, so I'll only start the exercise.

Write

\rho = \rho_0 \left( \frac{a_0}{a} \right)^3.

Use this in the Friedmann equation, integrate the Friedmann equation, and find \rho \left( t \right) and H \left( t \right). The value of a_0 is not needed to do this.