Newtonian and GR Friedman equation

[PAllen]

I think the most common English scientific term for what Dr. Stupid was referring to is recession velocity. It is, of course, a function of distance (in the standard foliation) for objects with no peculiar velocity; and it is not a relative velocity in the normal sense.

 

[DrStupid]

If you mean that, in order to know the actual function that gives density vs. time

Yes, that's what I mean.

you need to know distances but not positions

And that's why I do not see why I should use positions at all.

 

[PeterDonis]

I do not see why I should use positions at all.

If you're the only person that's going to look at your calculations, of course it makes no difference what method you use as long as your method gives the correct answers for observable quantities. But since coordinate charts are the standard tool for this job, if you want other people to understand what you're doing, it helps to be able to describe it in standard terms.

 

[DrStupid]

if you want other people to understand what you're doing, it helps to be able to describe it in standard terms.

In #24 I provided such a description but it seems that it wasn't very helpful and as I do not know how to integrate this equation I can't provide you with the corresponding solution. In the meanwhile I tried to start from Newton's law of gravitation without using gauss law but I stuck with the same problem:

A test mass i at position r_i within an infinite homogeneous mass distribution can be assumed to be located on the surface of a ball with the centre r₀, surrounded by an infinite large spherical shell. According to the shell theorem the force acting on the test mass depends on the mass M_i of the inner ball only and is independent from the shell outside. The resulting acceleration is

\ddot r_i = - G \cdot M_i \cdot \frac{{r_i - r_0 }}{{\left| {r_i - r_0 } \right|^3 }}

With

M_i = {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \left| {r_i - r_0 } \right|^3

this turns into the same equation as already posted above

\ddot r_i = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \left( {r_i - r_0 } \right) = g_0 - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot r_i

The only difference is that in this case g₀ is not some unknown integration constant but the result of the arbitrarily choice of the centre r₀. As an infinite mass distribution has no centre it is impossible to determine this constant except per definition. There simply is no absolute acceleration in such a configuration.

To solve this problem I need to switch to differences. In this case the unknown constant cancels out:

\Delta \ddot r_{i,j} = \ddot r_i - \ddot r_j = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \left( {r_i - r_j } \right) = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \Delta r_{i,j}

That's why I prefer differences over positions to solve this problem. If you want to use positions you may do this, but then you have to live with unknown or arbitrarily offsets and ugly differential equations. As I don't see any benefits I still don't think that this makes sense.

 

[PAllen]

I find myself doubting the validity of this line of reasoning (post #34). Since the center is arbitrary, both direction and magnitude of force is arbitrary. You then try to remove this via differences. But the if the approach is valid, since the center is arbitrary, it should work if I choose a different center for each point. Since this is obviously not the case, without further justification, I doubt the whole argument.

 

[DrStupid]

But the if the approach is valid, since the center is arbitrary, it should work if I choose a different center for each point.

Try this with r_i=r_j and you will see why choosing different centers in the same calculation is not allowed. In #34 g₀ replaces the integration constant in #24 and like the integration constant it needs to be normalized with the same condition for the whole calculation (e.g. by definition of the acceleration at a specific position).