Question regarding Friedmann equation

[Buzz Bloom]

It has occurred to me that the Friedmann equation

allows for a solution
H = H₀ = 0 .
This seems to say that independently of the value of the scale factor a, and the various Ωs, a static universe is possible. I am guessing that this solution is spurious and is a side effect of the derivation of the equation.

I am curious about how the derivation introduced this spurious solution of the cosmological form of the GR equations, and would much appreciate someone posting an explanation.

Regards,
Buzz

 

[PeterDonis]

allows for a solution
H = H₀ = 0 .

What makes you think that? You do realize that, if $H_0 = 0$, the LHS is undefined, right?

 

[ChrisVer]

how js H=0 a solution inependently of the value of a? Only if the Omegas are zero this can be true...

 

[Buzz Bloom]

Hi Chris:

Thanks for your post.

If H₀ = 0, then H = 0 no matter what values a or the Ωs have. I understand that H can also be zero with H₀ not zero for some combinations of value for a and the Ωs. Such a solution would correspond to a mathematically possible configuration of a GR universe, but H = H₀ = 0 does not.

Regards,
Buzz

 

[Buzz Bloom]

if H0=0, the LHS is undefined

Hi Peter:

Got it. Thank you. I have also seen the equation somewhere in the form
H = H₀ √ . . .
This form, together with another of my occasional senior moments, are the sources of the spurious solution.

Regards,
Buzz

 

[Orodruin]

If H is zero the critical density is zero and so there is nothing in the universe. You recover Minkowski space.

 

[ChrisVer]

Hi Chris:
If H₀ = 0, then H = 0 no matter what values a or the Ωs have. I understand that H can also be zero with H₀ not zero for some combinations of value for a and the Ωs. Such a solution would correspond to a mathematically possible configuration of a GR universe, but H = H₀ = 0 does not.

Well a's are not constant. In some point they can give an overall zero result and it's when the H=0 (acceleration halts for a moment).
The thing then is that if you want it to be always zero indepent of the value of a, the omegas to be zero. As for example the only way for:
ax^2+ bx+c =0 (always 0 independently of the value of x) would need a=b=c=0.

 

[PeterDonis]

If H is zero the critical density is zero

More precisely, if H is zero at all times then the critical density is zero.

 

[Chalnoth]

It has occurred to me that the Friedmann equation

View attachment 94379
allows for a solution
H = H₀ = 0 .
This seems to say that independently of the value of the scale factor a, and the various Ωs, a static universe is possible. I am guessing that this solution is spurious and is a side effect of the derivation of the equation.

I am curious about how the derivation introduced this spurious solution of the cosmological form of the GR equations, and would much appreciate someone posting an explanation.

Regards,
Buzz

If you permit this kind of division by zero, you can prove 1 = 2. All you've shown is that if you divide by zero, you can prove anything.

 

[Buzz Bloom]

If you permit this kind of division by zero, you can prove 1 = 2. All you've shown is that if you divide by zero, you can prove anything.
Hi Chalnth:

Sorry I wasn't clearer in my post #5.

What you said above is what I meant by "another of my occasional senior moments".

Regards,
Buzz