# I Friedmann equation

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1. Oct 15, 2017

### Das apashanka

while deriving the friedmann equation using Newtonian Mechanics the 2nd term of the r.h.s is coming to be 2^U/(r^2*a^2) where U is a constt,but it is replaced by -kc^2/(r^2*a^2)?

2. Oct 15, 2017

### Staff: Mentor

Which is only heuristic in any case. A proper derivation uses GR, not Newtonian mechanics.

Can you show more of your work? I don't understand what you're doing.

3. Oct 15, 2017

### Das apashanka

from the concept of kinetic energy and potential energy being constt for a test particle situated over a gravitational sphere having radius at time t to be a(t)r[o]

4. Oct 15, 2017

### Staff: Mentor

Sorry, this doesn't help. Please post some actual equations and reasoning, and please use the PF LaTeX feature. You can find help on that here:

https://www.physicsforums.com/help/latexhelp/

5. Oct 15, 2017

### Das apashanka

test particle in a gravitational mass M ,situated on the surface:
K.E+P.E=constt
1/2*(dR(t)/dt)^2-GM(t)/R(t)=U.....................(1)
U=constt,M(t),R(t)=mass and radius at time t
M(t)=p(t)*V(t),p(t) and V(t)=density and volume at time t
R(t)=a(t)*R0,a(t)=scale factor
V(t)=4/3*pi*a(t)^3*R0^3
putting R(t),M(t) in 1:
H(t)^2=8πGp(t)/3+2U/(a(t)^2*RO^2)
the 2nd term is replaced by -kc^2/(RO^2*a(t)^2)?

6. Oct 15, 2017

### Bandersnatch

It's hard to read (please, use LaTeX in the future), but it looks like you omitted the test particle mass term in both kinetic and potential energy.
Other than that, $kc^2$ is substituted for $-\frac {2U}{mR_0^2}$ in the final step.

Last edited: Oct 15, 2017
7. Oct 16, 2017

### Das apashanka

my question is that it is coming 2U/(..) but everywhere it is written -kc^2/(...)

8. Oct 16, 2017

### Bandersnatch

Yes, because in the last step you make a substitution for the constants in the last term, and call it curvature parameter k.

9. Oct 16, 2017

### Das apashanka

No I didnt make a substitution my question is why it is being substituted

10. Oct 16, 2017

### Das apashanka

will you please explain why is term of 2U is written in terms of kc^2

11. Oct 16, 2017

### Staff: Mentor

@Das apashanka , you marked this thread as "A", indicating a graduate level knowledge of the subject matter. The questions you are asking indicate that you don't have that background; accordingly, I have changed the thread level to "I".

12. Oct 16, 2017

### Bandersnatch

When one arrives at the first Friedmann equation:
$$H^2(t)=\frac{8πG}{3}\rho(t)+\frac{2U}{m R_0^2 a^2(t) }$$
in the first term on the r.h.s. we have some time-variable $\rho(t)$, and some universal constants. In the second term, we have a time-variable $a^2(t)$, and a bunch of parameters $\frac{2U}{m R_0^2}$ which are all time-independent. Whatever the total energy is, it is conserved throughout expansion. So is test particle mass (and it cancels out with itself in total energy anyway), and the comoving distance $R_0$ is likewise unchanging. So, for convenience, we gather all of these parameters into one constant, and call it $k$ (where $c^2$ is just a unit-conversion factor).

From the derivation one gets some intuitive understanding that the constant $k$ is related to the total energy - which as far as I understand is the main reason for using Newtonian derivation.
It's to give intuitive meaning to what pops up in the General Relativistic derivation.

Sorry, love. I was using 'you' to mean 'one', or 'we', or 'it's how it's done'. I did not mean you in particular. Just a figure of speech

13. Oct 17, 2017

### Das apashanka

Ok that's fine but why the constant k is taken to be the curvature?

14. Oct 17, 2017

### Das apashanka

no no nothing happened like that ,thanks for replying I mean for last paragraph

15. Oct 17, 2017

### Bandersnatch

I believe it's because that's what you get from the General Relativistic derivation.

16. Oct 17, 2017

### Staff: Mentor

As @Bandersnatch said, the only way to show this is to do the correct GR derivation. The Newtonian derivation, since it assumes flat space, cannot possibly tell you about spatial curvature.