Why is U replaced by -kc² in the Friedmann equation derivation?

[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)?

 

[PeterDonis]

while deriving the friedmann equation using Newtonian Mechanics

Which is only heuristic in any case. A proper derivation uses GR, not 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)?

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

 

[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]

 

[PeterDonis]

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]

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/

 

[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)?

 

[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.

 

[Das apashanka]

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

 

[Bandersnatch]

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

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

 

[Das apashanka]

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

 

[Das apashanka]

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

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

 

[PeterDonis]

@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".

 

[Bandersnatch]

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

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.

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

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

 

[Das apashanka]

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

 

[Das apashanka]

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

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

 

[Bandersnatch]

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

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

 

[PeterDonis]

why the constant k is taken to be the curvature?

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.