I Energy-momentum tensor and Friedmann Equations

[Diferansiyel]

Hi everyone,

I want to derive the Friedmann equations from Einstein Field Equations. However, I have a problem that stems from the energy-momentum tensor. I am also trying to keep track of $c^2$ terms.

FRW Metric:
$$ ds^2= -c^2dt^2 + a^2(t) \left( {\frac{dr^2}{1-kr^2} + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2} \right)$$

EFE:
$$ G_{\mu \nu} \equiv R_{\mu \nu }- \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$

I calculated metric dependent components which are Christoffel symbols, Riemann Curvature Tensor, Ricci Tensor & Scalar. What I've found is

$$ R_{tt} = -3 \frac{\ddot{a}}{a} $$

$$ R_{ii} = \dfrac{g_{ii}}{c^2a^2}( a \ddot{a} +2 \dot{a}^2 +2kc^2 ) $$

$$ R=g^{\mu \nu}R_{\mu \nu} = 6\left[\frac{\ddot{a}}{ac^2}+ \left(\frac{\dot{a}}{ac}\right)^2 +\frac{k}{a^2}\right] $$

For time-time components I want to obtain:
$$ \boxed{\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3c^2} \rho - \frac{kc^2}{a^2(t)}} $$
and for the space-space components:
$$ \boxed{\frac{2\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 = -\frac{8 \pi G }{c^2}p - \frac{kc^2}{a^2(t)}} $$
here $\rho$ is the energy density, not the mass density.

So in order to obtain the boxed equations, energy-momentum tensor $T_{\mu \nu}$ must be $diag(\rho c^2, p g_{ij})$ . However from the expression $T_{\mu \nu}=\left(\rho + \frac{p}{c^2}\right)u_\mu u_\nu +pg_{\mu \nu}$ I can't get what I want for $u^\alpha=(c,0,0,0)$ . Am I using wrong form of $T_{\mu \nu}$ or 4-velocity $u^\alpha$ ?

Thanks for your help,

K.

 

[Chalnoth]

Hi everyone,

I want to derive the Friedmann equations from Einstein Field Equations. However, I have a problem that stems from the energy-momentum tensor. I am also trying to keep track of $c^2$ terms.

FRW Metric:
$$ ds^2= -c^2dt^2 + a^2(t) \left( {\frac{dr^2}{1-kr^2} + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2} \right)$$

EFE:
$$ G_{\mu \nu} \equiv R_{\mu \nu }- \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$

I calculated metric dependent components which are Christoffel symbols, Riemann Curvature Tensor, Ricci Tensor & Scalar. What I've found is

$$ R_{tt} = -3 \frac{\ddot{a}}{a} $$

$$ R_{ii} = \dfrac{g_{ii}}{c^2a^2}( a \ddot{a} +2 \dot{a}^2 +2kc^2 ) $$

$$ R=g^{\mu \nu}R_{\mu \nu} = 6\left[\frac{\ddot{a}}{ac^2}+ \left(\frac{\dot{a}}{ac}\right)^2 +\frac{k}{a^2}\right] $$

For time-time components I want to obtain:
$$ \boxed{\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3c^2} \rho - \frac{kc^2}{a^2(t)}} $$
and for the space-space components:
$$ \boxed{\frac{2\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 = -\frac{8 \pi G }{c^2}p - \frac{kc^2}{a^2(t)}} $$
here $\rho$ is the energy density, not the mass density.

So in order to obtain the boxed equations, energy-momentum tensor $T_{\mu \nu}$ must be $diag(\rho c^2, p g_{ij})$ . However from the expression $T_{\mu \nu}=\left(\rho + \frac{p}{c^2}\right)u_\mu u_\nu +pg_{\mu \nu}$ I can't get what I want for $u^\alpha=(c,0,0,0)$ . Am I using wrong form of $T_{\mu \nu}$ or 4-velocity $u^\alpha$ ?

Thanks for your help,

K.

The correct form of $T_{\mu \nu}$ is certainly $diag(\rho c^2, p g_{ij})$ (though possibly with a negative sign somewhere depending upon the sign convention you're using), and that is the correct 4-velocity. My guess is that you've got a sign error.

 

[Diferansiyel]

The correct form of $T_{\mu \nu}$ is certainly $diag(\rho c^2, p g_{ij})$ (though possibly with a negative sign somewhere depending upon the sign convention you're using), and that is the correct 4-velocity. My guess is that you've got a sign error.

Thanks for the answer, I am using (-,+,+,+) convention. The problem is that for 00 component $(\rho + p/c^2) \times c^2 + p(-c^2)$ does not give $\rho c^2$

 

[Chalnoth]

Thanks for the answer, I am using (-,+,+,+) convention. The problem is that for 00 component $(\rho + p/c^2) \times c^2 + p(-c^2)$ does not give $\rho c^2$

Doesn't it?

Edit: Ahh, you have a $c^2$ discrepancy. I believe it should be $(\rho + p/c^2) \times c^2 + p(-1)$. This is one reason why physicists usually leave off the factors of $c$: being only unit conversion factors, they add no information and can cause errors.

 

[Diferansiyel]

Doesn't it?

Edit: Ahh, you have a $c^2$ discrepancy. I believe it should be $(\rho + p/c^2) \times c^2 + p(-1)$. This is one reason why physicists usually leave off the factors of $c$: being only unit conversion factors, they add no information and can cause errors.

Yes I agree with you, however in our case we kept $c$ in any term, so we should also keep $c$ in the metric.

The other possibility is that
$$ T_{\mu \nu} = (\rho + p)u_\mu u_\nu + pg_{\mu \nu} $$
by choosing the above expression as our energy-momentum tensor, then we can obtain the intended result.

 

[Chalnoth]

Yes I agree with you, however in our case we kept $c$ in any term, so we should also keep $c$ in the metric.

Yes, but the fact that you've got that extra $c^2$ in there means that you made a mistake. Probably some conflict in the convention of where to place the factors of $c$.

You can see that it's a mistake because it's not possible to add (or subtract) two parameters with different units. Imagine trying to add a position to a velocity, or a position to a time. It makes no sense.

In the above equations, $\rho$ has units of matter density, so $\rho c^2$ has units of energy density. The pressure $p$ also has units of energy density.

My guess is you made a mistake when raising/lowering the indices of the metric.

The other possibility is that
$$ T_{\mu \nu} = (\rho + p)u_\mu u_\nu + pg_{\mu \nu} $$
by choosing the above expression as our energy-momentum tensor, then we can obtain the intended result.

That only works if you use units where $c=1$. With your convention of using units where $c \neq 1$, you can't add a matter density to an energy density.

 

[Diferansiyel]

Yes, but the fact that you've got that extra $c^2$ in there means that you made a mistake. Probably some conflict in the convention of where to place the factors of $c$.

You can see that it's a mistake because it's not possible to add (or subtract) two parameters with different units. Imagine trying to add a position to a velocity, or a position to a time. It makes no sense.

In the above equations, $\rho$ has units of matter density, so $\rho c^2$ has units of energy density. The pressure $p$ also has units of energy density.

My guess is you made a mistake when raising/lowering the indices of the metric.That only works if you use units where $c=1$. With your convention of using units where $c \neq 1$, you can't add a matter density to an energy density.

I highlighted that "$\rho$ is the energy density, not the mass density" in my main post. Thus, there is no problem with $T_{\mu \nu} = (\rho + p)u_\mu u_\nu + pg_{\mu \nu}$ . On the other hand, this time $T_{00}$ is in the unit of energy density x c² = mass density x c⁴ which is meaningless.

 

[Chalnoth]

I highlighted that "$\rho$ is the energy density, not the mass density" in my main post. Thus, there is no problem with $T_{\mu \nu} = (\rho + p)u_\mu u_\nu + pg_{\mu \nu}$ . On the other hand, this time $T_{00}$ is in the unit of energy density x c² = mass density x c⁴ which is meaningless.

With that convention, sure, that works fine.

 

[rolling stone]

High to everybody!

With just one and half year delay (sorry), here is a way to derive the Friedmann equations keeping track of the “c” term.

Starting point is the fully contravariant energy-stress tensor of a perfect fluid in a Local Inertial Frame:

(1) T_Hij = diag[ ρ * c^2, p, p, p ]

where Hij means “High ij indexes”, ρ is the mass density, p is the pressure of the fluid.

In the LIF the metric tensor is:

(2) g_Hij = g_Lij = diag[ +1, -1, -1, -1 ]

and the contravariant 4-velocity tensor is:

(3) u_Hi = d/dτ [ c*t, x, y, z ] = γ [ c, dx/dt, dy/dt, dz/dt ],

where τ is the proper time and γ the relativistic contraction factor.

So, if the fluid and the LIF are at rest:

(4) u_Hi = [ c, 0, 0, 0 ]

If we leave the LIF and enter a generic frame, the energy-stress tensor becomes:

(6) T_Hij = (ρ – p/c^2)*(u_H0)^2 + p*g_Hij

Note that eq.6 becomes eq.1 if u_Hi is as in eq.4 and g_Hij as in eq.2 .

Now, if the new reference frame is the Friedmann Walker Robertson one

the fully covariant metric tensor is:

(7) g_Lij = diag[ + c^2, - a^2/(1- k * x^2), - (a^2 * x^2), - (a^2 * x^2 * sinθ^2) ]

the fully contravariant metric tensor is:

(8) g_Hij = diag[ +1/c^2, - (1- k * x^2)/a^2, - 1/(a^2 * x^2), - 1/(a^2 * x^2 * sinθ^2) ]

and the contravariant 4-velocity tensor is:

(9) u_Hi = d/dτ [ t, x, θ, φ] = γ [ 1, dx/dt, dθ/dt, dφ/dt ]

So, if the fluid and the FWR frame are at rest:

(10) u_Hi = [ 1, 0, 0, 0 ]

Thus in the FRW frame, from eq.6, by eq.10 and eq.8, the fully contravariant energy-stress tensor is:

(11) T_Hij = diag[ + ρ, + p*(1- k * x^2)/a^2, + p/(a^2 * x^2), + p/(a^2 * x^2 * sinθ^2)]

The trick to remove the distorting effect of the metric on the fully contravariant energy-stress tensor as in eq.11 is to consider its Mixed indexes tensor. Since:

(12) T_Mij = T_Hik * g_Lkj

from eq.12, eq.11 and eq.7 we obtain:

(13) T_Mij = diag[ + ρ*c^2, - p, - p , -p ]

Therefore the Einstein Field Equations to be solved (keeping track of the “c” term) are:

(14) G_Mij = (8 π G/ c^4)*T_Mij

The mixed Einstein tensor G_Mij in eq.14 may be computed using the open source wxMaxima algebra system app by, for instance, the following (.wmx) program:

kill(all)$
dim: 4$
array(g,dim,dim)$
g[1,1]: c^2$
g[1,2]: 0$
g[1,3]: 0$
g[1,4]: 0$
g[2,1]: 0$
g[2,2]: -(a(t)^2)/(1-k*x^2)$
g[2,3]: 0$
g[2,4]: 0$
g[3,1]: 0$
g[3,2]: 0$
g[3,3]: -(a(t)*x)^2$
g[3,4]: 0$
g[4,1]: 0$
g[4,2]: 0$
g[4,3]: 0$
g[4,4]: -(a(t)*x*sin(theta))^2$
gg: genmatrix(g,dim,dim)$
kill(ctensor)$
load(ctensor)$
lg: gg$
cmetric(false) $
ct_coords: [t,x,theta,phi]$
christof(false)$
ricci(false)$
R: scurvature()$
einstein(true)$

The output is:
(15) G_M00 = 3*k / a^2 + 3*(da/dt)^2 / a^2
and
(16) G_M11 = k / a^2 + (da/dt)^2 / (c^2 * a^2) + 2 (d^2 a / dt^2) / (c^2 * a)

From eq.15, eq.14 and eq.13, we obtain the equations:
(17) 3*k / a^2 + 3*(da/dt)^2/ a^2 = (8 π G/ c^4)* ρ*c^2
and
(18) k / a^2 + (da/dt)^2 / (c^2 * a^2) + 2 (d^2 a / dt^2) / (c^2 * a) = - (8 π G/ c^4)* p

which are the basis for the deduction of the Friedmann equations by keeping track of the “c” term.