# Energy-momentum tensor and Friedmann Equations

• I
• Diferansiyel
In summary: 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... unless you use that mysterious factor of ##c## that's floating around.
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.

Diferansiyel said:
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.

Chalnoth said:
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 ##

Diferansiyel said:
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.

Chalnoth said:
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.

Diferansiyel said:
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.

Diferansiyel said:
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
Chalnoth said:
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.

Diferansiyel said:
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.

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.

## 1. What is the Energy-Momentum Tensor?

The Energy-Momentum Tensor is a mathematical object used in the theory of general relativity to describe the distribution of energy and momentum in a given region of space. It has 4 components (one for each dimension of space-time) and is used to represent the energy, momentum, and pressure of matter and radiation in a given space.

## 2. How is the Energy-Momentum Tensor related to the Friedmann Equations?

The Energy-Momentum Tensor is an important component in the Friedmann Equations, which are a set of equations used to describe the expansion of the universe in the context of general relativity. The energy and momentum contained in the Energy-Momentum Tensor play a crucial role in determining the curvature and dynamics of space-time in the Friedmann Equations.

## 3. What are the implications of the Energy-Momentum Tensor for the universe's expansion?

The Energy-Momentum Tensor is a key factor in determining the overall energy content and behavior of the universe. It helps to determine the rate of expansion, the curvature of space-time, and the overall fate of the universe. Its precise values and distribution have important implications for the evolution and future of the universe.

## 4. How is the Energy-Momentum Tensor calculated or measured?

The Energy-Momentum Tensor can be calculated using the Einstein field equations in the theory of general relativity. In practical terms, it can be measured through observations of the effects of matter and energy on the curvature of space-time. This can include measurements of the universe's expansion rate, gravitational lensing, and other phenomena.

## 5. What are the main applications of the Energy-Momentum Tensor and Friedmann Equations?

The Energy-Momentum Tensor and Friedmann Equations are used in a variety of fields, including cosmology, astrophysics, and particle physics. They are essential for understanding the behavior of the universe on a large scale and for making predictions about its future. They are also important for testing and improving our understanding of gravity and the fundamental laws of physics.

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