Tips for deriving the friedmann equations

In summary, we discussed the use of the metric ds^{2}=c^{2}dt^{2}-a^{2}(t)[\frac{dr^{2}}{1-kr^{2}}+r^{2}(dθ^{2} + sin^2θd^{2}∅)] to derive the Christoffel symbols: \Gamma^{\sigma}_{\mu\nu}=\frac{1}{2}g^{\sigma\rho}(\partial_{\nu}g_{\rho\mu} +\partial_{\mu}g_{\rho\nu} -\partial_{\rho}g_{\mu\nu}). We found that for the specific case of \
  • #1
pleasehelpmeno
157
0
Hi I am using the metric [itex]ds^{2}=c^{2}dt^{2}-a^{2}(t)[\frac{dr^{2}}{1-kr^{2}}+r^{2}(dθ^{2} + sin^2θd^{2}∅)[/itex]]
and am subsequently trying to derive the christoffel symbols:
[itex]\Gamma^{\sigma}_{\mu\nu}=\frac{1}{2}g^{\sigma\rho}(\partial_{\nu}g_{\rho\mu} +\partial_{\mu}g_{\rho\nu} -\partial_{\rho}g_{\mu\nu})[/itex]

I am stuck with finding these and would like some help for instance why does [itex]\Gamma^{0}_{11}=\frac{a\dot{a}}{c(1-kr^{2}}[/itex] where does the c come from?

I understand that one substitutes in the numbers but when attempting this i don't fully understand the differential but for instance:
[itex] \partial_{0}=g_{11}[/itex] is this basically [itex] \frac{\partial}{\partial t}a^{2}\frac{dr^{2}}{1-kr^{2}}[/itex]? and if so are all differentials with numbers greater than 0 just = 0 then i.e what would [itex] \partial_{2}[/itex] be equal to?

I appreciate any help
 
Space news on Phys.org
  • #2
pleasehelpmeno said:
Hi I am using the metric [itex]ds^{2}=c^{2}dt^{2}-a^{2}(t)[\frac{dr^{2}}{1-kr^{2}}+r^{2}(dθ^{2} + sin^2θd^{2}∅)[/itex]]
and am subsequently trying to derive the christoffel symbols:
[itex]\Gamma^{\sigma}_{\mu\nu}=\frac{1}{2}g^{\sigma\rho}(\partial_{\nu}g_{\rho\mu} +\partial_{\mu}g_{\rho\nu} -\partial_{\rho}g_{\mu\nu})[/itex]

I am stuck with finding these and would like some help for instance why does [itex]\Gamma^{0}_{11}=\frac{a\dot{a}}{c(1-kr^{2}}[/itex] where does the c come from?

I understand that one substitutes in the numbers but when attempting this i don't fully understand the differential but for instance:
[itex] \partial_{0}=g_{11}[/itex] is this basically [itex] \frac{\partial}{\partial t}a^{2}\frac{dr^{2}}{1-kr^{2}}[/itex]? and if so are all differentials with numbers greater than 0 just = 0 then i.e what would [itex] \partial_{2}[/itex] be equal to?

I appreciate any help

Hi pleasehelpmeno,

Note: try \phi to get a phi symbol in LaTeX

The more general equation for the line element, meaning with any metric, is:$$ds^2 = g_{\mu\nu} dx^\mu dx^\nu$$Remember that there is an implied summation over repeated indices so that this is actually ##\sum_\mu \sum_\nu g_{\mu\nu} dx^\mu dx^\nu##. In your case, you are using a set of spherical spatial coordniates, so that a coordinate point in spacetime (x0, x1, x2, x3) is expressed as (t, r, θ, ϕ).

In your case, if you examine your line element, you notice that you have terms like dx0dx0 and dx1dx1 (which would be dt2 and dr2 respectively), but you have NO terms like dx0dx1 or dx2dx3 (which would be dtdr and dθdϕ respectively). What this means is that all of the cross terms in your metric are zero:$$g_{\mu\nu} = 0~\textrm{if}~\mu \neq \nu$$Another way to look at it: you can express your metric as a 4x4 matrix: gμν is the element in row mu and column nu of this matrix, where each of the indices mu and nu varies over the values (0,1,2,3). Then in THIS case, for THIS particular metric, all of the off-diagonal terms in this matrix are zero, leaving only diagonal terms. In fact, by inspection, we can see that the matrix is$$ g_{\mu \nu} = \left[ \begin{array}{cccc} c & 0 & 0 & 0 \\ 0 & -a^2/(1-kr^2) & 0 & 0\\ 0 & 0 & -a^2r^2 & 0 \\ 0 & 0 & 0 & -a^2r^2\sin^2(\theta)\end{array}\right]$$The metric tensor has the property that ##g^{\mu \nu} g_{\nu\sigma} = \delta^{\mu}_{\sigma}##, the Kronecker delta, which is basically the identity matrix, in matrix form. Since the metric is diagonal, it's easy to see that the matrix form of the version with raised indices is the same as the above matrix, except that each element on the diagonal is the reciprocal of the one in the version with lowered indices (so that when you multiply the two, you get a matrix with 1s on the diagonal and 0s elsewhere).

Now we can consider the Christoffel symbol that you were trying to compute. It's the one for which sigma = 0 and mu = nu = 1. Now, since we have a prefactor of ##g^{\sigma \rho}## multiplying the whole expression, and sigma = 0, it must be true that rho = 0 as well, so that we have ##g^{00}##. For all other values of rho, the metric vanishes: ##g^{0\rho} = 0~\textrm{for}~\rho \neq 0##. What this also means is that the first two terms inside the parentheses vanish as well, because mu and nu are 1, whereas rho is 0, and g01 = g10 = 0 (off-diagonal elements). We are left with the third term in the parentheses:$$\Gamma^0_{11} = \frac{1}{2}g^{00}(-\partial_\rho g_{\mu \nu})$$ $$ = \frac{1}{2}g^{00}(-\partial_0 g_{11})$$Now, the ∂'s are shorthand for partial derivatives with respect to the coordinates. In particular, ∂0 is a shorthand for ##\frac{\partial}{\partial x_0} = \frac{\partial}{\partial t}## Also, since g00 = c, it must be that g00 = 1/c (because of the statement in blue above). So we have:$$\Gamma^0_{11} = \frac{1}{2}g^{00}(-\partial_0 g_{11}) = \frac{1}{2c}\left[-\frac{\partial}{\partial t}\left( -\frac{a^2}{1-kr^2}\right)\right]$$Remember that a(t) is a function of time, whereas r, which is a coordinate, does not depend on t (coordinates are considered to be independent variables). So:$$\Gamma^0_{11} = \frac{1}{2c}\frac{1}{1-kr^2}\left(+\frac{\partial }{\partial t} (a^2) \right) = \frac{1}{2c(1-kr^2)}2a\frac{\partial a}{\partial t} = \frac{1}{2c(1-kr^2)}2a\dot{a}$$ $$ = \frac{a\dot{a}}{c(1-kr^2)}$$Does this make sense?
 
Last edited:
  • #3
yeah thanks but shouldn't it be [itex]c^{2}[/itex].
I think i see know so;
[itex] \partial_{0} = \partial_{t} [/itex]
[itex]\partial_{1} = \partial_{2}[/itex]
[itex]\partial_{2} = \partial_{\theta}[/itex]
[itex]\partial_{3} = \partial_{\phi}[/itex]
 
  • #4
pleasehelpmeno said:
yeah thanks but shouldn't it be [itex]c^{2}[/itex].

Yeah, good point, g00 should be c2. I am not sure if it is supposed to be c2 in the final answer for ##\Gamma_{11}^0## as well, or if I just missed a factor of c somewhere. However, you get the idea.
pleasehelpmeno said:
I think i see know so;
[itex] \partial_{0} = \partial_{t} [/itex]
[itex]\partial_{1} = \partial_{2}[/itex]
[itex]\partial_{2} = \partial_{\theta}[/itex]
[itex]\partial_{3} = \partial_{\phi}[/itex]

If you meant to write this: [itex]\partial_{1} = \partial_{r}[/itex], then yes. In general:$$\partial_\mu \equiv \frac{\partial}{\partial x_\mu}$$
 
  • #5
oh yes thx, yeah I am doing them now, might c equal 1 to avoid this problem lol.
 
  • #6
pleasehelpmeno said:
oh yes thx, yeah I am doing them now, might c equal 1 to avoid this problem lol.
In general, yeah, I'd just set c=1. You can figure out the factors of c you need later by examining the units (e.g. if you have a distance, but end up with units in seconds, you need to multiply by a factor of c).
 
  • #7
can i ask one more question about lowering indices is:
[itex]g_{\mu\nu}g_{\mu\nu}T^{\mu\nu} = T g_{\mu\nu} = T_{\mu\nu}[/itex]
and is:
[itex]g_{\mu\nu}g_{\mu\nu}u^{\mu} = g_{\mu\nu} u_{\nu} = u_{\mu}[/itex]
 
  • #8
pleasehelpmeno said:
can i ask one more question about lowering indices is:
[itex]g_{\mu\nu}g_{\mu\nu}T^{\mu\nu} = T g_{\mu\nu} = T_{\mu\nu}[/itex]
No. The operation [itex]g_{\mu\nu}T^{\mu\nu} = T[/itex] is a contraction over all indices, which reduces the tensor to a scalar, which removes much of the information of the tensor. Instead, you have to do the following operation:

[tex]g_{\mu\sigma}g_{\nu\tau}T^{\mu\nu} = T_{\sigma\tau}[/tex]

The main problem here is that you're using the same index too many times. The same goes for below:

pleasehelpmeno said:
and is:
[itex]g_{\mu\nu}g_{\mu\nu}u^{\mu} = g_{\mu\nu} u_{\nu} = u_{\mu}[/itex]
This should simple be:

[tex]g_{\mu\nu}u^{\mu} = u_{\nu}[/tex]
 

FAQ: Tips for deriving the friedmann equations

1. What are the Friedmann equations?

The Friedmann equations are a set of equations used in cosmology to describe the evolution of the universe. They were derived by the physicist Alexander Friedmann based on Einstein's theory of general relativity.

2. Why are the Friedmann equations important?

The Friedmann equations are important because they allow us to understand the dynamics of the universe and how it has evolved over time. They are also crucial in making predictions about the future of the universe.

3. What are some of the key tips for deriving the Friedmann equations?

Some key tips for deriving the Friedmann equations include understanding the principles of general relativity, using the correct mathematical framework, and considering various physical factors such as the density and curvature of the universe.

4. What are some common pitfalls when deriving the Friedmann equations?

Some common pitfalls when deriving the Friedmann equations include making incorrect assumptions, not considering all relevant physical factors, and making mathematical errors.

5. How are the Friedmann equations used in cosmology?

The Friedmann equations are used in cosmology to study the past, present, and future of the universe. They are used to understand the expansion of the universe, the formation of galaxies and other structures, and the distribution of matter and energy in the universe.

Similar threads

Replies
16
Views
853
Replies
6
Views
2K
Replies
8
Views
2K
Replies
1
Views
572
Replies
9
Views
826
Replies
27
Views
4K
Back
Top