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Tips for deriving the friedmann equations

  1. Jan 26, 2013 #1
    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 dont 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
     
  2. jcsd
  3. Jan 26, 2013 #2

    cepheid

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    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: Jan 26, 2013
  4. Jan 26, 2013 #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]
     
  5. Jan 26, 2013 #4

    cepheid

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    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.


    If you meant to write this: [itex]\partial_{1} = \partial_{r}[/itex], then yes. In general:$$\partial_\mu \equiv \frac{\partial}{\partial x_\mu}$$
     
  6. Jan 26, 2013 #5
    oh yes thx, yeah im doing them now, might c equal 1 to avoid this problem lol.
     
  7. Jan 27, 2013 #6

    Chalnoth

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    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).
     
  8. Jan 27, 2013 #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]
     
  9. Jan 27, 2013 #8

    Chalnoth

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    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:

    This should simple be:

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