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Form of spherically symmetric metric from formal analysis

  1. May 14, 2015 #1

    ShayanJ

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    Gold Member

    In most GR textbooks, the general form of a spherically symmetric metric is obtained by inspection which is acceptable. But in the textbook I'm reading, the author does that with a mathematical analysis just to illustrate the method. But I can't follow his calculations. In fact he omits much of the steps.

    At first he says that under rotations, Cartesian coordinates change by ##x^\mu \rightarrow x^\mu+\xi^\mu ## where ## \xi^0=0 \ , \ \xi^i=\epsilon^{ij}x_j ## and ## \epsilon^{ij}=-\epsilon^{ji} ## is a set of arbitrary infinitesimal constants. The change in functional form of the metric is given by ## \delta g^{\mu\nu}=-\xi^\sigma \partial_\sigma g^{\mu\nu}+g^{\mu \sigma}\partial_\sigma \xi^\nu+g^{\nu\sigma}\partial_\sigma \xi^\mu ## so if we set ## \delta g_{\mu\nu}=0 ##(yeah, he sets the covariant thing to zero!), we'll get ## \xi^\sigma \partial_\sigma g^{\mu\nu}=g^{\mu \sigma}\partial_\sigma \xi^\nu+g^{\nu\sigma}\partial_\sigma \xi^\mu##. Then by considering the several different values of ## \mu ## and ## \nu ##, he gets several equations which constrain the form of the metric.

    When he sets ## \mu=\nu=0##, he gets ## (\partial_i g_{00})x_j=(\partial_j g_{00})x_i ##. Which I'm not sure how he gets it.

    By setting ## \mu=0##, he gets ## g_{0i}\epsilon^i_{\ j}+(\partial g_{0j})\epsilon^{ik}x_k=0 ##. He says this indicates that ## g_{0i}dx^idt=a_1(r,t)x^idx_idt=f(r,t)dr dt ##(where ##a_1## and f are arbitrary functions. But I neither understand how he gets the equation nor how he gets that form of the metric from the equation.

    Then he assumes ## \mu,\nu \neq 0## which gives him ## g_{ik}\epsilon^k_j+g_{kj}\epsilon^k_i+(\partial_k g_{ij})\epsilon^{kl}x_l=0## . Then he says it shows that ##g_{ij}dx^idx^j=[a_2(r,t)\delta^{ij}+a_3(r,t)x^ix^j]dx_idx_j ## which again I don't get the whole thing!

    Any help is much appreciated!
    Thanks
     
  2. jcsd
  3. May 14, 2015 #2

    PeterDonis

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    Staff: Mentor

    I've never seen this done in Cartesian coordinates, so I can't really help much with unpacking this. But it might be an interesting exercise to take the known form of a spherically symmetric metric in spherical coordinates and transform it to Cartesian coordinates.

    Also, IIRC there should only be two unknown functions when all is said and done. What you're showing here has four, but two of them are in front of terms that it should be possible to eliminate by coordinate transformations (the ##dr dt## term and the ##x^i x^j## term).
     
  4. May 15, 2015 #3
    Hi Shyan,

    This is a rather unusual approach, to start with a cartesian coordinate system to describe spherical symmetric configuration. Not sure it is any more rigorous than the usual approach which rightaway starts from the spherical polar coordinates. Anyways, regarding your issues with the derivation, they are pretty straightforward. First as opposed to your upper indexed objects i.e. inverse metric, gμν I will be using the metric, gμv instead. In that case, under a coordinate change,
    xμ→xμμ
    the change of the metric is,
    δgμν= - ξρρgμν - gρνμξρ - gμρνξρ.

    Now, for SPATIAL rotations, i.e. ξ0=0,ξiij xj, the metric is unchanged (since it is symmetric under rotations/spherical symmetry) so the above change in metric should be zero. This leads to simplifications which we now will derive when we consider component by component. But before we dive in we should note that for spatial rotations, the ξ's are not functions of time, just space, i.e. time derivatives vanish, ∂0ξi=0.

    Now, lets look at the metric change vanishing condition for the 00 component of the metric, i.e. δg00=0. From the general expression we see that plugging in μ,ν=0, we get,

    δg00= - ξρρg00 - gρ00ξρ - g0ξρ.

    Recalling the time-independence, ∂0ξi=0, the last 2 terms vanish and noting that ξ0=0 and this rhs simplifies to,

    δg00 = - ξiig00.

    Further after plugging in ξiij xj, we get

    δg00 = - εij xjig00.

    So this is the change in the 00-component of the metric under spatial rotation. But then this is a symmetry and must vanish. Using antisymmetry of εij then we get,
    xij g00 - xji g00 = 0.
    Note that in the previous equation i and j were dummy indices and in this equation they are not. Since the dummy index is summed over all components, we use isotropy of cartesian coordinates to set EACH contribution of the sum to be zero i.e. for each pair i and j.

    Next, set μ=0,ν=i. We have again from the general expression for metric change, plugging μ=0,ν=i,

    δgμ0= - ξρρg0i - gρ0iξρ - g0ξρ.
    Since ξ0=0 and the ξi's are time-independent, the rhs again simplifies a lot. In particular the last term vanishes. We thus have now,
    δgμ0= - ξjjg0i - gj0iξj.
    Now plugin the special form, ξjjk xk, we further simplify to
    δgμ0= - εjk xkjg0i - gj0 εji.

    Now this variation also must vanish if rotations are symmetry, giving us,
    εji gj0 + εjk xkjg0i =0,

    which is same as,
    εjk ( δki gj0 + xkjg0i ) = 0.
    Although, this is a sum of over dummy indices j and k as before, isotropy implies this must hold for each pair j and k. So Setting the quantity inside the brackets to zero for each j and k, we have,
    δki gj0 + xkjg0i = 0

    The Kronecker delta on the lhs forces i to be same as k, and we have after taking trace of both sides
    gj0 = - 1/3 xij g0i.
    This also implies, after multiplying both sides by dxj dt ,
    gj0 dxj dt = - 1/3 xi ( dxjj g0i) dt.

    But I am not sure how to conclude that " xi ( dxjj g0i)" is proportional to δij, which is neccessary to get to the point where the rhs becomes drdt times function of r and t. This is getting too long, so I am writing the next part in the next message.
     
  5. May 15, 2015 #4
    So plugging μ=i, μν=j again the in the general formula for metric change, we get,

    δgij = - ξρρ gij - gρji ξρ - gj ξρ.

    As before the rhs simplifies because ξ0 = 0 for spatial rotations, and becomes,

    δgij = - ξkk gij - gkji ξk - gikj ξk.

    This change is zero as spatial rotation is a symmetry. So we set the rhs to zero, and at the same time we plug in ξk = εkl xl. This gives us,

    εkl xlk gij+ gkj εki + gik εkj.

    This is the intermediate step. Now I guess the next step is to reorganize this in the form,

    εkl (xlk gij + gkj δil + gik δjl) = 0.

    Again the antisymmetric in k and l part in the quantity in the brackets must be zero for each pair of k and l due to isotropy. However not sure this leads to the form you had for gijdxidxj.
     
  6. May 15, 2015 #5

    ShayanJ

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    Thanks Roy. You cleared up so much confusion and I appreciate the time you devoted to reply.
    I just have problem with one thing. You assumed that ## \epsilon^i_k=-\epsilon^k_i ##. But initially we only had ## \epsilon^{ij}=-\epsilon^{ji}##. Now if we contract with ## g_{jk}##, we'll get ## \epsilon^i_{\ \ k}=-\epsilon_k^{\ \ i} ##. How can we make sure that this is the same as what you assumed? I mean, sometimes some textbooks actually consider it important that which index you're getting up or down. This is something I never understood properly!
     
    Last edited: May 15, 2015
  7. May 15, 2015 #6
    Probably my bad notation but the subscript j is to the right of superscript i here in my convention, not right under i. But regardless, I think in both case you get antisymmetry whether they are both upstairs, or one up and on down, as long as you distinguish the top one from the bottom one by NOT placing one right under the other.
    I am not sure about that I do not know your original convention, but I thought is proper to define ε, so that you get the spatial rotation to work correctly i .e. I defined or arrived at the right form for ε, thru the relation,

    ξi= εijxj.

    I used this because I always keep coordinates with upper index and derivatives with lower index to reduce confusion.
     
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