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I Non-zero components of Riemann curvature tensor with Schwarzschild metric

  1. Nov 16, 2017 #1
    I was working out the components of the Riemann curvature tensor using the Schwarzschild metric a while back just as an exercise (I’m not a student, and Mathematica is expensive, so I don’t have access to any computing programs that can do it for me, and now that I’m thinking about it, does anyone know of any comparable but less expensive alternatives that I could use to do all the calculating for me?)
    Anyways, back to the question at hand: at some point after working all the components out I came to understand that the Schwarzschild metric represents a space that is Ricci-flat— this understanding did not come as a consequence of all my hard work, in fact when I looked back at my work, of course I noticed most of the components that contribute to the Ricci tensor cancel out nicely, except for 1 (well 2 if you count its antisymmetric twin). Of course I thought I made a mistake and rechecked my work, but I still got the same answer. Of course I’m still making a mistake, but I’m failing to see where. So now I’m calling on the aid of the PF, the society so proficient at ferreting out mistakes.
    The component in question has a “1” upstairs, and “010” downstairs (I believe I’m following convention when I assign the index “0” for the time direction and “1” for the radial direction). For the sake of brevity, from here on out I will exclude all of the qualifiers like “I calculated...” and “I think...”
    *Convention (just for this post): for Christoffel symbols (of the 2nd kind), “a/bc” will mean the Γ with an “a” upstairs and a “bc” downstairs. There shouldn’t be any risk of confusing 1/01 with, say, a fraction in the context of this post.
    So, the only non-zero Christoffel symbols this component is affected by are 1/00, 1/11, and 0/01, and we will also need to take the partial derivative of 1/00 with respect to x^1.
    Using geometric units, 1/00 = m(r-2m)/r^3 and its derivative is 2m(3m-r)/r^4. 0/01 = m/(r^2-2mr), and 1/11 = -0/01
    Our component of R = d(1/00)/dx^1 + (1/00)*(1/11) - (0/01)*(1/00)
    Or R = d(1/00)/dx^1 - 2*(1/00)*(0/01)
    Which means this component of R = -2m(r-2m)/r^4
    Now, I did not find any other non-zero components of R with a “0i0” downstairs for this to cancel with when contracting indices to get the Ricci tensor, and therefore I ended up with the “00” component of the Ricci tensor being non-zero. So where am I going wrong?
    p.s. sorry for the notation, I’m typing on an iPad.
     
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  3. Nov 16, 2017 #2

    robphy

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  4. Nov 16, 2017 #3

    pervect

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    Yes. Maxima is free, though it's annoying to use. But you can't beat the price. It's notation is confusing, though, and doesn't match the usual GR textbook, so there's a lot of room for confusion in interpreting the result.

    There's and old PF thread that's very helpful on using Maxima written by an ex-user (at least we haven't seen him around for a while). https://www.physicsforums.com/threads/brs-using-maxima-for-gtr-computations.378991/

    Can you communicate to us the basis vectors that you are calculating that component in? At the moment, we'd have to guess. Components are always cacluated according to some basis, the question is - what basis are you using?
     
  5. Nov 16, 2017 #4

    Ibix

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    Maxima - its ctensor library does components of tensors nicely. Try:
    Code (Text):

    load(ctensor);
    ct_coordsys(exteriorschwarzschild);
    cmetric();
    christof(mcs);
    riemann(true);
     
    As pervect says, the notation is a bit odd. Indices range from 1-4 instead of 0-3. It renders ##\Gamma^i_{jk}## as ##mcs_{j,k,i}##, and ##R^i{}_{jkl}## as ##riem_{j,k,l,i}##.
     
  6. Nov 16, 2017 #5

    robphy

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  7. Nov 17, 2017 #6
    My apologies, I’ll be honest, I had never even considered the fact that you would need to use different basis vectors depending on who’s doing the measuring. Pretty obvious now that I think about it. In fact, I believe your reply indirectly points out exactly where I made my mistake in one of my previous posts (man, you guys are good).

    I was using time + spherical coordinates that would be used by a static observer infinitely far away from the black hole/gravitating body (using geometric units).
     
  8. Nov 21, 2017 #7

    pervect

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    The numerical values you get for the components depend on your choice of basis. Among the more popular choices are coordinate bases and orthonormal bases. One is easier to calculate, the other provides more physical insight.
     
  9. Nov 21, 2017 #8

    PeterDonis

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    Note that any vacuum spacetime (i.e., any spacetime with zero stress-energy tensor--note that a nonzero cosmological constant counts as "stress-energy" for this purpose) must be Ricci flat, since if the RHS of the EFE is zero the LHS, the Einstein tensor, must be as well, and a vanishing Einstein tensor implies a vanishing Ricci tensor.
     
  10. Nov 21, 2017 #9

    pervect

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    I suspect some potential communication issues, I'm not sure what remarks I can make that will help, but I'll try and review some basics in the hope that it will clarify my concerns. Let's review. A vector space consists of abstract entites called vectors, which can be added together and multiplied by scalars.

    When we pick out a set of basis vectors at a point, we can represent an arbitrary vector as a weighted sum of the basis vectors. Weighted sums entail multiplying the basis vector by some scalar (one of the allowed operations), and adding these weighted vectors (that have been multiplied by some scalar) together, which is another of the allowed operations on vectors.

    So if we have a set of four basis vectors ##\vec{e_0}, \vec{e_1}, \vec{e_2}, \vec{e_3}## , a typical choice of nomenclature in General Relativity (often, the arrows are omitted, I'm including them in what is an attempt at clarity so we know what quantites are vectors) then we can write an arbitrary vector ##\vec{u}## as ##u^0 \, \vec{e_0} + u^1\, \vec{e_1} + u^2\, \vec{e_2} + u^3\, \vec{e_3}##. The set of four numbers u^0, u^1, u^2, u^3 are the components of the vector ##\vec{u}##.

    But we need to know what the basis vectors are, before the numbers make any sense. In a coordinate basis, the vectors are typically not of unit length. Given a metric tensor ##g_{ab}## and vectors ##\vec{u}## # with components ##u^\mu##, where ##\mu## is understood to take on the values (0,1,2,3), we can write the squared length of a vector in several ways. In what's called abstract notation, we write length##^2## = ##\vec{u} \cdot \vec{u}##,. In what's called component notation it's written as ##g_{\mu\nu} u^\mu u^\nu##, where g is the metric tensor as stated previously. This is shorthand for ##\sum_{\mu=0...3, \nu=0...3} \, g_{\mu\nu} u^\mu u^\nu## by the EInstein summation convention.

    In the Schwarzschild coordinate basis, which is what is what I loosely described as "easy to calculate", the squared length of the basis vector ## \vec{e_0}## is, by the formula above, ##g_{00}##, which has a value of something like (1-2GM/r), or perhaps (1-2GM/r)*c depending on the details of one's setup. This is typically not one, thus the squared length of ##\vec{e_0}## is not one and it is not a vector of unit length. Similar remarks apply to the other components, none of them necessarily have a unit length (though it can happen by accident in some cases). Working with basis vectors that are not unit length is often a source of confusion for those that aren't experienced with tensors. Converting from the coordinate basis (where the vectors don't have unit length) to an orthonormal basis (where the vectors do have unit length) may (and probably is) also an unfamiliar process. One or the other must be learned, though, for the mathematical representation to make any physical sense, to allow physical interpretation and understanding of the significance of the numbers we calculate.
     
  11. Nov 24, 2017 #10
    I understand the basics of what an orthonormal basis is, but you are correct in assuming I am unfamiliar with a general method to convert from a coordinate basis to an orthonormal one. How about I just explain exactly how I was going about calculating the components of the RC tensor, and you tell me if I was doing it all wrong from the start. For each component I used the equation R = dΓ - dΓ + ΓΓ - ΓΓ (I’m typing on my phone, so you’ll have to imagine there are indices all over the place, and by “d” I mean del with subscript— as in partial derivative with respect to x^(aforementioned subscript)). And I calculated the Christoffel symbols simply using the components of the metric (and their inverses), and for the metric, I used the Schwarzschild metric, where
    g00 = (1 - 2m/r)
    g11 = -(1 - 2m/r)^(-1)
    g22 = -r^2
    g33 = -(r^2)*(sin(x^2))^2
    Those numbers by the g’s should be subscripts, and by “x^2” I am of course indicating which coordinate of x, not meaning x squared.
    And that’s it. Using that exact metric I calculated the Christoffel symbols, and using those Christoffel symbols I calculated the RC tensor.
     
  12. Nov 24, 2017 #11

    pervect

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    It's hard to tell if this is all right with all the omissions, I'm assuming you're basically using the formula from the Wikipedia:

    $$
    R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma}
    - \partial_\nu\Gamma^\rho{}_{\mu\sigma}
    + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}
    - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}
    $$

    This gives you the Riemann in a coordinate basis.

    An incomplete list of a few of the nonzero components from the program I use
    $$R^t{}_{rtr} = \frac{2m}{r^2(r-2m)}$$
    $$R^t{}_{\theta t \theta} = -\frac{m}{r}$$
    $$R^t{}_{\phi t \phi} = -\frac{m}{r} \sin^2 \theta$$

    These three components are related to the tidal forces, though the relation is much plainer in an orthonormal basis.

    As far as converting the basis goes, if you're familiar with the notation, the basis vectors in the Schwarzschild metric are labelled ##\partial_t \, \partial_r \, \partial_\theta \, \partial_\phi##. The basis vectors are already orthogonal, so the only thing one has to do is adjust the lenghts. Which can be done by computing the square of the length of the vector, and introducing a new basis that is scaled so the new basis vectors are of unit length. You can also use the tensor transformation rules for changing basis, of course.

    The results are much simpler as well as being able to interpret. For the same three components

    $$\hat{R}_{\hat{t}\hat{r}\hat{t}\hat{r}} = \frac{2m}{r^3} \quad \hat{R}_{\hat{t} \hat{\theta} \hat{t} \hat{\theta}} = -\frac{m}{r^3} \quad \hat{R}_{\hat{t} \hat{\phi} \hat{t} \hat{\phi}} = -\frac{m}{r^3}$$

    The "hat" symbol ^ represents a unit vector, a convention one of my textbooks uses (but it's not universal).

    We can also note that

    $$\hat{R}_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = \hat{R}_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = \frac{m}{r^3} \quad \hat{R}_{\hat{\theta} {\phi} {\theta} {\phi} } = -\frac{2m}{r^3}$$

    since it's easier to write down the components.

    There are a lot of symmetries of the Riemann, so even the more complete second list ( in the orthonormal basis) doesn't explicitly list all of the nonzero componnets. One needs some of the symmetri relations like interchanging the first and second components ##R_{abcd} = -R_{bacd}## reverses the sign of R. Interchanging the last two components is also antisymmetric. And R_{abcd} = R{cdab}. See for isntance the wiki article "symmetries of the Riemann".
     
  13. Nov 25, 2017 #12
    Thanks for that clear and simple explanation, this gives me a better understanding of orthonormal bases. So to answer your question, I was using the coordinate basis, and yes, you assumed correctly, although I did not get it from Wikipedia, that formula is the one I was using. But back to my original question: where am I going wrong when I say ##R^{r}_{trt} = -\frac {2m(1-2m/r)} {r^3}?##
    1. ##Γ^r_{tt} = \frac {m(1-2m/r)} {r^2}##
    2. ##∂_rΓ^r_{tt} = \frac {2m(3m-r)} {r^4}##
    3. ##Γ^t_{tr} = \frac m {r^2 (1 - 2m/r)} = -Γ^r_{rr}##
    4. There are no other non-zero Γ's that ##R^{r}_{trt}## depends on
    So ##R^{r}_{trt} = ∂_rΓ^r_{tt} + Γ^r_{tt} Γ^r_{rr} - Γ^t_{tr} Γ^r_{tt} = -\frac {2m(1-2m/r)} {r^3}##
    5. I also calculated the other 3 ##R^μ_{tμt}##'s to be 0, so when contracting indices to get the Ricci tensor, I'm left with ##R_{tt} = R^{r}_{trt} ≠ 0##
    For reference:
    ##g_{00} = 1 - 2m/r##
    ##g_{11} = - (1 - 2m/r)^{-1}##
    ##g_{22} = - r^2##
    ##g_{33} = - r^2 sin^2(θ)##
    I've numbered the parts that I used in calculating this erroneous component of the Ricci tensor simply to make pointing out my mistake(s) more convenient.
     
  14. Nov 25, 2017 #13

    Ibix

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    I agree your ##R^r {}_{trt}##, but if I'm interpreting the Maxima output correctly, ##R^t {}_{ttt}=0## and ##R^\phi {}_{t\phi t}=R^\theta {}_{t\theta t}=-R^r {}_{trt}/2##. Obviously that cancels to give zero for the Ricci tt component.
     
  15. Nov 25, 2017 #14
    Yes. Wow. Thank you. It appears I just got sloppy and messed up on those 2 components and never questioned the initial zeros I got for them. The one thing I don’t re-examine too closely is where I made my mistake. Imagine that.
     
  16. Nov 25, 2017 #15

    Ibix

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    It's always the way. This is a major reason I like Maxima - it eliminates a whole class of stupid mistakes that I, at least, make all the time.
     
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