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A Interpretation of the derivative of the metric = 0 ?

  1. Dec 22, 2016 #1
    I am trying to learn GR, primarily from Wald. I understand that, given a metric, a unique covariant
    derivative is picked out which preserves inner products of vectors which are parallel transported.
    What I don't understand is the interpretation of the fact that, using this definition of the derivative,
    the covariant derivative of the metric tensor = 0.

    In my mind, the covariant derivative applied to any tensor is telling us something about
    how the tensor changes from point to point.
    Isn't saying that the covariant derivative of the metric tensor = 0 (everywhere, correct?) the same thing
    as saying the metric tensor is "constant" (in some sense) throughout spacetime?
    But, of course, it does not have to be constant. And wouldn't that imply constant curvature
    (or zero curvature) everywhere? This, of course, must be a wrong interpretation (and Riemann and Ricci
    tensors are clearly nonzero, and they more directly relate to curvature).

    So what is wrong with my interpretation of a derivative as a measure of change?
    thanks.
     
  2. jcsd
  3. Dec 22, 2016 #2

    andrewkirk

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    This stack exchange answer casts some light on it. The derivative is chosen from amongst the many available connections as the (provably unique) connection that gives zero when applied to the metric. Preserving inner products of parallel-transported vectors - the officially stated criterion for selection of the connection - is mathematically equivalent to preserving the metric.

    As you observe, the constancy of the derivative at zero does not entail constant curvature, because the metric tensor is not itself a curvature tensor.

    In the approach described on stack exchange the derivative is selected based on the metric. But it can also be done the other way around. IIRC, Lee does that in his book 'Riemannian Manifolds'. He defines various types of connection and then says that, given an affine connection on a manifold, a Riemannian metric is the unique symmetric 2-tensor that has a certain property in relation to the connection. I don't have the book here right now, so don't remember the property, but it's probably equivalent to giving a zero derivative of the tensor.

    Here's another hokey, folksy way of looking at it, that may or may not mean anything or have any validity: using a connection to differentiate a tensor is - in a very loose sense - measuring the rate of change of the tensor field relative to the connection. But since there is a unique relation between affine connections and their corresponding metric tensors, we could say that differentiating the metric is measuring the rate of change of the tensor field relative to itself. And that is something that one might intuitively expect to be zero.
     
  4. Dec 23, 2016 #3

    PeterDonis

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    No. One way to look at it, heuristically, is that the point of the covariant derivative is to tell how a vector or tensor changes "relative to the metric"; i.e., it is intended to filter out changes that are due to the way the geometry of spacetime changes, and focus on changes that are due to the way the vector or tensor itself changes. But of course all changes in the metric tensor are changes in the way the geometry of spacetime changes, so if you filter out those changes, you're left with zero change.
     
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