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Deriving the Christoffel Symbols

  1. Jul 19, 2012 #1
    Hi All,

    I am currently reading Menzel's "Mathematical Physics" and one part in particular confuses me. When he is introducing Riemannian Geometry he derives the Christoffel symbols almost out of thin air. He starts by differentiating a vector with respect to a coordinate system ∂[itex]_{i}[/itex]u[itex]^{j}[/itex]e[itex]_{j}[/itex]=(∂[itex]_{i}[/itex]u[itex]^{j}[/itex])e[itex]_{j}[/itex]+u[itex]^{j}[/itex](∂[itex]_{i}[/itex]e[itex]_{j}[/itex])

    he then focuses on this term (∂[itex]_{i}[/itex]e[itex]_{j}[/itex]) and simply says that whatever "it" is, it must depend on the basis vectors and should look like this
    [itex]\Gamma^{i}[/itex][itex]_{j}[/itex][itex]_{k}[/itex]e[itex]_{i}[/itex]

    and he says "these are the christoffel symbols of the (first? second?forgot...) kind". it all seems handwavy to me and I was wondering if someone could explain to me WHY it must depend on the basis and why it just happens to be an object with 3 indices (I do realize one of them is a dummy index, but still the other two perplex me). He explained none of this in his "derivation". (on a side not, it is a fabulous book still one of my favorites even though it is pretty dated, great read)
     
  2. jcsd
  3. Jul 19, 2012 #2
    Well, the e_i form a complete orthonormal set, and so ∂_ie_j, must be given by a linear superposition of the e_i. The result must however be rank-2, and therefore must be of the form Gamma^k_{ij} e_k.
     
  4. Jul 19, 2012 #3
    the basis isn't, in general, orthonormal. that much I am sure of
     
  5. Jul 19, 2012 #4
    Sorry, it doesn't matter if it is orthonormal or not..
    The crucial ingredient is completeness.
    (By the way, in case it isn't clear, apart from the dummy indices, the other two indices are present because the rank of the two objects must be the same.)
     
  6. Jul 19, 2012 #5
    ok, I understand why it depends on the basis now, but I still don't understand why it goes up in rank. One reason is that it isn't a tensor so it doesn't really have a rank.
     
  7. Jul 20, 2012 #6
    the number of free indices on both sides of any tensor equation must always be the same, even if the objects don't transform as tensors. Furthermore, the number of downstairs and upstairs indices should always match as well. By the way, even objects that don't transform as tensors have a rank, it's just that when you transform them you pick up extra terms which come from the non-tensor nature. In any case, the number of free indices is always preserved in every term of the equation. Also, there can be only two repeated dummy indices in any one term, one upstairs and one downstairs.
     
  8. Jul 20, 2012 #7
    Also, Gamma^k_{ij} has three indices because this is the only way of ensuring that the result is a linear superposition of ALL the e_k while preserving the number of free indices, (i,j) in this example,
    $$
    \partial_i {\bf e}_j = \Gamma^k_{ij}{\bf e}_k
    $$
     
  9. Jul 20, 2012 #8
    I see now, thank you :)
     
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