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Preservation of the infinitesimal element of length

  1. Nov 11, 2012 #1
    My two questions are related to the title. The problematic is: "How can we connect the coordinates ... xα... for α = 0, 1, 2 and 3 to the coordinates of the same object in another frame, say ...yλ for λ = 0, 1, 2, 3 in preserving the quantity η(x)αβ. dxα.dxβ = η(y)λμ. dyλ. dyμ?

    Can someone explain clearly:
    1°) why we have supposed that the coordinates transformations should absolutely be linear (and not, e.g. a Taylor's developpment including terms of highter degree)? -> Poincaré group;

    2°) why didn't we prefered to work with the relations proposed in 1869 by E.B. Christoffel for such problematic?

  2. jcsd
  3. Nov 12, 2012 #2
    It is supposed that the infinitesimals will satisfy the linear transformation i.e. the connection should be affine in character and that is where coefficient of affine connection comes into picture (Christoffel symbols).which is the key to parallel displacement without embedding suface into higher dimensional euclidean spaces.
  4. Nov 12, 2012 #3
    Can you please elaborate a little bit? Do you mean that non linear transformations would result in some deformations of space-time (the rubber) which are themselves not linear? I am sorry for such questioning which may sound stupid but I have the biggest difficulties to get a concrete representation of the maths I am doing myself...
    Last edited: Nov 12, 2012
  5. Nov 12, 2012 #4


    Staff: Mentor

    I'm a bit unclear on the context of your question. Are you just talking about Lorentz transformations in flat spacetime? Or are you talking about general coordinate transformations in curved spacetime?
  6. Nov 13, 2012 #5
    parallelism connection first used by levi-civita requires embedding of a surface into higher dimensional euclidean space but the concept used by weyl is based on an affine connection which even applied to spaces where metric element is not even defined.it is given in his book'space,time and matter'.The linearity is supposed to be for simplicity ,I don't know if non-linearity will impose some deformation which are not liable to any topological transformation.
  7. Nov 13, 2012 #6
    @ PeterDonis
    Answer: I am re-reading and studying the E.B. Christoffel's work concerning the preservation of homogeneous differential expressions of degree 2 (Über die Transformationen der homogenen Differenzialausdrücke zweites Grades: E.B. Christoffel, in Journal für die reine und angewandte Mathematik, (pp. 46 - 70), 03.01.1869). A fine analysis leads to the conclusion that the work concerns only symmetric expressions (ωαβ = ωβα). An immediate application to the preservation of the infisitesimal element of length defined within the GTR context is thus possible. I note that the Christoffel's symbols appear at this step and that...

    ... no notion of connection is introduced in that historical work. In extenso, at this step: the Christoffel's symbols [...] and the {...} are not interpreted as representing a connection.

    @andrien: thanks for re-insisting on the link "Levi-Civita connection - parallelism". I shall re-read again some historical documents with the hope to understand that link better (the question of the embedding).

    Otherwise, I think (do you think it is correct?) the Christoffel's work can a priori be re-used for non-symmetric differential expressions
  8. Nov 13, 2012 #7


    Staff: Mentor

    This doesn't directly answer the question I was asking, but it seems to imply that you are considering general coordinate transformations in curved spacetime. However, your use of the symbol [itex]\eta_{\alpha \beta}[/itex] instead of [itex]g_{\alpha \beta}[/itex] for the metric tensor suggests that you are only considering flat spacetime. So I'm still confused about the context of your question.

    I can't answer this since I'm not familiar enough with Christoffel's work in this area. However, I'm not sure why it's relevant to the question you posed in the OP, which talked about why we use linear coordinate transformations in relativity, not why the metric tensor is or is not symmetric.

    As far as the question of why we use linear transformations, I believe it's because those are the only ones that preserve geometric invariants. (The line element is only one of these, btw.) There's another thread currently active that touches on this topic:


    It's only talking about flat spacetime, though.

    As far as the question of why the metric tensor is symmetric, that follows from the use of the Levi-Civita connection. I believe there have been proposals to construct a theory like General Relativity but using a different connection, which would allow the metric tensor to be non-symmetric (since it's a rank-2 tensor, that basically means there would be a symmetric part and an antisymmetric part, the "torsion"--the Levi-Civita connection is "torsion-free"). My understanding is that all such proposals to date make exactly the same physical predictions as standard GR does, so I don't really see the point. But I am not up to date in this area. You might try Googling "gravity with torsion" or something similar and see what comes up.
  9. Nov 13, 2012 #8


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    I'm not terribly sure I understand BlackForest's notation, but I think he's starting out asking for a group of isometries, i.e. diffeomorphisms that preserve the metric, assuming that's what he means by the [itex]\eta_{\alpha\beta}[/itex]. And going on from there.

    I think the existence of such diffeomorphisms depends on the nature of the space itself, in particular it's Killing vectors. If you've got at least one Killing vector, you can write down at least one such diffeomoprphism.

    But if you don't have any Killing vectors at all, the answer to the "How can you" question becomes "You can't" rather than "how". So it might be good to start out with asking in general "can you" before asking "how", unless you're thinking specifically of flat space-times.

    As I recall from Wald (I didn't look it up) every one parameter group of diffeomorphisms is associated with a vector field, but only if the group of diffeomrphisms is an isometry is the vector field a Killing vector field. And you can think of the Killing Vector as the generator of the group, an infinitesimal translation (or rotation - or whatever) that generates the whole group.

    I haven't read Christoffel's original work, I just use the symbols, so I don't know how it applies to this problem.

    My intuition is saying to me that the isometries might be linear only in flat, homogeneous spaces, but I'm not sure it's right at this point.
  10. Nov 13, 2012 #9
    The OP title talks about preservation of infinitesimal length, and that suggests he is referring to infinitesimal isometries, so I would say the transformations would be linear regardless of the curvature of the manifold.
  11. Nov 13, 2012 #10


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    Do you mean the Christoffel symbols are symmetric from the use of the levi - civita connection? Given a riemannian manifold M there is a unique affine connection [itex]\triangledown [/itex], on M, such that [itex]\triangledown [/itex] is symmetric and compatible with the riemannian metric on M, part of which which leads to the christoffel symbols (the coefficients of this connection) being symmetric. The riemannian metric is symmetric by definition as it associates with each Tp(M) an inner product.
  12. Nov 13, 2012 #11


    Staff: Mentor

    I was assuming that the OP was asking about passive transformations, not active transformations. He uses the phrase "coordinates of the same object", which to me implies that we are transforming the description of a given object in one chart, into a description of the same object in another chart. You are describing isometries that "move one object to another", so to speak (by "moving" along an orbit of a Killing vector field). But I might be wrong; clarification from BlackForest would be welcome.
  13. Nov 13, 2012 #12


    Staff: Mentor

    They are, but I was referring to the *metric* that is preserved, not just the Christoffel symbols, which *are* the connection. However, I may have been mistaken about what using a different connection implies about the symmetry of the metric tensor; see below.

    Yes, agreed.

    Does an inner product require a symmetric metric tensor? The inner product simply contracts a vector with itself:

    [tex]v = g_{ab} v^a v^b[/tex]

    Write this out for a simple 2-dimensional space:

    [tex]v = g_{00} v^0 v^0 + g_{01} v^0 v^1 + g_{10} v^1 v^0 + g_{11} v^1 v^1 = g_{00} ( v^0 )^2 + ( g_{01} + g_{10} ) v^0 v^1 + g_{11} ( v^1 )^2[/tex]

    This will be well-defined and unique for any vector [itex]v[/itex] whether or not [itex]g_{ab}[/itex] is symmetric, because the "cross terms" are effectively symmetrized anyway. In other words, the inner product only "sees" the symmetric part of the metric, so the presence of a nonzero antisymmetric part doesn't affect it at all. At least, that's how it looks to me.
  14. Nov 13, 2012 #13


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    I don't disagree at all. I'm just saying that it is usually defined to be symmetric in the two vector inputs without regards to any connection. I don't disagree that in certain field theories the symmetric part is dropped but I just mean in standard riemannian geometry texts.
  15. Nov 14, 2012 #14
    Sorry for the time-delay but we are living along different "parallels"... Well first of all, thank you @all who have tried to understand my question. Yes PeterDonis you have understood it correctly. The same object observed from two different places. This was explaining the x-coordinates for the one and the y-coordinates for the other. The choice of η was certainly confusing for you all because we actually refer it automatically to the flat Minkowski metric. In my mind and because of the followed quest, this was a generic notation and it would have been a better choice to write g(x) and g(y). In fact, half conscient of the risk I have tried to induce this in your mind in writing η(x) and η(y). Ok, my error.

    So, the Christoffel's work is centered on the preservation of quantities like pαβ. xα. xβ where -when you read the details of the work- one must have pαβ = pβα. Now, except if I did misinterpret this work (of Christoffel), it is possible to re-write it in supposing that p is not automatically symmetric; i. e. p could be antisymmetric, implying pαβ + pβα = 0. This eventuality is reducing the differential expression to a sum on β of pββ. (xβ)2...

    It is probable that Christoffel's work concern real numbers but it could be meaning full to relook it for complex numbers and -why not- for quaternions... this would invalid the argument yα. yβ = yβ. yα... and make everything quite more complicated, offering a natural link with a non-commutative world...

    So I must work - Come later again.
  16. Nov 16, 2012 #15
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