Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Curved spacetime and imaginary coordinate

  1. Oct 18, 2009 #1
    In Misner, Thorne, Wheeler: "Gravitation" it is stated on that "no one has discovered a way to make an imaginary coordinate work in the general curved spacetime manifold" (p.51). Can anyone elaborate on this? Right now, I don't get why it wouldn't work and nothing more is said in the book.
     
  2. jcsd
  3. Oct 18, 2009 #2
    Either way, the "norm" of a four vector [itex](ct,x,y,z)[/itex] in flat spacetime is [itex]-ct^2 +x^2 +y^2 +z^2[/itex]. Some textbooks (I believe Marion-Thornton does this) try to "hide" the negative sign by instead saying that the dot product of four-vectors is as in Euclidean space, but the time component is imaginary. This of course is mathematically equivalent, but hides the very idea of a metric which we will need later in GR.

    Also, in flat spacetime we become complacent with the idea that a vector can be a 'displacement vector', instead of event coordinates being separate from the vector coordinates in the tangent space. Once in curved space this becomes obvious, and pushing the imaginary number onto the coordinates of the tangent space means even less then. Furthermore, the metric would start to have imaginary components, and the curvature and other things would need to be redone just to fix this kludge.

    So imaginary coordinates are not useful at all (they don't "work").
    I don't think they are claiming you literally cannot make changes in definitions to force such a substitution on the notation, but one would have to be acutely aware of this substitution to prevent problems ... to the point where you'd most likely be forced to just work it out normally and do the substitution afterwards (making that notation pointess).
     
  4. Oct 19, 2009 #3
    I think that's it. The whole point of imaginary time is to get rid of the negative sign from any of the metric components. Before considering curvature you need cross-terms in the metric, which corresponds to putting [itex]\sqrt{-1}[/itex] in some of the metric components if you use imaginary time. The cure is worse than the disease.
     
  5. Oct 19, 2009 #4
    Okay, so assigning imaginary values to a dimension in curved spacetime is not prinicpally impossible, in the sense that it would lead to inconsistencies, but it's rather that it's just highly impracticable and therefore of no use. Hope I got that right; thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Curved spacetime and imaginary coordinate
Loading...