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Real relativistic field and curvature?

  1. Dec 19, 2009 #1
    Say we have a real field that satisfies:

    E^2 = P^2 + m^2

    Assume spacetime is 4D. Assume the field is at rest and grab a single point of this field and slowly displace it a distance x. Just as an anchored string (string with an additional sideways restoring force) with fixed end points will have its length change when a point is displaced and just as a two dimensional "anchored" membrane will change its area when a single point is displaced can we say that a 3 dimensional "anchored" membrane will change its volume if a single point is displaced a distance x? When I say "anchored" membrane it is the real relativistic field I am thinking of.

    For small displacements, x, the change in volume is proportional to what power of x?

    Thanks for any suggestions on how to solve this.
    Last edited: Dec 19, 2009
  2. jcsd
  3. Dec 19, 2009 #2
    I'm thinking the change in volume will also depend on the radius of volume around the point that is displaced?

    Thanks for any help.
  4. Dec 19, 2009 #3
    And then we would like to let the point go and picture volume changing with time and space. But first things first.
  5. Dec 19, 2009 #4
    And then repeat all the above questions for a massless real field,

    E^2 = P^2
  6. Dec 20, 2009 #5
    I have to think more clearly about grabbing a point of a field that has a space dimension greater then one.

    So that the math does not blow up, for a field of three space dimensions we grab a small spherical shell of radius R, of the field. Now we can displace this shell a distance w in the tangent space of the field. Now I think we can try to calculate volume changes.

    the following function may be useful,

    D(r) = A*exp[-m*r]/r for r > or = R

    r is the radial distance from the center of the displaced shell, D(r) is the displacement of the field as a function of the radial coordinate r, and A is chosen so that:

    D(R) = A*exp[-m*R]/R = w

    Thanks for any help.
  7. Dec 21, 2009 #6
    If I did the math right the radial component of the laplacian operator in spherical coordinates when operating on D(r) gives a constant squared times D(r) for r > R.

    Thanks for any help!
  8. Dec 21, 2009 #7
    For a massless field use the function

    D(r) = A*exp[-m*r]/r for r > or = R

    and set m = 0
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