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Retarded Green Function in Curved Spacetime

  1. Sep 20, 2012 #1
    Hello,

    Can anyone explain to me the next settence, found here in section 1.4:

    "The causal structure of the Green’s functions is richer in curved spacetime: While in flat spacetime the retarded Green’s function has support only on the future light cone of [itex]x'[/itex], in curved spacetime its support extends inside the light cone as well;[itex]G^{\alpha}_{+\beta '}[/itex] is therefore nonzero when [itex]x\in I^{+}(x')[/itex], which denotes the chronological future of [itex]x'[/itex]. This property reflects the fact that in curved spacetime, electromagnetic waves propagate not just at the speed of light, but at all speeds smaller than or equal to the speed of light; the delay is caused by an interaction between the radiation and the spacetime curvature. A direct implication of this property is that the retarded potential at [itex]x[/itex] is now generated by the point charge during its entire history prior to the retarded time [itex]u[/itex] associated with [itex]x[/itex]: the potential depends on the particle’s state of motion for all times [itex]\tau \leq u[/itex]."

    I do understand the implications of such but why is there a tail term in the Green function?why do "electromagnetic waves propagate not just at the speed of light, but at all speeds smaller than or equal to the speed of light; the delay is caused by an interaction between the radiation and the spacetime curvature."

    Thank you
     
  2. jcsd
  3. Sep 20, 2012 #2

    Ben Niehoff

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    That's an odd, convoluted way of putting it.

    There is nothing too unusual about the Green function being nonzero inside the lightcone. This even happens in flat space in 2+1, 4+1, 6+1, etc. dimensions (cf. ripples on a 2-d surface, like a pond).

    I disagree with the paragraph's assertion that spacetime curvature alters the speed of a wave. The speed of propagation is still c, but the shape of the wave may change, including new wavefronts developing inside the lightcone.
     
  4. Sep 20, 2012 #3

    Bill_K

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    Yes, but please don't make it sound like ripples on a pond have anything to do with the wave equation! Water waves derive from a solution of an elliptic equation, in fact the three-dimensional Laplace's equation. And they have no "light cone" to be inside of - the maximum velocity of propagation of water waves is infinite.
     
  5. Sep 20, 2012 #4
    I'm still not following...Why is it possible in curved spacetime (or 2+1,etc dimensions) that a perturbation that starts at a point [itex]x'[/itex], and that perturbation travels at the speed of light, to influence a point [itex]x \in I^{+}(x')[/itex] (but [itex]x \notin J^{+}(x')\backslash I^{+}(x')[/itex])?

    Is there a good reference from where I can learn this?
     
    Last edited: Sep 20, 2012
  6. Sep 20, 2012 #5

    Bill_K

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    Could be wrong, but it looks to me from Eq (1.13) that the effect on the electromagnetic Green's function is due solely to Ricci curvature. That is, it doesn't happen in empty space.

    On the other hand, the corresponding equation (16.4) for the gravitational Green's function contains the Riemann tensor. In both cases, the equation is no longer just the wave equation, it's more like the Klein-Gordon equation, i.e. dispersive. Discontinuities will still propagate with velocity c, but low frequency waves will not, so that wave packets will tend to develop a tail.
     
  7. Sep 20, 2012 #6
    So after given some thought about it, what Bill_K said was that the wave changes form developing a tail due to the curvature and instead of a well localized pulse one gets more like a barrage of pulses, an afterglow. Well the final question is why?Why does curvature alongates a pulse...?


    From Ben Niehoff post I remembered that I've read a few months ago something about de Huygens principle and how it only applies to spacetimes with odd spatial dimensions.

    I've found then a good reference about the Huygen principle: here.

    It does explain the fact that at each point of the wave front acts as a source for another wave and for even spatial dimensions those wavelets are no longer a pulse in time but more like a wave that gradually looses intensity... It even talks about the same principle might be applied to curved spacetimes in which the curvature does have the same efect...

    The problem is that the proofs in this article are mathematical and it doesn't explain what's the mechanism behind this...So why only even spatial dimensions develop the tail?
     
  8. Sep 21, 2012 #7

    Ben Niehoff

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    Do you have a reference for that? I don't see how Laplace's equation can allow traveling wave solutions.
     
  9. Sep 21, 2012 #8

    Bill_K

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    The definitive book on the subject of water waves is by J J Stoker, and is available online here.

    In a nutshell, the motion of the fluid is irrotational, which implies that v is derivable from a velocity potential, v = ∇φ. Also incompressible,∇·v = 0, which implies that ∇2φ = 0.

    For small amplitude waves in water of infinite depth, the boundary condition at the surface z = 0 is φtt + gφz = 0, leading to wave solutions of the form φ = exp(ikx + kz - iωt) with ω2 = gk.
     
    Last edited: Sep 21, 2012
  10. Sep 21, 2012 #9

    Ben Niehoff

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    The fact that ikx and iωt have opposite signs inside the exponential would indicate that the full equations, including time, are hyperbolic. The spacelike part of the wave equation is of course Laplace's equation (or Helmholtz's equation).

    The Boussinesq equation for water waves is hyperbolic, for example:

    http://en.wikipedia.org/wiki/Boussinesq_approximation_(water_waves)
     
  11. Sep 21, 2012 #10

    Bill_K

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    Actually, doesn't it just indicate whether the wave is traveling to the left or right? In fact I originally had written +iωt and went back and changed it. :smile:

    Laplace's equation is the full equation, not an approximation to something hyperbolic. The velocity potential φ(x, t) is a solution of ∇2φ = 0, with arbitrary time dependence! Only the boundary condition constrains the time dependence.

    Regarding the boundary condition, there are approximation schemes for the two opposite cases: deep water and shallow water. ("Deep" and "shallow" means compared to the wavelength.) For deep water, the boundary condition is linear, and the dispersion formula I gave, ω2 = gk, applies to the simplest case of infinite depth. For shallow water (equivalently, "long waves"), the boundary condition is nonlinear. Boussinesq is an example of this. It's more complex, but also more interesting because the solutions include solitons.
     
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