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Naked Singularity ?

  1. Apr 15, 2007 #1
    Naked Singularity .... ?????


    How the Scaler Field collapse, Dust collapse and spherical gravitational collapse provide the possibility of existance of NAKED SINGULARITY?

    Also can any one explain me about this.....
    "In known examples of naked singularities for dust and perfect fluids, the redshift along outgoing geodesics emerging from the singularity is found to be infinite (when calculated for observers in the vacuum region). This could be interpreted to mean that no \information" is being transmitted from the naked singularity and could be yet another approach to preserving censorship. "
  2. jcsd
  3. Mar 5, 2011 #2
    Re: Naked Singularity .... ?????

    For the benefit of the non-scientist; how does an infinite redshift manifest itself?
    Last edited: Mar 5, 2011
  4. Mar 5, 2011 #3
    Re: Naked Singularity .... ?????

    Since the redshift of light emitted by an object is one plus the ratio of the change in wavelength divided by the wavelength at which it was emitted, infinite redshift requires that for any finite value of the emitted wavelength, the wavelength of the light received at infinity is infinitely long. This means that no light (or anything else emitted from the naked singularity) could escape (to infinity).
  5. Mar 5, 2011 #4
    Re: Naked Singularity .... ?????

    But it could escape to any distance short of infinity?

    How would that differ from saying it could escape to any distance, since it could never reach infinity?
  6. Mar 6, 2011 #5
    Re: Naked Singularity .... ?????

    The energy [tex]E[/tex] (not including relativistic effects) due to gravity and motion of an object (I will call "O", because it looks like a 2-d planet.) with a mass, [tex]m[/tex], moving at a velocity, [tex]v[/tex], with respect to a much heavier object (I will call "S", for Sun) of mass, [tex]M[/tex], a distance, [tex]r[/tex], away is (approximately): [tex]\frac{1}{2}mv^{2}-\frac{GM}{r}[/tex] where [tex]G[/tex] is the universal gravitational constant. If [tex]E[/tex] is greater than zero, then O has a hyperbolic trajectory, it will come near S once and never return. All of the properties of these objects are constant except for their relative velocity, [tex]v[/tex], and the distance between them, [tex]r[/tex]. So, one can solve for the velocity of O in terms of a bunch of constants and the distance between the two objects: [tex]v=\sqrt{\frac{2E}{m}+\frac{2GM}{mr}}[/tex]. Now, take the limit as [tex]r[/tex] goes to infinity. If [tex]E[/tex] is positive or zero there are no problems, the second term under the square root gets smaller and smaller until in the limit, it goes to zero. The value of [tex]v[/tex] in this limit tells you something about the shape of the trajectory O has. If the value is zero, the shape is a parabola, if it is greater than zero the shape is a hyperbola. If the value under the square root becomes negative (which can only happen (when the other constants are positive) if [tex]E[/tex] is negative), then O cannot escape S and will forever orbit it. The case is slightly different for redshift and light, since the local speed of light is constant. What happens instead is that the energy/frequency of the light decreases (the kinetic energy of light is not given by the same equation as I gave above for O, instead it is proportional to the frequency of the light, so the frequency decreases in order for [tex]E[/tex] to remain constant as [tex]r[/tex] increases). So, similar situations to those for O can happen for light. Instead of velocity the frequency of the light at a given distance can be solved for: [tex]E=h\nu{}-\frac{GM}{r}\Rightarrow{}\nu{}=\frac{E}{h}+\frac{GM}{hr}[/tex], where [tex]h[/tex] is planck's constant. So, similarly, if [tex]E[/tex] is positive, the light escapes S since as [tex]r[/tex] goes to infinity [tex]\nu{}[/tex] approaches a positive value. If [tex]E=0[/tex], then [tex]\nu{}[/tex] approaches zero as [tex]r[/tex] goes to infinity. This is the case of infinite redshift. What it means is that the farther one is from the naked singularity, the lower the frequency/energy of light one receives for a given frequency of light emitted.

    So, theoretically, yes light could escape to any distance short of infinity. However, practically, if an instrument that can detect given range of frequencies is placed far enough from the singularity that emits a given (non-infinite) range of frequencies of light, it will not be able to detect anything.
    Last edited: Mar 6, 2011
  7. Mar 6, 2011 #6
    Re: Naked Singularity .... ?????

    Thanks, IsometricPion. I would ask, wouldn't I!!! Now I shall have to try to find time to make sense of those equations. Its a good thing you added the last 3 lines, at least I shall know what I'm aiming for. :smile:
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