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Black hole evaporation

  1. May 12, 2008 #1
    in classical electromagnetic a charge can make energy. Now consider Howking radiation effect: if an anti-particle fall into the BH the volume of BH will reduce. But if that particle be a charged particle, this charge can make positive energy and increase the volume. Is that right?
  2. jcsd
  3. May 12, 2008 #2
    I am not sure what you are refering to, or if what you read confused you. An anti-particle has a positive mass. If an anti-particle falls into a BH, the mass of the BH increases.

    Independently of this, in Hawking's first calculations, one can interpret an anti-particle from the fluctuation of a pair as the one carrying negative energy (by definition), the particle being the one carrying positive energy (and this can be a positron BTW). This split near the horizon makes it possible for the particle to escape at infinity, whereas the antiparticle must keep falling. From an outside observer, the BH has emitted a particle. The particle-antiparticle definition is almost irrelevant in this context, except for the amount of energy carried and the possibility to escape at infinity, since most of those pairs are photons. Strictly speaking once again, a black hole will emit as much positrons as electrons (provided it is not charged. In fact, once it has emitted one electron, it is more likely to emit a positron). You can not go further in the interpretation : the process occurs at the quantum level and only the outside observer interpretation is a valid one that can be measured. Strictly speaking, there is no such thing as a negative-energy real (anti-)particle. What is observed is the emission of thermal radiation, not negative-energy particles.

    IMO, this discussion is not BtSM.
    Last edited: May 12, 2008
  4. May 12, 2008 #3


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    Staff Emeritus
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    I agree. I'm moving to General Physics: Doc Al or Zz may find a more suitable home for it.
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