Electron Falling in Kerr Metric: Release of 40% Rest Energy?

  • #1
jfy4
649
3
I have here a quote from Hartle's Gravity, page 321:

"The fraction of rest energy that can be released in making a transition from an unbound orbit far from an extremal black hole to the most bound innermost stable circular orbit is [itex](1-1/\sqrt{3})\approx 42\%[/itex]".

My question is about releasing an electron into a black hole in precisely this fashion. An electron is a fundamental particle, hence, cannot decay into other fundamental particles. Then how is the electron losing about 40% of its rest energy during this process?
 
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  • #2
I would guess that they're just expressing the loss of potential energy as a fraction of the rest energy.
 
  • #3
bcrowell said:
I would guess that they're just expressing the loss of potential energy as a fraction of the rest energy.

After re-reading over and over again, that seems right. This appears to be the key in that paragraph:

"The binding energy of any orbit is the difference between the energy of a particle at rest at infinity (including rest energy) and the energy of the same particle moving the orbit as measured from infinity. Since [itex]\mathbf{e}[/itex] is the energy measured from infinity per unit rest mass, the binding energy per unit rest mass is [itex]\mathbf{(1-e)}[/itex]. This is the fraction of rest energy that can be released in the process of gravitational binding".

So another way to say this is:

This is the amount of binding energy expressed as a fraction of the rest energy that can be released in the process of gravitational binding.

?
 
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