Electron falling in the Kerr metric

[jfy4]

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 (1-1/\sqrt{3})\approx 42\%".

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?

[bcrowell]

I would guess that they're just expressing the loss of potential energy as a fraction of the rest energy.

[jfy4]

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 \mathbf{e} is the energy measured from infinity per unit rest mass, the binding energy per unit rest mass is \mathbf{(1-e)}. 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.

?