In quantum field theory in flat spacetime, there are both positive and negative frequency solutions to the classical field equations, but upon quantization we get only positive energy particles. But in Hawking's original paper about Hawking radiation, it is stated that negative energy particles can exist inside a black hole:

Just outside the event horizon there will be virtual pairs of particles, one with negative energy and one with positive energy. The negative particle is in a region which is classically forbidden but it can tunnel through the event horizon to the region inside the black hole where the Killing vector which represents time translations is spacelike. In this region the particle can exist as a real particle with a timelike momentum vector even though its energy relative to infinity as measured by the time translation Killing vector is negative. The other particle of the pair, having a positive energy, can escape to infinity where it constitutes a part of the thermal emission

That is, Hawking is saying that the energy of a particle can be defined as $$E = p^\mu K_\mu$$ where $p^\mu$ is its four-momentum and $K^\mu$ is the time translation Killing vector. I can see how this works in Minkowski space, where $K = \partial_t$ and we get $E = p^0$ as expected.

But why is this the correct definition of energy? What kind of observer would measure $E$ to be the energy of the particle? Can this quantity be shown to be conserved? Why should we trust this equation when $K^\mu$ is not even timelike inside the black hole?

  • $\begingroup$ I don't think my answer was very good, so I'm going to delete it. To allow for a good answer, could we edit the question a little bit, and maybe split it up? It seems to me that you have three things going on here: (1) semiclassical gravity and Hawking radiation, (2) interpretation of energy for test particles in GR, and (3) interpretation of that energy inside the event horizon. It seems to me that 1 is peripheral to the question, while 2 and 3 are really separate questions. Could we edit this question to address either 2 or 3, and maybe start a different question for the other one? $\endgroup$
    – user4552
    Apr 10, 2018 at 16:52
  • $\begingroup$ If I don't hear back from knzhou and if I get some time within the next few days, I will write up a separate question more specifically on topic 2 and self-answer it. (Or we may already have a question on this topic -- I haven't searched yet.) $\endgroup$
    – user4552
    Apr 11, 2018 at 15:25
  • $\begingroup$ @BenCrowell Sorry, I’m on vacation now. I’d be happen with an answer to this question only about (2) or (3), and I really think they’re closely linked. Feel free to start other questions though! If you do, could you comment here so I can find them? $\endgroup$
    – knzhou
    Apr 12, 2018 at 13:12
  • $\begingroup$ related: physics.stackexchange.com/questions/218121/… $\endgroup$
    – user4552
    Apr 12, 2018 at 21:08
  • 1
    $\begingroup$ It seems to me that the various aspects of this question are now pretty thoroughly covered in gj255's answer to this question, as well as the answers to physics.stackexchange.com/questions/218121/… by Valter Moretti and me. $\endgroup$
    – user4552
    Apr 13, 2018 at 3:32

2 Answers 2


As far as I know, any observer (inertial or otherwise) with 4-velocity $U^\mu$ will measure the energy of a particle with 4-momentum $p^\mu$ to be $U^\mu p_\mu$. Since no observers have spacelike 4-velocities, no observer would measure the energy of a particle to be $K^\mu p_\mu$ inside the black hole. Trying to define the energy 'as measured at infinity' strikes me as dangerous: we can only measure local quantities in GR, so the only way to talk about an observer measuring something far away is to imagine some signal travelling between events. But this cannot happen if our particle is behind the horizon.

What I believe Hawking is trying to say is that it’s OK for the quantity $K^\mu p_\mu$ to be negative, precisely because it isn't energy. There is a brief moment, as the particle approaches the horizon, during which the particle has negative energy, which would be problematic were it not for the brevity of the moment (forgive the hand-waving). But once inside the black hole, this quantity doesn't correspond to energy, and so we no longer have a problem. Indeed, a typical observer inside the black hole would have 4-velocity something like $-\partial/\partial r$, and so would measure the particle's energy as $-p_r$, which is positive, since $p^r$ is negative.

Note: the quantity $K^\mu p_\mu$ for any Killing vector $K^\mu$ is conserved along the particle's geodesic.

I assume the mostly minus metric convention.


In addition to gj255 answer here is a visual aid, a light-cone diagram in Schwarzschild coordinates:modified image from arXiv:1401.1797

Schwarzschild coordinates here are convenient because timelike at infinity Killing vector field is especially simple here: $\partial_t$, so in the diagram it is a simple vertical translation.

So the contraction $E=p^\mu K_\mu$ in Schwarzschild coordinates is a t-component of 4-momentum (vertical component in the image). Since 4-momentum must lie within the future light-cone at a given point, outside the horizon only positive energies are possible (purple vectors in the image).

Inside the horizon, however, we may also have negative energies (as well as positive), since light-cones are now pointing sideways. So in image we have green vectors, representing negative energies.

But why is this the correct definition of energy?

This is a global definition of conserved quantity which we could identify with 'energy' because at spatial infinity the vector field $K_\mu$ is timelike. It is not the same as energy measured by a local observer, and inside the horizon there are (as you already guessed) no observers that would move along $K_\mu$.

Can this quantity be shown to be conserved?

Look at the answers to this question by Valter Moretti and Ben Crowell.


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