Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

With Hawking radiation, one half of virtual pair falls into horizon and this particle has negative energy.

What would an observer inside horizon observe when seeing negative particles ? How do these negative particles interact with ordinary matter ?

share|cite|improve this question
up vote 5 down vote accepted

The Schwarzschild solution for a neutral black hole is $$ c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right) $$ You see that the terms $dt^2$ and $dr^2$ are multiplied by $(1-r_s/r)$ or its inverse where $r$ is the radial coordinate and $r_s$ is a constant, the Schwarzschild radius.

An important subtlety of this $(1-r_s/r)$ is that it becomes negative for $r_s\lt r$; this is true for any black hole beneath its event horizon. That's why in the black hole interior, the changes of the coordinate $r$ are actually timelike and the changes of the coordinate $t$ are spacelike; the role of the space and time are interchanged in the interior relatively to the exterior!

When we say that the outgoing/infalling particles from the pair have positive/negative energy, we are talking about the energy that generates translations of the $t$ coordinate. However, the $t$ coordinate is really spacelike in the black hole interior so the $t$-component of the energy-momentum vector is interpreted as a spatial component of the energy-momentum vector by observers inside.

It means that from the internal observers' viewpoint who are capable of observing the infalling particle (and not all of them can!), it is just an ordinary particle with some value of the momentum $p_t$, which is a spatial component and may unsurprisingly be both positive and negative. There is certainly no "new kind of matter" occuring in general relativity right beneath the horizon. All the local physics obeys the same laws as it does outside the black hole.

This general assertion is a special example of the fact that the event horizon is a coordinate singularity, not an actual singularity. When one choose more appropriate, Minkowski-like coordinates for the region of the spacetime near the horizon, it looks almost flat – as seen from the fact that the Riemann curvature tensor has very small values, at least if the black hole is large enough. So an observer crossing the event horizon of a large enough black hole doesn't feel anything special at all. His life continues for some time before he approaches the singularity and this is where the curvature becomes intense and where he's inevitably killed by extreme phenomena. But life may continue fine near the horizon and even beneath it.

Note that the energy is the only conserved component of the energy-momentum vector on the Schwarzschild background – because this solution is time-translationally invariant but surely not space-translationally invariant. Due to the intense curvature caused by black hole, one must be careful and not interpret individual components of vectors "directly physically". We saw an example that what looks like a temporal component of a vector, namely energy, from the viewpoint of the observer at infinity, can really be a competely different, spatial, component from the viewpoint of coordinates appropriate for a different observer, one who is inside.

share|cite|improve this answer

Let me try to explain first what is understood in this context by a negative energy particle. Then we will see that it behaves exactly as a normal particle, and in fact it is normal.

When the pair of virtual particles is created, its total energy is $0$. If one of the particles moves faster, its kinetic energy is larger, and that particle is the one having positive energy (relative to the event horizon). The potential energies are initially the same, since the particles are created at the same position. The potential energy is negative for both particles, since you have to give energy to the particles to move them far from the black hole.

The only difference between the negative energy particle and the positive energy one is that the negative energy particle has small velocity (w.r.t. the event horizon), so that its rest plus kinetic energy is smaller than the modulus of its potential energy. If the positive energy particle moves fast enough, it may escape the black hole, but the negative energy particle will fall in the black hole.

According to the principle of equivalence, it doesn't matter if the reference frame is near the event horizon of a large black hole, or far from any strong gravitational field. Locally, the physics near the event horizon, and even inside the black hole, is the same as anywhere. So, the fact that the particle is very close to the source of gravity doesn'e make it behave differently. The fact that it has smaller kinetic energy also doesn't matter, since the kinetic energy depends on the reference frame. So, nothing will make the negative energy particle behave differently than a normal particle. In particular, it interacts like a normal particle.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.