If an electron falls into a black hole. How can the Heisenberg uncertainty principle hold? The electron has fallen into the singularity now so it has a well defined position which means that it doesn't have a well defined momentum? Furthermore, the electron can't have a well defined position in space because space eigenkets are unphysical. It's momentum must certainly cause it to move.

Another question, Can one calculate the amount of new mass(the relativistic mass) that the black hole acquire after quantum particles fall in the singularity?

Doesn't this mean that the electron can't be described by a wave packet at the singularity of the black hole? If we want quantum mechanics to be applicable inside a black hole,the wavefunction should leak outside right ?

  • $\begingroup$ I think you are assuming that an outside observer can see past the event horizon and that the observer can measure the position of the singularity precisely. $\endgroup$ – RedGrittyBrick Nov 25 '12 at 10:14
  • $\begingroup$ No , I'm not assuming that an outside observer can figure out what's going on inside the black hole . I just want to understand physics inside the event horizon that has nothing to do with the fact that anything inside can't be known to an outside observer . But , It seems that this is not true as in QM , the Wavefunction leaks outside so maybe information can leak to an outside observer $\endgroup$ – nabil Nov 25 '12 at 10:26
  • $\begingroup$ I'm assuming that the position of singularity can be known with arbitrary precision. $\endgroup$ – nabil Nov 25 '12 at 10:27
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    $\begingroup$ There is no it. The electron isn't there, it ceases to exist at the spacetime singularity. $\endgroup$ – Alfred Centauri Nov 25 '12 at 12:43
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    $\begingroup$ I should also say that trying to apply non-relativistic quantum mechanics in the neighborhood of a black hole is never going to work successfully. The difference in time directions in the neighborhood of a horizon gives rise to different notions of particle, and an electron field state near the horizon will not necessarily be seen as a pure single-particle state far from the horizon. non-relativistic QM is not set up to deal with these distinctions. In a curved background, you are stuck with QFT. $\endgroup$ – Jerry Schirmer Nov 26 '12 at 18:18

I guess you have in mind something like a Schwarzschild black hole? An electron falls into the event horizon, and the laws of general relativity deem that it will eventually make its way to the singularity. It's important to remember that the theorems which predict the existence of the singularity are predicated on classical i.e non-quantum general relativity. When Planck-scale quantum effects are modelled, it may not be appropriate to talk of singularities any more. There is no general agreement on how to handle this at the present time.

However, even if the Schwarzschild singularity does exist in the form predicted by GR, it is difficult to talk about application of the uncertainty principle there. The problem is that, at the singularity, time ends - the time experienced by the infalling electron comes to a stop. Lack of ability to perform time derivatives makes it rather problematic to talk of the electron's momentum at the singularity.


You take a fully classical object (pointlike black hole with definite position and momentum) and you make it interact with a fully quantum object (pointlike particle described by a wave function) at "0 distance" (in the singularity) and you are surprised that your reasoning does not make too much sense?

Well, this is a bit like putting a lion and a tiger in the same cage and expecting them to friendly scratch each other's back.

I am just surprised that people even try to argue with you by talking about external observers, event horizons, ect.

The point is this: either you describe the whole setup classically (classical black hole with a classical infalling particle ) and you get an unphysical , but formally consistent result, or you describe the whole thing quantum mechanically (quantum mechanical black hole interacting with quantum mechanical particle) and see what results you get. Unfortunately, quantum gravity has not been fully "discovered/explained" yet, so probably nobody can fully describe a quantum black hole very enough to give meaningful answers to your question.


To take a haphazard approach.

Let us begin with the idea of conservation ultimately information must be conserved without respect to any perticular measurements this limits what can occur to your "electron".

If I were to watch this electron hit the eh it would stop and any measurement I try to obtain would be significantly shifted if not pulled in itself but, for the sake of your scenario lets assume we receive a measurement at the moment of this measure ocuring you would be applying a force to the electron thereby altering its momentum. To elaborate the exact position could technically be specific but the momentum of course would be obscured (regardless of any bh/singularity present.

Now to the heart of the issue this electron would I believe we would all agree pass through the eh and proceed toward the inner singularity. Now things get obscured in a classical sense the electron can not exist as it does not "appear" to experiance time in a linear sense. To continue despite this we would soon realize the singularity itself does not possession space-like coordinates which will prevent any measure of location.

Momentum: logically to me at least this should be measurable but without physical markers ie coordinates to establish path or trajectory momentum becomes increasingly arbitrary you would also need to contend with the fact that time Osborne longer linear so any measurement would not have a reference to any other measurement and could not be repeated or confirmed with prediction.


One thing to remember is that the momentum used in the Heisenberg uncertainty relation, $\Delta q \, \Delta p \geq \hbar/2$, should be conjugate to the coordinates used.

In the Schwarzschild solution, $$d\tau^2 = \left(1-\frac{r_s}{r}\right) dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 d\Omega^2$$ with Lagrangian $$L = \sqrt{\left(1-\frac{r_s}{r}\right) \dot{t}^2 - \left(1-\frac{r_s}{r}\right)^{-1} \dot{r}^2 - r^2 \dot{\Omega}^2}$$ the momentum conjugate to $r$ is $$p_r = \frac{\partial L}{\partial \dot{r}} = \frac{1}{L}\left(1-\frac{r_s}{r}\right)^{-1}\dot{r}$$

Thus, the uncertainty relation in this case reads $$\Delta r \, \Delta \left(\frac{1}{L}\left(1-\frac{r_s}{r}\right)^{-1}\dot{r}\right) \geq \frac{\hbar}{2}$$

Even if we try to change to "Euclidean coordinates" (whatever that means in a curved space), we will have a complicated uncertainty relation that includes effects close to the event horizon.


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