Discussion:
SEKI
2018-04-30 13:39:35 UTC
Hawking radiation postulates that particles with negative energy
fall into a black hole.

Einstein's gravitational equation is presented in terms of linear
expression of energy-momentum tensor.
So, I suppose that particles with negative energy, if actually
present, are to be repelled by the black hole.

SEKI
Thomas 'PointedEars' Lahn
2018-06-11 12:37:52 UTC
Post by SEKI
Hawking radiation postulates that particles with negative energy
fall into a black hole.
That is only *one* possible, the *alternative* conceptualization of Hawking
radiation (HR), and it only applies for an observer far away from the event
horizon (EH) of the black hole (BH).

Usually HR is *locally* conceptualized as the result of pair production (PP)
from vacuum fluctuations (VF) near but outside the EH, where due to the BH
the virtual particles become real, one particle falls beyond the EH (“into
the black hole”) and the other one does not. So considering that the vacuum
energy from which the pair was created formerly contributed to the energy
density (ρ = dE∕dV) of the BH, the BH loses mass–energy (density) equivalent
to that of the escaping particle’s total energy (E) (density). As the
(assumed non-rotating, uncharged) BH’s Schwarzschild radius is rₛ = 2 G M∕c²
and E₀ = M c², if dE₀∕dt < 0, then dM∕dt < 0 → drₛ∕dt < 0: the BH shrinks.

It can also be conceptualized as PP from VF beyond the EH where one of the
particles tunnels through the EH (see “quantum tunneling effect”); the same
as above applies.

Post by SEKI
Einstein's gravitational equation is presented in terms of linear
expression of energy-momentum tensor.
That is an oversimplification. Instead, the _Einstein field equation*s*_
(EFEs) state that the curvature in a region of spacetime (ST) is equivalent
to the density and flux of energy and momentum in that region.

The EFEs are:

G_µν ≡ R_µν − (½ R − Λ) g_µν = 8πG∕c⁴ T_µν

“ST curvature” = “energy–momentum”

G_µν – Einstein tensor
R_µν – Ricci curvature tensor
R – Ricci curvature scalar R = tr_g R_µν
Λ – cosmological constant
g_µν – metric tensor
G – Newton’s gravitational constant
c − speed of light in vacuum
T_µν – (stress–)energy–momentum tensor

<https://en.wikipedia.org/wiki/Einstein_field_equations>
Post by SEKI
So, I suppose that particles with negative energy, if actually
present, are to be repelled by the black hole.
No, I do not think so. Contrary to popular belief, a BH is _not_ a massive
everything, but a *region* of *ST* where the ST curvature (STC) is that
large that “space falls faster than light” (Andrew J.S. Hamilton, see also ).

The total STC in the ST neighborhood of the negative-energy particle (NEP)
is the result of the STC equivalent to the energy–momentum of the NEP *and*
that of the BH *combined*. The energy−momentum density and flux, and
therefore the STC in the BH’s singularity is *infinitely* *large* (that is
why it is a *singularity*: all spacetime geodesics *end* there).

That is, even if the ρ_NEP of the NEP is *a little* negative, the ρ_BH in
the region beyond the EH is *infinitely more positive*. So the total scalar
STC where the NEP is is very much *positive*, and *all* geodesics for the
NEP *still* lead *towards* the singularity, “*into* the BH”.

[That is also why the evaporation of large BHs due to HR would take a
very long time, and does not even happen if the BH is large enough and
there are enough positive–energy particles in the vicinity to fall into
it: HR would still be emitted, but in total the BH will still *stay the
same* or *grow*.]

IOW, for us distant observers to observe the NEP to be repelled by the BH,
i.e. for the total scalar STC to be negative, the absolute of its negative
ρ_NEP would have to be *greater than* the absolute of the (positive) energy
density of the BH:

|ρ_NEP| > |ρ_BH|

IOW, the NEP would have to be a white hole that is greater than the BH.
Obviously, that is NOT so.

If that helps, you can think of the NEP near the EH like a canoe near a
waterfall: Sitting in the canoe, you can use all your strength (negative
energy) paddling against the downward current, but eventually you will still
go down the waterfall because the speed relative to the abyss that you can
achieve by paddling against the current is still less than the speed of the
downward current:

At the EH, space is falling at the speed of light in vacuum, c. A NEP
with non-zero mass m cannot even move away *at* c, much less *more* than that:

T = E − E₀ = m c² (1∕√(1 − u²∕c²) − 1)

lim T = +∞,
|u|→c

where T is the particle’s local kinetic energy and u is its local relative
(3-)velocity.

Post by SEKI
Yes, it is NOT. You are proceeding from false assumptions.

However, note that as long as we have no theory of quantum gravity¹, HR is
really only a semi-classical hypothesis: it is based on the *assumption*
that the laws of quantum mechanics apply *unchanged* in curved ST. AFAIK,
to date there is no evidence of HR, no evidence of evaporating BHs.

You should have asked this in <news:sci.physics.relativity> as it pertains
more to BHs and general relativity (GR) than particle physics.

_______
¹ <https://en.wikipedia.org/wiki/Quantum_gravity>
--
PointedEars