Black Hole "Traffic Jam" in GM Sgr
Jim Graber
I quote from http://science.nasa.gov/newhome/headlines/ast22sep99_1.htm
It says in part:
"In either case, the X-ray outburst is powered by
an incredibly fast fall down the steep gravitational
field of the compact body. At the end, matter from
the primary body either slams into a neutron star's
tougher-than-diamond surface, or is super-heated
in an incredible traffic jam at the event horizon
before it disappears into a black hole. The kinetic
energy from the superhot matter is turned into
X-rays."
Apparently GM Sgr could be a neutron star, but let's assume it's a
black hole. My question is "Where is this traffic jam?"
Is it really at the event horizon? If so wouldn't the Xrays
be redshifted to nothing?
So let me make my question a little more precise, and add a few
related questions:
If a blob of matter falls into a black hole, what happens to its
density? Its temperature? At what point is its radiation at a maximum?
TIA Jim Graber
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Share what you know. Learn what you don't.
J. Scott Miller
>
> Even more intriguing, after a singularity, what then? Can a singularity go
> on absorbing matter and energy ad infinitum? Or does it change? Into what?
>
> tim
Classical black hole theory indicates the singularity simply accumulates all
mass entering the event horizon. As the mass increases, the event horizon
moves farther from the singularity. With each solar mass of material it
absorbs in this way, the event horizon moves out 3 kilometers. The real trick
over long periods of time is whether the incoming matter can offset mass loss
via the Hawking radiation process. Left unchecked, this could eventually
cause the black hole to evaporate in a flurry of gamma rays.
--
J. Scott Miller, Program Coordinator Scott....@louisville.edu
Gheens Science Center and Rauch Planetarium
http://www.louisville.edu/planetarium
University of Louisville
John Baez
J. Scott Miller <Scott....@louisville.edu> wrote:
>The real trick
>over long periods of time is whether the incoming matter can offset mass loss
>via the Hawking radiation process. Left unchecked, this could eventually
>cause the black hole to evaporate in a flurry of gamma rays.
I hope people understand that when you say a "long time" you mean a
REALLY long time, at least dozens of orders of magnitude more than the
current age of the universe! I forget the exact figure - the point
is, Hawking radiation from a large black hole is predicted to be
incredibly feeble.
I was waiting for a real expert to answer the original question, but
since none has yet, let me put in my 2 cents:
I'd guess the reported "traffic jam" near this black hole is an
"accretion disk". We don't usually expect gas from nearby stars
to fall straight into a black hole, because it has angular momentum.
Instead, it should slowly spiral in, a bit like water going down the
bathtub drain. People have studied this sort of thing and found that
as it spirals in, the gas forms a flat spinning disk and get extremely hot,
emitting X-rays and the like. Jets of plasma may also shoot out from
the north and south poles of this accretion disk. In fact, huge jets
of this type have been observed coming out of the centers of many
galaxies. That's one reason most astrophysicists think these galaxies
have black holes at their centers. In some cases, we can also see a
small central object.
There's an object at the center of our galaxy called Sgr A*, which
is widely believed to be a black hole, but as far as I know it
doesn't have big jets of plasma coming out of it. I don't know
if this object is related to GM Sgr - other than the fact that they're
both in Sagittarius. Does anyone out there know?
Tom Roberts
> [...] The real trick
> over long periods of time is whether the incoming matter can offset mass loss
> via the Hawking radiation process. Left unchecked, this could eventually
> cause the black hole to evaporate in a flurry of gamma rays.
For a solar-mass black hole, the Hawking temperature is (IIRC) only
micro-degrees Kelvin, and it would take many tens or hundreds of
billions of years for such a black hole to radiate away its mass
down to 0, ASSUMING NOTHING fell into it. But as its temperature is
significantly below the CMBR temperature, that is a bad assumption
(CMBR radiation will fall in at a faster rate than Hawking radiation
will leave). And it is expected that such black holes will usually be
located near dense galactic centers, so additional matter will also
be infalling, almost surely at a rate FAR exceeding the CMBR infall.
Tom Roberts tjro...@lucent.com
Jim Graber
>
> There's an object at the center of our galaxy called Sgr A*, which
> is widely believed to be a black hole, but as far as I know it
> doesn't have big jets of plasma coming out of it. I don't know
> if this object is related to GM Sgr - other than the fact that they're
> both in Sagittarius. Does anyone out there know?
As far as I know, they're unrelated. GM Sgr is a little guy, maybe even
a neutron star. Jim
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Before you buy.
John Baez
Gerry Quinn <ger...@indigo.ie> wrote:
>There is no such thing as a singularity. General relativity may
>confidently be expected to represent a particular mathematical
>formalisation of the classical limit of a (so far unformulated) quantum
>gravity theory. Therefore, it will break down at the Planck scale (if
>not before).
This is an excellent point, but if you don't mind a pedantic addendum,
we don't know *for sure* that there's no such thing as a singularity.
Everyone hates infinities, including me, so everybody hopes that quantum
gravity or something will get rid of the singularities predicted by
general relativity. There are a lot of annoying things about these
singularities. However, we don't know any *logical inconsistency* that
would be caused by their existence, so it is ultimately an empirical
question whether they exist or not.
Of course, there are lots of impressive philosophical arguments
proving that a good physical theory cannot have singularities, some
going back all the way to the Greeks, but most of them boil down
to: "infinity sucks!"
Interestingly, in many toy models of quantum cosmology that have
been studied so far - e.g., minisuperspace models of Bianchi IX
cosmologies - quantum effects do *not* eliminate the big bang
singularity. Of course, these toy models are a far cry from a
full-fledged theory of quantum gravity. However, they have slightly
lessened our confidence that quantum gravity will cure the singularities
of general relativity: they have transformed this notion from a
universally accepted platitude into something worthy of study
and debate.
Keith Ramsay
writes:
|I forget the exact figure - the point
|is, Hawking radiation from a large black hole is predicted to be
|incredibly feeble.
According to Wald, (formula 14.3.8 of his textbook on GR) the
temperature of the radiation is 6*10^{-8} K for a black hole the mass
of the Sun, and inversely proportional to the mass. So if you want to
have any hope of evaporating a black hole of normal size, you have to
put it in a container cooled down to the nanokelvin level, at least
until the universe is itself that cold.
Keith Ramsay
Erik Max Francis
> For a solar-mass black hole, the Hawking temperature is (IIRC) only
> micro-degrees Kelvin, ...
~100 nK, actually.
> ... and it would take many tens or hundreds of
> billions of years for such a black hole to radiate away its mass
> down to 0, ASSUMING NOTHING fell into it.
~10^67 y, actually.
--
Erik Max Francis | icq 16063900 | whois mf303 | email m...@alcyone.com
Alcyone Systems | irc maxxon (efnet) | web http://www.alcyone.com/max/
San Jose, CA | languages en, eo | icbm 37 20 07 N 121 53 38 W
USA | Mon 1999 Sep 27 (59%/949) | &tSftDotIotE
__
/ \ Everyone wants to look good at his own funeral.
\__/ Louis Wu
(Greg Weeks)
: This is an excellent point, but if you don't mind a pedantic addendum,
: we don't know *for sure* that there's no such thing as a singularity.
The word "singularity" can be misleading, I think. What you actually have
is:
1. no upper bound on curvature (over space-time)
2. world-lines of finite proper length (duration)
Personally, I have a hard time imagining that #2 will ever go away. But if
it does, and if general relativity does hold at the macroscopic level, then
it seems that something *really interesting* must happen when you fall into
a black hole.
Greg
ba...@galaxy.ucr.edu
In article <wolfram-2609...@dialppp-7-69.rz.ruhr-uni-bochum.de>,
Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> wrote:
>So when Hawking radiation is active, even the infalling astronaut can not
>reach the horizon in a finite proper time, because in his proper time he
>should reach the black hole at the very moment it evaporates.
No, Hawking radiation does not prevent the freely falling astronaut
from reaching the event horizon in a finite amount of proper time -
in fact, for a largish black hole, the effect of the Hawking radiation
on the astronaut is negligible.
Joachim Draeger
I think at first we have to define what a singularity is, before we
start the discussion. Usually every breakdown of the strcuture of the
space time manifold is considered as singularity of space time. A
infinity is not needed for causing such a singularity. Consider as
example the space time of a cosmic string. There are other types of
singularities as well, e.g. naked singularities.
Perhaps even horizons can be considered as very mild forms of
singularities. The symmetry of neighbourhood is broken there. You can
reach point A from point B, but not point B from point A (e.g. event
horizon of a BH).
Wether we can get rid of singularities or at least special types of
them, depends on the underlying mathematical strcuture, too. There is
no problem with horizons if manifolds are used. If differential spaces
or similar strcutures are used instead of manifolds, even cosmic
strings can be handled without problems. The whole topic is connected
with the underlying mathematical framework.
BTW, I don't understand completely what is meant by #2 above. Do you
mean that world lines end because of 'holes' cut into the space-time
manifold because of infinities and so on? If this is so, it is a
relict of insufficient mathematical strcutures. You want to model
space time as manifold. Now there are some points (?) in space time,
which are inconsistent with the properties of a manifold. A naive
solution of this problem is very simple. You cut out these points, the
remaining space time is a manifold, and you are happy. Every infinity
or other singularity is contained in the boundary of the manifold, but
not in the manifold itself. This is sufficient for most
cases. However, if you want to explore the properties of the
singularities, you have to reconstruct them in one way or the
other. Of course, this is a nontrivial problem; for example, there is
no unique way of reconstruction, and in the general you will get many
different solutions for such a reconstruction. This is the main
reason, why so many generalizations of the manifold concept were
introduced in differential geometry. If it is possible to consider a
singularity as `normal` element of spacetime, there were no problems
of reconstructing the singularity anymore. One can do this, for
example, with the quasiregular singularities like cosmic
strings. However, because they are normal elements, no world line
crossing the singularities will end suddenly.
As a remark, of course you will have some crazy effects at the
singularity itself. Some models for cosmic strings predict a change
of space time dimension from 4 to 5, for example.
Joachim Draeger
John Baez
Daryl McCullough <da...@cogentex.com> wrote:
>My guess is that there is some critical radius (as measured using
>Schwarzchild coordinates) such that an astronaut falling from rest
>at a distance R_critical(M) towards a black hole of mass M will miss
>the singularity. If an astronaut starts falling from a distance
>closer than R_critical(M), then he will hit the singularity in
>finite coordinate time (as well as proper time).
This all sounds fine except for 2 things:
1) "finite coordinate time" doesn't mean anything until we describe
which coordinates we're using. Of course you address that point below,
but I have some qualms about what you say - more on that later.
2) By "the singularity" here, we really mean "the place where the
curvature gets really big, which would be singular if classical
general relativity were true" - since we don't know what quantum
effects or other effects might do here. The only reason I mention
this is that it's actually very interesting to wonder what happens
to the remains of someone who fell into a black hole when the black
hole finally evaporates.
>Since Schwarzchild coordinates are for an eternal black hole,
>you have to use some other coordinates to describe an evaporating
>black hole, but I think any sensible coordinate system generalizing
>Schwarzchild coordinates would have to have the property that
>*either* the infalling astronaut hits the singularity in finite time,
>just as the black hole evaporates, or else he escapes and lives
>to tell the tale.
Hmm. If we ignore black hole evaporation for a minute, and consider
an eternal nonrotating black hole in ordinary Schwarzschild coordinates,
then an infalling person takes an *infinite* amount of coordinate time
to reach the horizon. He then travels *backwards* for an infinite
amount of coordinate time until he hits the singularity. But this just
goes to show that Schwarzschild coordinates are really lousy for
understanding this issue! The infalling person only takes a finite
amount of *proper* time to hit the singularity. People only started
understanding black holes when they developed coordinate systems that
are better than the Schwarzschild coordinates.
You seem to be interpreting Wolfram Schroers' original comments differently
than I did. I thought he was saying something that I've often hear others
say, namely: "It takes an infinite amount of coordinate time for an person
to reach the horizon of a Schwarzschild black hole, so if the black hole
actually evaporates in a finite amount of time, the person will never have
time to cross the horizon". This of course is erroneous; it's a confusion
based on confusing coordinates. As you note, anyone who falls in reasonably
soon will get squashed at the singularity (or what passes for the
singularity) in a finite amount of proper time.
As usual, all of this becomes obvious with the help of a Penrose
diagram - see page 413 of Wald's "General Relativity".
But I could have misunderstood what he was saying - I had some trouble
understanding it.
Toby Bartels
>The word "singularity" can be misleading, I think. What you actually have
>is:
> 1. no upper bound on curvature (over space-time)
> 2. world-lines of finite proper length (duration)
>Personally, I have a hard time imagining that #2 will ever go away.
But #2 is the odious one.
You have an object which, in a finite amount of its own proper time,
ceases to exist. What happens to it? Where does it go???
I really have a hard time with this. I wonder if it's the same feeling
that makes people believe souls must exist after bodies die.
-- Toby
to...@ugcs.caltech.edu
Daryl McCullough
>Daryl McCullough <da...@cogentex.com> wrote:
>
>>My guess is that there is some critical radius (as measured using
>>Schwarzchild coordinates) such that an astronaut falling from rest
>>at a distance R_critical(M) towards a black hole of mass M will miss
>>the singularity. If an astronaut starts falling from a distance
>>closer than R_critical(M), then he will hit the singularity in
>>finite coordinate time (as well as proper time).
>
>This all sounds fine except for 2 things:
>
>1) "finite coordinate time" doesn't mean anything until we describe
>which coordinates we're using. Of course you address that point below,
>but I have some qualms about what you say - more on that later.
Well, I think that I can phrase it like this:
*Whatever* coordinate system you are using, if the hole is gone
at time T, then either the infalling astronaut is also gone at
time T, or else the astronaut escapes. There is no coordinate
system in which the hole disappears at time T, and yet the
astronaut hovers above some event horizon for the rest of
eternity. The original poster was proving this conclusion
via a reductio ad absurdum.
>Hmm. If we ignore black hole evaporation for a minute, and consider
>an eternal nonrotating black hole in ordinary Schwarzschild coordinates,
>then an infalling person takes an *infinite* amount of coordinate time
>to reach the horizon.
I know. The point is that that description can't hold for evaporating
black holes. No matter how misleading Schwarzchild coordinates are,
they are not going to tell you that the astronaut takes an infinite
amount of time to fall into the hole *and* the hole only lasts for
a finite amount of time.
>He then travels *backwards* for an infinite amount of coordinate
>time until he hits the singularity.
Oh, I didn't know that! (Warning: the next
two paragraphs are *not* discussing what
*really* happens, but are discussing how
they are described in a particular coordinate
system. I know that's an evil thing for a
physicist to do, but I'm not a physicist.
Physics for me is entertainment.)
I thought that the Schwarzchild coordinates
were just incomplete---that is, that there were events (namely
those after the infalling astronaut falls into the black hole)
that were simply not assigned any coordinates at all. (In the
same way that if you try to describe the x-y plane using the
coordinates z = 1/x, w = 1/y, then the origin (x=y=0) is given
no coordinates at all.)
What you seem to be saying is that the events after the
astronaut falls through the event horizon are described
in the Schwarzchild coordinates as happening in the opposite
order: The astronaut reaches the singularity in finite time
and then moves backwards up to the event horizon (taking
an infinite time to do so). In the Schwarzchild coordinates,
the infalling astronaut is in two different places at the
same time. I know you advise against taking coordinates
seriously, but this coordinate weirdness seems like fun
to me. It's as if the black hole were a perfect mirror,
and as the astronaut falls towards the event horizon,
there is a mirror image of the astronaut rising from
the center to meet him at the event horizon. I like that.
>But this just goes to show that Schwarzschild coordinates are
>really lousy for understanding this issue!
It depends on what you consider the issue to be. It seemed to
me that the original poster was wondering how to *reconcile*
two different descriptions of the same situation.
The situation is pretty straight-forward from the point of view
of the infalling astronaut---he just falls to his
death in a finite amount of time. But how do things look
from outside the event horizon?
>You seem to be interpreting Wolfram Schroers' original comments differently
>than I did. I thought he was saying something that I've often hear others
>say, namely: "It takes an infinite amount of coordinate time for an person
>to reach the horizon of a Schwarzschild black hole, so if the black hole
>actually evaporates in a finite amount of time, the person will never have
>time to cross the horizon". This of course is erroneous; it's a confusion
>based on confusing coordinates.
Rather than saying that Wolfram was making
an erroneous argument, I would say that he has correctly
shown, via reducto ad absurdum, that Schwarzchild coordinates
don't apply for evaporating black holes. For an evaporating
black hole, it is *not* the case (in any coordinate
system) that the astronaut takes forever to reach his
doom.
>But I could have misunderstood what he was saying - I had some trouble
>understanding it.
That's because you're too smart. You need an interpreter like me.
Daryl McCullough
CoGenTex, Inc.
Ithaca, NY
John Baez
Daryl McCullough <da...@cogentex.com> wrote:
>John Baez wrote:
>>If we ignore black hole evaporation for a minute, and consider
>>an eternal nonrotating black hole in ordinary Schwarzschild coordinates,
>>then an infalling person takes an *infinite* amount of coordinate time
>>to reach the horizon. He then travels *backwards* for an infinite amount
>>of coordinate time until he hits the singularity.
>Oh, I didn't know that! (Warning: the next
>two paragraphs are *not* discussing what
>*really* happens, but are discussing how
>they are described in a particular coordinate
>system. I know that's an evil thing for a
>physicist to do, but I'm not a physicist.
>Physics for me is entertainment.)
Yes, but there are are a lot of impressionable kids out there reading
this, and if we're not careful, they'll walk away thinking that black
holes have something to do with travelling backwards in time! That's
why I almost didn't mention this - I know *you* can handle it, but the
kids out there....
>I thought that the Schwarzchild coordinates
>were just incomplete---that is, that there were events (namely
>those after the infalling astronaut falls into the black hole)
>that were simply not assigned any coordinates at all. (In the
>same way that if you try to describe the x-y plane using the
>coordinates z = 1/x, w = 1/y, then the origin (x=y=0) is given
>no coordinates at all.)
>
>What you seem to be saying is that the events after the
>astronaut falls through the event horizon are described
>in the Schwarzchild coordinates as happening in the opposite
>order: The astronaut reaches the singularity in finite time
>and then moves backwards up to the event horizon (taking
>an infinite time to do so). In the Schwarzchild coordinates,
>the infalling astronaut is in two different places at the
>same time.
Right. Check out Figure 31.4 a) on page 835 of Misner, Thorne
and Wheeler's book "Gravitation". This compares the kooky
Schwarzschild coordinate description of someone falling into
a black hole with the more sensible description in Kruskal
coordinates.
>I know you advise against taking coordinates
>seriously, but this coordinate weirdness seems like fun
>to me.
Right: as long as you don't take it seriously, it's fun.
>It's as if the black hole were a perfect mirror,
>and as the astronaut falls towards the event horizon,
>there is a mirror image of the astronaut rising from
>the center to meet him at the event horizon.
Yup, exactly.
>I like that.
I don't, but I suppose it has a perverse charm. It just
goes to show that you can make stuff seem weird by using
weird coordinates.
Jim Graber
ba...@galaxy.ucr.edu (John Baez) wrote:
> In article <UhHH3.289$r46...@news.indigo.ie>,
> Gerry Quinn <ger...@indigo.ie> wrote:
>
> >There is no such thing as a singularity. General relativity may
> >confidently be expected to represent a particular mathematical
> >formalisation of the classical limit of a (so far unformulated)
quantum
> >gravity theory. Therefore, it will break down at the Planck scale
(if
> >not before).
>
> we don't know *for sure* that there's no such thing as a singularity.
>
quantum
> gravity or something will get rid of the singularities predicted by
> general relativity.
>
>2) By "the singularity" here, we really mean "the place where the
>curvature gets really big, which would be singular if classical
>general relativity were true" - since we don't know what quantum
>effects or other effects might do here. The only reason I mention
>this is that it's actually very interesting to wonder what happens
>to the remains of someone who fell into a black hole when the black
>hole finally evaporates.
>
inside black holes say at the Planck scale. My favorite model would be
to stop the collapse of the Oppenheimer Snyder collapse just before the
singularity develops. Since there is no singularity, the geodesics don't
stop. Instead, they pass through the center and exit on the other side.
But the event horizon is still there. Wouldn't this object be
simultaneously a black and a white hole, a sort of zebra hole, such as
Hawking is said to have briefly considered after discovering his black
hole radiation? If not, what would be the effects interior and exterior
if quantum gravity eliminated the singularities but not the event
horizon of black holes? TIA Jim Graber
John Baez
Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> wrote:
>That's what I thought, too. Schwarzschild coordinates simply don't cover
>the whole manifold, but there are other coordinate systems which do.
>So you may transform Schwarzschild coordinates to another system ONLY
>outside the horizon [...]
Whether Schwarzschild coordinates cover "the whole manifold" depends
on what you mean by "the whole manifold". They don't cover the whole
Kruskal extension, but they do cover the region inside the event
horizon where an infalling particle would be found. In Schwarzschild
coordinates, this is the region where r < 2m.
>Hawking radiation is (even for the distant observer) non-negligible on
>time scales of the order of the hole's lifetime. So why should it be
>negligible for the infalling astronaut?
One reason is that the infalling astronaut crosses the event horizon
and is gobbled up by the singularity long "before" the black hole
shrinks to a small size and starts emitting lots of Hawking
radiation. Of course, the word "before" can be confusing here,
since it depends on a coordinate system. But if you look at
a Penrose diagram of an evaporating black hole (see page 413
of Wald's book), you'll see that the worldline of an infalling
astronaut never comes anywhere near the regions of spacetime
that correspond to the late stages of black hole evaporation.
>Naively one would expect that the radiation increases as he
>approaches the horizon.
No! A person freely falling into a large black hole will
see no Hawking radiation whatsoever as he crosses the event
horizon. A person using rocket thrusters to hover over the
event horizon of a large black hole will see Hawking radiation.
The closer he hovers to the horizon, the higher the temperature
of the radiation he sees.
Remember, even someone accelerating in the vacuum in flat
Minkowski spacetime will see radiation: Unruh radiation. So
how much radiation you see depends on how much you accelerate,
not just where you are.
Ian Easson
Jim Graber (James S. Graber) <jgr...@my-deja.com> wrote in message
news:7tasjc$ea7$1...@nnrp1.deja.com...
> >
> Let's consider what happens if quantum gravity eliminates singularities
> inside black holes say at the Planck scale. My favorite model would be
> to stop the collapse of the Oppenheimer Snyder collapse just before the
> singularity develops. Since there is no singularity, the geodesics don't
> stop. Instead, they pass through the center and exit on the other side.
>
Just as there is no such thing as a (classical) trajectory in quantum
mechanics, there should be no such thing as a geodesic in any consistent
theory of quantum gravity. So the mental picture outlined above of
geodesics passing "through" the region that is classically singular makes no
sense.
Ian Easson
Ralph Hartley
> >approaches the horizon.
>
> No! A person freely falling into a large black hole will
> see no Hawking radiation whatsoever as he crosses the event
> horizon. A person using rocket thrusters to hover over the
> event horizon of a large black hole will see Hawking radiation.
> The closer he hovers to the horizon, the higher the temperature
> of the radiation he sees.
>
> Remember, even someone accelerating in the vacuum in flat
> Minkowski spacetime will see radiation: Unruh radiation. So
> how much radiation you see depends on how much you accelerate,
> not just where you are.
>
an observer falls toward the horizon then fires a rocket timed so that
his escape is after the hole has evaporated.
As I understand it now, what he sees is that on the way down nothing
unusual happens, in accordance with the principle that nothing special
happens (locally) at the horizon. It's after he starts accelerating
that he starts seeing radiation, with the black hole evaporating
behind him. Some or all of the radiation he sees, he might attribute
to his (very) high acceleration. Anyway, he sees a lot more radiation
than the distant observer, who sees the the same radiation red shifted
almost to nothing.
I assume that, as usual, all bets are off as to what an observer sees
while approaching the singularity instead of the horizon, but it might
involve infinite radiation fluxes as well as being crushed to zero
volume.
Ralph Hartley
har...@aic.nrl.navy.mil
Barry Adams
On 27 Sep 1999 14:24:26 -0500, ba...@galaxy.ucr.edu (John Baez) wrote:
>Interestingly, in many toy models of quantum cosmology that have
>been studied so far - e.g., minisuperspace models of Bianchi IX
>cosmologies - quantum effects do *not* eliminate the big bang
>singularity. Of course, these toy models are a far cry from a
>full-fledged theory of quantum gravity.
So what would a singularity look like in Quantum Gravity? are
their any ideas yet? I guess models with discrete space time
or extended fundamental objects don't contain singularities
anyway. String/Brane Theory have instead a single
string/brane containing the energy of everything that fall
info the black hole, unless there was a upper limit to
how mach enegry a single string or brane could hold
(never heard of a upper limit) in which case you'd have a lump of
maximal energy strings/branes. What would a singularity look
with Spin-Networks?
Barry Adams
Mark William Hopkins
Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> wrote:
>So when Hawking radiation is active, even the infalling astronaut can not
>reach the horizon in a finite proper time, because in his proper time he
>should reach the black hole at the very moment it evaporates.
In article <7srcv8$7tu$1...@rosencrantz.stcloudstate.edu> ba...@galaxy.ucr.edu writes:
>No, Hawking radiation does not prevent the freely falling astronaut
>from reaching the event horizon in a finite amount of proper time -
>in fact, for a largish black hole, the effect of the Hawking radiation
>on the astronaut is negligible.
It does not prevent the astronaut from getting there in a finite proper
time, but surely it prevents the black hole from BEING there when the
astronaut reaches the point where horizon was before the black hole
evaporated since it will have evaporated away be the time the astronaut
gets there!
John Baez
Barry Adams <bad...@adept77.demon.co.uk> wrote:
>On 27 Sep 1999 14:24:26 -0500, ba...@galaxy.ucr.edu (John Baez) wrote:
>>Interestingly, in many toy models of quantum cosmology that have
>>been studied so far - e.g., minisuperspace models of Bianchi IX
>>cosmologies - quantum effects do *not* eliminate the big bang
>>singularity. Of course, these toy models are a far cry from a
>>full-fledged theory of quantum gravity.
Once upon a time Jim Graber asked me some questions about this
remark, which I never got around to answering:
>Do these toy models eliminate black hole singularities?
The models I mentioned have nothing at all to say about black
holes. In a "minisuperspace model", you pick a class of solutions
of Einstein's equation depending on *finitely many parameters*, and
you just quantize the theory of those solutions. This reduces
the difficult problem of quantum gravity to an ordinary quantum
mechanics problem - with *finitely many degrees of freedom*.
The simplest example is to take the usual theory of a homogeneous
isotropic big bang/big crunch universe, and quantize the theory of
that. In this case the only degree of freedom is the radius of
the universe - so mathematically, this problem looks just like the
the quantum mechanics of a particle in a potential in one dimension!
The position of the particle corresponds to the radius of the universe.
It's easy to solve this problem, and it's a lot of fun - see below for i
what the answer looks like. But in this oversimplified model, there are
no galaxies, no black holes, no gravitational radiation - none of the
*inhomogeneities* that make the universe an interesting place! So
the model is not terribly realistic.
The Bianchi IX cosmologies are a bit more exciting, because these
are a class of *inhomogeneous, anisotropic* solutions of Einstein's
equations that have a big bang and big crunch. These solutions have
more degrees of freedom so the math becomes harder - but still just
finitely many degrees of freedom. In particular, there are no black
holes in these cosmologies. So when we quantize them, we learn nothing
about black holes. The main thing we *do* learn is that in these
models, the big bang/big crunch singularities are not "smoothed out"
by quantum effects. But it's not clear if this is a real result, or
just an artifact of an oversimplified model!
>Are there toy models which eliminate black hole singularities?
>(My impression is that some recent string theory models do this.)
>Are there good toy models which do not do this?
I'll let the string theory gurus say what string theory says about
this issue. You can certainly cook up minisuperspace models based
on the Schwarzschild solution. Voila! - a toy model of a "quantum
black hole". If you want it to rotate and have electric charge,
start from the Kerr solution. I think these models have singularities
much like the classical solutions on which they're based. But I'm
not an expert on this. And there's a reason why I'm not:
I sort of regard this minisuperspace stuff as a kind of game you play
when you don't feel up to *actually* studying quantum gravity. A
little bit is very helpful, but too much only makes you lose sight of
some important issues. Also, it's never clear whether the results
we get from minisuperspace models are qualitatively correct, or artifacts
of oversimplifying the problem.
Okay, back to Barry Adams:
>So what would a singularity look like in Quantum Gravity? are
>there any ideas yet? I guess models with discrete space time
>or extended fundamental objects don't contain singularities
>anyway. String/Brane Theory have instead a single
>string/brane containing the energy of everything that fall
>info the black hole, unless there was a upper limit to
>how mach enegry a single string or brane could hold
>(never heard of a upper limit) in which case you'd have a lump of
>maximal energy strings/branes. What would a singularity look
>with Spin-Networks?
I really wish I knew what the inside of a quantum black hole is
like in the spin network approach! But I don't. So far most
work has focused on the event horizon. I think I have a pretty
good idea what that's like. For slides of a talk about this, try:
http://math.ucr.edu/home/baez/black/black.html
and for a cute encapulated Postscript picture of a black hole a la
spin networks, try:
http://math.ucr.edu/home/baez/blackhole.eps
(By the way: if some smart person out there can turn this into a jpg
or gif file, I'd be really grateful! I'll put it on my website and
acknowledge you....)
And now, let me give you a cute example of a minisuperspace model
worked out in detail. But first, a completely irrelevant puzzle,
just to get your brain going:
Puzzle #8: Which 39-year-old female mathematician is rumored to be
secretly in charge of one of the world's largest countries?
For the answer, go to http://math.ucr.edu/home/baez/puzzles/8.html
.....................................................................
Also available at http://math.ucr.edu/home/baez/week6.html
February 20, 1993
This Week's Finds in Mathematical Physics (Week 6)
John Baez
1) Quantum cosmology, talk given at Texas/Pascos 1992 at Berkeley by
Alexander Vilenkin, preprint available in TeX form as gr-qc/9302016
This is, as Vilenkin notes, an elementary review of quantum cosmology.
It won't be news to anyone who has kept up on that subject (except
perhaps for a few speculations at the end), but for those who haven't
been following this stuff, like myself, it might be a good way to get
started.
Let's get warmed up....
Quantizing gravity is mighty hard. For one thing, there's the "problem
of time" - the lack of a distinguished time parameter in *classical*
general relativity means that the usual recipe for quantizing a
dynamical system - "represent time evolution by the unitary operators
exp(-iHt) on the Hilbert space of states, where t is the time and H, the
Hamiltonian, is a self-adjoint operator" - breaks down! As Wheeler so
picturesquely put it, in general relativity we have "many-fingered
time"; there are lots of ways of pushing a spacelike surface forwards in
time.
But if we simplify the heck out of the problem, we might make a little
progress. (This is a standard method in physics, and whether or not
it's really justified, it's often the only thing one can do!)
For one thing, note that in the big bang cosmology there is a
distinguished "rest frame" (or more precisely, field of timelike
vectors) given by the galaxies, if we discount their small random
motions. In reality these are maybe not so small, and maybe not so
random - such things as the "Virgo flow" show this - but we're talking
strictly theory here, okay? - so don't bother us with facts! So, if we
imagine that things go the way the simplest big bang models predict, the
galaxies just sit there like dots on a balloon that is being inflated,
defining a notion of "rest" at each point in spacetime. This gives
a corresponding notion of time, since one can measure time using
clocks that are at rest relative to the galaxies. Then, since we are
pretending the universe is completely homogeneous and isotropic - and
let's say it's a closed universe in the shape of a 3-sphere, to be
specific - the metric is given by
dt^2 - r(t)^2[(d psi)^2 + (sin psi)^2{(d theta)^2 + (sin theta)^2 (d phi)^2}]
What does all this mean? Here r(t) is the radius of the universe as a
function of time, the following stuff is just the usual metric on the
unit 3-sphere with hyperspherical coordinates psi, theta, phi
generalizing the standard coordinates on the 2-sphere we all learn in
college:
(d psi)^2 + (sin psi)^2{(d theta)^2 + (sin theta)^2 (d phi)^2}
and the fact that the metric on spacetime is dt^2 minus a bunch of stuff
reflects the fact that spacetime geometry is "Lorentzian," just as
in flat Minkowski space the metric is
dt^2 - dx^2 - dy^2 - dz^2.
The name of the game in this simple sort of big bang cosmology is thus
finding the function r(t)! To do this, of course, we need to see what
Einstein's equations reduce to in this special case, and since
Einstein's equations tell us how spacetime curves in response to the
stress-energy tensor, this will depend on what sort of matter we have
around. We are assuming that it's homogeneous and isotropic, whatever
it is, so it turns out that all we need to know is its density rho and
pressure P (which are functions of time). We get the equations
r''/r = -(4pi/3)(rho + 3P) (r')^2 = (8pi/3) rho r^2 - 1
Here primes denote differentiation with respect to t, and I'm using
units in which the gravitational constant and speed of light are equal to 1.
Let's simplify this even more. Let's assume our matter is "dust," which
is the technical term for zero pressure. We get two equations:
r''/r = -(4pi/3)rho (r')^2 = (8pi/3) rho r^2 - 1. (1)
Now let's take the second one, differentiate with respect to t,
2r'' r' = (8pi/3)(rho' r^2 + 2 rho r r')
plug in what the first equation said about r'',
-(8pi/3) rho r r' = (8pi/3)(rho' r^2 + 2 rho r r')
clear out the crud, and lo:
3 rho r' = - rho' r
or, more enlighteningly,
d(rho r^3)/dt = 0.
This is just "conservation of dust" - the dust density times the volume
of the universe is staying constant. This, by the way, is a special
case of the fact that Einstein's equations *automatically imply*
local conservation of energy (i.e., that the stress-energy tensor is
divergence-free).
Okay, so let's say rho r^3 = D, with D being the total amount of dust.
Then we can eliminate rho from equations (1) and get:
r'' = -4piD/3r^2 (r')^2 - (8pi/3) D/r = - 1 (2)
What does this mean? Well, the first one looks like it's saying
there's a force trying to make the universe collapse, and that the
strength of this force is proportional to 1/r^2. Sound vaguely
familiar? It's actually misleadingly simple - if we had put in
something besides dust it wouldn't work quite this way - but as long as we
don't take it too seriously, we can just think of this as gravity trying
to get the universe to collapse. And the second one looks like it's
saying that the "kinetic" energy proportional to (r')^2, plus the
"potential" energy proportional to -1/r, is constant! In other words,
we have a nice analogy between the big bang cosmology and a very
old-fashioned system, a classical particle in one dimension attracted to
the origin by a 1/r^2 force!
It's easy enough to solve this equation, and easier still to figure it out
qualitatively. The key thing is that since the total "energy" in
the second equation of (2) is negative, there won't be enough "energy"
for r to go to infinity, that is, there'll be a big bang and then a big
crunch. Here's r as a function of t, roughly:
| . .
| . .
r | . . Figure 1
| . .
| . .
|------------------------------------
t
What goes up, must come down! This curve, which I haven't drawn too
well, is just a cycloid, which is the curve traced out by a point on the
rim of rolling wheel. So, succumbing to romanticism momentarily we
could call this picture ONE TURN OF THE GREAT WHEEL OF TIME.... But
there is *no* reason to expect further turns, because the
differential equation simply becomes singular when r = 0. We may either
say it doesn't make sense to speak of "before the big bang" or "after
the big crunch" - or we can look for improved laws that avoid these
singularities. (I should repeat that we are dealing with unrealistic models
here, since for example there is no evidence that there is enough matter
around to "close the universe" and make this solution qualitatively
valid - it may well be that there's a big bang but no big crunch. In
this case, there's only one singularity to worry about, not two.)
People have certainly not been too ashamed to study the *quantum*
theory of this system (and souped-up variants) in an effort to get a
little insight into quantum gravity. We would expect that quantum
effects wouldn't matter much until the radius of the universe is very
small, but when it *is* very small they would matter a lot, and maybe -
one might hope - they would save the day, preventing the nasty
singularities. I'm not saying they DO - this is hotly debated - but
certainly some people hope they do. Of course, serious quantum gravity
should take into account the fact that geometry of spacetime has all
sorts of wiggles in it -it isn't just a symmetrical sphere. This may make
a vast difference in how things work out. (For example, the big crunch
would be a lot more exciting if there were lots of black holes around by
then.) The technical term for the space of all metrics on space is
"superspace" (sigh), and the toy models one gets by ignoring all but
finitely many degrees of freedom are called "minisuperspace" models.
Let's look at a simple minisuperspace model. The simplest thing
to try is to take the classical equations of motion (2) and try to
quantize them just like one would a particle in a potential. This is a
delicate business, by the way, because one can't just take some
classical equations of motion and quantize them in any routine way.
There are lots of methods of quantization, but all of them require a
certain amount of case-by-case finesse.
The idea of "canonical quantization" of a classical system with one
degree of freedom - like our big bang model above, where the one degree
of freedom is r - is to turn the "position" (that's r) into a
multiplication operator and the "momentum" (often that's something like
r', but watch out!) into a differentation operator, say -i hbar d/dr, so that
we get the "canonical commutation relations"
[-i hbar d/dr, r] = -i hbar.
We then take the formula for the energy, or Hamiltonian, in terms of
position and momentum, and plug in these operators, so that the
Hamiltonian becomes an operator. (Here various "operator-ordering"
problems can arise, because the position and momentum commuted in the
original classical system but not anymore!) To explain what I mean,
why don't I just do it!
So: I said that the formula
(r')^2 - (8pi/3) D/r = - 1 (3)
looks a lot like a formula of the form "kinetic energy plus potential
energy is constant". Of course, we could multiply the whole equation
by anything and get a valid equation, so it's not obvious that the
``right'' Hamiltonian is
(r')^2 - (8pi/3) D/r
or (adding 1 doesn't hurt)
(r')^2 - (8pi/3) D/r + 1
In fact, note that multiplying the Hamiltonian by some function of r
just amounts to reparametrizing time, which is perfectly fine in general
relativity. In fact, Vilenkin and other before him have decided it's
better to multiply the Hamiltonian above by r^2. Why? Well, it has to
do with figuring out what the right notion of "momentum" is
corresponding to the "position" r. Let's do that. We use the old
formula
p = dL/dq'
relating momentum to the Lagrangian, where for us the position, usually
called q, is really r.
The Lagrangian of general relativity is the "Ricci scalar" R - a measure of
curvature of the metric - and in the present problem it turns out to be
R = 6 (r''/r + (r')^2/r^2)
But we are reducing the full field theory problem down to a problem with
one degree of freedom, so our Lagrangian should be the above integrated
over the 3-sphere, which has volume 16 pi r^3/3, giving us
32pi (r'' r^2 + (r')^2 r)
However, the a'' is a nuisance, and we only use the integral of the
Lagrangian with respect to time (that's the action, which classically is
extremized to get the equations of motion), so let's do an integration
by parts, or in other words add a total divergence, to get the Lagrangian
L = -32pi (r')^2 r.
Differentiating with respect to r' we get the momentum "conjugate to r",
p = -64pi r'r.
Now I notice that Vilenkin uses as the momentum simply -r'r, somehow
sweeping the monstrous 64pi under the rug. I have the feeling that
this amounts to pushing this factor into the definition of hbar in the
canonical commutation relations. Since I was going to set hbar to 1 in
a minute anyway, this is okay (honest). So let's keep life simple and
use
p = -r'r.
Okay! Now here's the point, we want to exploit the analogy with good
old quantum mechanics, which typically has Hamiltonians containing
something like p^2. So let's take our preliminary Hamiltonian
(r')^2 - (8pi/3) D/r + 1
and multiply it by r^2, getting
H = p^2 - (8pi D/3)r + r^2.
Hey, what's this? A harmonic oscillator! (Slightly shifted by the
term proportional to r.) So the universe is just a harmonic
oscillator... I guess that's why they stressed that so much in all my
classes!
Actually, despite the fact that we are working with a very simple model
of quantum cosmology, it's not quite *that* simple. First of all,
recall our original classical equation, (3). This constrained the
energy to have a certain value. I.e., we are dealing not with a
Hamiltonian in the ordinary sense, but a "Hamiltonian constraint" -
typical of systems with time reparametrization invariance. So our
quantized equation says that the "wavefunction of the universe," psi(r),
must satisfy
H psi = 0.
Also, unlike the ordinary harmonic oscillator we have the requirement
that r > 0. In other word, we're working with a problem that's like a
harmonic oscillator and a "wall" that keeps r > 0. Think of a particle
in a potential like this:
|
| .
|
V(r) | Figure 2
| .
| .
|.-----.-------------
. r
Here V(r) = - (8pi D/3)r + r^2. The minimum of V is at r = 4 pi D/3
and the zeroes are at r = 0 and 8 pi D/3. Classically, a particle
with zero energy starting at r = 0 will roll to the right and make it
out to r = 4 pi D/3 before rolling back to r = 0. This is basically the
picture we had in Figure 1, except that we've reparametrized time so we
have simple harmonic motion instead of cycloid.
Quantum mechanically, however one must pick boundary conditions at r = 0
to make the problem well-defined!
This is where the fur begins to fly!! Hawking and Vilenkin have very
different ideas about what the right boundary conditions are. And note
that this is not a mere technical issue, since they determine the
wavefunction of the universe in this approach! I will not discuss this
since Vilenkin does so quite clearly, and if you understand what I have
written above you'll be in a decent position to understand him. I will
just note that Vilenkin, rather than working with a universe full of
"dust," considers a universe in which the dominant contribution to the
stress-energy tensor is the cosmological constant, that is, the negative
energy density of a "false vacuum", which believers in inflation (such
as Vilenkin) think powered the exponential growth of the universe at an
early stage. So his equations are slightly different from those above (and
are only meant to apply to the early history of the universe).
[Let me just interject a question to the experts if I may - since I've
written this long article primarily to educate myself. It would seem to
me that the equation H psi = 0 above would only have a normalizable
solution if the boundary conditions were fine-tuned! I.e., maybe the
equation H psi = 0 itself determines the boundary conditions! This
would be very nice; has anyone thought of this? It seems reasonable
because, with typical boundary conditions, the operator H above will
have pure point spectrum (only eigenvalues) and it would be rather
special for one of them to be 0, allowing a normalizable solution of
H psi = 0. Also, corrections and education of any sort are welcomed.
I would love to discuss this with some experts.]
Anyway, suppose we find some boundary conditions and calculate psi, the
"wavefunction of the universe." (I like repeating that phrase because
it sounds so momentous, despite the fact that we are working with a
laughably oversimplified toy model.) What then? What are the
implications for the man in the street?
Let me get quite vague at this point. Think of the radius of the
universe as analogous to a particle moving in the potential of Figure 2.
In the current state of affairs classical mechanics is an excellent
approximation, so it seems to trace out a classical trajectory. Of
course it is really obeying the laws of quantum mechanics, so the
trajectory is really a "wave packet" - technically, we use the WKB
approximation to see how the wave packet can seem like a classical
trajectory. But near the big bang or big crunch, quantum mechanics
matters a lot: there the potential is rapidly varying (in our simple
model it just becomes a "wall") and the wave packet may smear out
noticeably. (Think of how when you shoot an electron at a nucleus it
bounces off in an unpredictable direction - it's wavefunction just tells
you the *probability* that it'll go this way or that!) So some quantum
cosmologists have suggested that if there is a big crunch, the universe
will pop back out in a highly unpredictable, random kind of way!
I should note that Vilenkin has a very different picture. Since this
stuff makes large numbers of assumptions with very little supporting
evidence, it is science that's just on the brink of being mythology.
Still, it's very interesting.
John Baez
>The Bianchi IX cosmologies are a bit more exciting, because these
>are a class of *inhomogeneous, anisotropic* solutions of Einstein's
>equations that have a big bang and big crunch.
Whoops - these cosmologies are *homogeneous* but anisotropic!
Thanks go to Ted Bunn for catching this error.
>(By the way: if some smart person out there can turn this into a jpg
>or gif file, I'd be really grateful! I'll put it on my website and
>acknowledge you....)
Okay, we have some winners, please stop sending in entries!
The winners are Ted Bunn (first gif) and Wim van Dam (first jpg).
You can see their marvelous artwork at
http://math.ucr.edu/home/baez/blackhole.gif
and
Zachary Uram
> The winners are Ted Bunn (first gif) and Wim van Dam (first jpg).
> You can see their marvelous artwork at
>
> http://math.ucr.edu/home/baez/blackhole.gif
Wow what progam(s) did these guy use to make this? I want to be
able do this also :) BTW what does it mean when someone say "this
is STABLE solution to Einstein field equations"? Can you make
these equations work for any D? I remember seeing on NCSA side
D=2 and D=3 pages of equations I couldn't understand (yet! hehe).
Zach
________________________________________________________
ur...@cmu.edu
"Blessed are those who have not seen and yet have faith." - John 20:29
Harry Johnston
> To an external observer, it looks like the infalling person doesn't
> cross the event horizon until the very moment the black hole
> evaporates,
Actually, I suspect this isn't even true for an evaporating black
hole. My reasoning is as follows: suppose someone jumps in and
crosses the event horizon at a point when the event horizon radius (so
to speak) is 1km. One final photon is emitted exactly at the event
horizon.
Much, much later, the event horizon radius has shrunk to 500m. It's
still going to be a long time before the black hole evaporates, but
what happened to that final photon? It started life 1km "out" from
the singularity, a distance now well away from the event horizon, so
it is my guess that it has long ago gone on it's merry way and been
seen by any observer that might have been willing to wait for it.
I await comment or correction by the better informed ...
Harry.
---
Harry Johnston, om...@ihug.co.nz
One Earth, One World, One People
john baez
Jacques M. Mallah <jqm...@is2.nyu.edu> wrote regarding the
problem of time:
>What approaches to this problem are used by quantum
>gravitationalists, and why?
Yikes, what a huge question! As you'd expect with a famous unsolved
problem, *lots* of approaches have been tried - including ignoring
the issue and hoping it goes away. But it ain't gonna go away.
If you really want to know the answer to your question, try this:
Canonical Quantum Gravity and the Problem of Time, Chris J. Isham,
125 pages, preprint available as http://xxx.lanl.gov/abs/gr-qc/9210011
If you glance at the bibliography, you'll see why I'm not even gonna
*try* to give you a thorough answer.
>> I.e., we are dealing not with a Hamiltonian in the ordinary sense,
>> but a "Hamiltonian constraint" - typical of systems with time
>> reparametrization invariance. So our quantized equation says that
>> the "wavefunction of the universe," psi(r), must satisfy H psi = 0.
>Why?
Well, the concept of "the total energy of the universe" is not so simple
in general relativity, so the *meaning* of the Hamiltonian is a bit
different than you're used to. This is true already in *classical*
general relativity! It can't hurt to start by reading the FAQ:
http://math.ucr.edu/home/baez/physics/energy_gr.html
In particular, if you start by using general relativity to describe a
closed universe and use the standard recipe to work out the Hamiltonian,
you get an expression that you can show equals ZERO by using Einstein's
equation.
In other words: in general relativity, the Hamiltonian is a *constraint*.
>If it's an energy eigenstate, there would be no dynamics.
Ah, that's what you might think at first - that's why the Wheeler-DeWitt
equation is also called the "frozen formalism". But all the constraint
H = 0 really means is this: in the absence of a god-given best notion of
time, we are forced to think of a solution of Einstein's equation as
a geometry for *all of spacetime*. Nothing physical about this geometry
changes if we change coordinates. In particular, if we choose some time
coordinate t, and then "push t forwards" by defining t' = t + c, nothing
about the spacetime geometry is any different. Ultimately, this is why
we get the equation H = 0.
(Don't be surprised if it takes a while for this to sink in! This is
the big difference between "background-free physics" and conventional
physics where you fix the geometry of spacetime at the very beginning.)
So the trick is to extract dynamics from the frozen formalism, or
else to develop some other, "unfrozen", formalism.
>In quantum gravity that may no longer be a true or meaningful statement,
>but the toy model is just regular QM.
It's quantum mechanics for a system with a Hamiltonian *constraint*,
not the usual quantum mechanics where there's a Hamiltonian that can
take various different values.
>Or am I missing something? Would not a more reasonable requirement
>be <H> = 0?
No. If you ponder the quantization of constrained systems
long enough, you'll conclude that the constraint H = 0 should
really be imposed as an operator constraint H psi = 0, not just
a constraint on expectation values.
I'm glad you find this as strange as I did when I first learned
about it! I sympathize with you immensely.
Here are some more details....
Also available as http://math.ucr.edu/home/baez/week43.html
This Week's Finds in Mathematical Physics (Week 43)
John Baez
[....]
There are 3 basic steps in the "canonical quantization" of general
relativity. At each step there is a vector space of quantum states, but
only in the last do we really need a Hilbert space of states, since only
when we're done do we want to be able to compute expectation values of
observables, which takes an inner product.
In what follows I'll talk about the simplest situation, where we have
the *vacuum* Einstein equations
G = 0
where G is the "Einstein tensor" cooked up from the curvature of
spacetime. Say spacetime is of the form R x S, where R is the real
numbers (time) and S is a 3-dimensional manifold (space). We will think
of S as the "t = 0 slice" of R x S.
I) The first stage is to get the space of "kinematical states". In the
quantum mechanics of a point particle on the line, the space of
wavefunctions is a space of functions on the real line. Similarly, in
quantum gravity we naively expect kinematical states to be functions on
the space of Riemannian metrics on the 3-dimensional manifold S we're
taking to be "space". In the loop representation one does something a
bit more clever, but let's move on and then come back to that.
II) The second stage is getting the space of "diffeomorphism-invariant
states". In fact, Einstein's equations in coordinates look like
G_{mu nu} = 0
where the indices mu, nu range from 0 to 3. It's customary to work in
coordinates x_{mu} where x_0 is "time" and the other three coordinates
are the "space" coordinates on S. Then classically, the equations
G_{0 mu} = 0 serve as *constraints* on the initial data for Einstein's
equations, while the remaining equations describe time evolution. I.e.,
only for certain choices of a metric and its first time derivative at t
= 0 can we get a solution of Einstein's equations. In fact, G_{0 mu}
can be calculated knowing only the metric and its first time derivative
at t = 0, and the equations saying they are zero are the constraints
that this data must satisfy to get a solution of Einstein's equations.
Following the usual recipes of quantum theory, we want to turn these
constaints into *operators* on the kinematical Hilbert space of stage
I, and then demand that the states relevant for physics be annihilated
by these operators. The "diffeomorphism-invariant subspace" is the
subspace of the kinematical state space that is annihilated by the
constraints corresponding to G_{0i} where i = 1, 2, 3. Let us put off
for a moment why it's called what it is!
III) The third and final stage is getting the space of "physical states".
Here we look at the subspace of diffeomorphism-invariant states that are
also annihilated by the constaint corresponding to G_{00}. The equation
saying that our diffeomorphism-invariant state is annihilated by this
constraints is called the "Wheeler-DeWitt equation", and this is generally
regarded as the fundamental equation of quantum gravity.
Now, it should make some sense why we call the "physical states" what we
do. These are quantum states satisfying the quantum analogues of the
constraints that the *classical* initial data must satisfy to be initial
data for a solution of Einstein's equations. But why do we impose the
constraints G_{mu nu} = 0 in two separate stages, and call the states in
part II "diffeomorphism-invariant states"?
This is a very important question which gives quantum gravity much of
its curious character. In classical general relativity, G_{0i} not only
gives one of Einstein's equations, namely G_{0i} = 0, it also "generates
diffeomorphisms" of the 3-dimensional manifold S representing space. If
you don't quite know what this means, let me simply say that in
classical mechanics, observables give rise to one-parameter families of
symmetries. For example, momentum gives rise to spatial translations, while
energy (aka the Hamiltonian) gives rise to time translations. We say
that the observable "generates" the one-parameter family of symmetries.
This is (roughly) what I mean by saying that G_{0i} generates
diffeomorphisms of S. Similarly, G_{00} generates diffeomorphisms of
the spacetime R x S corresponding to time evolution.
A similar thing happens in quantum theory. BUT: in quantum theory, if a
state is annihilated by some observable, it implies that the state is
invariant under the one-parameter family of symmetries generated by that
observable. This is not true in classical mechanics. Indeed, it's
rather odd. But what it implies is that in step II we are really
restricting ourselves to kinematical states that are invariant under
diffeomorphisms of the spatial manifold S. This is why we call them
"diffeomorphism-invariant" states. Similarly, in step III we're further
restricting ourselves to states that are invariant under time evolution.
The final "physical states" are, at least heuristically, invariant under
ALL DIFFEOMORPHISMS OF SPACETIME. (So maybe the physical states are the
ones that really should be called "diffeomorphism-invariant" --- but
it's too late now.) While this may seem odd, all it really means is
that in the quantum theory of gravity --- at least when one does it this
way --- the physical states describe only those aspects of the world
that are independent of any choice of coordinate system. That has a
certain charm, philosophically speaking. It is, however, not something
physicists are used to.
The general scheme outlined above has been around ever since the
work of DeWitt:
1) Quantum theory of gravity, I-III by Bryce S. DeWitt, Phys. Rev. 160
(1967), 1113-1148, 162 (1967) 1195-1239, 1239-1256.
[....]
Vesselin G Gueorguiev
>... suppose someone jumps in and
> crosses the event horizon at a point when the event horizon radius (so
> to speak) is 1km. One final photon is emitted exactly at the event
> horizon.
>
> Much, much later, the event horizon radius has shrunk to 500m. It's
> still going to be a long time before the black hole evaporates, but
> what happened to that final photon? It started life 1km "out" from
> the singularity, a distance now well away from the event horizon, so
> it is my guess that it has long ago gone on it's merry way and been
> seen by any observer that might have been willing to wait for it.
Or may be not? The way I see it, there are three options for the photon:
1) It is trapped to circle at the even horizon, so it will follow its changes,
2) some fluctuations will get the photon into the BH, so it will fall in,
3) some fluctuations will get the photon out of the BH, your case above!
Pedro J. Hernández
1999101823...@charity.ucr.edu...
> In article <K47N3.45$t43....@typhoon.nyu.edu>,
> Jacques M. Mallah <jqm...@is2.nyu.edu> wrote regarding the
> problem of time:
> Well, the concept of "the total energy of the universe" is not so simpl=
e
> in general relativity, so the *meaning* of the Hamiltonian is a bit
> different than you're used to. This is true already in *classical*
> general relativity! It can't hurt to start by reading the FAQ:
>
> http://math.ucr.edu/home/baez/physics/energy_gr.html
Why can not one define the energy locally and then add all the energies a=
t
the same cosmic time in order to define the total energy of the universe =
(at
least in a closed universe)?.
cheers
Pedro J. Hern=E1ndez
ph...@correo.rcanaria.es
Jacques M. Mallah
> If you really want to know the answer to your question, try this:
>
> Canonical Quantum Gravity and the Problem of Time, Chris J. Isham,
> 125 pages, preprint available as http://xxx.lanl.gov/abs/gr-qc/9210011
Thanks. I can see that a lot of approaches have been tried,
though that doesn't mean all possibilities have been or that they have
been pursued with equal persistence. I read about the first 30 pages but
then it seemed to get bogged down in technicalities. If I wanted to get
technical I could do some work on my thesis instead of distracting myself
with this stuff :-(
Perhaps you could explain a bit about how the two major
approaches, canonical QG (as you use it) and string theory, deal with the
problem?
> In particular, if you start by using general relativity to describe a
> closed universe and use the standard recipe to work out the Hamiltonian,
> you get an expression that you can show equals ZERO by using Einstein's
> equation.
> In other words: in general relativity, the Hamiltonian is a *constraint*.
I couldn't help but notice the caveat 'closed'. I know SR is a
special case of GR, so I assume that H=0 is not the case for an open
universe. Could it be that in QG the only nontrivial universes are open?
How does the problem of time look for the open case? After all,
astronomers tell us that our universe will not recollapse.
> >If it's an energy eigenstate, there would be no dynamics.
>
> Ah, that's what you might think at first - that's why the Wheeler-DeWitt
> equation is also called the "frozen formalism". But all the constraint
> H = 0 really means is this: in the absence of a god-given best notion of
> time, we are forced to think of a solution of Einstein's equation as
> a geometry for *all of spacetime*.
But wouldn't that be true of a 4-geometry, not a 3-geometry? If
we had a psi[r(t)] or a psi(r/r_classical(t)) I wouldn't have so much of
a problem, but we just have a psi(r) (in the toy model), right? Is the
classical r(t) somehow hidden in the model?
> Nothing physical about this geometry
> changes if we change coordinates. In particular, if we choose some time
> coordinate t, and then "push t forwards" by defining t' = t + c, nothing
> about the spacetime geometry is any different. Ultimately, this is why
> we get the equation H = 0.
> (Don't be surprised if it takes a while for this to sink in! This is
> the big difference between "background-free physics" and conventional
> physics where you fix the geometry of spacetime at the very beginning.)
OK, but at this point I don't get it. Why is the above statement
not true in, for example, SR?
> So the trick is to extract dynamics from the frozen formalism, or
> else to develop some other, "unfrozen", formalism.
Right. How would one extract it from the frozen formalism?
> >Or am I missing something? Would not a more reasonable requirement
> >be <H> = 0?
>
> No. If you ponder the quantization of constrained systems
> long enough, you'll conclude that the constraint H = 0 should
> really be imposed as an operator constraint H psi = 0, not just
> a constraint on expectation values.
I see that for a regular constraint, but this case is a bit
unusual and it would seem worthwhile to try relaxing the contraint. Maybe
the case H ~= 0 is just singled out in the classical limit for a
macroscopic universe, for example. Given the large uncertainty in the age
of our universe, the required small uncertainty in H does not sound
unreasonable to me.
> I'm glad you find this as strange as I did when I first learned
> about it! I sympathize with you immensely.
Well, thanks for your help. Incidently, we have had an
aquaintance in common, Henry Flynt. I taught him some physics.
One reason I am interested in the problem of QG is that it is
relevant to interpretation of QM. As a computationalist I want some
physics that can implement computations. (Even though I still don't
fully know how to handle it even for classical mechanics, but that's
another story!) At least, I want to know what QG developments to root for
or against, if nothing else.
- - - - - - -
Jacques Mallah (jqm...@is2.nyu.edu)
Graduate Student / Many Worlder / Devil's Advocate
"I know what no one else knows" - 'Runaway Train', Soul Asylum
My URL: http://pages.nyu.edu/~jqm1584/
john baez
>... suppose someone jumps in and
> crosses the event horizon at a point when the event horizon radius (so
> to speak) is 1km. One final photon is emitted exactly at the event
> horizon.
A classical point particle at the event horizon moving outwards at the
speed of light would, by definition, stay right on the event horizon.
However, light has wavelike aspects; it's not made of classical point
particles. Thus we can't put some light "exactly at the event horizon".
We can only put it near the event horizon. As time passes, some of it
will fall into the black hole while the rest escapes.
(I'm avoiding the word "photon" because the definition of photons is
rather problematic in curved spacetime, and I don't want to get into
that here.)
John Baez
>John Baez wrote:
>>If we ignore black hole evaporation for a minute, and consider
>>an eternal nonrotating black hole in ordinary Schwarzschild coordinates,
>>then an infalling person takes an *infinite* amount of coordinate time
>>to reach the horizon. He then travels *backwards* for an infinite amount
>>of coordinate time until he hits the singularity.
>Oh, I didn't know that! (Warning: the next
>two paragraphs are *not* discussing what
>*really* happens, but are discussing how
>they are described in a particular coordinate
>system. I know that's an evil thing for a
>physicist to do, but I'm not a physicist.
>Physics for me is entertainment.)
Yes, but there are are a lot of impressionable kids out there reading
this, and if we're not careful, they'll walk away thinking that black
holes have something to do with travelling backwards in time! That's
why I almost didn't mention this - I know *you* can handle it, but the
kids out there....
>I thought that the Schwarzchild coordinates
>were just incomplete---that is, that there were events (namely
>those after the infalling astronaut falls into the black hole)
>that were simply not assigned any coordinates at all. (In the
>same way that if you try to describe the x-y plane using the
>coordinates z = 1/x, w = 1/y, then the origin (x=y=0) is given
>no coordinates at all.)
>
>What you seem to be saying is that the events after the
>astronaut falls through the event horizon are described
>in the Schwarzchild coordinates as happening in the opposite
>order: The astronaut reaches the singularity in finite time
>and then moves backwards up to the event horizon (taking
>an infinite time to do so). In the Schwarzchild coordinates,
>the infalling astronaut is in two different places at the
>same time.
Right. Check out Figure 31.4 a) on page 835 of Misner, Thorne
and Wheeler's book "Gravitation". This compares the kooky
Schwarzschild coordinate description of someone falling into
a black hole with the more sensible description in Kruskal
coordinates.
>I know you advise against taking coordinates
>seriously, but this coordinate weirdness seems like fun
>to me.
Right: as long as you don't take it seriously, it's fun.
>It's as if the black hole were a perfect mirror,
>and as the astronaut falls towards the event horizon,
>there is a mirror image of the astronaut rising from
>the center to meet him at the event horizon.
Yup, exactly.
John Baez
Pedro J. Hernández <ph...@correo.rcanaria.es> wrote:
>Why can not one define the energy locally and then add all the energies at
>the same cosmic time in order to define the total energy of the universe
>(at least in a closed universe)?.
You can, but this energy usually depends strongly on your choice of
time coordinate, and is usually not conserved. If your cosmology
is sufficiently symmetrical, like the Robertson-Friedman-Walker big
bang universe, there's a well-defined notion of "cosmic time" which
you can use as your time coordinate. However, the energy you define
this way will not be conserved!
Do you want to know why?
Then read the FAQ on "Is Energy Conserved in General Relativity"!
You can get this FAQ at various websites including:
http://math.ucr.edu/home/baez/physics/energy_gr.html
http://www.weburbia.demon.co.uk/physics/energy_gr.html
http://www.desy.de/user/projects/Physics/energy_gr.html
Jacques M. Mallah
ordinary QM occurred to me. Perhaps it is relevant.
Suppose you have a very long lived but unstable particle, but when
it decays it releases a lot of energy. Because it is long lived it has a
very well defined energy. This is placed in a system with a
clock/computer that is inert until the particle decays, but will be
powered by the energy from the decay. From the point of view of this
clock, once it is activated it knows how long it was since the decay, and
it sees normal dynamics and normal laws of physics, regardless of the
lifetime of the original particle.
But from a global viewpoint, what is happenning is that the system
is in a superposition of states in which the particle did not yet decay,
states in which the clock is running (low measure), and states in which
the free energy from the decay has long been used up. Slowly, the
amplitudes of these terms shift toward the final states. In the limit as
the lifetime of the particle approaches infinity, it approaches an energy
eigenstate and the dynamics become frozen, and the measure of states in
which the clock is running goes to zero.
This doesn't solve the problem of time, but it is still cool that
the clock sees no effect on its operations until the absolute limit of
an energy eigenstate is reached. Perhaps time, to it, is better defined
in terms of its clock states than in terms of actual time.
Wolfram Schroers
> In article <7tdao7$bkh$1...@rosencrantz.stcloudstate.edu>,
> ba...@galaxy.ucr.edu (John Baez) wrote:
> > >Naively one would expect that the radiation increases as he
> > >approaches the horizon.
> > No! A person freely falling into a large black hole will
> > see no Hawking radiation whatsoever as he crosses the event
> > horizon.
> However, I don't understand this point fully yet: If a person falls
> towards a black hole close to evaporation it might still evaporate before
> he reaches it [...] Where has its mass gone if it hasn't passed
> him in the form of radiation?
Apparently nobody answered to my original question. So I try to solve this
problem myself ...
If the infalling observer doesn't see any Hawking radiation, the hole's
mass will not vary and there is no problem. The same is true for an
orbiting observer - no acceleration and thus no change in mass. However,
the distant observer is accelerating (to hold his position) and thus he
will see a change in mass.
The only situation in which a problem occurs is when two observers meet
and compare their ideas of the hole's mass. This reminds me somehow of the
twin paradox - in this situation the amount of proper time the two persons
have experienced varies, because one has undergone some acceleration and
the other doesn't.
So let's consider a similar thought experiment, but this time in a curved
spacetime with a black hole sitting in the middle (well, it needn't even
be a black hole, just a huge mass would do the job equally well). Two
starships orbit at some distance and have devices to measure the mass at
any time. Now one starship starts its engines and slows down his
tangential velocity, but tries to keep his distance to the central body -
finally he will no longer be orbiting, but stay at a fixed position and
burn his fuel to hold his position. His collegue will pass him several
times, BUT one of them will see Hawking radiation, and the other one
won't. One is wasting his fuel, the other one isn't.
After some time, the accelerating pilot does again some manouver to return
to his orbit and aligns with his collegue to exchange data on the central
bodie's mass.
Well, what would they find? It may either be that they find different
results and the central mass has changed depending on the manouver of the
starships. Or it may be that the mass is the same as it has ever been and
the radiation one pilot has seen was merely due to his manouver, but had
no influence on the central body.
If you vote for the second option, then a black hole won't radiate away
due to the Hawking radiation - only an observer hovering at a fixed
position would see radiation which depends on the mass and his distance,
but this should not influence the hole's mass itself.
What do you think? Is this just a bunch of nonsense or does it make any sense?
______________________________________________________________________
Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> |
Institute for Theoretical Physics, BUGH Wuppertal |
Harry Johnston
>>... suppose someone jumps in and
>> crosses the event horizon at a point when the event horizon radius (so
>> to speak) is 1km. One final photon is emitted exactly at the event
>> horizon.
>
> A classical point particle at the event horizon moving outwards at the
> speed of light would, by definition, stay right on the event horizon.
Just to make things absolutely clear: this is also true when the event
horizon is shrinking due to black hole evaporation?
Harry.
---
Harry Johnston, om...@ihug.co.nz
One Earth, One World, One People
[Moderator's note: The definition of an "event horizon" becomes
murkier for an evaporating black hole. However, when people speak of
such a thing they are usually talking about a lightlike surface; so,
yes, this would still be true. -MM]
Oz
writes
>
>I don't, but I suppose it has a perverse charm. It just
>goes to show that you can make stuff seem weird by using
>weird coordinates.
A small question.
Whilst one can clearly pick any wierd co-ordinate system one might feel
the urge to do so I presume that some are more sensible than others.
I just wonder if the co-ordinates used really (in more sensible cases
anyway) reflect the viewpoint of a particular observer. So that the
Schwarzschild co-ordinates reflect that seen by a distant stationary
observer. If so what do the Kruskal co-ordinates represent? The
viewpoint of a freely falling observer? Or am I way off on a tangent
here (again).
--
Oz
Paul Arendt
>Wolfram Schroers wrote:
>> ba...@galaxy.ucr.edu (John Baez) wrote:
(someone uncited wrote:)
>> > >Naively one would expect that the radiation increases as he
>> > >approaches the horizon.
>> > No! A person freely falling into a large black hole will
>> > see no Hawking radiation whatsoever as he crosses the event
>> > horizon.
>[...]
>> However, I don't understand this point fully yet: If a person falls
>> towards a black hole close to evaporation it might still evaporate before
>> he reaches it [...] Where has its mass gone if it hasn't passed
>> him in the form of radiation?
This is a nifty paradox. Perhaps a discussion of the scales involved
can shed some light on its resolution.
Notice that John Baez said a "large" black hole. If a black hole is
close to evaporation, then its mass is of order the Planck mass!
Since the Planck mass is so small, the size of a black hole which
is near evaporation is also very small -- a few times the Planck
length (10^{-33} cm). So you don't want to use a "person" to fall
in! If you'd like to invoke the equivalence principle for a
freely-falling observer, the spatial extent of the observer's world
has to be smaller than the characteristic length scales for tidal
variations... really really really really really really small!
(Tidal forces at the event horizon go up as the mass of the hole
goes *down*, so you're more likely to survive -- for a short while
-- passing the event horizon of a supermassive hole than a small
hole.)
Because of the small mass and size, you're not likely to notice
the gravity due to a small black hole nearby (unless it directly
collides with you), any more than you're likely to notice the
gravity of a nearby grain of dust. (You might notice the Hawking
radiation, however!!)
And because of the short length scale for tidal forces, the
freely falling observer (who sees no Hawking radiation) has to
have a small spatial extent -- where "small" means compared to
the Planck mass! So the resolution of the paradox is going to
involve quantum effects of spacetime; Hawking radiation involves
quantum matter/energy fields, but classical GR spacetime.
Barry Adams
On 20 Oct 1999 09:32:04 GMT, ba...@math.ucr.edu (john baez) wrote:
>Ah, that's what you might think at first - that's why the Wheeler-DeWitt
>equation is also called the "frozen formalism". But all the constraint
>H = 0 really means is this: in the absence of a god-given best notion of
>time, we are forced to think of a solution of Einstein's equation as
>a geometry for *all of spacetime*. Nothing physical about this geometry
>changes if we change coordinates. In particular, if we choose some time
>coordinate t, and then "push t forwards" by defining t' = t + c, nothing
>about the spacetime geometry is any different. Ultimately, this is why
>we get the equation H = 0.
Ok fine, so we've solved H psi = 0 in the frozen formalism, and we
got how wave function of the universe. Now how we getting any
predictions/retrodictions out of it?
We presemably get eigenstates for each possible geometry of all
spacetime. So we got a big set of completely deterministic universes
but we don't ever get enough information to say which one were
in. Where's the prediction? where the dynamics?
I just read Julian Barbour's end of time, he insteads the eigenstates
are for relative positions of particles in space - not space time.
Then goes on to somehow introduce an notion of time for an
observer. But i've seen nothing to indicate how that time is defined,
or to say anything as to why we observe any consistant laws from
one moment to next in this model.
Barry Adams
Phillip Helbig
<vess...@baton.phys.lsu.edu> writes:
> If gravitons are the carriers of gravity, then in almost empty space one
> would expect no gravitons and no radiation around an observer at 'rest'.
> Now if this observer fires his thruster he would undergo an acceleration
> and thus he should see gravitons and 'other radiation' around him
> popping up from the vacuum. (note: the vacuum fluctuations should be
> described by Lorentz (Poincaré) invariant function in the 'rest'
> frame). Now consider an observer on the surface of a planet with the
> equivalent gravity.
>
> Should he observe the same 'other radiation'?
>
> To save the equivalence principle we assume that there will be the same=
> 'other radiation'. Therefore, the gravity seems to be more about the
> vacuum fluctuations of the 'other radiation' and how such fluctuations
> propagate. Humm!
Related is perhaps the following problem, which as far as I know has not
been resolved, or if so I don't understand the resolution.
Assume the equivalence principle is true. We know that accelerating
charges radiate. We do not observe charges at rest in a gravitational
field radiating. So can one look at a charge, see if it is radiating,
and determine, completely locally, if one is accelerating or in a
gravitational field?
Readers might want to read gr-qc/9303025 before following up! Comments
on this work most appreciated!
--
Phillip Helbig Email ......... p.he...@jb.man.ac.uk
University of Manchester Tel. ... +44 1477 571 321 (ext. 2635)
Jodrell Bank Observatory Fax ................ +44 1477 571 618
Macclesfield Telex ................ 36149 JODREL G
UK-Cheshire SK11 9DL Web ... http://www.jb.man.ac.uk/~pjh/
************************ currently working at *******************************
Kapteyn Instituut Email (above preferred) hel...@astro.rug.nl
Rijksuniversiteit Groningen Tel. ...................... +31 50 363 4067
Postbus 800 Fax ....................... +31 50 363 6100
NL-9700 AV Groningen
The Netherlands Web ... http://gladia.astro.rug.nl/~helbig/
My opinions are not necessarily those of either of the above institutes.
t...@rosencrantz.stcloudstate.edu
Phillip Helbig <p.he...@jb.man.ac.uk> wrote:
>Assume the equivalence principle is true. We know that accelerating
>charges radiate. We do not observe charges at rest in a gravitational
>field radiating. So can one look at a charge, see if it is radiating,
>and determine, completely locally, if one is accelerating or in a
>gravitational field?
This is a very interesting puzzle, in my opinion. Unfortunately,
I don't think it has a resolution, for a reason I'll suggest
at the end of this post.
By the way, I advise anyone who wants to think seriously about this
question to resolve at the outset to be scrupulously, ridiculously
precise in defining the terms. In particular, a specific, operational
definition of the verb "to radiate" is crucial. What experiment are
you going to do to answer the question, "Is it radiating?"
I mention this not solely to be pedantic (although as a professional
pedant I reserve the right to be pedantic at will), but because it's
actually relevant to the discussion. People frequently talk past each
other on this subject because they haven't decided exactly what they
mean by the terms. In particular, note that the most common
definition of "radiate" involves the electric and magnetic fields far
from the source (in, oddly enough, the "radiation zone"); for the
present discussion, as Phillip rightly observes, this won't do: a
local definition is required.
Such definitions do exist -- for instance, you can integrate the
Poynting vector over an infinitesimal sphere -- but you've got to be
extremely careful about how you set things up.
>Readers might want to read gr-qc/9303025 before following up! Comments
>on this work most appreciated!
I agree that people who want to think about this puzzle should read
the above paper. It's the one by Stephen Parrott, called something
like "Radiation from a uniformly accelerated charge and the
equivalence principle."
(Does anyone know if it was ever published, by the way? It seems to
have been revised last in 1994, but last time I did a search of the
published literature it hadn't ever appeared. Maybe there were
referee troubles. I think that'd be a shame, since, although I
disagree with his final conclusion, I do think it's a well-argued,
thought-provoking paper that's much better than lots of things that do
get published.)
Parrott's main point, as I recall it -- it has been a while since I
studied the paper closely -- is that it's possible to distinguish
between a uniformly accelerated charge and a charge at rest in a
uniform gravitational field because the force required to hold them up
will be different. Imagine little rocket engines attached to the two
charges to support them (accelerating the one and holding the other at
rest in the gravitational field). Since the accelerated charge is
radiating energy, there'll be a radiation reaction force back on it
(the argument goes), and its little rocket will have to burn more fuel
than the other rocket. This provides a way to distinguish, via a
purely local experiment, between the two cases.
In other words, Parrott's criterion (or at least a sufficient
condition) for "radiating" is "feeling a nonzero radiation reaction
force." At least, that's my understanding. I hope someone will
correct me if I'm misinterpreting him.
The problem is that we don't know for sure that it's true that the
accelerated charge feels a radiation reaction force. In order to
determine whether there is or not, we need an equation of motion for a
classical point charge. Unfortunately, there is no entirely
satisfactory such equation of motion; this is the dirty little secret
of classical electrodynamics. In fact, the closest thing we have to a
classical theory of charged point particles, namely the Lorentz-Dirac
equation, predicts that there is no radiation reaction force on a
uniformly accelerated charge, so there's no problem with the
equivalence principle.
Parrott rejects the Lorentz-Dirac equation, but he neither replaces it
with a specific alternative nor argues convincingly (to me, anyway)
that this specific prediction of the equation is necessarily wrong.
That's why I find his argument ultimately unconvincing.
There are various problems with the Lorentz-Dirac equation. It has
obviously unphysical "runaway" solutions, which is annoying. Worse
yet, as Parrott shows in another paper, there are some physically
reasonable situations for which the Lorentz-Dirac equation has *no*
physically acceptable solutions (all solutions are "runaways"). So
Parrott's skepticism about the applicability of the equation can
certainly be justified. However, I don't think he's argued
convincingly that it's necessarily invalid in the sorts of situations
that come up in the equivalence-principle discussion, or that, even
assuming it is invalid, the radiation reaction force is necessarily
nonzero for uniform acceleration.
I'm not saying that either of those claims is necessarily *untrue*,
merely that he hasn't shown that they *are* true. Since he's the
one making the controversial claim, I think the burden of proof
is on him in this case.
In fact, I don't know whether I want to assign a truth-value at all to
the statement that a uniformly accelerated point charge feels a
radiation reaction force. Before doing so, at the risk of being
Clinton-esque, I'd want to know what the meaning of "true" is. It
clearly doesn't mean "in agreement with physical reality," since there
are no classical point charges in physical reality (since the world
isn't classical). If there were a standard, generally-accepted theory
of classical point charges (preferably one that was a good
approximation to reality in the classical limit), then "true" could mean
"in agreement with the predictions of this theory," but unfortunately,
there is no known satisfactory theory of classical point charges, and
there may never be one.
It's entirely possible -- even likely, in my opinion -- that the
notion of a point charge and classical electrodynamics don't fit in
with each other well enough for a satisfactory theory to exist. If
that's true, then I don't think that a satisfactory answer can be
found to the question, "Is there a radiation reaction force on a
classical point charge?" If that's true, then there's no satisfactory
answer to the question, "Do classical point charges violate the
equivalence principle?" The equivalence principle is by its nature
local (meaning that we need to talk about point charges), so I suspect
we're just out of luck.
- -Ted
John Baez
Barry Adams <bad...@adept77.demon.co.uk> wrote:
>Ok fine, so we've solved H psi = 0 in the frozen formalism, and we
>got how wave function of the universe. Now how we getting any
>predictions/retrodictions out of it?
Well, if you know the wavefunction of the universe, you're smarter
than I am! The equation H psi = 0 presumably has lots of solutions,
and I don't know which one describes our universe. Hawking and Hartle
have proposed a specific solution as "the wavefunction of the universe",
but even if they're right, this is only the right wavefunction to use
*before you've measured anything* - i.e., it's analogous to what
probability theorists call the "a priori probability distribution",
the one you use when you don't know any better.
The Hartle-Hawking wavefunction is presumably a superposition of lots
of wavefunctions which all satisfy H psi = 0 but describe specific
"branches" of the universe (to borrow some terminology from the
many-worlds interpretation, without necessarily endorsing that
interpretation in detail). In some of these solutions psi there
is a high chance that a person named John Baez with blond hair
types this sentence. In others he has brown hair. In others
he's named Fred and types some other sentence. In most, he
doesn't exist at all. So unless you're doing cosmology, where the
existence of puny people like me doesn't much matter, to find
the relevant solution of the Wheeler-DeWitt equation you need to
work a bit! For example, you can take the Hartle-Hawking
solution and "condition" it. Here I'm using the language of
probability theory, but what I really mean is: project the
wavefunction down to the subspace of states where the observables
you've measured take the values you measured them to have.
Suppose you do this and get some wavefunction psi.
Once you do this, you can take any observable A and try to work
out the expectation value <psi, A psi>. This is how you make
predictions, just as in ordinary quantum mechanics!
>We presumably get eigenstates for each possible geometry of all
>spacetime.
Be careful. Geometrical observables won't all commute in quantum gravity,
so we can't find an eigenstate of *all* these observables simultaneously.
If we find an eigenstate of *some* of them, this will not be an
eigenstate of others. Nothing new here - just quantum mechanics.
>So we got a big set of completely deterministic universes
>but we don't ever get enough information to say which one were
>in.
First of all, there's nothing "completely deterministic" going
on here - for the reason explained in the previous paragraph.
If we know some things for sure in the wavefunction psi, others
will be uncertain.
Secondly, we need to put in the information about "what our
world is like" in the process of choosing the wavefunction psi
in the first place - as I described earlier.
>Where's the prediction? where the dynamics?
Don't worry, it's there - I sorta sketched just now how it goes.
In principle it makes sense; it's just the DETAILS of ACTUALLY
DOING these calculations that gets hard. For example: how do
we describe the observables? (See the old thread on sci.physics.
research about precisely that topic!)
John Baez
Jacques M. Mallah <jqm...@is2.nyu.edu> wrote:
> I'm still hoping John Baez will say more on the topic.
More? More?!?
Looking back on this thread, it seems my last post in response
to you never appeared. Is that true? Just in case, let me
post it again. Hopefully this will satisfy you - at least as
much as I can possibly satisfy you on a question that remains
one of the biggest unsolved problems in physics!
........................................................................
Newsgroups: sci.physics.research
Subject: Re: The problem of time
From: ba...@galaxy.ucr.edu (John Baez)
References: <7sb1gh$hbu$1...@nnrp1.deja.com> <LLLP3.46$Jc3....@typhoon.nyu.edu>
Organization: U. C. Riverside
In article <LLLP3.46$Jc3....@typhoon.nyu.edu>,
Jacques M. Mallah <jqm...@is2.nyu.edu> wrote:
>If I wanted to get technical I could do some work on my thesis instead
>of distracting myself with this stuff :-(
Yeah - thinking about quantum gravity is not the obvious best way to
take a relaxing break from your thesis!
>Perhaps you could explain a bit about how the two major
>approaches, canonical QG (as you use it) and string theory, deal with the
>problem?
For the most part, string theory deals with this issue by ignoring
it for now and hoping that it will eventually be solved. Of course
the string theorists wouldn't put it that way! If you read my recent
conversation with Jacques Distler here on sci.physics.research, you'll
see major differences of opinion on this issue.
A more precise and less unkind way to describe the string-theoretic
approach is as follows. Typically one starts by choosing a spacetime
manifold with a "background metric": a solution of Einstein's equations
(or more precisely, the supersymmetric analogue of Einstein's equations).
One then describes strings moving around in this background metric. If
this metric has a symmetry in the timelike direction, one can talk about
"time evolution" for strings just as one can in ordinary quantum mechanics
or ordinary quantum field theory on Minkowski spacetime.
Of course, people from the general relativity community find this
approach unsatisfactory. If quantum gravity is a theory in which the
metric is a quantum field of some sort, why should we start by choosing
a fixed "background metric" on spacetime? I think the best justification
of this approach is that 1) it gives interesting results, and 2) it's hard
to know what else to do.
Canonical quantum gravity deals with this issue by worrying about it
endlessly, trying lots of things that don't quite work, and never solving
it. The Hamiltonian is treated as a constraint. But it's not easy to
make this constraint into a well-defined operator on a well-defined
Hilbert space. Various ways to do this have been proposed in loop
quantum gravity - see for example:
http://vishnu.nirvana.phys.psu.edu/mog/mog8/node7.html
for my summary of Thiemann's approach. However, all these ways
have problems. Moreover, if you could formulate the Hamiltonian
constraint as an operator and find solutions (which represent
"quantum 4-geometries") you would still need to extract the dynamics
from the theory - to "thaw the frozen formalism", as it were.
On the bright side, in certain toy models like quantum gravity
in 3-dimensional spacetime, one can actually do all this stuff:
in particular, one can "thaw the frozen formalism" and see how
dynamics is contained in the solutions to the Hamiltonian constraint.
Personally, this issue has pushed me towards a version of loop
quantum gravity which deals with spacetime rather than space from
the very start: theories of this sort are called "spin foam models".
They are more related to the Feyman diagram approach to quantum
field theory than the Hamiltonian approach.
>> In particular, if you start by using general relativity to describe a
>> closed universe and use the standard recipe to work out the Hamiltonian,
>> you get an expression that you can show equals ZERO by using Einstein's
>> equation.
>>
>> In other words: in general relativity, the Hamiltonian is a *constraint*.
> I couldn't help but notice the caveat 'closed'. I know SR is a
>special case of GR so I assume that H=0 is not the case for an open
>universe.
Yes, a Hamiltonian can be defined for asymptotically flat solutions of
general relativity - see the physics FAQ:
http://math.ucr.edu/home/baez/energy_gr.html
This Hamiltonian is given as integral over the "sphere at spatial
infinity". I.e., it's a "boundary term" left over from doing an
integration by parts. When space is a closed manifold, there's no
boundary term when the equations of motion hold, so instead of a
Hamiltonian there's just a Hamiltonian constraint.
>Could it be that in QG the only nontrivial universes are open?
Who knows?
>How does the problem of time look for the open case?
One can certainly attempt to formulate a Hamiltonian (not just
a Hamiltonian constraint) for asymptotically flat quantum gravity.
I've written about this, and so has Thiemann, and so have other
people. This might be especially useful for studying things like
a small black hole in a big empty region of spacetime that can be
approximated as asymptotically flat.
>After all, astronomers tell us that our universe will not recollapse.
Right - but they don't say it's asymptotically flat.
It may be good to separate out issues of quantum cosmology from issues
of "local" physics. (Some people argue otherwise.)
>> >If it's an energy eigenstate, there would be no dynamics.
>> Ah, that's what you might think at first - that's why the Wheeler-DeWitt
>> equation is also called the "frozen formalism". But all the constraint
>> H = 0 really means is this: in the absence of a god-given best notion of
>> time, we are forced to think of a solution of Einstein's equation as
>> a geometry for *all of spacetime*.
> But wouldn't that be true of a 4-geometry, not a 3-geometry?
Ah, this is the sneaky part. Once you understand this issue, you'll
understand what all the fuss is about! (See below for a slight nudge
in the right direction.)
>> Nothing physical about this geometry
>> changes if we change coordinates. In particular, if we choose some time
>> coordinate t, and then "push t forwards" by defining t' = t + c, nothing
>> about the spacetime geometry is any different. Ultimately, this is why
>> we get the equation H = 0.
>> (Don't be surprised if it takes a while for this to sink in! This is
>> the big difference between "background-free physics" and conventional
>> physics where you fix the geometry of spacetime at the very beginning.)
> OK, but at this point I don't get it. Why is the above statement
>not true in, for example, SR?
There's a fundamental difference between "gauge symmetries" and ordinary
symmetries. In special relativity time translation is an ordinary
symmetry: it's generated by an observable called the Hamiltonian, and
this observable is not necessarily zero. In general relativity time
translation is a "gauge symmetry": it's generated by an observable,
there's also a constraint saying this observable is zero!
To understand this stuff, you have to understand the classical
mechanics of constrained systems. It takes a while to explain.
Briefly: if someone hands you a Lagrangian, you can try to work out
the corresponding Hamiltonian. But sometimes when you do this, you
find terms that consist of a Lagrange multiplier times a constraint.
These constraints generate "gauge symmetries" of your classical
mechanics problem.
If you do this to Maxwell's equations, you find that the usual
gauge transformations of the vector potential are gauge symmetries,
but time translation is not. If you do it to Einstein's equations,
you find that time translation is a gauge symmetry!
Of course, it takes a while to explain why this is such a big deal.
I had a huge long discussion about this on sci.physics.research
with Greg Weeks and other people. I'm sort of reluctant to repeat
such a long story, so I urge you to read what we said using Deja.com.
The thread was called "Re: string theory" and it happened around
4/16/1999.
>> So the trick is to extract dynamics from the frozen formalism, or
>> else to develop some other, "unfrozen", formalism.
> Right. How would one extract it from the frozen formalism?
Another huge question - it's a bit like "how do we end world hunger?"
I'm too tired to answer this one right now!
>> No. If you ponder the quantization of constrained systems
>> long enough, you'll conclude that the constraint H = 0 should
>> really be imposed as an operator constraint H psi = 0, not just
>> a constraint on expectation values.
> I see that for a regular constraint, but this case is a bit
>unusual and it would seem worthwhile to try relaxing the contraint. Maybe
>the case H ~= 0 is just singled out in the classical limit for a
>macroscopic universe, for example. Given the large uncertainty in the age
>of our universe, the required small uncertainty in H does not sound
>unreasonable to me.
Okay, write a paper about this! It will join the sea of papers where
people try all sorts of clever tricks to deal with the problem of time.
>> I'm glad you find this as strange as I did when I first learned
>> about it! I sympathize with you immensely.
> Well, thanks for your help. Incidently, we have had an
>acquaintance in common, Henry Flynt. I taught him some physics.
Huh! He's a strange and interesting guy - he was the one who
invented the term "conceptual art" (later shortened to "concept
art"), and he's a Marxist how makes his living by investing in
precious metals. I helped him learn the proof of Goedel's theorem,
and he helped me find a girlfriend. I haven't seen him for a long
time.
>At least, I want to know what QG developments to root for
>or against, if nothing else.
Root for me!
Raymond E. Rogers
>
> Related is perhaps the following problem, which as far as I know has not
> been resolved, or if so I don't understand the resolution.
>
> charges radiate. We do not observe charges at rest in a gravitational
> field radiating. So can one look at a charge, see if it is radiating,
> and determine, completely locally, if one is accelerating or in a
> gravitational field?
>
> on this work most appreciated!
I think you are right about the equivalence principle breaking down, but
it is supposed to break down for physical affects involving the second
derivatives of the metric tensor. It is a tangent space affirmation.
I have been working on proving that acceleration ( and hopefully
geodesic motion in a gravitational field with certain natural
configurations) produces an inline electric field along the accelerated
path, and that this electric field is exactly the "back reaction". Also
I hope to mathematically prove that this field together with the
classical incoming and outgoing radiation fields yield the classical
equations.
Actually this should be a general effect of Wave equations, and the
process might have other applications. Unfortunately my results so far,
while affirming, are lame and are causing me to relearn many things I
thought I understood.
Ray
[Moderator's note: Quoted text trimmed. -MM]
Wolfram Schroers
> >> > No! A person freely falling into a large black hole will
> >> > see no Hawking radiation whatsoever as he crosses the event
> >> > horizon.
> >> However, I don't understand this point fully yet: If a person falls
> >> towards a black hole close to evaporation it might still evaporate before
> >> he reaches it [...] Where has its mass gone if it hasn't passed
> >> him in the form of radiation?
> Notice that John Baez said a "large" black hole. If a black hole is
> close to evaporation, then its mass is of order the Planck mass!
> Since the Planck mass is so small, the size of a black hole which
> is near evaporation is also very small -- a few times the Planck
> length (10^{-33} cm). So you don't want to use a "person" to fall
> in! If you'd like to invoke the equivalence principle for a
> freely-falling observer, the spatial extent of the observer's world
> has to be smaller than the characteristic length scales for tidal
> variations... really really really really really really small!
Well, the black hole under consideration (no charge, no angular momentum)
is characterized by only one scale, its mass parameter. So the physical
processes should not change qualitatively if the mass gets smaller (at
least as long as we stay far above the Planck mass). Assuming the validity
of the calculations for a large black hole (MUCH larger than the Planck
scale), we still encounter similar paradoxes (cf. my recent post and
below). If it is true that Hawking radiation is just a frame-dependant
effect and does not really alter the black hole's mass, then a large black
hole can never become a Planck-mass object in the first place.
> And because of the short length scale for tidal forces, the
> freely falling observer (who sees no Hawking radiation) has to
> have a small spatial extent -- where "small" means compared to
> the Planck mass! So the resolution of the paradox is going to
> involve quantum effects of spacetime; Hawking radiation involves
> quantum matter/energy fields, but classical GR spacetime.
Of course, you are right, the argumentation is not applicable to
final-state black holes, but I can also construct variants of the paradox
which do not involve Planck-scale objects. You can find one in my recent
post, here is another one:
An observer hovers with a starship at a distance from a large massive
object (again it needn't be a black hole; any other large mass should do
the job equally well). He will see some Hawking radiation; how much,
depends on his distance - I think everybody agrees on this point so far.
Now an astronaut jumps out and falls freely towards the central body. If
it is true that he sees no Hawking radiation during his fall, then the
central bodies' mass will not change. However, if it was also true that
the body loses mass by Hawking radiation, then the observer in the
starship will measure a mass becoming smaller as time passes.
So again we encounter either the concept of a frame-dependant mass or a
body can not lose mass by Hawking radiation.
Vesselin G Gueorguiev
[...]
> A more precise and less unkind way to describe the string-theoretic
> approach is as follows. Typically one starts by choosing a spacetime
> manifold with a "background metric": a solution of Einstein's equations
> (or more precisely, the supersymmetric analogue of Einstein's equations).
> One then describes strings moving around in this background metric. If
> this metric has a symmetry in the timelike direction, one can talk about
> "time evolution" for strings just as one can in ordinary quantum mechanics
> or ordinary quantum field theory on Minkowski spacetime.
>
> Of course, people from the general relativity community find this
> approach unsatisfactory. If quantum gravity is a theory in which the
> metric is a quantum field of some sort, why should we start by choosing
> a fixed "background metric" on spacetime? I think the best justification
> of this approach is that 1) it gives interesting results, and 2) it's hard
> to know what else to do.
3) Almost everything we do we do it with mean fields.
[...]
> http://math.ucr.edu/home/baez/energy_gr.html
Is this link right? I can't find it.
[...]
> >After all, astronomers tell us that our universe will not recollapse.
>
> Right - but they don't say it's asymptotically flat.
Some of them say that it is FLAT, surprise! (of course you know that)
See Bahcall, Ostriker, Perlmutter, & Steinhardt,
"The Cosmic Triangle: Revealing the State of the Universe",
Science 284, 1481, 28 May 1999
Here is the abstract of a talk she gave at LSU, 30 Sep 1999:
Neta Bahcall, "The Cosmic Triangle: The Mass, Acceleration,
and Curvature of the Universe"
The Cosmic Triangle is a new way of representing the past, present, and
future status of the Universe. Our current location within the Cosmic
Triangle is determined by the answers to three fundamental questions:
How much matter is in the Universe? Is the expansion rate slowing down or
speeding up? And, is the Universe flat or curved? By combining recent
observations of distant supernovae, clusters of galaxies, and the cosmic
blackbody radiation, we find mounting evidence suggesting that we live in a
surprisingly lightweight Universe. The observations also suggest that our
Universe may be "flat", with some sort of "dark energy" opposing the
attractive force of gravity. The measurements and their implications for the
dark energy, the unseen dark matter that makes up much of the mass of the
Universe, and the observed expansion rate of the Universe will be discussed.
Getting two lines to intercept is easy (much more chance then to be parallel),
but getting a third line right through the intersection of the other two is
some what indicative of something special.
[...]
> There's a fundamental difference between "gauge symmetries" and ordinary
> symmetries. In special relativity time translation is an ordinary
> symmetry: it's generated by an observable called the Hamiltonian, and
> this observable is not necessarily zero. In general relativity time
> translation is a "gauge symmetry": it's generated by an observable,
> there's also a constraint saying this observable is zero!
[...]
> Of course, it takes a while to explain why this is such a big deal.
> I had a huge long discussion about this on sci.physics.research
> with Greg Weeks and other people. I'm sort of reluctant to repeat
> such a long story, so I urge you to read what we said using Deja.com.
> The thread was called "Re: string theory" and it happened around
> 4/16/1999.
[...]
> >> If you ponder the quantization of constrained systems
> >> long enough, you'll conclude that the constraint H = 0 should
> >> really be imposed as an operator constraint H psi = 0, not just
> >> a constraint on expectation values.
> >Maybe the case H ~= 0 is just singled out ...
I was just wondering, what if we have some mechanism of fluctuations
that let us briefly do something like this:
Psi_{E=0}-->Psi_{E!=0}-->Phi_{E=0}
thus we jump from one geometry to another by virtual excitation
to state Psi_{E!=0}.
Jacques M. Mallah
> Jacques M. Mallah <jqm...@is2.nyu.edu> wrote:
> > I'm still hoping John Baez will say more on the topic.
>
> to you never appeared. Is that true?
Yes.
> From: ba...@galaxy.ucr.edu (John Baez)
> Jacques M. Mallah <jqm...@is2.nyu.edu> wrote:
>
> >If I wanted to get technical I could do some work on my thesis instead
> >of distracting myself with this stuff :-(
>
> Yeah - thinking about quantum gravity is not the obvious best way to
> take a relaxing break from your thesis!
Now you tell me.
> >Perhaps you could explain a bit about how the two major
> >approaches, canonical QG (as you use it) and string theory, deal with the
> >problem?
>
> A more precise and less unkind way to describe the string-theoretic
> approach is as follows. Typically one starts by choosing a spacetime
> manifold with a "background metric": a solution of Einstein's equations
> (or more precisely, the supersymmetric analogue of Einstein's equations).
> One then describes strings moving around in this background metric. If
> this metric has a symmetry in the timelike direction, one can talk about
> "time evolution" for strings just as one can in ordinary quantum mechanics
> or ordinary quantum field theory on Minkowski spacetime.
Then the background metric, I take it, is not affected by what the
strings do. That doesn't make too much sense to me because you said it's
a solution of Einstein's equations, which depend on the mass distribution.
I could understand if it just played a role like that of the static metric
in the 'flat interpretation of GR' thread, but I thought that had trouble
dealing with with black hole formation.
> Of course, people from the general relativity community find this
> approach unsatisfactory. If quantum gravity is a theory in which the
> metric is a quantum field of some sort, why should we start by choosing
> a fixed "background metric" on spacetime? I think the best justification
> of this approach is that 1) it gives interesting results, and 2) it's hard
> to know what else to do.
I think what you are saying is that, in the string approach, the
background metric is fixed but there is also a variable metric field. If
so, I don't see anything wrong with that except that I thought it could
not handle black hole formation. (Of course, I wouldn't mind if GR was
modified on the macroscopic level to eliminate black holes, but I don't
think string theory does that.)
> Canonical quantum gravity deals with this issue by worrying about it
> endlessly, trying lots of things that don't quite work, and never solving
> it. The Hamiltonian is treated as a constraint. But it's not easy to
> make this constraint into a well-defined operator on a well-defined
> Hilbert space. [...]
>
> Moreover, if you could formulate the Hamiltonian
> constraint as an operator and find solutions (which represent
> "quantum 4-geometries") you would still need to extract the dynamics
> from the theory - to "thaw the frozen formalism", as it were.
>
> On the bright side, in certain toy models like quantum gravity
> in 3-dimensional spacetime, one can actually do all this stuff:
> in particular, one can "thaw the frozen formalism" and see how
> dynamics is contained in the solutions to the Hamiltonian constraint.
Why didn't you say so before! So, how does it work in 2+1
spacetime? I suppose, in principle, it would work about the same way in
3+1 but be harder to carry out.
> Personally, this issue has pushed me towards a version of loop
> quantum gravity which deals with spacetime rather than space from
> the very start: theories of this sort are called "spin foam models".
> They are more related to the Feyman diagram approach to quantum
> field theory than the Hamiltonian approach.
Are you saying that you consider psi as a functional on
4-geometries?
> a Hamiltonian can be defined for asymptotically flat solutions of
> general relativity - see the physics FAQ:
> http://math.ucr.edu/home/baez/energy_gr.html
http://math.ucr.edu/home/baez/physics/energy_gr.html
> One can certainly attempt to formulate a Hamiltonian (not just
> a Hamiltonian constraint) for asymptotically flat quantum gravity.
> I've written about this, and so has Thiemann, and so have other
> people. This might be especially useful for studying things like
> a small black hole in a big empty region of spacetime that can be
> approximated as asymptotically flat.
>
> >After all, astronomers tell us that our universe will not recollapse.
>
> Right - but they don't say it's asymptotically flat.
But what about the (hyperbolic) non-flat open case? From the
above it looks like any open case would do.
> It may be good to separate out issues of quantum cosmology from issues
> of "local" physics. (Some people argue otherwise.)
Well, eventually it must all be put in a single theory.
> There's a fundamental difference between "gauge symmetries" and ordinary
> symmetries. In special relativity time translation is an ordinary
> symmetry: it's generated by an observable called the Hamiltonian, and
> this observable is not necessarily zero. In general relativity time
> translation is a "gauge symmetry": it's generated by an observable,
> there's also a constraint saying this observable is zero!
So in GR in asymptotically flat spacetime, it is not a "gauge
symmetry"? That sounds strange.
> To understand this stuff, you have to understand the classical
> mechanics of constrained systems. It takes a while to explain.
> Briefly: if someone hands you a Lagrangian, you can try to work out
> the corresponding Hamiltonian. But sometimes when you do this, you
> find terms that consist of a Lagrange multiplier times a constraint.
> These constraints generate "gauge symmetries" of your classical
> mechanics problem.
OK, I should take a closer look at that.
> Of course, it takes a while to explain why this is such a big deal.
> I had a huge long discussion about this on sci.physics.research
> with Greg Weeks and other people. I'm sort of reluctant to repeat
> such a long story, so I urge you to read what we said using Deja.com.
> The thread was called "Re: string theory" and it happened around
> 4/16/1999.
I looked at some of it (I think).
I don't see why it's a big deal though. I know gauge symmetries
help with renormalization in QFT, and I know classical GR has them. For
these reasons, I think, they are popular. What I don't know is why people
have grown so attached to them. I see no philosphical reason to favor
them. In fact, I would be rather pleased if the gauge symmetry is broken
because it would horrify both GR people and particle physics people :-)
> >> No. If you ponder the quantization of constrained systems
> >> long enough, you'll conclude that the constraint H = 0 should
> >> really be imposed as an operator constraint H psi = 0, not just
> >> a constraint on expectation values.
>
> > I see that for a regular constraint, but this case is a bit
> >unusual and it would seem worthwhile to try relaxing the contraint. Maybe
> >the case H ~= 0 is just singled out in the classical limit for a
> >macroscopic universe, for example. Given the large uncertainty in the age
> >of our universe, the required small uncertainty in H does not sound
> >unreasonable to me.
>
> Okay, write a paper about this! It will join the sea of papers where
> people try all sorts of clever tricks to deal with the problem of time.
Maybe I'll try, after I finish my thesis.
Actually I don't know if H =/=0 would be so bad even in classical
cosmology. Perhaps you know, or can help me find out: is it possible the
model with the GR Hamiltonian H but without the constraint, violating
Einstein's equations, could mimic the effects (as far as astronomers can
tell) of a cosmological constant and/or dark matter?
(Correct me if I'm wrong, but I think that would mean violating
the time-related portion of the equations but keeping the other equations,
and allowing a greater range of initial conditions. And breaking that
nasty diffeomorphism invariance!)
> >At least, I want to know what QG developments to root for
> >or against, if nothing else.
>
> Root for me!
I have my own reasons for being interested in QG (as I see it as a
subset of the more important problem of interpretation of QM), but of
course I do root for you - if you succeed in solving the problems of
quantum gravity, you'll be able to explain it!
Vesselin G Gueorguiev
>
> Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> wrote:
> >Wolfram Schroers wrote:
> >> ba...@galaxy.ucr.edu (John Baez) wrote:
> (someone uncited wrote:)
> >> > >Naively one would expect that the radiation increases as he
> >> > >approaches the horizon.
> >> > see no Hawking radiation whatsoever as he crosses the event
> >> > horizon.
[...]
> Notice that John Baez said a "large" black hole. If a black hole is
> close to evaporation, then its mass is of order the Planck mass!
[...]
Somewhat different but related thing.
If gravitons are the carriers of gravity, then in almost empty space
one would expect no gravitons and no radiation around an observer at
'rest'. Now if this observer fires his thruster he would undergo an
acceleration and thus he should see gravitons and 'other radiation'
around him popping up from the vacuum. (note: the vacuum fluctuations
should be described by Lorentz (Poincar ) invariant function in the
'rest' frame). Now consider an observer on the surface of a planet
with the equivalent gravity.
Should he observe the same 'other radiation'?
To save the equivalence principle we assume that there will be the
same 'other radiation'. Therefore, the gravity seems to be more about
Zachary Uram
and Spin J. Is there no more properties? How can one tell if the
black hole has these properities and how does one measure them
for a given black hole? Is this spin the same as angular
momentum?
John Baez
<t...@rosencrantz.stcloudstate.edu> wrote:
>In article <wolfram-0711...@dialppp-1-46.rz.ruhr-uni-bochum.de>,
>Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> wrote:
>>An observer hovers with a starship at a distance from a large massive
>>object (again it needn't be a black hole; any other large mass should do
>>the job equally well). He will see some Hawking radiation;
Not according to the calculations people have done.
>My understanding is that you don't get
>Hawking radiation unless you've got a horizon. I found this fact so
>astonishing and strange when someone (probably Baez) said it that it
>stuck in my head.
Yeah, it's surprising. But then, Hawking radiation is surprising
in the first place! I don't really understand this fact myself, but I was
assured that it's true by Bill Unruh, who is one of the real experts on
Hawking radiation, Rindler radiation, and the like. We can ask in various
situations whether such radiation exists or not - let's call all these forms
of radiation "acceleration radiation", for lack of a better name. And we
can make the question operational by asking whether an atom in various
situations will or will not be excited by acceleration radiation. Unruh
told me:
1) An atom sitting on the earth will not be excited by acceleration
radiation.
2) An atom freely falling near the earth will (with some very small
probability) be excited by acceleration radiation.
3) An atom at constant distance from a black hole will (with some
small probability) be excited by acceleration radiation.
4) An atom freely falling near the horizon of a black hole will not
be excited be acceleration radiation (in the approximation where it's
right at the horizon).
5) An atom freely falling or at rest far from a black hole will
(with some small probability) be excited by acceleration radiation.
>>Now an astronaut jumps out and falls freely towards the central body. If
>>it is true that he sees no Hawking radiation during his fall, then the
>>central bodies' mass will not change.
>I'm not sure I believe this. Actually, I'm not sure I know exactly
>what it means.
I guess it's a bit ambiguous in the first place, but I don't know
any good way to make it into into a true statement, so I'd just say
it's false.
I think one needs to reason very carefully about these situations
if one wants to get the right answers - raw intuition can easily lead
us astray, since quantum field theory on curved spacetime is not
something we have direct experience of.
Chris Hillman
On 12 Nov 1999, Zachary Uram wrote:
> I read that black holes only have 3 properties: Mass M, Charge Q
> and Spin J.
You are thinking of the unicity theorem about the Kerr-Newman spacetime.
> Is there no more properties?
Real astrophysical objects may not be axially symmetric, or may have a
magnetic dipole moment. On the theoretical side, one can speculate about
spherically symmetric static solutions featuring kind of scalar field, or
Yang-Mills field. The unicity result you are thinking of does not apply
in these situations, although there are some more general unicity results
(see Frolov & Novikov, Black Hole Physics).
Here are some simple examples of static axisymmetric vacuums other than
the Schwarzschild vacuum (which is static spherically symmetric, a
stronger property):
Chazy-Curzon:
ds^2 = f^2 [-dt^2 + g^2 (dz^2 + dr^2)] + exp(-2m/sqrt(z^2+r^2)) r^2 du^2
-infty < t, z < infty, r > 0 , -pi < u < pi
f = exp(-m/sqrt(z^2+r^2)
g = exp(m^2 r^2/(2 (z^2+r^2)^2)
Bardeen-Horowitz:
ds^2 = f^2 [-(1+r^2) dt^2 + dr^2/(1+r^2) + du^2] + sin u (dv+rdt)^2/f^2
-infty < t < infty, r > 0, 0 < u < pi, -pi < v < pi
f = sqrt(3+cos(2u))/2
Hopefully I haven't mistyped anything. (In both cases, d/dt and d/du are
Killing vectors, if you know what that means: hence the phrase "static
axisymmetric".)
Also, be aware that the Kerr-Newman generalization of the Kerr solution
has been further generalized by many other authors to a five or six
parameter family which includes many of the most interesting known exact
solutions in gtr.
The solutions I've mentioned all have singularities. Here's a vacuum
solution which is, in a slightly different sense, cylindrically symmetric,
but which doesn't have any singularities:
ds^2 = 2 f dw du + f^2 du^2 + cosh^2 u dx^2 + sinh^2 u dy^2, (*)
-infty < t < infty, u > 0, -infty < x < infty, -pi < y < pi
f = sqrt(cosh(2u))/k
Compare H^3 in cylindrical coordinates:
ds^2 = du^2 + cosh^2 u dx^2 + sinh^2 u dy^2, (**)
u > 0, -infty < x < infty, -pi < y < pi
An easy exercise is to check that the line element I gave above really
does describe H^3. Use this parameterization of H^3 as a hypersurface in
E^(1,3) with signature -+++:
[ cosh u cosh x ]
[ cosh u sinh x ]
[ sinh u cos y ]
[ sinh u sin y ]
Compute the three tangent vectors d/du, d/dx, d/dy, and then compute the
various inner products in E^(1,3). The result is the induced metric (**).
This should be visualized via stereographic projection to E^3:
[x1] 1 [ cosh u sinh x ]
[x2] = ----------------- [ sinh u cos y ]
[x3] 1 + cosh u cosh x [ sinh u sin y ]
In the solution (*), the cylinders are the static wavefronts of a
"plane-fronted" gravitational wave. A harder exercise is to integrate the
geodesic equations, which can be done in closed form. This shows that all
test particles move past these static wavefronts at the local speed of
light, as you'd expect. Typically, they spiral from u = infinity into u =
0 and then back out again.
Chris Hillman
Home Page: http://www.math.washington.edu/~hillman/personal.html
t...@rosencrantz.stcloudstate.edu
Zachary Uram <zu...@andrew.cmu.edu> wrote:
>I read that black holes only have 3 properties: Mass M, Charge Q
Yes. According to standard theory, these are the only observable
properties of a black hole. This is what people mean when they
say (quite cryptically IMO) that "black holes have no hair."
>How can one tell if the
>black hole has these properities and how does one measure them
>for a given black hole?
Let's take them one at a time. I'll give a simplified procedure for
measuring each of the three quantities, and include some slightly more
detailed descriptions as parenthetical comments labeled "Fine print."
Mass: You can measure the mass of a black hole in a number of
different ways. For instance, you can put a satellite in a large
circular orbit around the black hole, measure the period and radius of
the orbit, and use Kepler's third law.
(Fine print: Here "large" means "many times the Schwarzschild radius."
Otherwise, Newtonian gravity, which is really what Kepler's law is,
isn't an accurate approximation. You can't actually measure the
radius of the orbit, since the center of the orbit is at the location
of the black hole; to find the radius, you measure the circumference
and divide by 2 pi.)
Charge: If the black hole is not rotating significantly, You can
measure the charge by measuring the electric field at a point a known
distance r from the black hole and applying Coulomb's law.
(Fine print: Again, r is the circumference of a circle passing
through the point, divided by 2pi.)
If the black hole has significant angular momentum, then strictly
speaking Coulomb's law doesn't apply (because the situation's not
spherically symmetric). You can still use Gauss's law, though.
Measure the E field over a closed surface enclosing the black hole.
(A sphere is probably simplest, but any shape is OK.) Integrate the
normal component of E over this surface, and Gauss's law tells you the
electric charge just as it would in ordinary electrostatics.
(Fine print: Using Coulomb's law would probably be OK if you did the
experiment with a large value of r: even for rotating black holes, at
large distances the electric field is dominated by the monopole term,
which is described by Coulomb's law. Actually, pretty much the same
comments apply to the mass measurement.)
Spin: This is a bit more complicated than the others. There are some
general-relativistic effects that go by names like "frame dragging"
and "gravitomagnetism" that are important in describing rotating black
holes (or other rotating massive objects). I may be misremembering
the details here, but doubtless an expert will pop up to correct me if
I get things too terribly wrong.
Roughly speaking, objects that "try to stay still" outside a rotating
black hole get "pulled around" in the direction of rotation. For
instance, if you put an object in a circular orbit that takes it over
the axis of rotation of the black hole (i.e., not an equatorial
orbit), the orbit will precess. If you put two objects in equatorial
orbits, one going with the direction of rotation of the black hole and
one going against it, they'll have different orbital periods.
Finally, if you put a gyroscope in orbit, its rotation axis will
precess.
This last one is supposed to be tested in Earth orbit some time soon.
For a relatively low-mass, slow-spinning body like the Earth, the
effect should be pretty weak, but people say they'll be able to
measure it.
These effects depend on how much angular momentum the black hole has,
so by measuring them you can work out the angular momentum.
>Is this spin the same as angular
>momentum?
Yes. In fact, it's probably better to call it angular momentum rather
than spin. The term "spin" has connotations of quantum mechanics,
whereas these effects need not (and for large black holes almost
certainly *do* not) have anything to do with quantum mechanics.
-Ted
Vesselin G Gueorguiev
[...what is the mass of the central body...]
This is an interesting question, I guess some astronomers could
tell better then me how to measure it.
>
> There happen, by chance, to be a bunch of observers standing on
> a sphere centered on the black hole. They measure the surface
One thing, it is not obvious to me how to assure these observers
as on a sphere without measuring the radius.
> area of their sphere and use it to figure out the sphere's radius.
> In other words, they measure their Schwarzschild-r coordinate.
> Assume that the black hole is evaporating slowly enough that they
> can do this measurement and compare notes in a time short enough
> that the geometry hasn't changed significantly [1]. At the same
> time as they make this measurement, our hapless astronaut falls
> past. They yell out to him the value of r they've just determined.
> At that instant, he measures the tidal force on him and uses the
> standard Schwarzschild formula for tidal forces to figure out
> the mass of the black hole.
May I suggest to our brave explorer to do two consecutive measurements
of the tidal force on him at r1 and r2, then deduce the mass as
as the "difference" in the observed tidal force, so we have:
Result1=TidalForce(r1,M)
Result2=TidalForce(r2,M)
s=DistanceBetween(r1,r2)
> That sounds quite complicated, but I honestly can't think of any
> simpler experiment the astronaut can do that would yield a number that
> could reasonably be called a measurement of the black hole's mass.
Me too.
Toby Bartels
>I read that black holes only have 3 properties: Mass M, Charge Q
>and Spin J. Is there no more properties? How can one tell if the
>black hole has these properities and how does one measure them
>for a given black hole? Is this spin the same as angular
>momentum?
The last question first: yes. J is the hole's angular momentum.
The theorem you cite is often expressed as "Black holes have no hair.",
meaning black holes have no structure, only gross characteristics.
For example, there is no way to tell if a black hole is the result
of a collapse of a bunch of neutrons, or of a bunch of neutrinos.
If Q = 0, you know it's wasn't a bunch of protons or a bunch of electrons,
but it might have been a bunch of protons *and* electrons.
You can't look inside a black hole to see what makes it up.
Black holes are not made up of stuff; they're just there.
Black holes have position, speed, energy, and momentum,
but different observers will measure different values,
so the only classical, observer independent values are M, Q, and J.
You measure M by measuring the strength of the curvature the hole causes,
you measure J by measuring the strength of frame dragging effects,
and you measure Q by measuring the electrostatic force it causes.
If quark colours could be separated,
a black hole might be produced by the collapse of only red quarks;
then you could measure a hole's redness, blueness, and greenness
by measuring the strength of the strong force it produced.
Of course this is not a part of the classical theory.
Even classically, there is temperature T from statistical mechanics.
But, since a hole has no structure, it has no entropy, so T = 0.
Quantumly, we know a black hole actually has non0 temperature,
but Hawking's calculation shows T is a function of M, Q, and J,
so they are still the only *mathematically independent* characteristics.
-- Toby
to...@ugcs.caltech.edu
Toby Bartels
>Assume the equivalence principle is true. We know that accelerating
>charges radiate. We do not observe charges at rest in a gravitational
>field radiating. So can one look at a charge, see if it is radiating,
>and determine, completely locally, if one is accelerating or in a
>gravitational field?
Do we observe charges at rest in a gravitational field radiating
if we are ourselves in free fall, not at rest in the field?
-- Toby
to...@ugcs.caltech.edu
Keith Ramsay
t...@rosencrantz.stcloudstate.edu writes:
|It's entirely possible -- even likely, in my opinion -- that the
|notion of a point charge and classical electrodynamics don't fit in
|with each other well enough for a satisfactory theory to exist.
I read described somewhere the approach of taking some field theory
which has bound distributions of charge of positive size, and then
taking a limit where the size goes to zero. Does this break down in
some known way, or has it just not been made to work?
|If
|that's true, then I don't think that a satisfactory answer can be
|found to the question, "Is there a radiation reaction force on a
|classical point charge?" If that's true, then there's no satisfactory
|answer to the question, "Do classical point charges violate the
|equivalence principle?" The equivalence principle is by its nature
|local (meaning that we need to talk about point charges), so I suspect
|we're just out of luck.
If so, it seems like we should at least be able to analyze a
"small-sized particle" approximation to both principles.
Keith Ramsay
Keith Ramsay
ba...@galaxy.ucr.edu (John Baez) writes:
|1) An atom sitting on the earth will not be excited by acceleration
|radiation.
|2) An atom freely falling near the earth will (with some very small
|probability) be excited by acceleration radiation.
|3) An atom at constant distance from a black hole will (with some
|small probability) be excited by acceleration radiation.
|4) An atom freely falling near the horizon of a black hole will not
|be excited be acceleration radiation (in the approximation where it's
|right at the horizon).
|5) An atom freely falling or at rest far from a black hole will
|(with some small probability) be excited by acceleration radiation.
So in some sense, the predicted situation around the Earth isn't the
same as the situation at the same distance from a black hole of the
same mass, even though the classical solution to GR is the same away
from the central mass?
Lightlike geodesics near the Earth don't seem to converge or diverge
very much as you go into the past, and even if the Earth were very
much more compact, they would converge to only a limited degree right
near the center.
In a nonrotating black hole solution, it seems to me one of the more
striking features of having a horizon, is that there are past-directed
lightlike geodesics converging on the horizon. Light emitted at
various distances just outside the horizon and directed outward
spreads out in space, with the time required to leave being very
sensitive to exactly how close to the horizon it was.
I don't see offhand what the situation is with other past-directed
geodesics, whether a lot of them converge on the horizon (and back to
the event at which the horizon formed?), or whether this feature
depends upon having the geodesic lined up in a precisely radial
direction, say. Would the ones directed close enough to the hole trace
back to a near orbit of the hole, which slowly escapes to where you
are?
Incidentally, can one predict from which direction the Hawking
radiation would appear to be coming from?
We need some nonlocal feature to explain why the situations differ,
and differ in such a sharp way, with all the situations up to just
short of having a horizon differing from one in which there is a
horizon. For me, this sort of "lensing in reverse" property would do
a lot to make it more plausible. If in the classical setting, what you
see as you look around in many directions would be remnants of
radiation from one event, or at least to a relatively localized piece
of space-time, it seems to me intuitively plausible that this would be
the relevant global feature of the space-time you are in, to explain
also why in the quantum case the EM field is in an unusual state. I
just don't know whether it's actually relevant.
Keith Ramsay
Vesselin G Gueorguiev
> I have an old preprint (10/97) gr-qc/9710084 which discusses the
> application of the pilot wave interpretation to the minisuperspace (MSS)
> model. This paper claims that trajectories can be recovered which look
> like the classical ones for macroscopic universes, but does not seem to
> discuss the Hamiltonian constraint, which is not imposed. The
> wavefunction in initially a Gaussian in the radius.
If the Hamiltonian constraint is not imposed how do you know that it is
satisfied by the state? (May be I should have a look at this paper)
> So at least for MSS models, it looks like the constraint does not
> need to be imposed, and it will be approximately true in the classical
> limit as long as the wavefunction is chosen suitably.
Well this "suitably" sounds a little too vague.
Vesselin G Gueorguiev
> >and spin J.
> The theorem you cite is often expressed as "Black holes have no hair.",
> meaning black holes have no structure, only gross characteristics.
> For example, there is no way to tell if a black hole is the result
> of a collapse of a bunch of neutrons, or of a bunch of neutrinos.
> If Q = 0, you know it's wasn't a bunch of protons or a bunch of electrons,
> but it might have been a bunch of protons *and* electrons.
> You can't look inside a black hole to see what makes it up.
Seems reasonable.
> Black holes are not made up of stuff; they're just there.
Well we don't know that yet! However, your post made me rethink my
previous post where I was adding isospin (hyper charge) to the list
of things that may characterize a black hole. [*]
[...]
> If quark colours could be separated,
> a black hole might be produced by the collapse of only red quarks;
> then you could measure a hole's redness, blueness, and greenness
> by measuring the strength of the strong force it produced.
> Of course this is not a part of the classical theory.
[*] Notice that M & Q are related to interactions with infinite range,
in some sense we may expect that properties of short range interactions
could not characterize BH in general. Thus we may not be able to measure
characteristics related to the strong and weak interaction.
We may even have some problem with measuring Q as well, because if we have
Q inside the BH how would we know about it! No photons get out of the BH,
and non of our photons gets back when we probe the BH. I really wonder,
how do we measure Q?
We can measure M from curvature of space and my guess for measuring J
is related to the frame dragging effects (which I don't understand yet).
Vesselin G Gueorguiev
John Baez wrote:
>
> In article <80hji5$48h$1...@rosencrantz.stcloudstate.edu>,
> <t...@rosencrantz.stcloudstate.edu> wrote:
>
> >In article <wolfram-0711...@dialppp-1-46.rz.ruhr-uni-bochum.de>,
> >Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> wrote:
>
> >>An observer hovers with a starship at a distance from a large massive
> >>object (again it needn't be a black hole; any other large mass should do
> >>the job equally well). He will see some Hawking radiation;
What is Hawking radiation anyway? [1]
> Not according to the calculations people have done.
>
> >My understanding is that you don't get
> >Hawking radiation unless you've got a horizon.
So my understanding is that since we have "large massive object
(again it needn't be a black hole) ...." we don't see Hawking radiation
since we are missing its main generation component, the BH horizon [1].
[...]
> Hawking radiation, Rindler radiation, and the like. We can ask in various
> situations whether such radiation exists or not - let's call all these forms
> of radiation "acceleration radiation", for lack of a better name.
I have no idea what is Hawking radiation, and Rindler radiation,
but at least for Hawking radiation, I guess, it is the outcome from the
Hawking process [1] which really needs BH horizon.
Then only through the equivalence principle I can make some "connection"
with something which we may call "acceleration radiation" [2].
> And we can make the question operational by asking whether an atom in
> various situations will or will not be excited by acceleration radiation.
> Unruh told me:
>
> 1) An atom sitting on the earth will not be excited by acceleration
> radiation.
acceleration radiation from where?
> 2) An atom freely falling near the earth will (with some very small
> probability) be excited by acceleration radiation.
I really didn't get this.
> 3) An atom at constant distance from a black hole will (with some
> small probability) be excited by acceleration radiation.
> 4) An atom freely falling near the horizon of a black hole will not
> be excited be acceleration radiation (in the approximation where it's
> right at the horizon).
> 5) An atom freely falling or at rest far from a black hole will
> (with some small probability) be excited by acceleration radiation.
I have been reading 3,4,5 over and over, and it seems that an atom outside
of the horizon will always be excited by 'acceleration radiation'
(with some small probability). We can think of any motion as composed of
short free falls and brief frizzes at constant distances. It is like
jumps between different free fall trajectories. We really need energy to
perform these jumps (something to hold us at constant distance).
If I have understood correctly, we can say that an atom will not
be excited if it is freely falling, since if it were, then it wouldn't
be freely falling, but a freely falling atom have natural chance of
spontaneous excitation which we can blame on some "acceleration radiation".
[1] Hawking process as described by Schutz on p.305 involves the BH
horizon and the Heisenberg uncertainty principle Delta(E)Delta(t)>=hbar.
So, near the BH horizon we get some *short* time violation of the
energy conservation by Delta(E), it may be a pair of photons or
neutrinos, or something else. Then in some cases we get one of the
created particles moving toward the BH horizon and some times even
pass the horizon, when this happen the other particle can go free.[*]
Thus we can have a process that generates some particles in vicinity
of the BH horizon, I guess these particles are part of the Hawking
radiation.
[2] I would thing of something called "acceleration radiation" as
something due to object moving with acceleration. For example,
a charged particle moving with acceleration.
[*] I don't quite understand how it is possible to get these particles
free, this would be an obvious violation of the energy conservation for
times much longer then Delta(t), and even more, since it has happen
near the BH horizon this 'free' particles will need infinite amount of
time to get to some distant observer far, far away from the BH. Thus no
way of interacting and observing them. Even more, what I see then is a BH
getting more mass, thus increasing in radius, but by no mean 'evaporating'!
John Baez
Keith Ramsay <kra...@aol.commangled> wrote:
>In article <80kfbh$f...@charity.ucr.edu>,
>ba...@galaxy.ucr.edu (John Baez) writes:
>|1) An atom sitting on the earth will not be excited by acceleration
>|radiation.
>|2) An atom freely falling near the earth will (with some very small
>|probability) be excited by acceleration radiation.
>|3) An atom at constant distance from a black hole will (with some
>|small probability) be excited by acceleration radiation.
>|4) An atom freely falling near the horizon of a black hole will not
>|be excited be acceleration radiation (in the approximation where it's
>|right at the horizon).
>|5) An atom freely falling or at rest far from a black hole will
>|(with some small probability) be excited by acceleration radiation.
>So in some sense, the predicted situation around the Earth isn't the
>same as the situation at the same distance from a black hole of the
>same mass, even though the classical solution to GR is the same away
>from the central mass?
Right. Very crudely speaking, the radiation in case 3) "comes from the
event horizon". I sort of understand how this case works, but I don't
understand case 1), which is actually more complicated. This means I
I don't really understand the whole business well enough to give wise
advice about it. So I won't say much....
>Incidentally, can one predict from which direction the Hawking
>radiation would appear to be coming from?
From the black hole.
>We need some nonlocal feature to explain why the situations differ,
>and differ in such a sharp way, with all the situations up to just
>short of having a horizon differing from one in which there is a
>horizon.
Right. The usual calculation in case 3) is quite global in nature
and relies crucially on the existence of a horizon. Similarly
for case 6), which I forgot to mention:
5) An atom moving at constant velocity in flat empty space will
not be excited by acceleration radiation.
6) An atom accelerating at a constant rate as measured in its
own instantaneous rest frame will (with some probability) be
excited by acceleration radiation.
John Baez
Jacques M. Mallah <jqm...@is2.nyu.edu> wrote:
>John Baez (ba...@galaxy.ucr.edu) wrote:
>> A more precise and less unkind way to describe the string-theoretic
>> approach is as follows. Typically one starts by choosing a spacetime
>> manifold with a "background metric": a solution of Einstein's equations
>> (or more precisely, the supersymmetric analogue of Einstein's equations).
>> One then describes strings moving around in this background metric. If
>> this metric has a symmetry in the timelike direction, one can talk about
>> "time evolution" for strings just as one can in ordinary quantum mechanics
>> or ordinary quantum field theory on Minkowski spacetime.
> Then the background metric, I take it, is not affected by what the
>strings do.
Right.
>That doesn't make too much sense to me because you said it's
>a solution of Einstein's equations, which depend on the mass distribution.
The background is actually a solution of the *vacuum* Einstein equations
(or more precisely, their supersymmetric analogue). The idea behind
this approach is that we're treating the strings *perturbatively*.
It's a lot like old-fashioned perturbative quantum gravity, where we fix
a solution g_0 of the vacuum Einstein equations and write the metric as
g_0 plus a small perturbation:
g = g_0 + h
and then quantize only the h field. It's a sensible thing to try.
The main problem with it is that to any order in perturbation theory,
the h field will propagate within the lightcones defined by the background
metric g_0. In other words, no signals travel faster than light, but
"faster than light" is defined using the background metric g_0, *not*
the actual metric, g. This is only a small problem if g and g_0 are
close, but it's a big problem if they're not close.
This is one reason I don't do string theory - I can't stomach the idea
of fixing an unphysical background metric at the start, and letting that
determine how causality works. Someday somebody will come up with a
better approach and then maybe I'll do string theory.
>I could understand if it just played a role like that of the static metric
>in the 'flat interpretation of GR' thread, but I thought that had trouble
>dealing with with black hole formation.
Right, it's exactly the same business. Starting with the Minkowski
space as your background, you're never gonna see a radically different
causal structure on spacetime, like an event horizon. To study an
event horizon in this approach you have to put it in from the start,
when you choose your background metric.
>> Of course, people from the general relativity community find this
>> approach unsatisfactory. If quantum gravity is a theory in which the
>> metric is a quantum field of some sort, why should we start by choosing
>> a fixed "background metric" on spacetime? I think the best justification
>> of this approach is that 1) it gives interesting results, and 2) it's hard
>> to know what else to do.
> I think what you are saying is that, in the string approach, the
>background metric is fixed but there is also a variable metric field. If
>so, I don't see anything wrong with that except that I thought it could
>not handle black hole formation.
It can't. More generally, it can't address any situation where you're
interested in how the "variable metric" defines a different causal
structure from the "fixed background metric". E.g., when you have a
quantum state that's a superposition of two states with drastically
different metrics.
Of course, if you talk to a string theorist about these issues, you'll
get answers that sound very different from mine. You'll typically have
to press them very, very hard to get them to admit I'm right. If you
aren't careful, they may just throw enough jargon at you that you'll feel
stupid and give up. :-)
>> On the bright side, in certain toy models like quantum gravity
>> in 3-dimensional spacetime, one can actually do all this stuff:
>> in particular, one can "thaw the frozen formalism" and see how
>> dynamics is contained in the solutions to the Hamiltonian constraint.
> Why didn't you say so before!
I guess I was too busy blabbing about *other* things.
>So, how does it work in 2+1 spacetime?
Try:
Steven Carlip, Quantum Gravity in 2+1 Dimensions, Cambridge
University Press, 1998.
>I suppose, in principle, it would work about the same way in
>3+1 but be harder to carry out.
No, it's very different in 2+1 dimensions, because general
relativity in 2+1 dimensions has no local degrees of freedom: all
solutions of the vacuum Einstein equations are *flat* in this case.
>> Personally, this issue has pushed me towards a version of loop
>> quantum gravity which deals with spacetime rather than space from
>> the very start: theories of this sort are called "spin foam models".
>> They are more related to the Feyman diagram approach to quantum
>> field theory than the Hamiltonian approach.
> Are you saying that you consider psi as a functional on
>4-geometries?
No, I'm saying that spin foams are like higher-dimensional analogues
of Feynman diagrams, and that we compute transition amplitudes by
summing over spin foams. (But the meaning of "transition amplitude"
is subtler, thanks to the problem of time.)
About energy in general relativity.....
> But what about the (hyperbolic) non-flat open case? From the
>above it looks like any open case would do.
You can *define* total energy in any reasonably nice open universe,
and it'll usually be nonzero, but I think it's usually conserved only
when you have time translation symmetry at spacelike infinity - since
it's symmetries that give conservation laws.
I'm not 100% sure about this - I'd need to scribble a bit to be sure,
and I'm too busy now.
>> It may be good to separate out issues of quantum cosmology from issues
>> of "local" physics. (Some people argue otherwise.)
> Well, eventually it must all be put in a single theory.
Yup. But eventually I'll be dead.
>> There's a fundamental difference between "gauge symmetries" and ordinary
>> symmetries. In special relativity time translation is an ordinary
>> symmetry: it's generated by an observable called the Hamiltonian, and
>> this observable is not necessarily zero. In general relativity time
>> translation is a "gauge symmetry": it's generated by an observable,
>> there's also a constraint saying this observable is zero!
> So in GR in asymptotically flat spacetime, it is not a "gauge
>symmetry"? That sounds strange.
Strange but true! An extremely important fact, too. It's just
another way of saying that there's a nonzero Hamiltonian in
asymptotically flat spacetime.
> Actually I don't know if H =/=0 would be so bad even in classical
>cosmology. Perhaps you know, or can help me find out: is it possible the
>model with the GR Hamiltonian H but without the constraint, violating
>Einstein's equations, could mimic the effects (as far as astronomers can
>tell) of a cosmological constant and/or dark matter?
I don't know.
> (Correct me if I'm wrong, but I think that would mean violating
>the time-related portion of the equations but keeping the other equations,
>and allowing a greater range of initial conditions. And breaking that
>nasty diffeomorphism invariance!)
The Hamiltonian constraint is the 00 component of Einstein's equation
G_{ab} = 8 pi k T_{ab}
If you want to throw out this component, you either have to throw out
all the rest or destroy diffeomorphism-invariance, since diffeomorphisms
mix the different components. You may think diffeomorphism invariance
is "nasty", but I don't! It's one of the basic ideas behind Einstein's
equation, and if you throw it out, it's not clear how much is left.
At the very least, this is a major project. But personally, I recoil
in horror from this whole idea, just as if you asked me to take Maxwell's
equations and throw out the one that says
partial_x E_y - partial_y E_x = - d B_x / dt
while keeping all the rest. If you want to go this route, I can't
help you.
Vesselin G Gueorguiev
> I read that black holes only have 3 properties: Mass M, Charge Q
> and Spin J. Is there no more properties?
I am not sure but isospin (hyper charge) should get into this list too.
> How can one tell if the
> black hole has these properties and how does one measure them
> for a given black hole?
I leave this one for the experts.
> Is this spin the same as angular momentum?
Yes, although I am not sure how one defines total angular momentum
in this case.
[Moderator's note: isospin and hypercharge should *not* be on the
list. - jb]
Ralph E. Frost
>
> In article <807eii$u92$1...@rosencrantz.stcloudstate.edu>,
> t...@rosencrantz.stcloudstate.edu writes:
>
> If so, it seems like we should at least be able to analyze a
> "small-sized particle" approximation to both principles.
Do you mean a specific structure or pattern of particles or a composite?
--
Best regards,
Ralph E. Frost
http://www.dcwi.com/~refrost/index.htm ..Feeling is believing..
Keith Ramsay
to...@ugcs.caltech.edu (Toby Bartels) writes:
|You measure M by measuring the strength of the curvature the hole causes,
|you measure J by measuring the strength of frame dragging effects,
|and you measure Q by measuring the electrostatic force it causes.
How about measuring J by dropping objects of known angular momenta
(generally opposite the direction of the angular momentum of the black
hole, as detected by frame dragging effects) into the hole until the
frame dragging effect goes away? That way it doesn't depend upon
knowing details.
Likewise, measure the charge by dropping opposite charges into it
until it stops having an electric field, and measure the mass by
adding mass to it until the effects of its gravitational field match
that of a standard "benchmark" hole, created by inducing a known
mass of ordinary matter to collapse. :-)
Keith Ramsay
Vesselin G Gueorguiev
Keith Ramsay wrote:
>
> In article <80kfbh$f...@charity.ucr.edu>,
> ba...@galaxy.ucr.edu (John Baez) writes:
> |1) An atom sitting on the earth will not be excited by acceleration
> |radiation.
> |2) An atom freely falling near the earth will (with some very small
> |probability) be excited by acceleration radiation.
> |3) An atom at constant distance from a black hole will (with some
> |small probability) be excited by acceleration radiation.
> |4) An atom freely falling near the horizon of a black hole will not
> |be excited be acceleration radiation (in the approximation where it's
> |right at the horizon).
> |5) An atom freely falling or at rest far from a black hole will
> |(with some small probability) be excited by acceleration radiation.
>
> So in some sense, the predicted situation around the Earth isn't the
> same as the situation at the same distance from a black hole of the
> same mass, even though the classical solution to GR is the same away
> from the central mass?
This is strange, very strange! Any arguments other then #1-#5 above that
can make this less contra-intuitive?
John Baez
Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>Things are getting more and more confusing....
Good - that's usually a sign that one is about to learn
something interesting!
>What is Hawking radiation anyway? [1]
It's the radiation emitted by the event horizon of a black hole.
For a nice explanation of how it works, see Wald's "General Relativity".
For a less technical explanation, see the FAQ that I wrote:
http://math.ucr.edu/home/baez/physics/hawking.html
>> >My understanding is that you don't get
>> >Hawking radiation unless you've got a horizon.
>So my understanding is that since we have "large massive object
>(again it needn't be a black hole) ...." we don't see Hawking radiation
>since we are missing its main generation component, the BH horizon [1].
That sounds right - but I don't understand this stuff as well
as I should, so don't completely trust me.
>> Hawking radiation, Rindler radiation, and the like. We can ask in various
>> situations whether such radiation exists or not - let's call all these forms
>> of radiation "acceleration radiation", for lack of a better name.
> I have no idea what is Hawking radiation, and Rindler radiation, [...]
I just explained what Hawking radiation is - though not how it works.
Rindler radiation is similar, but simpler: it's the radiation seen by
an observer accelerating through flat Minkowski spacetime at a constant
rate (as measured in his own rest frame at any given moment). Here
the "horizon" responsible for the radiation is the "Rindler horizon":
the surface separating the region of spacetime that this accelerating
observer can see someday from the region he cannot ever see.
>I would think of something called "acceleration radiation" as
>something due to object moving with acceleration. For example,
>a charged particle moving with acceleration.
No - I am NOT using "acceleration radiation" to mean the radiation
emitted by an accelerating charged particle! I'm using it to mean
the radiation seen by an observer accelerating through empty space.
This is the quantum-field-theoretic effect that we are discussing
in this thread. The simplest examples are Hawking radiation and
Rindler radiation.
>> 1) An atom sitting on the earth will not be excited by acceleration
>> radiation.
>acceleration radiation from where?
Huh? I'm saying there ISN'T ANY acceleration radiation in this
situation. I'm just phrasing it in operational terms, by saying
that if you have an atom, it won't be excited.
>> 2) An atom freely falling near the earth will (with some very small
>> probability) be excited by acceleration radiation.
>I really didn't get this.
This is similar to Rindler radiation, but more subtle. I don't
understand it very well, but I assume it works a lot like Rindler
radiation.
>We can think of any motion as composed of
>short free falls and brief frizzes at constant distances.
What's a "frizz"? Something like a "moment of infinite acceleration"?
Toby Bartels
>[1] Hawking process as described by Schutz on p.305 involves the BH
>horizon and the Heisenberg uncertainty principle Delta(E)Delta(t)>=hbar.
Perhaps you heard the collective grown that erupted when
all the regulars on s.p.r happened (by chance) to read that at the same time.
The explanation you cite is a very crude explanation of Hawking radiation,
and it really doesn't begin to get at what HR *really* is.
Unfortunately, I don't understand it well enough myself
to give a more appropriate but equally simple explantion.
The best I can do is point to Wald's book
Quantum Field Theory in Curved Spacetime and ...,
but this book requires a good handle on QFT in *flat* spacetime to start.
Nevertheless, I'll work with it and correct the following.
>[*] I don't quite understand how it is possible to get these particles
>free, this would be an obvious violation of the energy conservation for
>times much longer then Delta(t), and even more, since it has happen
>near the BH horizon this 'free' particles will need infinite amount of
>time to get to some distant observer far, far away from the BH. Thus no
>way of interacting and observing them. Even more, what I see then is a BH
>getting more mass, thus increasing in radius, but by no mean 'evaporating'!
You have 3 problems here: violating energy conservation,
the infinite time needed, and the BH growing, not shrinking.
The first and last problems cancel each other out.
When the captured particle is absorbed by the BH,
energy conservation requires the BH to *lose* energy
in order to cancel the positive energy of the escaped particle.
Because the BH loses energy, it loses mass and evaporates.
If it seems strange that the BH loses energy by absorbing a particle,
blame Schutz for his confusing explanation.
(An explanation along Schutz's lines but involving *virtual* particles
would make this seem more reasonable but would still miss the point of HR.)
As for the middle problem, remember that the pair creation
happened *near* the EH, not *at* it.
So, only a finite (albeit very long) time is required
for the escaped particle to be detected by you and me.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
>
> In article <382CEE9A...@phys.lsu.edu>,
>
> >> There happen, by chance, to be a bunch of observers standing on
> >> a sphere centered on the black hole. They measure the surface
>
> >One thing, it is not obvious to me how to assure these observers
> >as on a sphere without measuring the radius.
>
> (meaning that it doesn't change significantly on timescales comparable
> to the radius of the sphere) and spherically symmetric. I think this
> problem can be solved in this case. Just have them all measure the
> surface gravity (rate at which they have to burn rocket fuel to hold
> themselves still) at their locations. If they all get the same
> answer, they're on a sphere.
OK I can buy this, seems as a good way to define it.
> >May I suggest to our brave explorer to do two consecutive measurements
> >of the tidal force on him at r1 and r2, then deduce the mass as
> >as the "difference" in the observed tidal force, so we have:
> >
> >Result1=TidalForce(r1,M)
> >Result2=TidalForce(r2,M)
> >s=DistanceBetween(r1,r2)
>
> I'm sorry, but I don't understand this.
No wonder, I don't understand it myself :). I mean, I don't know the
details involved in such measurements, and whether it is really plausible.
So, let assume that our brave explorer have a way to extract a number
from some kind of measurement related to the tidal force. For example,
the eccentricity of an ellipsoid (initially we have a sphere but due to
tidal force it becomes an ellipsoid). Then we compare the results of
two measurements, first measurement done at point 1, and second measurement
done at point 2, let the distance between these points be s.
Now we use our theory to derive what is the functional relation between
the measured eccentricity, mass M of the BH and position of the
measurement r, I call this Result=TidalForce(r,M). Now we have two
measurements, together with our theory that tells us how to obtain
the functional form Result=TidalForce(r,M), and how physical distance s
is obtained in the particular coordinate system where these events have
coordinates r1 and r2. This gives us ground for fitting procedure that
will give M (I am not expecting simple function for Result=TidalForce(r,M)
in an ideal situation we would be able to solve for M).
One more thing, I am talking about two measurements since r is not
really measurable it is a coordinate of some event (our measurement).
What is really measurable is 's' the distance between two events.
Thus we really need the above system of three equations since r1, r2 and
M are unknown or arbitrary till we solve all these together.
P.S. We can use something like a perfect sphere with liquid on its surface
and measure the level change in the tide on one of its sides (far or near to
the BH).
Toby Bartels
>Similarly for case 6), which I forgot to mention:
>5) An atom moving at constant velocity in flat empty space will
>not be excited by acceleration radiation.
>6) An atom accelerating at a constant rate as measured in its
>own instantaneous rest frame will (with some probability) be
>excited by acceleration radiation.
For purposes of discussing the 7 cases John has given,
I think we should all agree now to count the above
as cases 6 and 7, respectively, since 5 is already taken.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
[...]
> Likewise, measure the charge by dropping opposite charges into it
> until it stops having an electric field, and measure the mass by
> adding mass to it until the effects of its gravitational field match
> that of a standard "benchmark" hole, created by inducing a known
> mass of ordinary matter to collapse. :-)
What if our "benchmark" is smaller than the object we study? Are you going
to take mass off? :-)
The other two (Q & J) are good examples of nice ZERO output experiments.
Paul Colby
news:80k8d0$o...@gap.cco.caltech.edu...
quantized amplitudes multiplying classical solutions of the field equations
(the normal modes). So one could ask two seperate questions. First, does the
charge radiate classically. Classically, a charge radiates if its far field
falls off like 1/r at large distance from the source. Is this true for a
charge at rest on the surface of a massive body? I think not because of
energy conservations. The second question is does the charge radiate quantum
mechanically. By this I mean will an atom in the far field have a finite
prbability of absorbing a quanta from the stationary charge. I think the
answer to this is also no since the amplitude is proportional to the
classical far field.
As far as the free fall observer is concerned the answer may well be a
matter of degree. If he wisses by this charge fast enough and close enough,
he may well see some radiation.
Regards
Paul Colby
Vesselin G Gueorguiev
> >horizon and the Heisenberg uncertainty principle Delta(E)Delta(t)>=hbar.
>
> Perhaps you heard the collective grown that erupted when
> all the regulars on s.p.r happened (by chance) to read that at the same time.
Oh that is what it was, I thought it was from the stadium near by :)
> The explanation you cite is a very crude explanation of Hawking radiation,
> and it really doesn't begin to get at what HR *really* is.
> Unfortunately, I don't understand it well enough myself
> to give a more appropriate but equally simple explantion.
> The best I can do is point to Wald's book
> Quantum Field Theory in Curved Spacetime and ...,
> but this book requires a good handle on QFT in *flat* spacetime to start.
> Nevertheless, I'll work with it and correct the following.
I am thankful for your consideration and I may really get to read
Wald's (I hope I have some good basis in QFT), but for now I have it marked.
> >[*] I don't quite understand how it is possible to get these particles
> >free, this would be an obvious violation of the energy conservation for
> >times much longer then Delta(t), and even more, since it has happen
> >near the BH horizon this 'free' particles will need infinite amount of
> >time to get to some distant observer far, far away from the BH. Thus no
> >way of interacting and observing them. Even more, what I see then is a BH
> >getting more mass, thus increasing in radius, but by no mean 'evaporating'!
>
> You have 3 problems here: violating energy conservation,
> the infinite time needed, and the BH growing, not shrinking.
Yes, these are the three things that were troubling me in the
crude explanation I cited. (I wonder why do they allow printing things
that confuse people, I am almost ready to bet that there is no way to
overcome 1 & 3 if you don't change the mechanism. For example,
using Bogoliubov transformations as in my resent post based on JB's FAQ)
However, I would like to present another line of thoughts I had
due to your message. [see the end]
> The first and last problems cancel each other out.
> When the captured particle is absorbed by the BH,
> energy conservation requires the BH to *lose* energy
Hold on! *lose* energy how? The only way I see to close
the loop of virtual particles would need to get them together
which has become impossible due to the presence of the EH.
> in order to cancel the positive energy of the escaped particle.
> Because the BH loses energy, it loses mass and evaporates.
> If it seems strange that the BH loses energy by absorbing a particle,
> blame Schutz for his confusing explanation.
I do!
> (An explanation along Schutz's lines but involving *virtual* particles
> would make this seem more reasonable but would still miss the point of HR.)
I thought his explanation _was involving_ *virtual* particles since what else
is a pair of particle-antiparticle popping up from the vacuum to spend
short (Dealta(t)) vacation near the EH.
> As for the middle problem, remember that the pair creation
> happened *near* the EH, not *at* it.
> So, only a finite (albeit very long) time is required
> for the escaped particle to be detected by you and me.
Yes I do keep this in mind, but I was thinking that the intensity of some
particular "spectral lines" will be proportional to 'how close' to the EH the
pair was created and thus how much detectable the radiation would be; and
since I am thinking about detecting then the age of the universe is also
important in estimating whether and when such radiation would be observable.
Now, in attempt to put your explanation above in my own words through my
'compiler' I came up with the following thoughts:
Notice that nothing leaves the BH, except may be gravitons that manage to get
out somehow. For example, consider bunch of photons, we usually don't have them
interact with each other (there is some QED effects), thus we cannot confine
them by the interaction they carry, but we do confine photons via gravity
or multiple interactions with matter. Thus we expect some similar situation
for the case of gravitons, everything interacts with them but they can not
be confine by the interaction they carry, this means that we assume somewhat
different gravity-matter coupling from gravity-gravity coupling which let some
gravitons escape the BH. Note, GR is non-linear theory and as such we expect
quite significant gravity-gravity coupling, but since there is no more
interactions we better assume that gravitons are confined in a way similar to
the photons, that is, via multiple interactions with matter.
Now we have one of this smart gravitons that gets out of the BH behind the EH.
There this graviton interacts with matter or suddenly decides to turn himself
into few other particles. So we got the BH to lose energy, and we got some
particles near the EH. Now some of these particles can get back into the BH,
but the rest will form the BH radiation.
I think that this mechanism solves all of the three problem I had before,
energy is conserved, the BH is losing energy and thus evaporates, and there
is no real concern where exactly this radiation is formed it can be as far
as needed from the EH provided sufficiently energetic graviton has gone behind
the EH.
Jacques M. Mallah
> This is one reason I don't do string theory - I can't stomach the idea
> of fixing an unphysical background metric at the start, and letting that
> determine how causality works. Someday somebody will come up with a
> better approach and then maybe I'll do string theory.
I see. String theorists do believe that eventually, it will be
possible to study a non-perturbative string like theory. So I suppose it
is premature to ask how string theory deals with the problem of time,
since the current techniques are not likely to be retained in more
advanced string theories (if there will be such).
> Jacques M. Mallah <jqm...@is2.nyu.edu> wrote:
> >I could understand if it just played a role like that of the static metric
> >in the 'flat interpretation of GR' thread, but I thought that had trouble
> >dealing with with black hole formation.
>
> Right, it's exactly the same business. Starting with the Minkowski
> space as your background, you're never gonna see a radically different
> causal structure on spacetime, like an event horizon. To study an
> event horizon in this approach you have to put it in from the start,
> when you choose your background metric.
OK.
From what I understood, it blows up in situations where a BH would
normally form. Is that so? I would not be concerned if it just started
giving different predictions from GR, but of course it must not get
ill-defined and singular or that is a problem.
> > I think what you are saying is that, in the string approach, the
> >background metric is fixed but there is also a variable metric field. If
> >so, I don't see anything wrong with that except that I thought it could
> >not handle black hole formation.
>
> It can't. More generally, it can't address any situation where you're
> interested in how the "variable metric" defines a different causal
> structure from the "fixed background metric". E.g., when you have a
> quantum state that's a superposition of two states with drastically
> different metrics.
I wonder if the problem could be handled using a countable number
of background metrics, each with a different horizon structure. The
wavefunction could be in a superposition of such states and transitions
could occur quantum mechanically.
> Steven Carlip, Quantum Gravity in 2+1 Dimensions, Cambridge
> University Press, 1998.
>
> >I suppose, in principle, it would work about the same way in
> >3+1 but be harder to carry out.
>
> No, it's very different in 2+1 dimensions, because general
> relativity in 2+1 dimensions has no local degrees of freedom: all
> solutions of the vacuum Einstein equations are *flat* in this case.
OK, not that I really understand that. You did say, though, that
it does have a 'frozen formalism' and it can be 'thawed'. I may take a
look at that book.
> >> Personally, this issue has pushed me towards a version of loop
> >> quantum gravity which deals with spacetime rather than space from
> >> the very start: theories of this sort are called "spin foam models".
> >> They are more related to the Feyman diagram approach to quantum
> >> field theory than the Hamiltonian approach.
>
> > Are you saying that you consider psi as a functional on
> >4-geometries?
>
> No, I'm saying that spin foams are like higher-dimensional analogues
> of Feynman diagrams, and that we compute transition amplitudes by
> summing over spin foams. (But the meaning of "transition amplitude"
> is subtler, thanks to the problem of time.)
So, psi is still a functional on 3-geometries (and the fields on
them) in your approach?
Feynmann diagrams are just another notation for perturbation
theory, as far as I know. Is that true of your approach?
Oh, by the way ... just what is the meaning of 'transition
amplitude' in your approach? (I bet you knew I would ask that:) That is
what interests me more than the mathematical details.
> About energy in general relativity.....
>
> > But what about the (hyperbolic) non-flat open case? From the
> >above it looks like any open case would do.
>
> You can *define* total energy in any reasonably nice open universe,
> and it'll usually be nonzero, but I think it's usually conserved only
> when you have time translation symmetry at spacelike infinity - since
> it's symmetries that give conservation laws.
>
> I'm not 100% sure about this - I'd need to scribble a bit to be sure,
> and I'm too busy now.
I'd be interested to know. I'm not sure what it would mean for
the Hamiltonian not to be conserved, unless it depends explicitly on time.
> > So in GR in asymptotically flat spacetime, it is not a "gauge
> >symmetry"? That sounds strange.
>
> Strange but true! An extremely important fact, too. It's just
> another way of saying that there's a nonzero Hamiltonian in
> asymptotically flat spacetime.
OK. I suppose it is also more than that, or we would not need two
ways to say the same thing. Why is gauge symmetry so important?
> > Actually I don't know if H =/=0 would be so bad even in classical
> >cosmology. Perhaps you know, or can help me find out: is it possible the
> >model with the GR Hamiltonian H but without the constraint, violating
> >Einstein's equations, could mimic the effects (as far as astronomers can
> >tell) of a cosmological constant and/or dark matter?
>
> I don't know.
I should check the literature, because if no one has looked at the
question, I'd be proud to be the first to ask it.
> > (Correct me if I'm wrong, but I think that would mean violating
> >the time-related portion of the equations but keeping the other equations,
> >and allowing a greater range of initial conditions. And breaking that
> >nasty diffeomorphism invariance!)
>
> The Hamiltonian constraint is the 00 component of Einstein's equation
> G_{ab} = 8 pi k T_{ab}
>
> If you want to throw out this component, you either have to throw out
> all the rest or destroy diffeomorphism-invariance, since diffeomorphisms
> mix the different components. You may think diffeomorphism invariance
> is "nasty", but I don't! It's one of the basic ideas behind Einstein's
> equation, and if you throw it out, it's not clear how much is left.
Hopefully, what's left is a Hamiltonian, at least in a closed
universe. Just let H=E instead of H=0.
I think diffeomorphisms of space would remain symmetric, but
spacetime diffeomorphisms would not.
> At the very least, this is a major project.
Drat.
> But personally, I recoil
> in horror from this whole idea, just as if you asked me to take Maxwell's
> equations and throw out the one that says
> partial_x E_y - partial_y E_x = - d B_x / dt
> while keeping all the rest.
Why? In that case (assuming 'throw out' means 'change' since a
complete set of equations of motion is still needed) all that would mean
is, for example, breaking rotational invariance. If there existed a
'problem of direction' with technical and conceptual problems plaguing
E&M, it would be perfectly natural to look at models with broken
rotational invariance. (BTW, I think it was this year, there were some
astronomers who alleged that the speed of light depended on direction and
they made the news. The allegation was then refuted by other astronomers.)
Look, I don't think the suggestion I made is the solution either.
But I _do_ think it's worth checking out. It's true that Occam's razor
favors symmetry. So our Bayesian prior probability should be higher for
the more symmetric model, but only by a finite factor. That factor can be
overcome if the more complicated lack of symmetry is compensated by making
other problems less complicated to solve.
That's similar to what happenned when experimental evidence forced
people to overcome their prejudices and stop ignoring that fact that the
symmetry of parity is broken.
Besides, at this point in my career, I'd rather be known as a guy
who was wrong, than not to be known at all.
> If you want to go this route, I can't help you.
Oh, you don't have to be so modest.
Thanks for all the help so far, and I think I speak for all
readers of the sci.physics hierarchy in saying that.
- - - - - - -
Jacques Mallah (jqm...@is2.nyu.edu)
Graduate Student / Many Worlder / Devil's Advocate
"I know what no one else knows" - 'Runaway Train', Soul Asylum
My URL: http://pages.nyu.edu/~jqm1584/
Toby Bartels
>We may even have some problem with measuring Q as well, because if we have
>Q inside the BH how would we know about it! No photons get out of the BH,
>and non of our photons gets back when we probe the BH. I really wonder,
>how do we measure Q?
No gravitons get out either, yet we can measure M.
The reason in each case is that the collapsing matter
has left the metric and EM fields outside the BH warped by its passage.
That is why we can measure M and Q.
Hypercharge, on the other hand, doesn't do this,
which is why hypercharge is part of the hair a BH doesn't have.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
I think the above #5 and the previous:
5) An atom freely falling or at rest far from a black hole will
(with some small probability) be excited by acceleration radiation.
are equivalent. Aren't they?
Vesselin G Gueorguiev
[...]
> >What is Hawking radiation anyway? [1]
>
> It's the radiation emitted by the event horizon of a black hole.
> For a nice explanation of how it works, see Wald's "General Relativity".
Aah, yea. I just took MWT from our department to read it for some year
or two, hopefully I will read it before I finish my Ph.D. and now what,
go read Wald's .... Probably, I should read the source:
Stephen W. Hawking, Particle creation by black holes,
Commun. Math. Phys. 43 (1975), 199-220.
> For a less technical explanation, see the FAQ that I wrote:
>
> http://math.ucr.edu/home/baez/physics/hawking.html
Well, this is better, I looked at the FAQ, it's easy - just click,
click and voilˆ you got what you what (almost!). Too bad, Hawking's
paper is not on the web.
So to summarize and check if I have got the main point.
For flat space the vacuum state is much better defined than in the
case of curved space. For asymptotically flat space we can
define some reasonable vacuum states at t=+-oo for observer at
spacial infinity. Now to get from the far past vacuum state to
the far future vacuum state we consider a bunch of vacuum states
between that are obtained via Bogoliubov transformations. These
Bogoliubov transformations seem to be very much like the one
in Hartree-Fock-Bogoliubov method in nuclear physics, we have
a bunch of particles and we write down 'new vacuum' (the ground state)
as some combination of particle & antiparticle operators acting on
the "old vacuum" which looks like popping up particle-antiparticle pairs
from the vacuum.
Now the BH comes along and some of our particle-antiparticle pairs
happen to be close to the EH, and thus some particles get sucked
into the BH while their partners go 'wild'.
Is that better? I hope so! This way I am not confused by
the Heisenberg's principle and violation of the energy conservation.
This explanation makes irrelevant the question haw long it will take
to observe the radiation, at least until we have really solved for the
Bogoliubov transformations. And the problem I was having before with
the BH evaporation is also gone for the same reason.
[...]
> >> Hawking radiation, Rindler radiation, ..."acceleration radiation"...
[...]
> Rindler radiation is similar, but simpler: it's the radiation seen by
> an observer accelerating through flat Minkowski spacetime at a constant
> rate (as measured in his own rest frame at any given moment). Here
> the "horizon" responsible for the radiation is the "Rindler horizon":
> the surface separating the region of spacetime that this accelerating
> observer can see someday from the region he cannot ever see.
Yup, it seems easier. It looks like 'you get half the story' the
other half you would never get its behind the curtains, but I guess
transforming the vacuum in some way as determined by the acceleration
of our observer should converge to the BH case.
[...]
> I'm using it ["acceleration radiation"] to mean
> the radiation seen by an observer accelerating through empty space.
> This is the quantum-field-theoretic effect that we are discussing
> in this thread. The simplest examples are Hawking radiation and
> Rindler radiation.
Thanks for the explanation.
>
> >> 1) An atom sitting on the earth will not be excited by acceleration
> >> radiation.
>
> >acceleration radiation from where?
>
> Huh? I'm saying there ISN'T ANY acceleration radiation in this
> situation. I'm just phrasing it in operational terms, by saying
> that if you have an atom, it won't be excited.
Yes, this is one of these situations when I get confused when we talk
about something which is not present in the situation we are talking about!
I mean talking about radiation in situation where we don't have radiation.
Thanks for the clarification.
>
> >> 2) An atom freely falling near the earth will (with some very small
> >> probability) be excited by acceleration radiation.
>
> >I really didn't get this.
>
> This is similar to Rindler radiation, but more subtle. I don't
> understand it very well, but I assume it works a lot like Rindler
> radiation.
You mean I can use the equivalence principle and consider Rindler
radiation for observer that is moving with acceleration like he was
pulled by the Earth, right?
> >We can think of any motion as composed of
> >short free falls and brief frizzes at constant distances.
>
> What's a "frizz"? Something like a "moment of infinite acceleration"?
In presence of gravity 'frizzing' at constant distance (but not time)
is the same as firing your rocket's thruster to fight the pull of gravity.
Thus I was combining #3 & #5:
> 3) An atom at constant distance from a black hole will (with some
> small probability) be excited by acceleration radiation.
> 4) An atom freely falling near the horizon of a black hole will not
> be excited be acceleration radiation (in the approximation where it's
> right at the horizon).
> 5) An atom freely falling or at rest far from a black hole will
> (with some small probability) be excited by acceleration radiation.
to formulate the claim:
I have been reading 3,4,5 over and over, and it seems that an atom outside
of the horizon will always be excited by 'acceleration radiation'
(with some small probability). We can think of any motion as composed of
short free falls and brief frizzes at constant distances. It is like
jumps between different free fall trajectories. We really need energy to
perform these jumps (something to hold us at constant distance).
and then I went to some kind of mumbo jumbo bla bla about
freely falling particle that is not freely falling if it has been exited...
I guess this are kind of hallucinations one gets when staying up till 3 am,
I better stop now before I get into some mumbo jumbo bla bla, it is
only 2:15 am, but seems I have gotten into it already :)
John N. White
>>|3) An atom at constant distance from a black hole will (with some
>>|small probability) be excited by acceleration radiation.
Does this mean that if I suspended a probe near the event horizon
it would see the horizon as a hot blackbody, and the probe would
be bathed in radiation from this blackbody?
>>|4) An atom freely falling near the horizon of a black hole will not
>>|be excited be acceleration radiation (in the approximation where it's
>>|right at the horizon).
Does this mean that if I dropped a snowball into the black hole
it would pass through the horizon without encountering any
radiation?
What does the probe see as the snowball falls past it? Does
the snowball seem to fall through the intense bath of thermal
radiation without being affected in any way?
Keith Ramsay
"Ralph E. Frost" <ref...@dcwi.com> writes:
|> If so, it seems like we should at least be able to analyze a
|> "small-sized particle" approximation to both principles.
|
|Do you mean a specific structure or pattern of particles or a composite?
No.
Keith Ramsay
Vesselin G Gueorguiev
>
> Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>
> >We may even have some problem with measuring Q as well, because if we have
> >Q inside the BH how would we know about it! No photons get out of the BH,
> >and non of our photons gets back when we probe the BH. I really wonder,
> >how do we measure Q?
>
> No gravitons get out either, yet we can measure M.
In another post I was discussing this, and argued that a quantum nature
of gravity does not require it, and that it is very possible that
some gravitons gets out because locally we have flat Minkowsky space
till a graviton shows up near by.
> The reason in each case is that the collapsing matter
> has left the metric and EM fields outside the BH warped by its passage.
OK, I agree with that argument.
> That is why we can measure M and Q.
Oups, if we have been lest with metric and EM fields (waves) we will have
dF=0, d*F=0 & G=0 thus we see the waves but not the charges that made them.
Of course we can extend our solutions as far as possible till we get to
the region of space-time where dF=0, d*F=*J & G=T then we can deduce Q & M.
> Hypercharge, on the other hand, doesn't do this,
that is related to the localization of the generated waves, as I said in [*]:
[*] Notice that M & Q are related to interactions with infinite range,
in some sense we may expect that properties of short range interactions
could not characterize BH in general. Thus we may not be able to measure
characteristics related to the strong and weak interaction.
Gravity & EM waves can spread to infinity while the W+-,Z & the gluons
have a finite range to spread around their sources.
Wolfram Schroers
quantities in my previous posts. So before continuing I should better try
to understand what happens - at least according to currently accepted
theories - at various situations.
In article <80kfbh$f...@charity.ucr.edu>, ba...@galaxy.ucr.edu (John Baez) wrote:
> 1) An atom sitting on the earth will not be excited by acceleration
> radiation.
> 2) An atom freely falling near the earth will (with some very small
> probability) be excited by acceleration radiation.
> 3) An atom at constant distance from a black hole will (with some
> small probability) be excited by acceleration radiation.
> 4) An atom freely falling near the horizon of a black hole will not
> be excited be acceleration radiation (in the approximation where it's
> right at the horizon).
> 5) An atom freely falling or at rest far from a black hole will
> (with some small probability) be excited by acceleration radiation.
Later 6) and 7) were added:
>6) An atom moving at constant velocity in flat empty space will
>not be excited by acceleration radiation.
>7) An atom accelerating at a constant rate as measured in its
>own instantaneous rest frame will (with some probability) be
>excited by acceleration radiation.
I assume point 2 and point 7 are somehow related - point 2 seems to be (by
use of the equivalence principle) a special case of point 7, although the
acceleration is not constant. Is this correct? Furthermore is there some
relation between 3 and 7?
In article <80s51k$2rd$1...@rosencrantz.stcloudstate.edu>, John Baez wrote:
> Rindler radiation is similar, but simpler: it's the radiation seen by
> an observer accelerating through flat Minkowski spacetime at a constant
> rate (as measured in his own rest frame at any given moment). Here
> the "horizon" responsible for the radiation is the "Rindler horizon":
> the surface separating the region of spacetime that this accelerating
> observer can see someday from the region he cannot ever see.
I don't understand if this is just a coincidence or if "Rindler radiation"
is based on a similar mechanism as Hawking radiation. If so, can the
"horizon" shrink like the event horizon of a black hole can shrink? If the
answer to this question is "no" (which would appear plausible to me), why
do people then believe that Hawking radiation causes the event horizon of
a black hole to shrink?
I do not quite understand points 4 and 5. Apparently they involve
different approximations, do they? According to 4 an atom right at the
horizon will not be excited, but according to 5 it may be excited at a
"large distance". Is the probability of excitation in case 5 equal for a
freely falling atom and an atom at rest equal? When I moved to
"intermediate distances" (which is supposed to mean the region where the
two approximations 4 and 5 are not valid), would it be reasonable to
assume that the probability of being excited decreases for the freely
falling atom and the probability for the stationary atom increases? Or is
this behavior completely unknown?
If it is the case that it is unknown so far, why is everybody so sure that
the mass loss a black hole suffers in the proper time of an infalling
observer is negligible compared to the total mass?
______________________________________________________________________
Wolfram Schroers <wol...@theorie.physik.uni-wuppertal.de> |
Institute for Theoretical Physics, BUGH Wuppertal |
Toby Bartels
>Toby Bartels <to...@ugcs.caltech.edu> wrote:
>>John Baez <ba...@galaxy.ucr.edu> wrote:
>>>5) An atom moving at constant velocity in flat empty space will
>>>not be excited by acceleration radiation.
>I think the above #5 and the previous:
> 5) An atom freely falling or at rest far from a black hole will
> (with some small probability) be excited by acceleration radiation.
>are equivalent. Aren't they?
The new (5) is in flat space; the old (5) is in curved space.
The new (5) has no excitation; the old (5) has some excitation.
Both the premisses and the conclusions are different.
-- Toby
to...@ugcs.caltech.edu
Toby Bartels
>Toby Bartels <to...@ugcs.caltech.edu> wrote:
>In another post I was discussing this, and argued that a quantum nature
>of gravity does not require it, and that it is very possible that
>some gravitons gets out because locally we have flat Minkowski space
>till a graviton shows up near by.
There are gravitons near by, but they can't be coming out of the BH,
because gravitons travel at speed c, too slow to escape a BH.
I should say here that gravitons are quite speculative,
but at least there is a standard speculation on their speed: c.
(Also, how much do they even make sense around a BH?
That's where the perturbative approach to QG is least reasonable.)
>Oops, if we have been left with metric and EM fields (waves) we will have
>dF=0, d*F=0 & G=0 thus we see the waves but not the charges that made them.
Yet we may deduce the existence of the charges.
Another poster was explicit in the case of Q:
integrate the electric flux through a surface around the BH
and use Gauss's law. You can measure M similarly.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
Toby Bartels wrote:
>
> Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>
> >Toby Bartels <to...@ugcs.caltech.edu> wrote:
>
> >In another post I was discussing this, and argued that a quantum nature
> >of gravity does not require it, and that it is very possible that
> >some gravitons gets out because locally we have flat Minkowski space
> >till a graviton shows up near by.
>
> There are gravitons near by, but they can't be coming out of the BH,
> because gravitons travel at speed c, too slow to escape a BH.
It does not matter how slow or fast are they. My point is that they are
the carriers of the interaction and that if we feel the curvature of
spacetime it is due to particles interacting with these gravitons.
Now try to catch some one who is running as fast as you do!
No way except if you are on a collusion curse, so most gravitons will not
escape but there are some that got pretty good run that may let them out.
> I should say here that gravitons are quite speculative,
We do believe in some kind of a particle that is supposed to carry the
interaction, don't we.
> but at least there is a standard speculation on their speed: c.
> (Also, how much do they even make sense around a BH?
> That's where the perturbative approach to QG is least reasonable.)
These are problems along the way to constructing rigorous QG.
> >Oops, if we have been left with metric and EM fields (waves) we will have
> >dF=0, d*F=0 & G=0 thus we see the waves but not the charges that made them.
>
> Yet we may deduce the existence of the charges.
> Another poster was explicit in the case of Q:
> integrate the electric flux through a surface around the BH
> and use Gauss's law. You can measure M similarly.
I agree with that. Since it is a classical theory result it probably is
going to survive the the quantization. However, when we integrate the flux
we do noting more but counting interaction carriers though the surface
around the BH. If no carriers get out how could you count them?
Toby Bartels
>I am thankful for your consideration and I may really get to read
>Wald's (I hope I have some good basis in QFT), but for now I have it marked.
Note that 2 books by Wald have been cited in this thread.
General Relativity is about general relativity in general
and contains a section on the Hawking effect.
I haven't read that section, so I can't comment on it.
Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics
starts with a presentation of quantum field theory in flat spacetime,
generalises it to curved spacetime, derives the Hawking effect,
and then discusses the ramifications for thermodynamics.
It's a small book but is essentially all about the Hawking effect.
>Toby Bartels <to...@ugcs.caltech.edu> wrote:
>>(An explanation along Schutz's lines but involving *virtual* particles
>>would make this seem more reasonable but would still miss the point of HR.)
>I thought his explanation _was involving_ *virtual* particles since what else
>is a pair of particle-antiparticle popping up from the vacuum to spend
>short (Dealta(t)) vacation near the EH.
Yes, you're right; I should have said
"involving the other pop science explanation for virtual particles".
You've read the FAQ entry on Hawking radiation;
now read question (3) from the entry on virtual particles
(<http://math.ucr.edu/home/baez/physics/virtual_particles.html>).
Notice how the last nonparenthetical paragraph in that question
refers to the energy of the system being borrowed for a brief time.
This is what Schutz meant when he talked of the uncertainty principle.
But the middle paragraph talks about the energy of the virtual state
having the same energy as the initial and finial states
despite that being kinematically impossible from a classical POV.
This is what I was thinking of; as the last paragraph notes,
this is the usual way of talking about Feynman diagrammes.
In the rest of that FAQ entry, you can thus see virtual particles
while violate the usual rule m^2 = E^2 - p^2 (with c = 1).
This rule is satisfied by all *real* particles,
but not the virtual particles (called "off shell" if they violate it).
So, in the Hawking radiation case,
energy conservation requires one particle to have negative energy,
which is quite possible for virtual particles.
The positive energy particle becomes real and is observed,
while the negative energy particle enters the black hole.
As it has negative energy, the BH loses mass.
Problem solved.
>>As for the middle problem, remember that the pair creation
>>happened *near* the EH, not *at* it.
>>So, only a finite (albeit very long) time is required
>>for the escaped particle to be detected by you and me.
>Yes I do keep this in mind, but I was thinking that the intensity of some
>particular "spectral lines" will be proportional to 'how close' to the EH the
>pair was created and thus how much detectable the radiation would be; and
>since I am thinking about detecting then the age of the universe is also
>important in estimating whether and when such radiation would be observable.
I think you'd have to make a calculation to know for sure
whether this intensity/distance correlation is strong enough
to make the escaped particles practically undetectable or not.
And you can't make a calculation using this explanation,
because it's not a good explanation.
Rest assured that *Hawking's* calculation shows the radiation is detectable.
>Notice that nothing leaves the BH, except maybe gravitons that manage to get
>out somehow.
Gravitons travel at the speed of light, not faster,
so they don't get out any more than photons do.
The gravitational effect of the BH is *not* caused
by gravitons leaving the BH to pull on us;
it's caused by the warping of spacetime that
the collapsing matter left behind when it collapsed.
I'm not sure how to explain that in terms of gravitons,
since I don't know how to make it a perturbation of a flat solution
(which is the context where talk of gravitons makes the most sense),
but maybe someone else can. In any case, however,
gravitions are usually supposed to travel at c,
and nothing travelling at c (or slower) can escape a BH.
Since your mechanism depends on gravitons' escaping the BH,
I won't comment on it except to say this:
as GR is nonlinear, gravitons directly interact with themselves,
even though photons interact with themselves only indirectly.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
> In article <80kfbh$f...@charity.ucr.edu>, ba...@galaxy.ucr.edu (John Baez) wrote:
> > 1) An atom sitting on the earth will not be excited by acceleration
> > radiation.
> > 2) An atom freely falling near the earth will (with some very small
> > probability) be excited by acceleration radiation.
> > 3) An atom at constant distance from a black hole will (with some
> > small probability) be excited by acceleration radiation.
> > 4) An atom freely falling near the horizon of a black hole will not
> > be excited be acceleration radiation (in the approximation where it's
> > right at the horizon).
> > 5) An atom freely falling or at rest far from a black hole will
> > (with some small probability) be excited by acceleration radiation.
> Later 6) and 7) were added:
> >6) An atom moving at constant velocity in flat empty space will
> >not be excited by acceleration radiation.
> >7) An atom accelerating at a constant rate as measured in its
> >own instantaneous rest frame will (with some probability) be
> >excited by acceleration radiation.
>
> I assume point 2 and point 7 are somehow related - point 2 seems to be (by
> use of the equivalence principle) a special case of point 7, although the
> acceleration is not constant. Is this correct?
I think, NO. If you are freely falling near the earth your instruments will
not detect any acceleration, while in 7 this acceleration is measured to
differ from zero.
>Furthermore is there some relation between 3 and 7?
I guess so. If there is a black hole, your natural state is to freely fall
into it, but to stay at constant distance from a black hole means to use
a rocket engine and thus producing acceleration which is what one has in 7.
>
> In article <80s51k$2rd$1...@rosencrantz.stcloudstate.edu>, John Baez wrote:
> > Rindler radiation is similar, but simpler: it's the radiation seen by
> > an observer accelerating through flat Minkowski spacetime at a constant
> > rate (as measured in his own rest frame at any given moment). Here
> > the "horizon" responsible for the radiation is the "Rindler horizon":
> > the surface separating the region of spacetime that this accelerating
> > observer can see someday from the region he cannot ever see.
>
> I don't understand if this is just a coincidence or if "Rindler radiation"
> is based on a similar mechanism as Hawking radiation. If so, can the
> "horizon" shrink like the event horizon of a black hole can shrink? If the
> answer to this question is "no" (which would appear plausible to me), why
> do people then believe that Hawking radiation causes the event horizon of
> a black hole to shrink?
I don't know, but there is one important difference between Rindler horizon
and Black hole horizon. The Black hole horizon is a closed surface when the
Rindler horizon is not.
[... can't answer ..]
Jim Heckman
to...@ugcs.caltech.edu (Toby Bartels) wrote:
>
> If quark colours could be separated, a black hole might be produced
> by the collapse of only red quarks; then you could measure a hole's
> redness, blueness, and greenness by measuring the strength of the
> strong force it produced. Of course this is not a part of the classical
> theory.
I presume this is due to the infinite range (gluons are 0-mass quanta) of
the QCD color force, correct?
> Even classically, there is temperature T from statistical mechanics.
> But, since a hole has no structure, it has no entropy, so T = 0.
> Quantumly, we know a black hole actually has non0 temperature,
> but Hawking's calculation shows T is a function of M, Q, and J,
> so they are still the only *mathematically independent* characteristics.
>
> -- Toby
Does Hawking's T really depend on Q? I could have sworn I read
somewhere that T is proportional to the 'surface area' of the BH
(meaning the surface area of the event horizon, I presume?), which I
can see depending on M and *maybe* J -- but Q? If T, which you say is
a quantum effect, *does* depend on Q independently of M and J, why
does it not also depend on the QCD color charge (redness, etc.)? Did
Hawking make the ansatz that there are no long-range forces besides
gravity and EM? Would we even know *how* T would depend on color
for an SU(3) force?
--
~~ Jim Heckman ~~
-- "As I understand it, your actions have ensured that you will never
see Daniel again." -- Larissa, a witch-woman of the Lowlands.
-- "*Everything* is mutable." -- Destruction of the Endless
Sent via Deja.com http://www.deja.com/
Before you buy.
Vesselin G Gueorguiev
t...@rosencrantz.stcloudstate.edu wrote:
[...]
> This last one is supposed to be tested in Earth orbit some time soon.
> For a relatively low-mass, slow-spinning body like the Earth, the
> effect should be pretty weak, but people say they'll be able to
> measure it.
This is Gravity Probe B, it has been postponed till october 2000, see
the news section in Nature 402, the first issue for November, if I remember
it right.
I was surprised to learn that this is the longest project at NASA.
Toby Bartels
>Toby Bartels <to...@ugcs.caltech.edu> wrote:
>>There are gravitons near by, but they can't be coming out of the BH,
>>because gravitons travel at speed c, too slow to escape a BH.
>It does not matter how slow or fast are they. My point is that they are
>the carriers of the interaction and that if we feel the curvature of
>spacetime it is due to particles interacting with these gravitons.
>Now try to catch some one who is running as fast as you do!
>No way except if you are on a collusion course, so most gravitons will not
>escape but there are some that got pretty good run that may let them out.
I don't see how a "pretty good run" will let anything out.
If it travels less than c (that is it's not a tachyon),
then the causal structure of the BH forbids it from getting out.
Not most of them, but all. How do you get around that?
>>I should say here that gravitons are quite speculative,
>We do believe in some kind of a particle that is supposed to carry the
>interaction, don't we.
Personally, I don't. I gather string theorists do.
Classically, there is no such particle. Quantumly, we simply don't know.
I don't mind talking about them, however.
>>Also, how much do they even make sense around a BH?
>>That's where the perturbative approach to QG is least reasonable.
>These are problems along the way to constructing rigorous QG.
Forget rigour.
The idea of gravitons comes from a perturbative approach
where the difference between the metric and the Minkowski metric,
a field often called "h", is treated as a spin 2 quantum field.
The theory is not renormalisable,
but otherwise I understand you can get some results out of it.
But I don't see how you can get a BH metric
as a perturbation of the Minkowski metric.
If you do, tell me how -- I won't require rigour!
>>Yet we may deduce the existence of the charges.
>>Another poster was explicit in the case of Q:
>>integrate the electric flux through a surface around the BH
>>and use Gauss's law. You can measure M similarly.
>I agree with that. Since it is a classical theory result it probably is
>going to survive the the quantization. However, when we integrate the flux
>we do nothing more but counting interaction carriers though the surface
>around the BH. If no carriers get out how could you count them?
Are you suggesting that measuring the electric flux through a surface
involves counting the photons that pass through that surface?
That doesn't sound right to me.
More fundamentally, the photons and the gravitons should be analogous.
Since we measure Q without photons getting out of a BH
(and they don't, by definition of "black"),
we can measure M without gravitons getting out.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
Toby Bartels wrote:
>
> Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>
[...]
Thanks again for the brief note on the Wald's books. Still, no time.
By the way I am planing on buying a laptop and reading the newsgroup
in the library, so I can just walk few steps, take a book, look at it,
and get more to talk about. Is there any one who does it this way?
If so, I will be happy to here advises on this.
> >Toby Bartels <to...@ugcs.caltech.edu> wrote:
>
> >>(An explanation along Schutz's lines but involving *virtual* particles
> >>would make this seem more reasonable but would still miss the point of HR.)
>
> >I thought his explanation _was involving_ *virtual* particles since what else
> >is a pair of particle-antiparticle popping up from the vacuum to spend
> >short (Dealta(t)) vacation near the EH.
>
> Yes, you're right; I should have said
> "involving the other pop science explanation for virtual particles".
Could you briefly state what is "the other pop science explanation"?
Do you mean paragraph 2 & paragraph 3 in virtual particles FAQ?
> You've read the FAQ entry on Hawking radiation;
I did, it was quickly to get it on the screen (I have the FAQ on my hard disk)
and it was short and clear.
> now read question (3) from the entry on virtual particles
> (<http://math.ucr.edu/home/baez/physics/virtual_particles.html>).
This file was for quite a while on my desktop, but I didn't get to read it all.
However, you asked me to jump and read pretty short part of it, which is
noting new to me, but some other readers may need to read more on this.
> Notice how the last nonparenthetical paragraph in that question
> refers to the energy of the system being borrowed for a brief time.
Yup.
> This is what Schutz meant when he talked of the uncertainty principle.
Yup.
> But the middle paragraph talks about the energy of the virtual state
> having the same energy as the initial and finial states
> despite that being kinematically impossible from a classical POV.
> This is what I was thinking of; as the last paragraph notes,
> this is the usual way of talking about Feynman diagrammes.
There is no much difference for me since I consider
'energy is somehow "borrowed"' as equivalent to 'intermediate state'
which explains how to borrow energy.
> In the rest of that FAQ entry, you can thus see virtual particles
> while violate the usual rule m^2 = E^2 - p^2 (with c = 1).
> This rule is satisfied by all *real* particles,
> but not the virtual particles (called "off shell" if they violate it).
Yup.
> So, in the Hawking radiation case,
> energy conservation requires one particle to have negative energy,
> which is quite possible for virtual particles.
Could try to sell it, but!
> The positive energy particle becomes real and is observed,
> while the negative energy particle enters the black hole.
> As it has negative energy, the BH loses mass.
but, but, why only the negative energy particle get into the BH?
> Problem solved.
No it is not, till you answer: why only the negative energy particle
go into the BH?
And if I am not mistaken negative energy particle is the same as
positive energy antiparticle, but with opposite momentum which is
in this case pointing out of the black hole. So it is the particle
which is getting out :)
> >>As for the middle problem, remember that the pair creation
> >>happened *near* the EH, not *at* it.
> >>So, only a finite (albeit very long) time is required
> >>for the escaped particle to be detected by you and me.
>
> >Yes I do keep this in mind, but I was thinking that the intensity of some
> >particular "spectral lines" will be proportional to 'how close' to the EH the
> >pair was created and thus how much detectable the radiation would be; and
> >since I am thinking about detecting then the age of the universe is also
> >important in estimating whether and when such radiation would be observable.
>
> I think you'd have to make a calculation to know for sure
> whether this intensity/distance correlation is strong enough
> to make the escaped particles practically undetectable or not.
Yes it needs to be calculated, but when and how?
> And you can't make a calculation using this explanation,
> because it's not a good explanation.
I assume you are referring to the virtual particle explanation, yes I
never liked, and now I don't like it even more.
> Rest assured that *Hawking's* calculation shows the radiation is detectable.
hum! I am not following this. Could you try again?
>
> >Notice that nothing leaves the BH, except maybe gravitons that manage to get
> >out somehow.
>
> Gravitons travel at the speed of light, not faster,
> so they don't get out any more than photons do.
Yes, you are right! I realized this a little while after I sent the post.
> The gravitational effect of the BH is *not* caused
> by gravitons leaving the BH to pull on us;
> it's caused by the warping of spacetime that
> the collapsing matter left behind when it collapsed.
This is certainly true for the region occupied by the collapsing matter
through all times, but an observer is supposed to be far enough of such
regions to avoid accidental involvement in the collapsing process.
As a result, our observer detects 'gravitons' - gravitational waves
coming from collapsing matter. However, if we claim that we are
observing a BH eating matter, thus EM radiation from the matter as it
gets into the BH, we should also detect gravitational waves from the
BH itself, and this waves *are not* wave due to formation they are
due to already formed BH.
> I'm not sure how to explain that in terms of gravitons,
intuitively it should be the same as detecting EM waves.
> since I don't know how to make it a perturbation of a flat solution
> (which is the context where talk of gravitons makes the most sense),
> but maybe someone else can. In any case, however,
> gravitions are usually supposed to travel at c,
> and nothing travelling at c (or slower) can escape a BH.
That is due to the classical theory, the quantum theory should
have something similar to a tunneling mechanism, which will
let some things to escape, but few will get far enough.
Those that will get far enough are related to long range interactions,
thus photons and gravitons, all other will have too small range to
make it to the detectors of a good ('sane') observer.
> Since your mechanism depends on gravitons' escaping the BH,
> I won't comment on it except to say this:
> as GR is nonlinear, gravitons directly interact with themselves,
> even though photons interact with themselves only indirectly.
I did mentioned this as a note when I was discussing this stuff.
Now, I am thankful to you, since due to this discussion, I think
the tunneling mechanism should be very important.
Vesselin G Gueorguiev
>
> >>There are gravitons near by, but they can't be coming out of the BH,
> >>because gravitons travel at speed c, too slow to escape a BH.
>
> >It does not matter how slow or fast are they. My point is that they are
> >the carriers of the interaction and that if we feel the curvature of
> >spacetime it is due to particles interacting with these gravitons.
> >Now try to catch some one who is running as fast as you do!
> >No way except if you are on a collusion course, so most gravitons will not
> >escape but there are some that got pretty good run that may let them out.
>
> I don't see how a "pretty good run" will let anything out.
> If it travels less than c (that is it's not a tachyon),
When I said 'not matter how slow or fast are they' I didn't mean less then
c, but to emphasizes that they have finite velocity and that the
important thing is :
___they are the carriers of the interaction___
These are the keepers who make everything stay (classically) in the BH.
> then the causal structure of the BH forbids it from getting out.
___classicaly__
> Not most of them, but all. How do you get around that?
__tunneling___
>
> >>I should say here that gravitons are quite speculative,
>
> >We do believe in some kind of a particle that is supposed to carry the
> >interaction, don't we.
>
> Personally, I don't. I gather string theorists do.
> Classically, there is no such particle. Quantumly, we simply don't know.
> I don't mind talking about them, however.
Every one has the right to its own opinion. The fun part begins when
we try to explain them to each other!
> >>Also, how much do they even make sense around a BH?
> >>That's where the perturbative approach to QG is least reasonable.
>
> >These are problems along the way to constructing rigorous QG.
>
> Forget rigour.
> The idea of gravitons comes from a perturbative approach
I would say it come naturally due to analogy with all we see around us,
particles interacting with other particle ....
The perturbative approach is how we try to make the idea of a graviton
more precise.
> where the difference between the metric and the Minkowski metric,
> a field often called "h", is treated as a spin 2 quantum field.
I think, it was mentioned few time that one can do perturbative approach
about any background metric not necessarily the Minkowski metric.
> The theory is not renormalisable,
> but otherwise I understand you can get some results out of it.
> But I don't see how you can get a BH metric
> as a perturbation of the Minkowski metric.
> If you do, tell me how -- I won't require rigour!
You are right there is no way since it will not be a perturbation.
However, I would start with BH metric and study the perturbations of it.
Know some one who would pay me to try this?
Don't forget that all we do now are perturbations to mean field theories.
> >>Yet we may deduce the existence of the charges.
> >>Another poster was explicit in the case of Q:
> >>integrate the electric flux through a surface around the BH
> >>and use Gauss's law. You can measure M similarly.
>
> >I agree with that. Since it is a classical theory result it probably is
> >going to survive the the quantization. However, when we integrate the flux
> >we do nothing more but counting interaction carriers though the surface
> >around the BH. If no carriers get out how could you count them?
>
> Are you suggesting that measuring the electric flux through a surface
> involves counting the photons that pass through that surface?
> That doesn't sound right to me.
What else are we detecting with our instruments?
> More fundamentally, the photons and the gravitons should be analogous.
Yes and I guess, photon can tunnel as well.
> Since we measure Q without photons getting out of a BH
> (and they don't, by definition of "black"),
> we can measure M without gravitons getting out.
Well, considering that tunneling may supply the mechanism for
things to get out of the BH are you still thinking that
there is no photons & gravitons getting out of a BH?
If I try to get in your shoes, but keep the gravitons as messengers,
I will see many gravitons & photons coming from the collapsing matter.
I wouldn't also see many as well, they will be outside my 'light cone',
if the universe is not going to collapse. Right now I can't see how all
this particles should get arranged so that the effect will give me
'information' about the BH. Certainly a completely deterministic
description can do this and classical GR is such a theory, but quantumly,
I don't see how, may be boundary effects, but I don't want universe
with boundary since this requires something behind the boundary.
John Baez
John N. White <jn...@mindspring.com> wrote:
>John Baez wrote in message <80q3k1$r...@charity.ucr.edu>...
>>>|3) An atom at constant distance from a black hole will (with some
>>>|small probability) be excited by acceleration radiation.
>Does this mean that if I suspended a probe near the event horizon
>it would see the horizon as a hot blackbody, and the probe would
>be bathed in radiation from this blackbody?
Right.
>>>|4) An atom freely falling near the horizon of a black hole will not
>>>|be excited be acceleration radiation (in the approximation where it's
>>>|right at the horizon).
>Does this mean that if I dropped a snowball into the black hole
>it would pass through the horizon without encountering any
>radiation?
Right *at* the horizon, yes.
>What does the probe see as the snowball falls past it?
A snowball.
>Does
>the snowball seem to fall through the intense bath of thermal
>radiation without being affected in any way?
Good question!
I don't know what the probe thinks about the snowball, but I think
I know what the snowball thinks about the probe. It thinks the
probe is rapidly accelerating through a region of space containing
no radiation. Since the probe can detect electromagnetic radiation,
it contains charged particles. And since these particles are rapidly
accelerating, they interact with vacuum fluctuations of the
electromagnetic field in a way that excites the probe - mimicking the
effect of a bath of thermal radiation!
I'm sure some similar analysis explains what the probe thinks about
the snowball. These issues have been studied in great detail by
people like Bill Unruh (who told me the various facts listed in the
post you cite). It turns out to be easiest to understand a lot of
these issues if you forget about black holes and consider objects
accelerating through flat spacetime. Unruh showed that an observer
accelerating through flat spacetime will see thermal radiation with
a temperature proportional to his acceleration. A non-accelerating
observer can explain away this effect as described above. But I
never thought to ask what the accelerating observer thinks about the
non-accelerating one! Maybe I'll ask Unruh. But maybe I should
try to guess the answer first.
By the way, in a previous post I used the phrase "Rindler radiation"
when I really meant "Unruh radiation". Rindler invented a nice
coordinate system in which an accelerating observer in flat spacetime
would appear to be at rest. Unruh was the one who showed that an
accelerating observer in flat spacetime would see blackbody radiation
with a temperature proportional to his acceleration.
(The constant of proportionality is so small that the existence of
Unruh radiation has not yet been experimentally tested. But Unruh's
calculations only depend on ordinary quantum field theory, so they
are uncontroversial.)
John Baez
Toby Bartels <to...@ugcs.caltech.edu> wrote:
>Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>>Notice that nothing leaves the BH, except maybe gravitons that manage to get
>>out somehow.
>Gravitons travel at the speed of light, not faster,
>so they don't get out any more than photons do.
>The gravitational effect of the BH is *not* caused
>by gravitons leaving the BH to pull on us;
>it's caused by the warping of spacetime that
>the collapsing matter left behind when it collapsed.
>I'm not sure how to explain that in terms of gravitons,
>since I don't know how to make it a perturbation of a flat solution
>(which is the context where talk of gravitons makes the most sense),
>but maybe someone else can.
It's a terrible, terrible thing to worry about how gravitons
get out of a black hole before you already have a crystal-clear
understanding of general relativity and quantum field theory.
It leads to endless confusion, angst, and frustration. So if
you don't already understand general relativity and quantum field
theory, stop reading here.
If you insist on explaining the gravitational attraction of
masses in terms of gravitons, you must do so in terms of
*virtual* gravitons, not real ones. And if you want to
understand this stuff, it's best to start with electromagnetism
before tackling gravity. In quantum electrodynamics, the
electrostatic attraction or repulsion of charges can be explained
in terms of *virtual* photons, not real photons. The details of
how this works are a bit tricky - for a nice introduction, see
Matt McIrvin's website. But here's one thing that's easy to
understand. Virtual particles aren't required to satisfy
E^2 - p^2 = m^2 the way real ones are. This means that their
energy-momentum vector can be spacelike, like a tachyon's would
be. In very crude and sloppy terms - I know I'm going to regret
saying this! - you can think of them as being able to go "faster
than light". In some sense, this explains how virtual photons can
"get out" of a charged black hole and create its electrostatic
attraction or repulsion to other charged objects. It also explains
how virtual gravitons can "get out" of a black hole and create
its gravitational attraction to other massive objects. It also
explains why Tom van Flandern is under the persistent misimpression
that gravity travels faster than light, violating the basic tenets
of relativity. It's subtle! In a certain sloppy sense virtual
particles can go "faster than light" - but you can't actually use
them to communicate faster than light! So the earth acts like its
getting pulled to where the sun *is*, not where it *was*, but if
someone suddenly blew up the sun, there would be no way to
instantaneously detect this by observing the motion of the earth.
Similarly, after you've fallen into a black hole, you can't use
virtual photons or gravitons to communicate to someone outside.
John Baez
Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>Toby Bartels wrote:
>> The idea of gravitons comes from a perturbative approach
>> where the difference between the metric and the Minkowski metric,
>> a field often called "h", is treated as a spin 2 quantum field.
>> The theory is not renormalisable,
>> but otherwise I understand you can get some results out of it.
>> But I don't see how you can get a BH metric
>> as a perturbation of the Minkowski metric.
>> If you do, tell me how -- I won't require rigour!
>You are right there is no way since it will not be a perturbation.
>However, I would start with BH metric and study the perturbations of it.
>Know some one who would pay me to try this?
Maybe - but only after you've read some of the hundreds of papers
where people tried already!
Toby is right, it's awful to study a black hole as a perturbation of
Minkowski space. Vesselin is right, you can linearize Einstein's
equations around the Schwarzschild solution, quantize the resulting
linear equations, and get a free field theory of "gravitons" running
around a black hole. Then you can try to quantize the nonlinear
terms and get "graviton interactions", just like in ordinary particle
physics. But unfortunately, just like when you perturb around Minkowski
space, the resulting theory is nonrenormalizable. Presumably this is the
math gods' way of saying we're doing something dumb.
>Don't forget that all we do now are perturbations to mean field theories.
That's an exaggeration.
John Baez
> If quark colours could be separated, a black hole might be produced
> by the collapse of only red quarks; then you could measure a hole's
> redness, blueness, and greenness by measuring the strength of the
> strong force it produced. Of course this is not a part of the classical
> theory.
I don't know what "this" refers to, so I'm not 100% sure what you're
saying, but it's actually rather tricky to define the total redness of
a black hole, even using classical chromodynamics. I'll eventually
explain why....
We've got two issues to worry about here: 1) how *classical* chromodynamics
might interact with general relativity to give a static black hole hair,
and 2) how *quantum* chromodynamics might do so. (By "hair" I mean
measurable properties in addition to mass, charge and angular momentum.)
The first issue is easy to study mathematically in a rigorous way,
while the second one is not. Unfortunately, the first issue is not
relevant to our world, while the second one is.
Let me digress a moment and think about that last point. It seems to
me that in our world, whenever chromodynamics is important,
quantum effects are also important. In this respect chromodynamics
is very different from electrodynamics. It's an interesting
question why this is true. As far as I can tell, the answer is
"confinement": the strong force gets stronger at long distance scales,
so that all stable large objects are uncharged with respect to
the strong force. By "large" I mean roughly larger than the proton
radius - about a femtometer, i.e. 10^{-15} meters. So we can say
roughly that chromodynamics is only directly relevant at length scales
less than a femtometer. And presumably at these length scales we always,
in practice, need to use quantum mechanics to understand anything.
(The last sentence isn't true *in principle*, since very massive objects
can act classical even at very small length scales. So I guess the answer
also depends on the fact that the lightest quarks are fairly light!)
(Another caveat: of course the "spinoffs" of chromodynamics are
important at length scales far exceeding a femtometer - for example, you
can't understand a neutron star without understanding neutron/neutron
interactions pretty well, and these trace their way back to chromodynamics.
But I don't think anyone uses chromodynamics *directly* when modelling
neutron stars. Now, if someone found a quark star, that might be a
different matter....)
Okay, enough digressions! Would classical chromodynamics give black
holes hair? I think not. Unlike classical electrodynamics, classical
chromodynamics is a nonabelian gauge theory, so the simple vacuum Maxwell
equation
d*F = 0
gets repaced by
d*F + [A,*F] = 0.
In electrodynamics we can use d*F = 0 to conclude that the integral of
the electric field through a surface surrounding the black hole doesn't
depend on which surface we use (given that our black hole is a
solution of the vacuum Einstein-Maxwell equations). This quantity is
then the black hole's charge. This quantity is gauge-invariant. And by
choosing a big sphere surrounding the black hole we can even arrange to
measure its charge from far away.
In chromodynamics this argument fails. For starters, no matter
what surface we choose, in chromodynamics the integral of the chromoelectric
field will not be gauge-invariant! - and thus not a legitimate observable.
Moreover, the integral depends on the surface we choose.
In other words: if we pick a surface around the black hole, we can define
its total colors as integrals of various components of the chromoelectric
field. (Actually, since this field takes values in su(3) and has 8
components, what we really get are the total "red/bluenss", "red/greenness",
"blue/greenness" and other things like that!) But if do a gauge
transformation that vary in a complicated way over our surface, the total
colors, defined this way, will change in a messy way. You can't even figure
out the new totals using the old ones! In this situation folks usually
say we don't have an observable.
So: classical chromodynamics doesn't give black holes colored hair!
The same is true of quantum chromodynamics, for all the same reasons,
plus the extra reason of confinement. But one can't study the quantum
case as rigorously as the classical case.
Patrick Labelle
> >If T, which you say is
> >a quantum effect, *does* depend on Q independently of M and J, why
> >does it not also depend on the QCD color charge (redness, etc.)? Did
> >Hawking make the ansatz that there are no long-range forces besides
> >gravity and EM? Would we even know *how* T would depend on color
> >for an SU(3) force?
This is a question which I've been asking myself for a while.
Why are black holes completely specified by giving only the mass,
angular momentum and mass. What about the other quantum
numbers associated with local gauge invariance? In the standard
model, the electric charge is a linear combination of the weak
charge and the hypercharge...how does this fit in the picture?
And of course, what about color...
Did Israel and the others only take into account QED?
Charles Torre
> ...no matter
> what surface we choose, in chromodynamics the integral of the chromoelectric
> field will not be gauge-invariant! - and thus not a legitimate observable.
[snip]
> So: classical chromodynamics doesn't give black holes colored hair!
I can think of two escape routes from this conclusion. Of
course, other considerations may rule out these escapes.
(1) If the black hole solution of the coupled
Einstein-Yang-Mills equations admits a covariantly constant
section of the SU(3) principal bundle, then I think you can
construct a gauge invariant surface integral.
(2) Even if the solution does not allow for constant
sections, it may do so asymptotically. Then one might be
able to define a sensible color via a surface integral
"at infinity", much as we do with energy-momentum and
angular momentum for the gravitational field.
Anyone know if/why these scenarios cannot occur?
Charles Torre
John Baez
Toby Bartels <to...@ugcs.caltech.edu> wrote:
>Jim Heckman <jhec...@my-deja.com> wrote:
>>Does Hawking's T really depend on Q? I could have sworn I read
>>somewhere that T is proportional to the 'surface area' of the BH
>>(meaning the surface area of the event horizon, I presume?), which I
>>can see depending on M and *maybe* J -- but Q?
>For a Schwarzschild BH, T is proportional to the surface *gravity* K
>(not surface *area* A; that's supposedly the entropy S),
>and K is proportional to 1/M. But I know it depends on J
>for rotating BHs. And on Q for charged ones? I don't know.
The following chart is good to remember:
BLACK HOLES THERMODYNAMICS
black hole mass M energy E
event horizon area A entropy S
surface gravity K temperature T
FIRST LAW: dM = K dA / 8 pi + work dE = T dS + work
SECOND LAW: A increases S increases
THIRD LAW: can't get K = 0 can't get T = 0
But if you actually want to get the exact numbers right, you need
this:
M = E
A / 4 = S
K / 2 pi = T
All this is in units where hbar = c = G = k = 1.
>From this we see that:
1) The black hole temperature depends only on its *surface gravity* -
lots of spacetime curvature at the event horizon makes for hot Hawking
radiation.
2) The black hole entropy depends only on its surface *area* - we can
imagine information living on the horizon, at 1/4 of a nit per square
Planck length. (A nit is the natural logarithm version of a bit;
there are ln 2 bits per nit.)
But to figure out if the black hole temperature depends on charge,
we need the actual formula for, say, the surface area in terms of M,
J and Q:
A = 4 pi ( [M + sqrt(M^2 - Q^2 - J^2/M^2)]^2 + J^2/M^2 )
Yuck! I could never remember that; I looked it up. Anyway,
this shows that A depends on all three of M, J, and Q. The
temperature, which is proportional to dM/dA, also depends on all
three.
>>Did Hawking make the ansatz that there are no long-range forces besides
>>gravity and EM?
Yes. For example, if there was another field satisfying Maxwell's equations
in addition to electromagnetism, black holes would also have that kind of
"charge".
>Would we even know *how* T would depend on color for an SU(3) force?
Yes; as long as we're talking SU(3) Yang-Mills theory - i.e., the usual
theory of the strong force - we can figure out how it affects the behavior
of black holes, and the answer is: black holes DON'T have color, thanks to
confinement, and thus their temperature does NOT depend on color.
Vesselin G Gueorguiev
>
I meant everything has to stay in the BH. That is classical result.
A tunneling process makes possible for things to get out of the BH.
>
> >Toby Bartels <to...@ugcs.caltech.edu> wrote:
>
> >>then the causal structure of the BH forbids it from getting out.
> >>Not most of them, but all. How do you get around that?
>
> >tunneling
>
> But even tunneling cannot violate the causal structure of spacetime.
> In flat Minkowski space, things don't tunnel faster than light.
Yes, but my impression was that you were asking how things get out,
and I hope, I have given one possible mechanism that let things out.
As far as thing go about the interactions, John Baez has posted an
article pointing toward the virtual particles. This sounds very
reasonable to me but I still don't understand the mechanism.
[...]
> >I think, it was mentioned few time that one can do perturbative approach
> >about any background metric not necessarily the Minkowski metric.
>
> True, but if the background metric is itself curved,
> then this curvature will exist in the absence of all perturbations
> and hence in the absence of all gravitons.
> Thus, the gravitons will not explain all the curvature.
> In your discussion, you want the gravitions to produce the curvature,
> analogously to the way the photons produce the EM field,
> so that measurement of the BH's mass involves measuring gravitons.
> That works only if the perturbation is around a flat metric.
In the same article John Baez was explaining this too. I don't understand
how things really work in these cases but I have to admit that his
explanations are very attractive to me, I should have probably talked
about virtual gravitons in this discussion, but since I don't understand
how they really do the work I can just refer to John's post.
>
> >>Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>
> >>>However, when we integrate the flux
> >>>we do nothing more but counting interaction carriers though the surface
> >>>around the BH. If no carriers get out how could you count them?
>
> >>Are you suggesting that measuring the electric flux through a surface
> >>involves counting the photons that pass through that surface?
> >>That doesn't sound right to me.
>
> >What else are we detecting with our instruments?
>
> I agree we're measuring photons,
> but I don't think we're measuring the number that pass through the surface.
> After all, the number of photons passing through the surface is unitless,
> while the electric flux through the surface is length^{-2}
> (in units where c and hbar = 1).
Yep, but this is just related to the geometrical property of the
counter we use, which gives the flux as number of particles per unit
area of our detector per unit time.
Vesselin G Gueorguiev
> >masses in terms of gravitons, you must do so in terms of
> >*virtual* gravitons, not real ones.
>
> And so Vesselin is right that gravitons are escaping the BH,
> not because they're tunneling, but because they're virtual.
> So, Vesselin, allow me to go back and see what place this controversy holds
> in the previous parts of our discussion.
Actually, I didn't think about the virtual particles. I was counting on
real particles because of the problems one gets into when trying to
use the virtual particles. (You have explained your vision of these
problems at the end of the post)
> Let's see, in one part it was a problem with the photon/graviton analogy;
> you said we could measure M because gravitons escaped,
> but how could we measure Q when no photons escape?
My analogy was based on the fact that photons cannot bound themselves
in space, but they can help to bound changed particles since such
particles are subject to EM interaction whose carriers are the photons.
Now, I just substitute photon -> graviton and EM -> gravitational
which gives:
-gravitons- cannot bound themselves in space, (they actually can)
^^^^^^^^^^^^ assumed by the analogy
but they can help to bound 'changed' particles since such (all)
particles are subject to -gravitational- interaction whose carriers
are the -gravitons-.
> But John reminds us *virtual* photons can escape, so problem solved.
Yes using virtual particles it seems like the problem is solved.
Unfortunately, I don't understand how exactly the virtual particles do
their job in transmitting the interaction.
> In the other part of the thread, it's more serious.
> You propose this as a method of Hawking radiaton:
>
> :Now we have one of these smart gravitons that get out of the BH behind the EH.
> :There this graviton interacts with matter or suddenly decides to turn himself
> :into few other particles. So we got the BH to lose energy, and we got some
> :particles near the EH. Now some of these particles can get back into the BH,
> :but the rest will form the BH radiation.
>
> Since virtual particles of other sorts are also escaping the BH,
> the method needn't be limited to gravitons.
A really wanted to be limited to gravitons since I am presuming real
gravitons to avoid the problem that you are describing bellow.
> However, I still don't think it works as an explanation.
For virtual particles as explanation of Hawking radiation, YES!
It doesn't work for me too.
However, using virtual particles to explain why you can measure M & Q
for a BH it seems OK, I wish I understand this better.
> As virtual particles can as easily have negative energy as positive,
> on average the virtual particles' escapes won't change the BH's energy.
> Furthermore, I don't think they can ever become real,
> because this would provide a means of communication.
>
[...]
john baez
Jim Heckman <jhec...@my-deja.com> wrote:
>Thanks for saving the day, as usual, John!
I'm not sure I completely saved the day. If you read Charles Torre's post,
and my reply to that, you'll see I overlooked a bunch of important issues -
at least when it comes to *classical* chromodynamics.
However, if you're interested in honest physics, as opposed to mathematical
physics, what really matters is *quantum* chromodynamics. And in this case,
I'm quite sure that confinement steps in to save the day.
>You seem to be saying that a non-abelian gauge field can't really have
>a conserved "classical" charge, at least not a gauge-invariant one. But
>I thought one of the hallmarks of such a field is that its quanta carry
>charge, too, unlike the electrically neutral photon. (Am I mistaken here?)
It's hard for me to completely agree or disagree with you here. So
let me just say some stuff that's true.
Say we're talking about classical Yang-Mills theory, possibly together
with other fields. In this case we loosely say one of our fields is
"charged" if it transforms nontrivially under constant gauge transformations.
This will be true if and only if the field interacts with the Yang-Mills
gauge field. So: is the gauge field *itself* charged? The answer is yes
if and only if the gauge group is nonabelian.
Another way to put this is: the Yang-Mills equations are nonlinear
if and only if the gauge group is nonabelian. These nonlinearities
describe "self-interactions" of the gauge field.
Okay, great. But note: none of this so far says anything about the
notion of "charge" in the sense of some kind of number you compute
knowing your fields - much less the notion of "conserved charge" or
"gauge-invariant conserved charge"! In electromagnetism, we can
compute the total charge of the universe by computing the electric
flux through a large sphere and then taking the limit as the radius
of this sphere goes to infinity. Of course the limit may or may
not exist, but if it does, we get a "charge" which is gauge-invariant
and also, under reasonable assumptions, conserved.
In the post to which you're replying, I described how this doesn't
work quite so nicely when your gauge group is nonabelian: the
electric flux through a sphere is not gauge-invariant. In fact
it transforms in a thoroughly disgusting way! Charles Torre then
pointed out that the *limit as the radius of the sphere goes to
infinity* could, under reasonable conditions, still transform in
a simple way under gauge transformations that approach a constant
at infinity. Then, in my reply to Torre, I pondered the issue a bit
further and wound up agreeing with him... though by now it's getting
subtle enough that I feel I should be pretty cautious with further
pronouncements.
>So... since gravitons carry energy="gravitational charge", do we then
>expect its gauge field to be non-abelian? (I seem to recall that General
>Covariance is essentially the gauge transformations of gravitation, and
>that the non-linearity of Einstein's field equations is the manifestation of
>gravitational self-interaction.)
Now you're making it even HARDER for me to agree or disagree! First of
all, let's leave "gravitons" out of this discussion if possible: gravitons
are a concept from perturbative quantum gravity, while the "black holes
have no hair" theorems concern nonperturbative classical gravity. Secondly,
general relativity is not a Yang-Mills theory and does not have a "gauge
group" in quite the same sense, so none of what I said above is directly
relevant to gravity. Third, there's an important sense in which gravity
does NOT carry energy, and an important sense in which it DOES, and I'm
not sure I have the energy myself to describe these senses for you right
now - if you're interested, you might start with the FAQ:
http://math.ucr.edu/home/baez/physics/energy_gr.html
On the bright side, even though general relativity is not a Yang-Mills
theory, there are a number of ways of describing it as a kind of gauge
theory, and then its gauge group is nonabelian, so there is a kind of
analogy between general relativity and Yang-Mills theory.
>So... if gravitation is non-abelian, how do we then get
>gauge-independent (=coordinate-independent?) classical M and J from
>it? Why does your argument against QCD-color hair not work here?
Okay, this question actually makes quite a bit of sense despite all
my caveats above, so I'll be nice and answer it instead of complaining
further.
Briefly, the answer is that there was a loophole in my argument,
and general relativity slips through this loophole. Charles Torre
pointed out the loophole and noted how general relativity slips through
it, giving us "mass" and "angular momentum" as observables you can
measure for black holes in asymptotically flat spacetime. In my
reply to his post, I started thinking about to what extent classical
Yang-Mills theory can also slip through this loophole.