Re: Acceleration of charge
- From: vanep@xxxxxxx
- Date: 28 Feb 2006 17:10:10 -0800
sal wrote:
On Mon, 27 Feb 2006 04:24:52 -0800, vanep wrote:
Bilge wrote:
sal:
>On Wed, 22 Feb 2006 08:31:54 +0000, Bilge wrote:
>
>> vanep@xxxxxxx:
>>
>> >> But this suggests that the virtual pair, localized as it is in
>> >> its own free-falling coordinate system, should also be either
>> >> outside or inside ... not one particle out, one particle in.
>> >
>> >That is similar to what Thorne was saying.
>>
>>
>> Diagramatically, then you are picturing a closed loop that looks
>> like,
>>
>> e- ^ r
>> -->-+ |
>> | | +--> t
>> -->-
>> e+
>>
>> But, recall, that in a feynman diagram, positrons are depicted as
>> negative energy electrons propagating backwards in time, so you can
>> draw the same loop as,
>>
>> -->-
>> | |
>> -<--
>>
>> (In a feynman diagram, _all_ lines are particle lines by convention.
>> The difference between a particle and anti-particle in a feynman
>> diagram is the direction in time the arrow points, and the time axis
>> is generally the horizontal axis, which is backwards from the usual
>> spacetime diagram. So, for example, the following diagram
>> represents electron compton scattering:
>>
>>
>>
>> ~ ~
>> ~ ~
>> +-->--+
>> / \
>> / \
>>
>>
>> because the arrow on the particle line line points in the positive t
>> direction. (actually, there is a second in which the photon lines
>> are crossed). If the arrow pointed the other way, it would represent
>> compton scattering by a positron. If you rotate it 90 degrees so
>> that the arrow points along the spatial direction and the solid
>> lines are on the left, you have pair anihilation - the incoming
>> electron emits a pair of photons and scatters backwards in time.
>> Rotated the other way, you have pair creation.
>>
>> So, applying that to pair creation at the horizon, you would picture
>> the process as an electron falling into the hole, emitting a virtual
>> photon, then scattering backward in time and destroying _itself_
>> outside the horizon by emitting a photon. Since a particle which
>> propagates forward in time falls into the black hole, it will cross
>> the horizon the other way by propagating backward in time as an
>> anti-particle. Try to figure out the plot from the compton
>> scattering diagram and the closed loop will make more sense. It's
>> completely equivalent to having the virtual pair fall across the
>> horizon.
>
>(It's sometimes said that the only dumb question is the one you didn't
>ask. Let's test that a bit.)
>
>How literally can I take the notion of the electron propagating
>forward in time across the horizon, and then propagating back across
>horizon, going backward in time?
In my personal opinion, you might as well take it as seriously as you
take space and time. Which is to say, I seriously doubt that space and
time mean a whole lot at the quantum level.
>If that's literally correct then it seems to lead to a problem. From
>the point of view of an outside observer, it takes infinite time for
>something to cross the horizon going in.
Well, that is also something of a misstatement, since an observer will
certainly be unable to see the infalling object after a finite time
unless the observer is at infinity. While an observer can't see the
object crossing the horizon, the observer will see the infalling object
disappear as the light gets redshifted into the hawking background.
> If the electron does that,
>and then re-crosses going "the other way", it would seem that the
>visible radiation on the outside would be paired with "negative
>radiation" on the inside _but_ the "negative" radiation should happen
>an infinite time _later_ than the external "positive" radiation.
I don't understand that statement.
It seems that Sal is trying to follow the event from a remote coordinate
system.
Yes, exactly.
It makes more sense to analyse the event from the proper frame of
the virtual pair, the virtual pair event depicted in your Feynman
diagrams, and do the analyse to see if your conjecture makes sense.
Again, what I was trying to arrive at was a notion of _when_, viewed from
remote coordinates, the "negative" photon actually makes it inside the
event horizon. Since the horizon shrinks at that moment, that event
has observable consequences for a _distant_ observer.
I know, for some problems, the issue for how much coordinate time
elapses for an object to fall into a Schwarzschild black hole figures
prominently, such as Hawkings recent comments on resolving the
information paradox, but Schwarzschild coordinates only exist for r >
2M. For an astrophyicist reviewing accretion disk data for a suspected
black hole candidate he might be looking for matter falling off the
disk and into the hole. A signature would be radiation emmitted from
the falling object decreasing in energy for decreasing r and 'wink out'
at the event horizon. The Schwarzschild energy per unit mass for a
radially falling object is a constant along the radial wordline
E/m = ( 1 - 2M/r ) dt/dTau = 1
Observed in the coordinate frame it wouldn't be a constant along the
radial worldline and would correspond with the data observed by the
astrpphysicist.
E/m = ( 1 - 2M/r ) dt
Lets say we drop a beacon probe, from r = 200 M, which emits a pulse of
light as it falls to the hole. Find the integral for the radial speed
of light dr/dt = ( 1 - 2M/r )
int (rdr) / (r - 2M )
dt = r + 2M ln ( r - 2M )
dt = [ r_1 - r_2 ] + 2M ln [ (r_1 - 2M) / (r_2 - 2M) ]
Now find the definite integral from say r = 2.5M to r = 200M
dt = 209.5 m / c = .0772 second
It blows up for r = 2M because 2M isn't a valid coordinate in the
Schwarzschild geometry.
It seems to me that any conjecture the remote observer makes using a
remote coordinate r = 2M, in the Schwarzschild geometry, doesn't make
much physics sense.
For Bilge's idea: I think he might be saying that the event described
by the Feynman diagram is only subject to uncertainty limitiations and
we don't even need to define a relativistic coordinate system. Just a
guess. I'm interested to see what Bilge comes up with. This subject is
fascinating. It's considered an important element in the search for
quantum gravity.
Since we believe (or anyway Stephen Hawking says he believes) that the
entire black hole can potentially shrink away to nothing in finite time as
viewed by a distant observer, the time when the photon actually crosses
the horizon, as computed by that self-same distant observer using
Schwarzschild coordinates, must also be finite.
You may have seen this but it's a good discussion on your query.
http://www.mathpages.com/rr/s7-02/7-02.htm
Bruce
>The consequence would be that, while black holes could radiate, they
>could never (observably) evaporate as a result, because the
>evaporation is due to the hole swallowing negative energy -- and that
>won't happen until the end of time.
Since the singularity in a schwarzschild black hole is spacelike,
time ends at the singularity and the singularity occupies all of space.
--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org
*** Free account sponsored by SecureIX.com ***
*** Encrypt your Internet usage with a free VPN account from http://www.SecureIX.com ***
.
- Prev by Date: Re: My Quests on Learning Lorentz Transformation
- Next by Date: Re: Why is the Speed Of Light Constant?
- Previous by thread: Re: Why is the Speed Of Light Constant?
- Next by thread: Re: Acceleration of charge
- Index(es):
Relevant Pages
|
