[long] Perpetual Motion vs. Reactionless Drive (was: What is IT? ...)
Captain Button
[ quoted paragraphs reformatted ]
Wild-eyed conspiracy theorists insist that on Fri, 12 Jan 2001 12:17:50 -0500,
Michael D. Ward <mw...@ms.uky.edu> wrote:
> Kyle Haight wrote:
>> In article <3A5F264F...@ms.uky.edu>,
>> Michael D. Ward <mw...@ms.uky.edu> wrote:
>> >
>> >I don't see the connection between reactionless drives and perpetual
>> >motion machines. I thought a perpetual motion machine always produced
>> >more energy that it consumed and a reactionless drive produced motion
>> >without inertia (but might require huge amounts of energy to do it.)
>>
>> Well, in principle if you have a perpetual motion machine you can
>> produce an engine that isn't bound by the rocket equation, which is the
>> characteristic of a reactionless engine that makes it so desirable.
> I thought that what made a reactionless drive desirable was that it
> wouldn't require reaction mass. How would the existance of a perputual
> motion machine allow you to achieve spaceflight without it?
Perpetual motion machines and reactionless drives are indeed two
different things.
Both are impossible according to current science, and both would be
useful for spaceship propulsion, though for different reasons.
They can get confused because many SF stories have ships propelled
by a device/system that is both a PMM and an RD. This is because
you really need both if you want to plausibly get spaceships
which are fast, maneuverable, and able to stay out for long
periods of time without having to be constantly obsessed with
where the nearest fuel/reaction mass source is.
A PMM produces energy out of nothing, basically [1]. So you can use
one to power your spaceship. This eliminates the need for nuclear
reactors or other such power source, which are likely to be massive
and may also require large amounts of massive fuel.
An RD turns energy into motion (kinetic energy) in one direction
only [2]. So you can use it to propel your spaceship, as long
as you have a source of energy to power it with. This eliminates the
need for reaction mass, something that is thrown in the opposite
direction from the direction you want the ship to move.
Note that many kinds of non-PMM, non-RD drives combine fuel and
reaction mass in some fashion. Chemical rockets burn a chemical
fuel with an oxidizer, and then shoot the combustion products out
the rocket nozzle as reaction mass. A hypothetical fusion rocket
might use hydrogen as fuel, and then shoot the helium produced out
as reaction mass.
But I digress.
If you have a PMM but no RD, you can use energy from the PMM to
propel your reaction mass, for example by heating it up and letting
the hot gas or plasma out through a rocket nozzle. Another example
might be using the PMM to power an electromagnetic mass driver to throw
chunks of reaction mass.
This will improve ship performance by eliminating the need for
fuel/batteries/reactors, but your ship will eventually run out of
reaction mass, so you have to be sure to get back to a base or other
reaction mass source before that happens.
If you have a RD but no PMM, you use the energy from your power plant
(batteries, diesel engine, nuclear reactor, etc.) to propel your ship.
But you are limited by the amount of energy available from it. A
perfectly efficient RD will mean that every joule of energy from
the power plant will be turned into one joule of kinetic energy in
the desired direction, according to the formula KE = 1/2 mv^2.
But to get up to really useful speeds for space travel takes a
*lot* of energy. Especially if you want to get up to a large
fraction of the speed of light.
This will improve ship performance by eliminating the need for reaction
mass, but your ship's power plant will eventually run out of charge/fuel
/fissionables/fusionables/etc, so you have to be sure to get back to a
base or other fuel (or whatever) source before that happens.
If you have both a PMM and an RD, your ship has even better performance
and can basically cruise around accelerating and decelerating freely
without worrying about where the next base is. The only limit on
duration of travel is running out of food, or the need to stop for
equipment maintenance, or shore leave to keep the crew from getting
space cafard [4].
Lots of SF writers have ships powered by PMMs and propelled by RDs
because the sad truth is that space is really really big, and it
is very hard get anywhere fast without both of them.
The other sad truth is that both PMMs and RDs are impossible,
according to the best science now available.
[1] Violating the Law of Conservation of Mass/Energy - "Matter and
energy cannot be created or destroyed, only changed into different
forms or each other. [3]
[2] Violating the Law of Conservation of Momentum, aka
Newton's Third Law of Motion - "For every action there is an equal
and opposite reaction." [3]
[3] I may not be getting the precise names and phrasing right here,
but that is the essence.
[4] From a very old Star Trek novel. Basically your crew gets bored
and goes stir-crazy and starts running amok.
--
"You may have trouble getting permission to aero or lithobrake
asteroids on Earth." - James Nicoll
Captain Button - [ but...@io.com ]
Paul F. Dietz
> A PMM produces energy out of nothing, basically [1]. So you can use
> one to power your spaceship. This eliminates the need for nuclear
> reactors or other such power source, which are likely to be massive
> and may also require large amounts of massive fuel.
>
> An RD turns energy into motion (kinetic energy) in one direction
> only [2]. So you can use it to propel your spaceship, as long
> as you have a source of energy to power it with. This eliminates the
> need for reaction mass, something that is thrown in the opposite
> direction from the direction you want the ship to move.
In fact, a reactionless drive that works in any reference frame
is necessarily also a PMM of the first kind, since the change
in kinetic energy of the vehicle depends on the reference
frame.
Paul
Captain Button
Just goes to show that there's no such thing as breaking just
one law of physics. One crack in the dike, and all science
collapses into anarchy....
J.B. Moreno
-snip-
> Perpetual motion machines and reactionless drives are indeed two
> different things.
Absolutely true.
> Both are impossible according to current science, and both would be
> useful for spaceship propulsion, though for different reasons.
PMM are indeed deemed impossible, but RD are not -- reactionless drives
are not inertialess drives. They simply require that you get your
motion in some other way than throwing something out the back. A solar
sail is a reactionless drive.
--
JBM
"Moebius strippers only show you their back side." -- Unknown
Anton Sherwood
: -- reactionless drives are not inertialess drives. They simply
: require that you get your motion in some other way than throwing
: something out the back. A solar sail is a reactionless drive.
Does it not throw light out the back?
--
Anton Sherwood -- br0...@p0b0x.com -- http://ogre.nu/
Paul F. Dietz
> PMM are indeed deemed impossible, but RD are not -- reactionless drives
> are not inertialess drives. They simply require that you get your
> motion in some other way than throwing something out the back. A solar
> sail is a reactionless drive.
This is not right. Solar sails still conserve momentum,
transfering momentum to/from the vehicle from/to something
else (sunlight, in this case.)
Reactionless drives (as usually defined) do not conserve
momentum. I presume the 'reactionless' name comes from
the english phrasing of Newton's Third Law (conservation
of momentum.)
Paul
Matthew Austern
> Perpetual motion machines and reactionless drives are indeed two
> different things.
>
> Both are impossible according to current science, and both would be
> useful for spaceship propulsion, though for different reasons.
Although in some sense they're impossible for the same reason.
Perpetual motion machines violate conservation of energy, and
reactionless drives violate conservation of momentum. But in
a relativistic framework, energy is just the 0 component of
the momentum 4-vector.
Marc Lombart
Moreno) wrote:
>, but RD are not -- reactionless drives
>are not inertialess drives. They simply require that you get your
>motion in some other way than throwing something out the back.
No, I think you are mistaking the terms reaction and reaction
mass. Reactionless drives are not simply reaction massless. If that
were the only requirement, then we could say we have reactionless
drives now. How much reaction mass does your bicycle throw backwards
to propel yourself forward? It's not a reactionless drive because the
tires of the bicycle react against the ground.
Magnetic sails would work the same, reacting against the
magnetic fields of the planets and the solar system to acquire
Delta-V, but though it would not throw any mass backwards, still not a
reactionless drive.
The Dean Drive(sic) is supposed to provide delta-V without any
reaction between the machine and the universe outside.
--
Marc el Kato
mailto:master...@netzero.net
ICQ UIN: 3337155
Please, reply either to the group or via eMail, not both.
Mark Atwood
>
> Reactionless drives (as usually defined) do not conserve momentum.
> I presume the 'reactionless' name comes from the english phrasing of
> Newton's Third Law (conservation of momentum.)
Is it possible to break "conservation of momentum" without breaking
"conservation of mass/energy"?
--
Mark Atwood | I'm wearing black only until I find something darker.
m...@pobox.com |
http://www.pobox.com/~mra
Paul F. Dietz
> Is it possible to break "conservation of momentum" without breaking
> "conservation of mass/energy"?
Um, no, I don't think so.
Paul
Dan Swartzendruber
> "Paul F. Dietz" <di...@interaccess.com> writes:
> >
> > Reactionless drives (as usually defined) do not conserve momentum.
> > I presume the 'reactionless' name comes from the english phrasing of
> > Newton's Third Law (conservation of momentum.)
>
> Is it possible to break "conservation of momentum" without breaking
> "conservation of mass/energy"?
Interesting question. On a quasi-related note, I recall awhile back a thread about statis
fields, and some of the more knowledgeable people (physics-wise) asserted that a stasis field
can't be perfectly rigid, but I don't recall anyone explaining why. I did some thinking about
it, and the only explanation I can think of is that if a bobble were perfectly rigid, I could
push on one end and have a force manifest at the other at an effective velocity faster than C.
If that's not it, I'm at a loss :) Anyway, if that *is* it, I was playing around with a
couple of ideas for bobbles that could be perfectly rigid (unfortunately, they require other
forms of handwaving, which I'd like to see critiqued). Option #1: the bobble is rigid, but
any momentum imparted to it is not manifested until it's legal to do so. e.g. if light will
travel N distance in M time, then if you try to push a bobble of diameter N, it won't start
moving until M time has elapsed (or maybe an infinitesimally larger time?) Option #2: bobbles
do not move, so you can't push them. This one is uglier, since it seems to me that that
requires some sort of stationary preferred frame of reference?
Dan Swartzendruber
Why not? Not being wise-ass, just curious. If I take something moving with X kinetic energy
in one direction and instantaneously reverse its direction, how is mass-energy conservation
violated?
Paul F. Dietz
> Why not? Not being wise-ass, just curious. If I take something
> moving with X kinetic energy in one direction and instantaneously
> reverse its direction, how is mass-energy conservation violated?
Look at it in a different reference frame.
Paul
Dan Swartzendruber
I was afraid you'd say that :)
Timothy Little
>
>Is it possible to break "conservation of momentum" without breaking
>"conservation of mass/energy"?
No, at least not in either Newtonian or Einsteinian physics. In
Newtonian, non-conservation of momentum alone in one frame breaks
conservation of energy in all others. In Einsteinian, conservation of
momentum and energy are not separate laws -- they are expressions of a
single law in different coordinates.
- Tim
Erik Max Francis
> Interesting question. On a quasi-related note, I recall awhile back a
> thread about statis
> fields, and some of the more knowledgeable people (physics-wise)
> asserted that a stasis field
> can't be perfectly rigid, but I don't recall anyone explaining why. I
> did some thinking about
> it, and the only explanation I can think of is that if a bobble were
> perfectly rigid, I could
> push on one end and have a force manifest at the other at an effective
> velocity faster than C.
> If that's not it, I'm at a loss :)
Yes, that's pretty much it. Relativity prohibits perfectly rigid
bodies. In an extreme case, ask yourself what would happen if you
dipped a stasis field half-in, half-out of a black hole's event horizon.
It's _got_ to come out broken.
But then stasis fields seem to violate physical laws left and right, so
that's not really surprising.
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ Love is like war: easy to begin but very hard to stop.
\__/ H.L. Mencken
The laws list / http://www.alcyone.com/max/physics/laws/
Laws, rules, principles, effects, paradoxes, etc. in physics.
J.B. Moreno
> J.B. Moreno <pl...@newsreaders.com> writes
> : -- reactionless drives are not inertialess drives. They simply
> : require that you get your motion in some other way than throwing
> : something out the back. A solar sail is a reactionless drive.
>
> Does it not throw light out the back?
Light bounces off of it, and get's thrown back, it's not something that
is gather (ram scoop) or carried (conventional fuel) by the ship and
then thrown out.
J.B. Moreno
> "J.B. Moreno" wrote:
>
> > PMM are indeed deemed impossible, but RD are not -- reactionless drives
> > are not inertialess drives. They simply require that you get your
> > motion in some other way than throwing something out the back. A solar
> > sail is a reactionless drive.
>
> This is not right. Solar sails still conserve momentum,
> transfering momentum to/from the vehicle from/to something
> else (sunlight, in this case.)
Right.
> Reactionless drives (as usually defined) do not conserve
> momentum. I presume the 'reactionless' name comes from
> the english phrasing of Newton's Third Law (conservation
> of momentum.)
You're thinking of inertialess drives (ala Doc Smith) not reactionless
drives, which are generally more narrowly defined as ships that don't
use some variation of rockets to move.
In SF this is frequently done via some type of gravity control (i.e. KK
ships in the Flinx series, where an artificial black hole is constantly
being created directly in front of the ship, gravity then pulls them
towards it and because the ship has moved the black hole is created
again a little further away).
barnacle
>Anton Sherwood <an...@home.ogre.nu> wrote:
>
>> J.B. Moreno <pl...@newsreaders.com> writes
>> : -- reactionless drives are not inertialess drives. They simply
>> : require that you get your motion in some other way than throwing
>> : something out the back. A solar sail is a reactionless drive.
>>
>> Does it not throw light out the back?
>
>Light bounces off of it, and get's thrown back, it's not something that
>is gather (ram scoop) or carried (conventional fuel) by the ship and
>then thrown out.
>
But always radial to the sun?
--
I have a quantum car. Every time I look at the speedometer I get lost...
barnacle
http://www.nailed-barnacle.co.uk
Paul F. Dietz
> > Reactionless drives (as usually defined) do not conserve
> > momentum. I presume the 'reactionless' name comes from
> > the english phrasing of Newton's Third Law (conservation
> > of momentum.)
>
> You're thinking of inertialess drives (ala Doc Smith) not reactionless
> drives, which are generally more narrowly defined as ships that don't
> use some variation of rockets to move.
No, I'm thinking of reactionless drives.
By your definition, my station wagon has a reactionless drive.
Paul
John Schilling
>On Sat, 13 Jan 2001 00:23:52 -0500, pl...@newsreaders.com (J.B.
>Moreno) wrote:
>>, but RD are not -- reactionless drives
>>are not inertialess drives. They simply require that you get your
>>motion in some other way than throwing something out the back.
> No, I think you are mistaking the terms reaction and reaction
>mass. Reactionless drives are not simply reaction massless. If that
>were the only requirement, then we could say we have reactionless
>drives now. How much reaction mass does your bicycle throw backwards
>to propel yourself forward?
5.9742 x 10^24 kilograms. Next question?
--
*John Schilling * "Anything worth doing, *
*Member:AIAA,NRA,ACLU,SAS,LP * is worth doing for money" *
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William December Starr
Erik Max Francis <m...@alcyone.com> said:
> Yes, that's pretty much it. Relativity prohibits perfectly rigid
> bodies. In an extreme case, ask yourself what would happen if you
> dipped a stasis field half-in, half-out of a black hole's event
> horizon. It's _got_ to come out broken.
Why does it have to come out at all?
-- William December Starr <wds...@panix.com>
Erik Max Francis
> Why does it have to come out at all?
If one end is being held outside of the horizon, then that end isn't
subject to being pulled in. If things are so arranged so that if, no
matter what, a bit of the stasis field is inside the horizon then all of
it falls in, then you're right back to the case of a perfectly rigid
body.
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ If you are afraid of loneliness, do not marry.
\__/ Anton Chekhov
Official Omega page / http://www.alcyone.com/max/projects/omega/
The official distribution page for the popular Roguelike, Omega.
Riboflavin
>In article <3A60FB32...@alcyone.com>,
>Erik Max Francis <m...@alcyone.com> said:
>> Yes, that's pretty much it. Relativity prohibits perfectly rigid
>> bodies. In an extreme case, ask yourself what would happen if you
>> dipped a stasis field half-in, half-out of a black hole's event
>> horizon. It's _got_ to come out broken.
>
>Why does it have to come out at all?
>
and the other into a different one.
--
Kevin Allegood ribotr...@mindspring.pants.com
Remove the pants from my email address to reply
"Life may have no meaning. Or even worse, it may
have a meaning of which I disapprove." -- Ashleigh Brilliant
Captain Button
> William December Starr wrote in message <93tep7$1i9$1...@panix2.panix.com>...
>>In article <3A60FB32...@alcyone.com>,
>>Erik Max Francis <m...@alcyone.com> said:
>>> Yes, that's pretty much it. Relativity prohibits perfectly rigid
>>> bodies. In an extreme case, ask yourself what would happen if you
>>> dipped a stasis field half-in, half-out of a black hole's event
>>> horizon. It's _got_ to come out broken.
>>
>>Why does it have to come out at all?
>>
> To cut to the chase, take a stasis field, stick one end into one black hole
> and the other into a different one.
As an alternative, take a 4 light year long pole, put it in a
stasis field, and point it at Alpha Centauri. Push it forward
and pull it back to send morse code messages giving the current
market rates on resubliminated thiotimoline. Since it is perfectly
rigid, the message travels FTL.
Wait and see if you are arrested by the ISEC or assassinated by
enraged relativistic physicists.
Zack Weinberg
>William December Starr wrote in message <93tep7$1i9$1...@panix2.panix.com>...
>>In article <3A60FB32...@alcyone.com>,
>>Erik Max Francis <m...@alcyone.com> said:
>>> Yes, that's pretty much it. Relativity prohibits perfectly rigid
>>> bodies. In an extreme case, ask yourself what would happen if you
>>> dipped a stasis field half-in, half-out of a black hole's event
>>> horizon. It's _got_ to come out broken.
>>
>>Why does it have to come out at all?
>>
>To cut to the chase, take a stasis field, stick one end into one black hole
>and the other into a different one.
Nah, the black holes would just get dragged together.
zw
Mark Atwood
> >
> To cut to the chase, take a stasis field, stick one end into one black hole
> and the other into a different one.
Truely, hitting the unbreakable shield with the unstoppable force.
Frank
To clarify: suppose you're moving at the same speed, in the same
direction, as the object X. This means you're in the same reference
frame. As far as you're concerned, object X has no kinetic energy;
it's motionless relative to you.
Now suppose the reactionless drive gets switched on, so that to some
other observer it appears that object X has reversed direction. To
you, it appears that object X suddenly acquired a whole bunch of
kinetic energy from nowhere.
Does that help?
ste...@pcrm.win.tue.nl
But even in a Newtonian setting, a reactionless drive is a
PPM. Consider a spaceship that goes from speed 0 to speed V
by using its reactionless drive. Let M be the mass of the spaceship,
then it needs U=MV^2/2 energy to extract from its batteries.
So it extracts an amount U of energy from its energy source.
Now consider an observer who passes by with speed V in the
opposite direction. He sees the spaceship going from V
to 2V, e.g. a kinetic energy increase of
M(2V)^2/2 - MV^2/2 = 3/2 * MV^2 = 3*U. However, there is still
only an amount U of energy extracted from the spaceship's energy
source. So the observer sees a production 2*U energy from nothing.
Note that if the spaceship was an ordinary rocket, then the
observer would see a decrease in the kinetic energy of the
spaceship's exhaust gasses, which would make up for the
apparent energy gain.
Stephan
--
ir. Stephan H.M.J. Houben
tel. +31-40-2474358 / +31-40-2743497
e-mail: step...@win.tue.nl
Charles R Martin
I'm going to tread carefully here since I've been beating people up about
statistics and my physics is sadly deficient... but some of the RD ideas, or
similar -- in particular, Woodruff's capacitor gimmick -- depend on the truth
of the hitherto untested Mach notion for where inertia comes from, which is
that it's basically the vector sum of all the gravitational forces from
everything *else* in the universe. Thus a "reactionless" drive in this sense
is actually reacting against the mass of the whole universe.
Now, it appears to me that this would also impose a preferred frame on the
universe, which would be interesting. (If there *were* a universal frame,
several of Einstein's gedankenexperimente suddly fall apart....) But I'm not
making this up, just reporting.
ste...@pcrm.win.tue.nl
>I'm going to tread carefully here since I've been beating people up about
>statistics and my physics is sadly deficient... but some of the RD ideas, or
>similar -- in particular, Woodruff's capacitor gimmick -- depend on the truth
>of the hitherto untested Mach notion for where inertia comes from, which is
>that it's basically the vector sum of all the gravitational forces from
>everything *else* in the universe.
But how would this mesh with general relativity?
In GR, gravity isn't really a "force" at all.
And even in a Newtonian setting, I cannot see how this
would be the source of inertia. In an infinite universe, the
vector sum would not even be well-defined. In a finite universe,
it might be well-defined but then all objects on
"small" scales would experience the same acceleration, e.g.
they would be in free-fall and so locally you wouldn't be
able to notice anything of this force. However, inertia
is a local phenomenon.
A better way to understand inertia is to require that
energy must be conserved for *all* frames of reference.
(Because we suppose there is no special FOR.)
Then conservation of impulse rolls out automatically.
>Thus a "reactionless" drive in this sense
>is actually reacting against the mass of the whole universe.
This would require to act on the whole universe instantaneously,
e.g. FTL. Which opens up a whole other can o' worms.
>Now, it appears to me that this would also impose a preferred frame on the
>universe, which would be interesting.
Yes, as always in FTL a Special Provision (such as a special FOR)
needs to be made. If you have a special FOR, then throw inertia
away be specifying that only in the special FOR, energy needs to
be conserved.
>(If there *were* a universal frame,
>several of Einstein's gedankenexperimente suddly fall apart....)
No kidding! The basic premise of relativity is that there is
*not* such a FOR. The rest follows from that in the obvious[1]
way.
Stephan
[1] Although it might not be obvious at first.
Erik Max Francis
> I'm going to tread carefully here since I've been beating people up
> about
> statistics and my physics is sadly deficient... but some of the RD
> ideas, or
> similar -- in particular, Woodruff's capacitor gimmick -- depend on
> the truth
> of the hitherto untested Mach notion for where inertia comes from,
> which is
> that it's basically the vector sum of all the gravitational forces
> from
> everything *else* in the universe. Thus a "reactionless" drive in
> this sense
> is actually reacting against the mass of the whole universe.
Note that although Einstein was inspired by Mach's work, general
relativity is not inherently Machian; it doesn't conform to Mach's
principle (the jist of which is what you describe here).
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ No one should have to dance backward all their lives.
\__/ Jill Ruckelshaus
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A physics reference.
John Schilling
>William December Starr wrote:
>> Why does it have to come out at all?
>If one end is being held outside of the horizon, then that end isn't
>subject to being pulled in. If things are so arranged so that if, no
>matter what, a bit of the stasis field is inside the horizon then all of
>it falls in, then you're right back to the case of a perfectly rigid
>body.
Well, yes, that follows logically from the rules of stasis fields and
black holes. The part of the stasis field inside the event horizon is
subject to being pulled in by gravity. The part of the stasis field
outside the event horizon is subject to being pulled in by being
unbreakably attacked to something that is being irresistably pulled in
to a black hole. Therefore, the whole mess ends up being pulled in to
thhe black hole. You'll have to work a bit harder to set up any sort
of irresistable force/immovable object paradox here.
Not a *lot* harder, mind you. Stasis fields are silly.
Charles R Martin
>
> On Mon, 15 Jan 2001 07:32:01 -0700, Charles R Martin <crma...@indra.com> wrote:
> >I'm going to tread carefully here since I've been beating people up about
> >statistics and my physics is sadly deficient... but some of the RD ideas, or
> >similar -- in particular, Woodruff's capacitor gimmick -- depend on the truth
> >of the hitherto untested Mach notion for where inertia comes from, which is
> >that it's basically the vector sum of all the gravitational forces from
> >everything *else* in the universe.
>
> But how would this mesh with general relativity?
> In GR, gravity isn't really a "force" at all.
>
> And even in a Newtonian setting, I cannot see how this
> would be the source of inertia. In an infinite universe, the
> vector sum would not even be well-defined. In a finite universe,
> it might be well-defined but then all objects on
> "small" scales would experience the same acceleration, e.g.
> they would be in free-fall and so locally you wouldn't be
> able to notice anything of this force. However, inertia
> is a local phenomenon.
Don't expect me to have a good answer here (I'm primarily a logician, I only
do continuous math at all 'cause I use queueing theory a lot) but -- while
it's certainly controversial -- Mach's principal is taken seriously.
http://einstein.stanford.edu/gen_int/story_of_gpb/box3.html
http://chaos.fullerton.edu/Woodward.html
http://chaos.fullerton.edu/~jimw/nasa-pap/
http://www.treasure-troves.com/physics/MachsPrinciple.html
http://www.amazon.com/exec/obidos/ASIN/0817638237/ericstreasuretroA/104-8311831-9452748
... it all sounds pretty crazed, but _Fundamentals of Physics Letters_ is the
real thing....
>
> A better way to understand inertia is to require that
> energy must be conserved for *all* frames of reference.
> (Because we suppose there is no special FOR.)
> Then conservation of impulse rolls out automatically.
>
> >Thus a "reactionless" drive in this sense
> >is actually reacting against the mass of the whole universe.
>
> This would require to act on the whole universe instantaneously,
> e.g. FTL. Which opens up a whole other can o' worms.
Not necessarily -- how would we know how long it takes for hitting a golf ball
here to twiddle the Lesser Magellanic Cloud?
Erik Max Francis
> Well, yes, that follows logically from the rules of stasis fields and
> black holes. The part of the stasis field inside the event horizon is
> subject to being pulled in by gravity. The part of the stasis field
> outside the event horizon is subject to being pulled in by being
> unbreakably attacked to something that is being irresistably pulled in
> to a black hole. Therefore, the whole mess ends up being pulled in to
> thhe black hole. You'll have to work a bit harder to set up any sort
> of irresistable force/immovable object paradox here.
I think you're missing my point. Relativity prohibits infinitely rigid
bodies, and so stasis fields are out. You can't start with the
assumption that stasis fields are infinitely rigid to conclude that the
whole thing will get sucked into a black hole; that's assuming what you
set out to show. (The general _you_, not you specifically.)
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ Woman was God's _second_ mistake.
\__/ Friedrich Nietzsche
Interstelen / http://www.interstelen.com/
A multiplayer, strategic, turn-based Web game on an interstellar scale.
mstemper - psc . com
>As an alternative, take a 4 light year long pole, put it in a
>stasis field, and point it at Alpha Centauri. Push it forward
>and pull it back to send morse code messages giving the current
>market rates on resubliminated thiotimoline. Since it is perfectly
>rigid, the message travels FTL.
>
>Wait and see if you are arrested by the ISEC or assassinated by
>enraged relativistic physicists.
... who never get invited to parties of that sort.
--
Michael F. Stemper
#include <Standard_Disclaimer>
2 + 2 = 5, for sufficiently large values of 2
Charles R Martin
>
> Charles R Martin wrote:
>
> > I'm going to tread carefully here since I've been beating people up
> > about
> > statistics and my physics is sadly deficient... but some of the RD
> > ideas, or
> > similar -- in particular, Woodruff's capacitor gimmick -- depend on
> > the truth
> > of the hitherto untested Mach notion for where inertia comes from,
> > which is
> > that it's basically the vector sum of all the gravitational forces
> > from
> > everything *else* in the universe. Thus a "reactionless" drive in
> > this sense
> > is actually reacting against the mass of the whole universe.
>
> Note that although Einstein was inspired by Mach's work, general
> relativity is not inherently Machian; it doesn't conform to Mach's
> principle (the jist of which is what you describe here).
As I say, I'm not the one to try and defend Mach's Principle, but the stuff I
just posted is very interesting at least.
pervect
"Charles R Martin" <crma...@indra.com> wrote in message
news:3A6309E1...@indra.com...
> ste...@pcrm.win.tue.nl wrote:
> I'm going to tread carefully here since I've been beating people up about
> statistics and my physics is sadly deficient... but some of the RD ideas,
or
> similar -- in particular, Woodruff's capacitor gimmick -- depend on the
truth
> of the hitherto untested Mach notion for where inertia comes from, which
is
> that it's basically the vector sum of all the gravitational forces from
> everything *else* in the universe. Thus a "reactionless" drive in this
sense
> is actually reacting against the mass of the whole universe.
>
> Now, it appears to me that this would also impose a preferred frame on the
> universe, which would be interesting. (If there *were* a universal frame,
> several of Einstein's gedankenexperimente suddly fall apart....) But I'm
not
> making this up, just reporting.
Well, for whatever it's worth, General Relativity isn't a machian theory.
Mach's principle was influential in forming the theory, but when all is said
and done the theory isn't Machian.
One of the ways this is demonstrated is by the way that rotation in GR is
absolute.
Of course, this doesn't say much about what a Machian theory would be like,
if one existed, it just points out that our current theory doesn't qualify
as being Machian.
Charles R Martin
From the stuff in what I've read, it appears that GR can only be consistent
with a Machian theory if you do something weird with time as well -- like
eliminate it.
On the other hand, Woodward's experiments *seem* to confirm Machian
predictions. John Cramer's supposed to be making an independent test with
results next month.
Jim Davies
>Erik Max Francis <m...@alcyone.com> writes:
>
>>William December Starr wrote:
>
>>> Why does it have to come out at all?
>
>>If one end is being held outside of the horizon, then that end isn't
>>subject to being pulled in. If things are so arranged so that if, no
>>matter what, a bit of the stasis field is inside the horizon then all of
>>it falls in, then you're right back to the case of a perfectly rigid
>>body.
>
>
>Well, yes, that follows logically from the rules of stasis fields and
>black holes. The part of the stasis field inside the event horizon is
>subject to being pulled in by gravity. The part of the stasis field
>outside the event horizon is subject to being pulled in by being
>unbreakably attacked to something that is being irresistably pulled in
>to a black hole. Therefore, the whole mess ends up being pulled in to
>thhe black hole. You'll have to work a bit harder to set up any sort
>of irresistable force/immovable object paradox here.
>
>Not a *lot* harder, mind you. Stasis fields are silly.
But why does the *outside* of the stasis field have to remain rigid?
So long as the *inside* isn't affected, we can stretch, twist, warp,
bend and tie knots in the *outside*.
Provides you don't snip it in half or make a torus in it, the inside
can remain unaffected. It's not as though there's anything odd about
warping space, after all.
So when you start to pull it out, you just pull and pull and get more
and more stasis field (it may get thinner like a rubber band, it may
not) but the other end stays in the hole.
I don't see that a stasis field need have any particular inherent
rigidity at all; provided you can restore the original [energy]
conditions when you switch it off, there's no conflict. All this
suggests is that it may require energy to bend it.
-
Jim Davies
----------
Mind your manners, son! I've got a tall pointy hat!
Blaine Manyluk
<3a635d7a...@news.powernet.co.uk>...
> But why does the *outside* of the stasis field have to remain rigid?
> So long as the *inside* isn't affected, we can stretch, twist, warp,
> bend and tie knots in the *outside*.
>
> Provides you don't snip it in half or make a torus in it, the inside
> can remain unaffected. It's not as though there's anything odd about
> warping space, after all.
Are you suggesting that the inside of a stasis field may be in a
different space than the outside? Can the interior be larger in
volume?
(After watching an episode of _Doctor Who_...)
======================================================================
[To reply, remove the S's from my address, and change the R's to N's.]
An infinite number of monkeys on an infinite number of typewriters
will eventually come up with a good _Voyager_ script.
Gareth Wilson
>
> On the other hand, Woodward's experiments *seem* to confirm Machian
> predictions. John Cramer's supposed to be making an independent test with
> results next month.
Cite? (I mean this in the "wow, cool, where can I get more information?" sense,
not the "Yeah, _right_" sense).
--
~~~~~~~~~~~~~~~
Gareth Wilson
Christchurch
New Zealand
~~~~~~~~~~~~~~
Geoffrey A. Landis
> > is necessarily also a PMM of the first kind, since the change
> > in kinetic energy of the vehicle depends on the reference
> > frame.
Nope. If negative mass is not forbidden, then you can produce a drive
which has zero net momentum (and zero net energy) regardless of velocity.
Negative energy has a lot of other bizarre consequences; for example, it
tends to make the entire universe unstable :( ; but it's not logically forbidden.
--
Geoffrey A. Landis
author of MARS CROSSING
http://www.sff.net/people/geoffrey.landis
Geoffrey A. Landis
>
> Ah, but there is ALREADY a way to convert energy into thrust. Light
> (and radio waves, etc.) have momentum but no "rest mass". At considerable
> energy cost, you can produce thrust without throwing away anything
> except photons.
It's not reactionless because you are reacting against the photons.
Rest mass is irrelevant.
"J.B. Moreno" wrote:
> In SF this is frequently done via some type of gravity control (i.e. KK
> ships in the Flinx series, where an artificial black hole is constantly
> being created directly in front of the ship, gravity then pulls them
> towards it and because the ship has moved the black hole is created
> again a little further away).
note that this violates conservation of energy *and* conservation of
momentum; it's a "pull yourself up by grabbing your shoelaces and
pulling" type drive.
Erik Max Francis
> "J.B. Moreno" wrote:
>
> > In SF this is frequently done via some type of gravity control (i.e.
> > KK
> > ships in the Flinx series, where an artificial black hole is
> > constantly
> > being created directly in front of the ship, gravity then pulls them
> > towards it and because the ship has moved the black hole is created
> > again a little further away).
>
> note that this violates conservation of energy *and* conservation of
> momentum; it's a "pull yourself up by grabbing your shoelaces and
> pulling" type drive.
Yep. To choose a more mundane example, it's like sitting on an office
chair, holding a fan, and pointing at you, and expecting to zip around
the office. The fan's expelled air may be applying a force on you, but
you're applying an equal but opposite force on the fan since you're
holding it ...
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ History is a set of lies agreed upon.
\__/ Napoleon Bonaparte
Crank Dot Net / http://www.crank.net/
Cranks, crackpots, kooks, & loons on the Net.
Captain Button
[ snip ]
> It's not reactionless because you are reacting against the photons.
> Rest mass is irrelevant.
You are Neutrino of Borg and I claim my five pounds.
:-)}
Paul F. Dietz
> Nope. If negative mass is not forbidden, then you can produce a drive
> which has zero net momentum (and zero net energy) regardless of velocity.
>
> Negative energy has a lot of other bizarre consequences; for example, it
> tends to make the entire universe unstable :( ; but it's not logically forbidden.
I agree, but I would not call that reactionless,
since it doesn't violate conservation of momentum.
Paul
Jim Davies
>Jim Davies <j...@aaargh.YouCanKeepThisBit.org> wrote in article
><3a635d7a...@news.powernet.co.uk>...
>
>> But why does the *outside* of the stasis field have to remain rigid?
>> So long as the *inside* isn't affected, we can stretch, twist, warp,
>> bend and tie knots in the *outside*.
>>
>> Provides you don't snip it in half or make a torus in it, the inside
>> can remain unaffected. It's not as though there's anything odd about
>> warping space, after all.
>
>Are you suggesting that the inside of a stasis field may be in a
>different space than the outside? Can the interior be larger in
>volume?
>
>(After watching an episode of _Doctor Who_...)
If we're assuming that time passes infinitely slowly inside the field,
you can do anything you like to it until you restore it to its
original shape, as it won't actually make any difference.
It depends on what "hair" it has; scalar quantities like mass and
charge are trivial unless you have to consider their distribution as
well. Charge distribution is a non-event if the field is
superconducting; maybe there can be an equivalent for mass too?
Mark Atwood
>
> It depends on what "hair" it has; scalar quantities like mass and
> charge are trivial unless you have to consider their distribution as
> well. Charge distribution is a non-event if the field is
> superconducting; maybe there can be an equivalent for mass too?
Would that make a bobble appear of have a moment of intertia of zero?
That is, it acts like a point mass. Or does instead the mass inside
act like it's uniformly distributed?
Could you even tell? The surface is utterly uniform, and implied to be
frictionless. Is "orientation" even definable for a bobble? What
happens if you enclose a gyroscope? Or say you Confine something that
is not in balance. (One side is not in mass balance with the other.)
When the Confinment ends, will the contents have the same orientation?
Will it have rotated so that the contents *are* in balance wrt local
gravity/acceleration? Would it be random?
Do the conditions inside the bobble when it's formed, and/or events
that stress the outside affect the the duration of confinement, or the
energy or processing requirements neccessary to create it?
What happens when you expose it to intense graviational fields? Do
gravity waves go thru it?
If I had a bobble generator in a lab, these are questions I would be
trying to answer. That's a point that I had a hard time suspending my
disbelief over, that the Peace Authority didn't have a couple of high
security research labs hammering away at these questions.
Matthew Austern
> j...@aaargh.YouCanKeepThisBit.org (Jim Davies) writes:
> >
> > It depends on what "hair" it has; scalar quantities like mass and
> > charge are trivial unless you have to consider their distribution as
> > well. Charge distribution is a non-event if the field is
> > superconducting; maybe there can be an equivalent for mass too?
>
> Would that make a bobble appear of have a moment of intertia of zero?
> That is, it acts like a point mass. Or does instead the mass inside
> act like it's uniformly distributed?
If either of those is the case, it would be easy to use bobbles to
violate conservation of energy.
William December Starr
Erik Max Francis <m...@alcyone.com> said (to John Schilling):
> I think you're missing my point. Relativity prohibits infinitely
> rigid bodies, and so stasis fields are out. You can't start with
> the assumption that stasis fields are infinitely rigid to conclude
> that the whole thing will get sucked into a black hole; that's
> assuming what you set out to show. (The general _you_, not you
> specifically.)
My point though, which may be sort of collateral to the discussion
going on between you and John Schilling, was that if you dipped a
stasis field half-in, half-out of a black hole's event horizon, I
can't see how "It's _got_ to come out broken."
"_If_ it comes out, it's got to come out broken" I can see, but not
the flat statement without the conditional.
-- William December Starr <wds...@panix.com>
Mark Atwood
> Mark Atwood <m...@pobox.com> writes:
> >
> > Would that make a bobble appear of have a moment of intertia of zero?
> > That is, it acts like a point mass. Or does instead the mass inside
> > act like it's uniformly distributed?
>
> If either of those is the case, it would be easy to use bobbles to
> violate conservation of energy.
How?
Matthew Austern
> Matthew Austern <aus...@research.att.com> writes:
> > Mark Atwood <m...@pobox.com> writes:
> > >
> > > Would that make a bobble appear of have a moment of intertia of zero?
> > > That is, it acts like a point mass. Or does instead the mass inside
> > > act like it's uniformly distributed?
> >
> > If either of those is the case, it would be easy to use bobbles to
> > violate conservation of energy.
>
> How?
Put it in a nonuniform gravitational field. If an irregularly shaped
object is in a nonuniform graviational field, its potential energy
depends on it orientation. So the steps are:
(1) start with the object in its minimum-potential orientation.
(2) bobble it.
(3) rotate the bobble until it's in the orientation that,
if the object were unbobbled, would be maximum potential.
(4) unbobble it.
By hypothesis (i.e. that the bobble behaves as a point mass or a
uniformly distributed mass), step 3 takes zero energy. You're
gaining potential energy for no work.
Richard D. Latham
Well, maybe not. After all, we do know it takes energy to create a
bobble, right ?
--
#include <disclaimer.std> /* I don't speak for IBM ... */
/* Heck, I don't even speak for myself */
/* Don't believe me ? Ask my wife :-) */
Richard D. Latham lat...@us.ibm.com
Kyle Haight
Matthew Austern <aus...@research.att.com> wrote:
>
> (1) start with the object in its minimum-potential orientation.
> (2) bobble it.
> (3) rotate the bobble until it's in the orientation that,
> if the object were unbobbled, would be maximum potential.
> (4) unbobble it.
>
>By hypothesis (i.e. that the bobble behaves as a point mass or a
>uniformly distributed mass), step 3 takes zero energy. You're
>gaining potential energy for no work.
Considering that bobbles are perfectly spherical and totally frictionless,
how do you propose to accomplish (3)?
--
Kyle Haight
kha...@alumni.ucsd.edu
Leonard Erickson
Except that "by definition" you *can't* rotate the bobble, since it's a
mathematically perfect sphere. Nothing to hold onto, nor any way to mark
it to tell if your efforts to rotate it have had any effect.
--
Leonard Erickson (aka Shadow)
sha...@krypton.rain.com <--preferred
leo...@qiclab.scn.rain.com <--last resort
Mark Atwood
> Mark Atwood <m...@pobox.com> writes:
> > Matthew Austern <aus...@research.att.com> writes:
> > > Mark Atwood <m...@pobox.com> writes:
> > > >
> > > > Would that make a bobble appear of have a moment of intertia of zero?
> > > > That is, it acts like a point mass. Or does instead the mass inside
> > > > act like it's uniformly distributed?
> > >
> > > If either of those is the case, it would be easy to use bobbles to
> > > violate conservation of energy.
> >
> > How?
> Put it in a nonuniform gravitational field. If an irregularly shaped
> object is in a nonuniform graviational field, its potential energy
> depends on it orientation. So the steps are:
> (1) start with the object in its minimum-potential orientation.
> (2) bobble it.
> (3) rotate the bobble until it's in the orientation that,
> if the object were unbobbled, would be maximum potential.
> (4) unbobble it.
Neat.
Of course, the next questions I had in my list was "Is orientation
defined for a bobble?", and of course, you also have the problem of
just *how* do you rotate a perfectly uniform perfectly frictionless
sphere!?
Perhaps when it unbobbles, the contents presents the orientation
that preserves the potential energy.
Lance Purple
Leonard Erickson <sha...@krypton.rain.com> wrote:
>Matthew Austern wrote:
>> Put it in a nonuniform gravitational field. If an irregularly shaped
>> object is in a nonuniform graviational field, its potential energy
>> depends on it orientation. So the steps are:
>> (1) start with the object in its minimum-potential orientation.
>> (2) bobble it.
>> (3) rotate the bobble until it's in the orientation that,
>> if the object were unbobbled, would be maximum potential.
>> (4) unbobble it.
>>
>> By hypothesis (i.e. that the bobble behaves as a point mass or a
>> uniformly distributed mass), step 3 takes zero energy. You're
>> gaining potential energy for no work.
>
>Except that "by definition" you *can't* rotate the bobble, since it's a
>mathematically perfect sphere. Nothing to hold onto, nor any way to mark
>it to tell if your efforts to rotate it have had any effect.
OK, rotate the nonuniform field around it. For example, move a huge
pile of lead weights from one side of the bobble to the opposite side.
If the bobble doesn't have multipole "hair", then it won't rotate 180
degrees in response, and the total potential energy won't be conserved.
--
,---------------------------------------,
/ Lance Purple (lpurple at io dot com) /
'---------------------------------------'
George William Herbert
Mmm. Rotate the frictionless perfectly spherical ball. That's good.
There are several cases:
* Bobbles appear as point masses
* Bobbles appear as constant density
* Bobbles have the same moments of inertia as the bobbled system did
If either 1 or 2 is true then rotation is impossible; since the
surface is frictionless then there will be no tangential forces,
only normal to the surface, and for a perfect sphere no normal
force will create a moment. Orientation won't change for any reason
while the bobble's active. This would actually be somewhat inconvenient
as you'd have a significant chance of coming out of one on your head
rather than on your feet, with a large chunk of ground above you.
Ouch.
There are actually a large number of problems with trying to interpret
bobbles literally and perfectly in physics. This set is one of them.
Another is "perfect sphere in what reference frame?".
-george william herbert
gher...@retro.com
Timothy Little
>> By hypothesis (i.e. that the bobble behaves as a point mass or a
>> uniformly distributed mass), step 3 takes zero energy. You're
>> gaining potential energy for no work.
>
>Well, maybe not. After all, we do know it takes energy to create a
>bobble, right ?
If the bobble-creation "knows" how much energy you're going to
extract, sure. Otherwise, it will always have to take at least mc^2
to create a bobble "just in case", since you can get near this much
energy by appropriate manipulation near a black hole.
- Tim
Mark Atwood
>
> OK, rotate the nonuniform field around it. For example, move a huge
> pile of lead weights from one side of the bobble to the opposite side.
> If the bobble doesn't have multipole "hair", then it won't rotate 180
> degrees in response, and the total potential energy won't be conserved.
What do you mean "move to the opposite side"? You can't move anything
around in the inside, pretty much by definition. You can't stick stuff
to the outside. You can't hang stuff off of it, and you can't pile
things on top of it.
Mark Atwood
>
> If the bobble-creation "knows" how much energy you're going to
> extract, sure. Otherwise, it will always have to take at least mc^2
> to create a bobble "just in case", since you can get near this much
> energy by appropriate manipulation near a black hole.
I was actually having wierd thoughts the other day that maybe a bobble
"talks back in time" along it's t "length". I mean, we know it takes a
semi-random amount of energy to create it...
If a bobble is going to be near a black hole sometimes in it's life,
then it just "happens" to require a LOT of ergs/computrons at it's
moment of creation.
Timothy Little
>Matthew Austern wrote:
>> (1) start with the object in its minimum-potential orientation.
>> (2) bobble it.
>> (3) rotate the bobble until it's in the orientation that,
>> if the object were unbobbled, would be maximum potential.
>
>Except that "by definition" you *can't* rotate the bobble, since it's a
>mathematically perfect sphere. Nothing to hold onto, nor any way to mark
>it to tell if your efforts to rotate it have had any effect.
(3a) Move the bobble around the gravity source so that if it preserves
its orientation, it will have maximum potential. (This can be done
for arbitrarily little energy)
Thought experiment: make a bobble with a heavy object just inside the
edge. Measure gravitational attraction between bobble and test
objects outside.
If the gravitation is non-uniform, then you have a way to interact
with the inside of bobbles.
If it is uniform, then you can break conservation of energy, momentum,
and/or angular momentum. If you bobble an off-centre object and let
it fall, there may be no way the bobble can decay that conserves all
three laws.
For example, consider an object moving at 1 km/s away from the Earth,
on a line passing some distance from Earth's centre. It has some
finite angular momentum with respect to the Earth. Suppose it is
bobbled by a sphere with centre moving directly away from Earth
(i.e. the object is off-center inside the bobble). Now, that's a
violation of angular momentum to begin with, but maybe we can assume
that a bobble has some intrinsic "spin" or something to balance it.
Now, the bobble is moving radially away with less than escape velocity
and so will eventually stop. When it does so, it has zero angular
momentum with respect to the Earth. If the bobble dissipates now,
there is no possible orientation the contents can adopt to conserve
angular momentum.
By similar means, you can easily construct situations that necessarily
violate either or both of energy and momentum conservation.
- Tim
Timothy Little
>Perhaps when it unbobbles, the contents presents the orientation that
>preserves the potential energy.
That's not always possible. In fact, if it does so it will in general
necessarily violate conservation of momentum and/or angular momentum.
- Tim
Robert Shaw
"George William Herbert" <gher...@gw.retro.com> wrote
>
> There are actually a large number of problems with trying to interpret
> bobbles literally and perfectly in physics. This set is one of them.
> Another is "perfect sphere in what reference frame?".
>
smaller than the radii of curvature of space, such a frame can be defined.
--
Matter is fundamentally lazy:- It always takes the path of least effort
Matter is fundamentally stupid:- It tries every other path first.
That is the heart of physics - The rest is details.- Robert Shaw
Timothy Little
>I was actually having wierd thoughts the other day that maybe a bobble
>"talks back in time" along it's t "length". I mean, we know it takes a
>semi-random amount of energy to create it...
>
>If a bobble is going to be near a black hole sometimes in it's life,
>then it just "happens" to require a LOT of ergs/computrons at it's
>moment of creation.
Especially if time really is 'frozen' inside -- it makes a lot of
sense to have to pay the whole energy bill 'up front', since from the
point of view of the bobble's interior, the formation and dissipation
are the same moment in time.
I leave it as an exercise for the reader to use this to break
causality :)
- Tim
Jens Kilian
> (3) rotate the bobble until it's in the orientation that,
> if the object were unbobbled, would be maximum potential.
Yes, but *can* you rotate a bobble?
--
mailto:j...@acm.org phone:+49-7031-464-7698 (TELNET 778-7698)
http://www.bawue.de/~jjk/ fax:+49-7031-464-7351
PGP: 06 04 1C 35 7B DC 1F 26 As the air to a bird, or the sea to a fish,
0x555DA8B5 BB A2 F0 66 77 75 E1 08 so is contempt to the contemptible. [Blake]
Lance Purple
>lpu...@fnord.io.com (Lance Purple) writes:
>> OK, rotate the nonuniform field around it. For example, move a huge
>> pile of lead weights from one side of the bobble to the opposite side.
>> If the bobble doesn't have multipole "hair", then it won't rotate 180
>> degrees in response, and the total potential energy won't be conserved.
>
>What do you mean "move to the opposite side"? You can't move anything
>around in the inside, pretty much by definition. You can't stick stuff
>to the outside. You can't hang stuff off of it, and you can't pile
>things on top of it.
No, no, no. In the lab, set up a gigantic pile of bricks, and a small
pendulum on a stand. The pendulum will lean ever-so-slightly towards
the bricks, like in the Cavendish experiment. Now bobble the pendulum
but not the bricks. While the pendulum apparatus is hidden inside the
bobble, move the bricks. What happens when you unbobble the pendulum ?
(a) the pendulum now leans toward the new location of the bricks,
because the bobble preserves multipole "hair" information.
(b) the pendulum still leans toward the old location of the bricks,
(and gently begins to swing towards the new location), which
means that the bobble violates conservation of energy...
George William Herbert
>"George William Herbert" <gher...@gw.retro.com> wrote
>> There are actually a large number of problems with trying to interpret
>> bobbles literally and perfectly in physics. This set is one of them.
>> Another is "perfect sphere in what reference frame?".
>
>In the bobbles own rest-frame. As long as the radius of the bobble is much
>smaller than the radii of curvature of space, such a frame can be defined.
Ahbut...
If bobbles aren't necessarily perfectly spherical in all reference
frames then you *can* rotate a bobble, because an ellipsoid is not
geometrically set up such that all normal rays to the surface are
through the centroid, unlike a sphere. Pushing on it at a point
halfway from the equator to pole will cause moment and rotation...
-george william herbert
gher...@retro.com
Mark Atwood
> "George William Herbert" <gher...@gw.retro.com> wrote
> >
> > There are actually a large number of problems with trying to interpret
> > bobbles literally and perfectly in physics. This set is one of them.
> > Another is "perfect sphere in what reference frame?".
> >
> In the bobbles own rest-frame. As long as the radius of the bobble is much
> smaller than the radii of curvature of space, such a frame can be defined.
Could that be maybe the source of the processing / energy requirements
and tradeoff in making a bobble? You're trying to force a small bit of
space to be "flat".
Mark Atwood
>
> No, no, no. In the lab, set up a gigantic pile of bricks, and a small
> pendulum on a stand. The pendulum will lean ever-so-slightly towards
> the bricks, like in the Cavendish experiment. Now bobble the pendulum
> but not the bricks. While the pendulum apparatus is hidden inside the
> bobble, move the bricks. What happens when you unbobble the pendulum ?
>
> (a) the pendulum now leans toward the new location of the bricks,
> because the bobble preserves multipole "hair" information.
>
> (b) the pendulum still leans toward the old location of the bricks,
> (and gently begins to swing towards the new location), which
> means that the bobble violates conservation of energy...
Close readings of the books imply (b). If it was (a), then the bobbles
would rotate towards even the most minute change in the lowest
gravitational potential, and thus, ferex, the buildings in the Korlov
city would be at random orientations towards each other after each
flicker.
Erik Max Francis
> My point though, which may be sort of collateral to the discussion
> going on between you and John Schilling, was that if you dipped a
> stasis field half-in, half-out of a black hole's event horizon, I
> can't see how "It's _got_ to come out broken."
>
> "_If_ it comes out, it's got to come out broken" I can see, but not
> the flat statement without the conditional.
My point was that relativity prohibits infinitely rigid bodies. If you
dip a stasis field half-in and half-out of an event horizon and pull it
back and it's not broken, then that's due to infinite rigidity and
violates relativity.
But if, as you seem to suggest, the stasis field _necessarily_ cannot be
brought back out and all of it must fall into the black hole if any of
it is dipped in, then that's another example of infinite rigidity and is
just as much in violation of relativity.
If I dip a stasis field half-in a horizon and the stasis field _cannot_
break, and so all of the stasis field must get sucked in, then you're
right back to talking about a case of infinite rigidity. See what I'm
getting at?
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ The great floodgates of the wonder-world swung open.
\__/ Herman Melville
The laws list / http://www.alcyone.com/max/physics/laws/
Laws, rules, principles, effects, paradoxes, etc. in physics.
Mark Lanett
which is used to create the bobble is going into reducing the angular
momentum to 0, which will take some time; when the bobble opens, that energy
is used to *try* to restore the original angular momentum, which would also
take time, but if it can't because the context is different, well then
normal physics can handle that.
~mark
"Timothy Little" <t...@freeman.little-possums.net> wrote in message
news:slrn96g45...@freeman.little-possums.net...
Leonard Erickson
And what if the contents are spinning when the bobble collapses?
> By similar means, you can easily construct situations that necessarily
> violate either or both of energy and momentum conservation.
There's evidence that a bobble has the momentum of whatever got bobbled.
So that one is easy enough to deal with.
How do you pull of violating conservation of energy?
Timothy Little
>Timothy Little wrote:
>> Now, the bobble is moving radially away with less than escape velocity
>> and so will eventually stop. When it does so, it has zero angular
>> momentum with respect to the Earth. If the bobble dissipates now,
>> there is no possible orientation the contents can adopt to conserve
>> angular momentum.
>And what if the contents are spinning when the bobble collapses?
Then you've immediately got non-conservation of momentum. Remember,
in this situation the bobble contains only a single off-centre mass.
It would also imply that by temporarily fiddling with a bobble while
it is intact, you can destroy the contents when it collapses (do the
"angular momentum" trick for a suitably high "exit spin").
>> By similar means, you can easily construct situations that
>> necessarily violate either or both of energy and momentum
>> conservation.
>There's evidence that a bobble has the momentum of whatever got
>bobbled. So that one is easy enough to deal with.
By this, I meant that if you fix one, you lose the others. e.g. your
attempt to fix conservation of angular momentum broke conservation of
momentum (and energy). I did not mean that you could always construct
a single situation in which all three were broken.
As you say, you can fix conservation of momentum alone (so long as
spacetime is flat in the neighbourhood of the bobble). Just don't
expect to be able to conserve the other two as well while retaining
the given properties of bobbles.
- Tim
Robert Shaw
"Timothy Little" <t...@freeman.little-possums.net> wrote > Leonard Erickson
<sha...@krypton.rain.com> wrote:
>
> Thought experiment: make a bobble with a heavy object just inside the
> edge. Measure gravitational attraction between bobble and test
> objects outside.
>
> If the gravitation is non-uniform, then you have a way to interact
> with the inside of bobbles.
>
> If it is uniform, then you can break conservation of energy, momentum,
> and/or angular momentum. If you bobble an off-centre object and let
> it fall, there may be no way the bobble can decay that conserves all
> three laws.
>
> For example, consider an object moving at 1 km/s away from the Earth,
> on a line passing some distance from Earth's centre. It has some
> finite angular momentum with respect to the Earth. Suppose it is
> bobbled by a sphere with centre moving directly away from Earth
> (i.e. the object is off-center inside the bobble). Now, that's a
> violation of angular momentum to begin with, but maybe we can assume
> that a bobble has some intrinsic "spin" or something to balance it.
Yes, assume the bobble inherits the angular momentum, mass, and
charge of its contents. The multipole moments get radiated away,
as with the formation of black holes. This lets bobble formation
be consistent with conservation laws.
>
> Now, the bobble is moving radially away with less than escape velocity
> and so will eventually stop. When it does so, it has zero angular
> momentum with respect to the Earth. If the bobble dissipates now,
> there is no possible orientation the contents can adopt to conserve
> angular momentum.
>
No. The bobble was formed with zero orbital angular momentum,
and has zero orbital angular momentum at all points on its orbit.
The spin of the bobble also stays constant. When the bobble
evaporates the contents have zero net momentum, and the
proper potential energy, but are spinning at a suitable rate to
conserve angular momentum.
The relevant quantities can be conserved during the lifetime of
the bobble, without causing problems, so since the bobble can
form without violating conservation, if evaporation is a time
reversal of formation, it can evaporate without violating
conservation.
Robert Shaw
"Erik Max Francis" <m...@alcyone.com> wrote
>
> My point was that relativity prohibits infinitely rigid bodies. If you
> dip a stasis field half-in and half-out of an event horizon and pull it
> back and it's not broken, then that's due to infinite rigidity and
> violates relativity.
It would require FTL propaation of stresses in the bobble's surface,
but that isn't matter or normal space. There's no reason for it to obey
the conditions on the stress tensor that normally rule out FTL.
Observers can't fly through the bobble but for a observer at the
transmitter to see the signal arrive before it was sent, they have to
be moving towards the receiver, through the sphere filled by the
bobble, which isn't possible.
It doesn't violate casuality to send signals outside the light cone,
as long as they can't entire your own past light cone. There
may be arrangements that allow this, even allowing for paths
being blocked by the bobble and the requirement for transmitters
and receivers to be stationary with respect to each other, but even then
constructing such a system would be implausible.
>
> But if, as you seem to suggest, the stasis field _necessarily_ cannot be
> brought back out and all of it must fall into the black hole if any of
> it is dipped in, then that's another example of infinite rigidity and is
> just as much in violation of relativity.
>
> If I dip a stasis field half-in a horizon and the stasis field _cannot_
> break, and so all of the stasis field must get sucked in, then you're
> right back to talking about a case of infinite rigidity. See what I'm
> getting at?
Vinge does suggest that his bobbles can be broken by strong
tidal forces, when he talks about sending them through a
wormhole. If bobbles can only be stretched and broken by such
extreme gravitational forces. not normally available in the solar
system, it would be understandable if that qualification was
normally forgotten.
Robert Shaw
> >
> >In the bobbles own rest-frame. As long as the radius of the bobble is
much
> >smaller than the radii of curvature of space, such a frame can be
defined.
>
> Ahbut...
>
> If bobbles aren't necessarily perfectly spherical in all reference
> frames then you *can* rotate a bobble, because an ellipsoid is not
> geometrically set up such that all normal rays to the surface are
> through the centroid, unlike a sphere. Pushing on it at a point
> halfway from the equator to pole will cause moment and rotation...
Except that, if you are touching the bobble, your rest frame is locally
identical with the bobble's, and it looks like a sphere to you, so it
can't be rotated.
Robert Shaw
"Timothy Little" <t...@freeman.little-possums.net> wrote
> Leonard Erickson <sha...@krypton.rain.com> wrote:
>
> >Timothy Little wrote:
> >> Now, the bobble is moving radially away with less than escape velocity
> >> and so will eventually stop. When it does so, it has zero angular
> >> momentum with respect to the Earth. If the bobble dissipates now,
> >> there is no possible orientation the contents can adopt to conserve
> >> angular momentum.
>
> >And what if the contents are spinning when the bobble collapses?
>
> Then you've immediately got non-conservation of momentum. Remember,
> in this situation the bobble contains only a single off-centre mass.
No. The mass can be spinning and still have zero total momentum
Classically, we can give the Moon arbitary angular momentum
without changing its linear momentum.
Jim Davies
>Mark Atwood <m...@pobox.com> writes:
>
>> Matthew Austern <aus...@research.att.com> writes:
>> > Mark Atwood <m...@pobox.com> writes:
>> > >
>> > > Would that make a bobble appear of have a moment of intertia of zero?
>> > > That is, it acts like a point mass. Or does instead the mass inside
>> > > act like it's uniformly distributed?
>> >
>> > If either of those is the case, it would be easy to use bobbles to
>> > violate conservation of energy.
>>
>> How?
>
>Put it in a nonuniform gravitational field. If an irregularly shaped
>object is in a nonuniform graviational field, its potential energy
>depends on it orientation. So the steps are:
> (1) start with the object in its minimum-potential orientation.
> (2) bobble it.
> (3) rotate the bobble until it's in the orientation that,
> if the object were unbobbled, would be maximum potential.
> (4) unbobble it.
>
>By hypothesis (i.e. that the bobble behaves as a point mass or a
>uniformly distributed mass), step 3 takes zero energy. You're
>gaining potential energy for no work.
This argument assumes that a) it takes no energy to unbobble it and
that b) it takes no energy to alter the bobble's local environment.
Violating (b) probably means that the bobble does have hair, of
course.
This discussion has been slightly muddied by confusing "stasis field
in general" with "bobble" (which is a very particular implementation
of a stasis field). Most notably, a bobble is spherical whereas a
generic field might not be.
-
Jim Davies
----------
Mind your manners, son! I've got a tall pointy hat!
George William Herbert
>> >In the bobbles own rest-frame. As long as the radius of the bobble is
>much
>> >smaller than the radii of curvature of space, such a frame can be
>defined.
>>
>> Ahbut...
>> If bobbles aren't necessarily perfectly spherical in all reference
>> frames then you *can* rotate a bobble, because an ellipsoid is not
>> geometrically set up such that all normal rays to the surface are
>> through the centroid, unlike a sphere. Pushing on it at a point
>> halfway from the equator to pole will cause moment and rotation...
>
>Except that, if you are touching the bobble, your rest frame is locally
>identical with the bobble's, and it looks like a sphere to you, so it
>can't be rotated.
You took "pushing on it" too literally.
Ok. I bobble a washing machine with a 2 meter diameter bobble,
hop on my spaceship and take off and accellerate to 0.9 C.
I then release the bobble. We fly through an asteroid belt,
and a small rock hits the bobble halfway from its central
axis to the equator (in either reference frame).
In my frame, the bobble's a sphere. No moment, no rotation
happens. Not that I should be able to tell anyways, but it
*should* not rotate it.
In the frame of the rock, it hit an ellipsoid offcenter
and that contact produces a moment, so the ellipsoid should
start rotating. Now, in the rock's reference frame, that
rotation will not be geometrically apparent, but it should
fundamentally lead to a rotation due to the moment.
It appears to me that this poses a paradox. If we assume the
combination of perfectly rigid body and zero-friction surface,
then you end up with relatavistically inconsistent solutions
to momentum transfers... in some reference frames you get
a linear recoil and in others you get that along with angular
momentum as well. Unless i'm missing my math somewhere the
classical bobble is thus inconsistent with relativity.
-george william herbert
gher...@retro.com
George William Herbert
>William December Starr wrote:
>> My point though, which may be sort of collateral to the discussion
>> going on between you and John Schilling, was that if you dipped a
>> stasis field half-in, half-out of a black hole's event horizon, I
>> can't see how "It's _got_ to come out broken."
>>
>> "_If_ it comes out, it's got to come out broken" I can see, but not
>> the flat statement without the conditional.
>
>My point was that relativity prohibits infinitely rigid bodies. If you
>dip a stasis field half-in and half-out of an event horizon and pull it
>back and it's not broken, then that's due to infinite rigidity and
>violates relativity.
>
>But if, as you seem to suggest, the stasis field _necessarily_ cannot be
>brought back out and all of it must fall into the black hole if any of
>it is dipped in, then that's another example of infinite rigidity and is
>just as much in violation of relativity.
>
>If I dip a stasis field half-in a horizon and the stasis field _cannot_
>break, and so all of the stasis field must get sucked in, then you're
>right back to talking about a case of infinite rigidity. See what I'm
>getting at?
One thing to be careful about here is not applying the sort of "flat space"
nonrelatavistic interpretation of event horizon to a bobble. A black
hole is not a flat region of space with a spherical boundary beyond
which nothing comes back. It's complexly curved space, with a large
number of significant effects due to its curvature. This is evidenced
by the various orbital trajectory zones, etc.
You can't just walk up to the event horizon and dip something (a bobble
or anything else) in. By the time you approach it, you're already in a
region of space where all sub-speed-of-light geodesics intersect the
event horizon. So the reality is, you can't take the bobble down into
a region where you can do a half-in experiment without all of it
falling in.
If we make the assumption that bobbles are always spherical in their
local reference frame, then we have two situations: bobble is small
compared to curvature of space near it, and bobble is not small
compared to curvature of space near it (i.e., curvature over its
radius is significant, and "the bobble's local reference frame"
ceases to be flat enough to define "sphere" consistently, and you
have to look at more complex descrptions of its shape and properties
combined with the shape and properties of local space).
If it's small compared to local curvature you can probably safely treat
the bobble as a point mass (cough and appologies to the rest of the thread)
in terms of its orbital characteristics.
If not, things get Ugly.
-george william herbert
gher...@retro.com
Timothy Little
>The relevant quantities can be conserved during the lifetime of
>the bobble, without causing problems, so since the bobble can
>form without violating conservation, if evaporation is a time
>reversal of formation, it can evaporate without violating
>conservation.
However, it does violate causality. Real advanced waves, etc.
The time-reversal of bobble formation is having various forms of
radiation converge on the bobble from "infinity", restoring its
multipole moments.
- Tim
Timothy Little
>
>No. The mass can be spinning and still have zero total momentum
Only if the axis of spin does not pass through the bobble's centre.
i.e. The state of the contents is not preserved, which seems to
violate the basic axiom of bobbles.
- Tim
Robert Shaw
"Timothy Little" <t...@freeman.little-possums.net> wrote
the advanced waves.
The advanced waves contain exactly the same information
as the retarded waves; namely the multipole moments, so
if the advanced waves are either nonexistent or unmeasurable
outside the future light cone of the moment of the bobble's
creation, when the retarded waves are zero, then they can't be
used for transmitting information backwards.
It would be as if the retarded waves had bounced off a spherical
mirror, though the actual mechanism would have to be more
subtle.
Robert Shaw
> hop on my spaceship and take off and accellerate to 0.9 C.
> I then release the bobble. We fly through an asteroid belt,
> and a small rock hits the bobble halfway from its central
> axis to the equator (in either reference frame).
>
> In my frame, the bobble's a sphere. No moment, no rotation
> happens. Not that I should be able to tell anyways, but it
> *should* not rotate it.
>
> In the frame of the rock, it hit an ellipsoid offcenter
No, at the instant of contact the rock must be stationary
with respect to the bobble, since the bobble is perfectly
rigid and impenetrable.
Robert Shaw
"Timothy Little" <t...@freeman.little-possums.net> wrote
spinning. External observers see the contents spinning at
the appropriate speed. Bobbled observers see the universe
spinning around them, but are initially in a locally inertial
frame. Masses can produce similar effects by frame dragging,
but without anything to sustain it the phenomena would
presumably end in a burst of gravitional waves.
Pekka P. Pirinen
On 20 Jan 2001 13:10:56 -0800, gher...@gw.retro.com (George William
Herbert) wrote:
[Re an object striking a bobble off-center at relativistic speed]
> It appears to me that this poses a paradox. If we assume the
> combination of perfectly rigid body and zero-friction surface,
> then you end up with relatavistically inconsistent solutions
> to momentum transfers... in some reference frames you get
> a linear recoil and in others you get that along with angular
> momentum as well.
That doesn't sound right to me. Surely friction can be ignored without
breaking relativity, and collision with a perfectly rigid body is just
the boundary case of an ordinary elastic collision. Even if the
collision was elastic, in the absence of friction the force would normal
to the surface, i.e., radial in the sphere's FoR. So the same paradox
should arise with ordinary spheres.
I suspect the resolution is that relativistic kinematics gives us a
different answer from the classical, and the sphere will actually get
some angular momentum, but I've forgotten too much of my physics to work
it out.
> Unless i'm missing my math somewhere the
> classical bobble is thus inconsistent with relativity.
It seems there are plenty of ways in which they are, starting from
perfect rigidity.
--
Pekka P. Pirinen
There may be no candidates you want to vote for, but there are
certainly some you want to vote *against*.
Timothy Little
>The advanced waves contain exactly the same information
>as the retarded waves; namely the multipole moments,
Not quite true. They contain multipole moments appropriate to where
the bobble will be when they converge at some time in the future.
This transmits information outside the light cone of any perturbation
the bobble might experience between its creation and dissipation.
- Tim
Timothy Little
>The other oprion is that the space where the bobble was is
>spinning. External observers see the contents spinning at the
>appropriate speed. Bobbled observers see the universe spinning around
>them, but are initially in a locally inertial frame. Masses can
>produce similar effects by frame dragging, but without anything to
>sustain it the phenomena would presumably end in a burst of
>gravitional waves.
That would be one hell of a burst! That would so badly violate
conservation of energy that it overwhelms any trivial concerns like a
few J-s of angular momentum non-conservation. As well as probably
destroying the contents through tidal disruption.
- Tim
Timothy Little
[...relativistic bobble striking an asteroid...]
>I suspect the resolution is that relativistic kinematics gives us a
>different answer from the classical, and the sphere will actually get
>some angular momentum,
No. In the asteroid's frame of reference, the bobble is indeed an
ellipsoid, and it strikes off-centre. However, the forces on the
bobble are normal to the surface in the *bobble's* frame of reference,
not the asteroid's. And in the bobble's frame of reference, the
surface normal passes through its centre.
In fact, the situation is rather similar to a packet of light
reflecting off a mirror of finite mass -- the angle of incidence is
equal to the angle of relection only in certain frames.
- Tim
George William Herbert
>"George William Herbert" <gher...@gw.retro.com> wrote
>> Ok. I bobble a washing machine with a 2 meter diameter bobble,
>> hop on my spaceship and take off and accellerate to 0.9 C.
>> I then release the bobble. We fly through an asteroid belt,
>> and a small rock hits the bobble halfway from its central
>> axis to the equator (in either reference frame).
>>
>> In my frame, the bobble's a sphere. No moment, no rotation
>> happens. Not that I should be able to tell anyways, but it
>> *should* not rotate it.
>>
>> In the frame of the rock, it hit an ellipsoid offcenter
>
>No, at the instant of contact the rock must be stationary
>with respect to the bobble, since the bobble is perfectly
>rigid and impenetrable.
No, no... this is not a two-body perfectly rigid collision.
The Bobble is perfectly rigid; the rock is not.
-george william herbert
gher...@retro.com
George William Herbert
Yeah. The key here is that the bobble surface boundary condition...
that there are no tangental forces, only normal... changes the
physics specifically in relatavistic terms. If there are only
normal forces, then the effects of a collision (or any other
force such as beam of light) depend on what "normal" is in a
particular frame of reference. In the Bobble's FOR, normal
is always through the centroid and there's never a moment.
In an accellerated FOR, anything not exactly on the poles
or equator of the flattened ellipsoid is *not* through the
centroid, and all hell breaks loose in the dynamics of the
situation as a result.
-george william herbert
gher...@retro.com
Robert Shaw
"Timothy Little" <t...@freeman.little-possums.net> wrote
> Robert Shaw <Rob...@shavian.fsnet.co.uk> wrote:
> >The advanced waves contain exactly the same information
> >as the retarded waves; namely the multipole moments,
>
> Not quite true. They contain multipole moments appropriate to where
> the bobble will be when they converge at some time in the future.
However that information can be deduced from the initial momentum,
energy and spin of the bobble.
> This transmits information outside the light cone of any perturbation
> the bobble might experience between its creation and dissipation.
>
If the advanced waves can't be measured outside the lightcone this
isn't a problem. The perturbation can theoretically be deduced from
information about the initial mass, velocity etc of the bobble contained
in retarded waves (which get to any point in the light cone first) so
the advanced waves don't transmit any new information.
If they were detectable outside the lightcone that would break
casuality, but they needn't be.
Robert Shaw
> Robert Shaw <Rob...@shavian.fsnet.co.uk> wrote:
> >Masses can
> >produce similar effects by frame dragging, but without anything to
> >sustain it the phenomena would presumably end in a burst of
> >gravitional waves.
>
> That would be one hell of a burst! That would so badly violate
> conservation of energy
Why? Gravity isn't part of the stress tensor. The energy density
of gravitational waves is zero. It is possible to emit gravitional waves
of arbitary amplitude but with zero energy, momentum and angular
momentum which won't violate conservation laws.
This doesn't stop them appearing to carry away energy sometimes,
but that's because in GR energy is not globally conserved, only
locally, except in special frames.
> that it overwhelms any trivial concerns like a
> few J-s of angular momentum non-conservation. As well as probably
> destroying the contents through tidal disruption.
>
Gravity interacts only weakly with matter. It's possible the waves would
be too weak to have noticable affect under normal conditions.
Robert Shaw
> >
> >No, at the instant of contact the rock must be stationary
> >with respect to the bobble, since the bobble is perfectly
> >rigid and impenetrable.
>
> No, no... this is not a two-body perfectly rigid collision.
> The Bobble is perfectly rigid; the rock is not.
>
If the rock is moving towards the bobble at the moment of contact
it would have to penetrate it. If it moving away at that moment
it would have to have pentrated it. It can't have any velocity
normal to the bobble surface.
It can have tranverse velocity but its trajectory can't be
non-tangential without passing through the bobble.
George William Herbert
>"George William Herbert" <gher...@gw.retro.com> wrote
>> Robert Shaw <Rob...@shavian.fsnet.co.uk> wrote:
>> >No, at the instant of contact the rock must be stationary
>> >with respect to the bobble, since the bobble is perfectly
>> >rigid and impenetrable.
>>
>> No, no... this is not a two-body perfectly rigid collision.
>> The Bobble is perfectly rigid; the rock is not.
>>
>
>If the rock is moving towards the bobble at the moment of contact
>it would have to penetrate it. If it moving away at that moment
>it would have to have pentrated it. It can't have any velocity
>normal to the bobble surface.
>It can have tranverse velocity but its trajectory can't be
>non-tangential without passing through the bobble.
Are you under the misaprehension that atoms and subatomic particles
are small rigid bodies?
Let me repeat myself: the Bobble is perfectly rigid, the rock is not.
-george william herbert
gher...@retro.com
Timothy Little
>Hrm? Are you SURE? If so, can you give an example where
>the angle of incidence doesn't equal the angle of reflectance
>in any frame?
I think angles of incidence always equal angle of reflection in *some*
frame, otherwise you call it something else like absorption or
refraction.
> Is this for flat mirrors?
Yes.
> Is the rebound due to light momentum the issue
When I posted, I had thought "yes". Now I realise I was wrong.
Consider a normal "heavy" mirror at rest, and a photon with an angle
of incidence of 45 degrees. In the rest frame of the mirror, the
reflected photon is at right angles to the incident photon.
Now, consider a 0.1c Lorentz boost in the direction of the incident
photon. In this reference frame, the reflected photon is no longer at
right angles to the incident photon -- it is at an angle of about
84.26 degrees from the original path. Now, this isn't the whole
story, because the mirror is no longer at a 45 degree angle in this
reference frame -- you need to apply the Lorentz transformations to
its dimensions, too. It is instead at an angle of about 45.14 degrees
in the boosted frame. From this, you can see that the angle of
reflection is 50.60 degrees from the mirror's surface in this frame,
not equal to the angle of incidence.
- Tim
Timothy Little
>> Not quite true. They contain multipole moments appropriate to where
>> the bobble will be when they converge at some time in the future.
>
>However that information can be deduced from the initial momentum,
>energy and spin of the bobble.
You can't deduce where the bobble will be when they converge. Suppose
the bobble is formed at day 0. It is perturbed at day 150, and
dissipates at day 200.
Now, the retarded waves travel out along the futureward light cone,
and we can ignore them. The advanced waves materialise at a distance
of about 100 light-days from the initial location of the bobble at day
100, just as the retarded waves pass that point. They converge on the
future location of the bobble's dissipation.
Now, there is a station at a distance of 60 light-days. It sees the
retarded waves go out on day 60, and notes where they came from. On
day 140 it sees the advanced waves converging and notes where they are
converging to. That location depends upon the perturbation, which is
still 10 days into the future and 60 light-days away. We have an FTL
communication system.
Give that station a bobble to play with as well, and we have all the
ingredients necessary for causality violation.
>If the advanced waves can't be measured outside the lightcone this
>isn't a problem.
Unfortunately it is, as I've shown above.
> The perturbation can theoretically be deduced from information about
>the initial mass, velocity etc of the bobble contained in retarded
>waves
No, the perturbation depends upon who shoves it in the interim and
how. In theory, this information may be contained in some sort of
retarded waves, but only in the sense that the future is predictable
from the present.
- Tim
Timothy Little
>
>> That would be one hell of a burst! That would so badly violate
>> conservation of energy
>
>Why? Gravity isn't part of the stress tensor. The energy density
>of gravitational waves is zero.
They do carry energy, even without having them in the stress tensor.
>This doesn't stop them appearing to carry away energy sometimes, but
>that's because in GR energy is not globally conserved, only locally,
>except in special frames.
That depends upon your definition of "energy". If you call it "the 00
component of the stress tensor", then yes. However, it's not the most
useful definition since gravitational waves can impart energy to
matter, and so energy is not even locally conserved if you use this
definition.
Any definition of "energy" useful for conservation has to consider
gravitational waves to carry some.
>Gravity interacts only weakly with matter. It's possible the waves
>would be too weak to have noticable affect under normal conditions.
You're not talking about trivial things like gravity radiation from
binary neutron star collisions here, you're talking about the
transient behaviour of a section of spacetime metric that is truly
warped.
- Tim