Speed Limits
Ray Dueland
> A basic result of special relativity is that there
>is no way a person or human-like autamaton can
>transmit information faster than the speed of light.
How fast do the affects of the forces travel, e.g. if the sun were
to disappear, would the earth stop orbitting immediately or 8
minutes later? If I turn my electromagnet off, when do magnetic
objects stop being attracted, etc.
--
Ray Dueland
{amdahl|clout|masscomp|nucsrl|sun|tellab5}!laidbak!rayd
Martin Francis Ryba
>In article <16...@hub.ucsb.edu> fla...@sbphy.ucsb.edu writes:
>> A basic result of special relativity is that there
>>is no way a person or human-like autamaton can
>>transmit information faster than the speed of light.
>How fast do the affects of the forces travel, e.g. if the sun were
>to disappear, would the earth stop orbitting immediately or 8
>minutes later? If I turn my electromagnet off, when do magnetic
>objects stop being attracted, etc.
In a word: all known forces also are transmitted at or below the speed
of light. If the sun blew up, it would take 8 minutes for us to know.
When supernova 1987A was seen, it was 150,000 years old.
Gravity is also transmitted at the speed of light, according to GR.
I'm not sure it has been experimentally verified, but I think so.
Cheers.
Marty Ryba (slave physics grad student)
They don't care if I exist, let alone what my opinions are!
Mary Holstege
>Gravity is also transmitted at the speed of light, according to GR.
This raises a question that has bugged me on and off. Maybe one of you physics
types can answer it for me: If forces are carried by particles (photons,
etc.), which is the current theory (right?), and these particles can't exceed
the speed of light, then how does a black hole exert any gravitational
influence? How does gravity "get out" of a black hole? Or doesn't it? Or is
there some weird quantum effect?
Not being a physicist I am probably being hopelessly naive here, but what about
it? In words of few syllables, please. Email is fine.
Confusedly,
-- Mary
Hols...@polya.stanford.edu
ARPA: holstege%po...@score.stanford.edu
BITNET: holstege%po...@STANFORD.BITNET
UUCP: {arpa gateways, decwrl, sun, hplabs, rutgers}!polya.stanford.edu!holstege
Carl Ellison
>In article <82...@phoenix.Princeton.EDU> mfr...@phoenix.Princeton.EDU (Martin Francis Ryba) writes:
>>I'm not sure it has been experimentally verified, but I think so.
When I last asked that question of a working physicist (in pre-net days)
I was told that the orbit of Mercury verified against General Relativity
established this.
Is this right?
--Carl Ellison UUCP:: c...@cloud9.Stratus.COM
SNail:: Stratus Computer; 55 Fairbanks Blvd.; Marlborough MA 01752
Disclaimer:: (of course)
Ethan Tecumseh Vishniac
> In article <11...@s.ms.uky.edu>, se...@ms.uky.edu (Sean Casey) writes:
> >In article <82...@phoenix.Princeton.EDU> mfr...@phoenix.Princeton.EDU (Martin Francis Ryba) writes:
> >>Gravity is also transmitted at the speed of light, according to GR.
> >>I'm not sure it has been experimentally verified, but I think so.
> > I would be REALLY interested in finding out if it has...
>
> When I last asked that question of a working physicist (in pre-net days)
> I was told that the orbit of Mercury verified against General Relativity
> established this.
This constitutes indirect proof. The theory of General Relativity
includes gravitational effects propagating at the speed of light.
Any competing theory I'm familiar with also does (although I'm sure
one could cook up some extremely contrived exception). General
Relativity has been tested in the limit of small deviations from
Newtonian gravity in a number of ways, including the precession of
the perihelion of Mercury's orbit. So far it works.
The only strong field test conducted so far has involved the timing
of the binary pulsar system discovered by Hulse and Taylor. The parameters
of the binary system are completely determined by the observations and
so it is possible to make an unambiguous determination of whether or not
the orbit is decaying at a rate consistent with the emission of gravitational
radiation at the rate predicted by GR. It is.
--
I'm not afraid of dying Ethan Vishniac, Dept of Astronomy, Univ. of Texas
I just don't want to be {charm,ut-sally,emx,noao}!utastro!ethan
there when it happens. (arpanet) et...@astro.AS.UTEXAS.EDU
- Woody Allen (bitnet) ethan%astro.as....@CUNYVM.CUNY.EDU
These must be my opinions. Who else would bother?
Steven Den Beste
> >In article <82...@phoenix.Princeton.EDU> mfr...@phoenix.Princeton.EDU (Martin Francis Ryba) writes:
> >>Gravity is also transmitted at the speed of light, according to GR.
> >>I'm not sure it has been experimentally verified, but I think so.
> > I would be REALLY interested in finding out if it has...
>
> When I last asked that question of a working physicist (in pre-net days)
> I was told that the orbit of Mercury verified against General Relativity
> established this.
>
> Is this right?
>
I went down to Portland State University once and dug up a grad student and
asked about this. (Oregon, not Maine.)
I was informed that the question was meaningless. It is theoretically
impossible to separate out the gravitational effects of one body from those of
all the other bodies. This was consistent with GR in as much as gravity isn't a
force and thus doesn't have a particle to mediate it. Instead, gravity is part
of the geometry of the universe - it doesn't travel, it just is.
[...and no, I didn't understand that answer either.]
Steven C. Den Beste, BBN Communications Corp., Cambridge MA
denb...@bbn.com(ARPA/CSNET/UUCP) harvard!bbn.com!denbeste(UUCP)
The Polymath
}to disappear, would the earth stop orbitting immediately or 8
}minutes later? If I turn my electromagnet off, when do magnetic
}objects stop being attracted, etc.
So far as I know, speed of light is still the limit. To an independent
observer, the Earth will keep orbiting until it "knows" the Sun has
vanished.
There's even a corresponding effect in the non-relativistic world, or so
said my high school physics teacher. He described the following
experiment, which, he said, was actually performed:
Suspend a weight from the ceiling by a length of primacord(sp?).
(Primacord is a high explosive, woven into a rope. Among other
characteristics, it burns faster than the speed of sound in
primacord).
Point a very high-speed movie camera at the weight and light the
primacord at the ceiling.
Result: The weight will not begin to fall until the primacord burns
down to it.
In fact, the weight doesn't "know" the rope has been cut until the
shockwave of released tension travels down the rope to it. The use of
primacord is simply to make the effect easier to photograph.
--
The Polymath (aka: Jerry Hollombe, holl...@ttidca.tti.com) Illegitimati Nil
Citicorp(+)TTI Carborundum
3100 Ocean Park Blvd. (213) 452-9191, x2483
Santa Monica, CA 90405 {csun|philabs|psivax}!ttidca!hollombe
Sean Casey
>I'm not sure it has been experimentally verified, but I think so.
I would be REALLY interested in finding out if it has...
Sean
--
*** Sean Casey se...@ms.uky.edu, se...@ukma.bitnet
*** Quid, me vexari? {backbone|rutgers|uunet}!ukma!sean
*** ``BITNET: slower than a speeding mountain...''
Greg Hennessy
#I went down to Portland State University once and dug up a grad student and
#asked about this. (Oregon, not Maine.)
# [Explanation deleted]
#[...and no, I didn't understand that answer either.]
#
It doesn't sound like your grad student understood it either.
There is no known particle to mediate the influence of gravity,
although many Grand Unification Theories propose a graviton which is
analogous to a photon. While the increase in the perihelion shift of
Mercury is a test of GR, it is not a test of the speed of gravity
waves. There was a paper in Nature (I think) that shows if you assume
that the gravity wave from SuperNova 1987A was emitted at the same
time as the increase in brightness of the SN, then the gravity wave
travelled at the speed of light within one part in 10 to the minus 9,
(this number is from memory). If people want, I will try to find the
paper.
-Greg Hennessy, University of Virginia
USPS Mail: Astronomy Department, Charlottesville, VA 22903-2475 USA
Internet: gs...@virginia.edu
UUCP: ...!uunet!virginia!gsh7w
Paul Rodman
>In article <82...@phoenix.Princeton.EDU> mfr...@phoenix.Princeton.EDU (Martin Francis Ryba) writes:
>>Gravity is also transmitted at the speed of light, according to GR.
>>I'm not sure it has been experimentally verified, but I think so.
>
>I would be REALLY interested in finding out if it has...
Observations of a binary pulsar system have shown that the energy of the
orbit is indeed being lost at the rate GR expects from emmission of
gravitational waves. [The precession of the major axis of the orbit
was on the order of 1 degree per year (! vs mercury's 40 seconds per
century!) due to the large masses and small orbits => large velocities
involved.]
Paul K. Rodman
rod...@Multflow.com
Kevin W. Williams
> In article <23...@laidbak.UUCP> ra...@laidbak.UUCP (Ray Dueland) writes:
> Gravity is also transmitted at the speed of light, according to GR.
> I'm not sure it has been experimentally verified, but I think so.
It has been. The orbits of the various planets and their interactions
with each other would vary depending on whether gravity was instantaneously
transmitted or whether it travels at c. I once read a calculation of the
sun's speed and direction of travel based on the slightly different sun
positions that the different planets orbit around. It jived well with other
methods of calculation.
Kevin Wayne Williams
UUCP : ...!ames!ncar!noao!asuvax!gtephx!williamsk
Remember : Brute force has an elegance all its own.
Hans Aberg
This is indeed so, at least in the interpretation in terms of some
perturbations Einstein did while working on general relativity. There
is a comment in the book of Misner, Thorne and Wheeler about this, but
I do not remember where.
Hans Aberg, Mathematics
ab...@math.rutgers.edu
Paul Schinder
>waves. There was a paper in Nature (I think) that shows if you assume
>that the gravity wave from SuperNova 1987A was emitted at the same
>time as the increase in brightness of the SN, then the gravity wave
>travelled at the speed of light within one part in 10 to the minus 9,
>(this number is from memory). If people want, I will try to find the
>paper.
>
>-Greg Hennessy, University of Virginia
> USPS Mail: Astronomy Department, Charlottesville, VA 22903-2475 USA
> Internet: gs...@virginia.edu
> UUCP: ...!uunet!virginia!gsh7w
What gravity wave? It is my understanding that *no* current gravity wave
detector could have detected SN1987A even if a large fraction of the mass
of the star was converted into gravity waves. (Is this something like the
Mont Blanc neutrinos?)
--------
Paul J. Schinder
Department of Astronomy, Cornell Univ.
schi...@astrosun.tn.cornell.edu
--
Paul J. Schinder
Department of Astronomy, Cornell Univ.
schi...@tcgould.tn.cornell.edu
James Preston
>I went down to Portland State University once and dug up a grad student and
>asked about this. (Oregon, not Maine.)
>
>I was informed that the question was meaningless. It is theoretically
>impossible to separate out the gravitational effects of one body from those of
>all the other bodies. This was consistent with GR in as much as gravity isn't a
>force and thus doesn't have a particle to mediate it. Instead, gravity is part
>of the geometry of the universe - it doesn't travel, it just is.
>
>[...and no, I didn't understand that answer either.]
The way it is usually explained is by making an analogy to two-dimensions:
Think of "space" as a huge sheet of rubber stretched out and the stars and
planets as billiard (or whatever) balls sitting on it. Each ball makes an
indentation in the rubber. The larger the ball, the deeper the indentation.
Anything in space--other balls, light waves, etc.--is by definition constrained
to stay on the rubber. Thus "gravity" is simply an effect of the shape of
space rather than a force transmitted by particles. When a moving ball or
light wave gets near another ball, it's path curves because it follows the
curve of the rubber caused by the other ball.
So now the question "If the sun vanished, how soon would we know it?" becomes
"How fast does space (the rubber sheet) spring back to its natural shape?"
This is, of course, left as an exercise for the reader. :-)
--James Preston
Greg Hennessy
#
#What gravity wave? It is my understanding that *no* current gravity wave
#detector could have detected SN1987A even if a large fraction of the mass
#of the star was converted into gravity waves. (Is this something like the
#Mont Blanc neutrinos?)
A group reported that a room temperature gravity wave detector went
"ding" around the same time as the supernova. I will try to find the
paper this week.
jim sullivan
But, according to what I think I remember learning (quite a
disclaimer I must say ;^) a gravitational field behaves
somewhat like an electric field in that if you move the
electron, the wave of the disturbance in the electric field
moves out at the speed of light. Gravity would work in a
similiar manner with a 'field' which, when a disturbance
occured (nova, death-star attack, etc...) to a gravitational
source, the change in the gravitational field would move out
at the speed of light. So, if the sun were to disappear
right now, we could not detect a change in the orbit
of the earth for about the 8 minutes it takes light to
get here from the sun.
So, as a star collapses and the field strengthens, the change
in the field would move at the speed of light. When the
field reached a point of where the acceleration was that of
light, no more information could cross that 'event horizon'
including a change in the gravitational field.
My question is: If matter falls into the hole, how would the
event horizon grow as Hawkins states in his book if the
information of the increase in the gravitational field could
not cross the event horizon?
Jim Sullivan
j...@nih-csl.dcrt.nih.gov
Paul Rodman
>I went down to Portland State University once and dug up a grad student and
>asked about this. (Oregon, not Maine.)
>
>I was informed that the question was meaningless. It is theoretically
>impossible to separate out the gravitational effects of one body from those of
>all the other bodies. This was consistent with GR in as much as gravity isn't a
>force and thus doesn't have a particle to mediate it. Instead, gravity is part
>of the geometry of the universe - it doesn't travel, it just is.
>
Wrong.
A system such as the binary pulsar system emits gravity "waves" . The
field is *changing* not constant. These changes propagate at c, it is
thought.
This emmission of waves steals energy from the orbit, hence it gets smaller,
and the orbital period decreases.
Weber-type gizmos look for gravity waves from more cataclysmic events.
(two black holes orbiting each other 10 times/second ?) They need
pretty vigorous events to get a signal right now...
Gravity *IS* thought have a particle to medeiate it, everyone hopes
someday to integrate GR with QM.....Hawkings interesting observations
about black hole evaporation point a tiny bit in this direction.
Take that grad student you "dug up" and re-bury him....:-)
Cheers,
Paul K. Rodman
rod...@Multiflow.com
Alan Lovejoy
This seems to imply that the speed-of-light limitation can be conceptualized
as a limit on the rate at which change in the curvature of space can propogate.
An object travelling faster than c would require propogating changes in the
curvature of space at a rate greater than c, because of the gravitational
effect of the object's mass.
So if a large body of mass were to instantaneously "disappear," there would be
a spherical boundary radiating away from the location of the disappeared object
at the speed of light. Outside this boundary, the curvature of space reflects
the presence of a large mass at the center of the sphere. Inside the boundary,
the curvature of space does not reflect the presence of this mass. Note that
at the instant that this spherical boundary passes an observer, he detects
an *instantaneous* change in the curvature of space. The observer's graph of
the curvature of space as a function of time would show a straight vertical
line at the instant he was passed by the boundary (x: time, y: curvature;
y = f(x)).
Space may not permit such discontinuities in curvatare to exist, regardless
of propogation speed. Of course, this is just another way of saying that
mass may not be able to just "disappear." But this cannot be proven, only
disproven.
Alan Lovejoy; alan@pdn; 813-530-2211; AT&T Paradyne: 8550 Ulmerton, Largo, FL.
Disclaimer: I do not speak for AT&T Paradyne. They do not speak for me.
_________________________Design Flaws Travel In Herds_________________________
Motto: If nanomachines will be able to reconstruct you, YOU AREN'T DEAD YET.
Ken Arromdee
>Think of "space" as a huge sheet of rubber stretched out and the stars and
>planets as billiard (or whatever) balls sitting on it. Each ball makes an
>indentation in the rubber. The larger the ball, the deeper the indentation.
I have seen this analogy a lot of times. It seems like about the worst
possible analogy to make, since the balls make indentations because of
gravity, so you are using gravity to explain gravity.
--
"Do you know what this is????" "No, what?" "I don't know either..."
-- Who said it, what story?
Kenneth Arromdee (UUCP: ....!jhunix!ins_akaa; BITNET: g49i0188@jhuvm;
INTERNET: arro...@crabcake.cs.jhu.edu) (please, no mail to arrom@aplcen)
Chris Phoenix
>In article <7...@key.COM> j...@penguin.key.COM (James Preston) writes:
>>So now the question "If the sun vanished, how soon would we know it?" becomes
>>"How fast does space (the rubber sheet) spring back to its natural shape?"
>>This is, of course, left as an exercise for the reader. :-)
>So if a large body of mass were to instantaneously "disappear," there would be
>a spherical boundary radiating away from the location of the disappeared object
>at the speed of light.
There seems to be another problem with this picture. There is a related
question: Do gravity waves have mass? If not, how can they have energy?
If so, how can they escape from a black hole?
The problem: The picture is that the gravity waves, like everything else,
travels along the sheet. But a black hole dips infinitely far into the
sheet. This means that any change in the gravity wave would take an infinite
time to get out. So a black hole could disappear *and we'd never know it.*
Other support for this point of view: It's been claimed that gravity can't
move faster than light. Then it, too, can't escape from a black hole, right?
Obviously, something is wrong. Can anyone explain it so that a quantum
klutz can understand it?
--
Chris Phoenix | "I was afraid of worms! Worms, Roxanne!"
cpho...@csli.Stanford.EDU | "More input! More input!"
Usenet: The real source of the "Tastes Great" vs. "Less Filling" debate.
Disclaimer: Don't mind me, I'm just a student!
----- Will Allen -----
How do these gravaty wave detectors y'all are talking about differentiate
beteen cosmic fluxuations in gravety and possible shifting of mass within
the earth? What would be the relative magnitudes of these two types of
events?
. . .Will
Vaso Bovan
Has the existence of tachyons been ruled out on theoretical grounds ?
Blair P. Houghton
>>The way it is usually explained is by making an analogy to two-dimensions:
>>Think of "space" as a huge sheet of rubber stretched out and the stars and
>>planets as billiard (or whatever) balls sitting on it. Each ball makes an
>>indentation in the rubber. The larger the ball, the deeper the indentation.
>
>I have seen this analogy a lot of times. It seems like about the worst
>possible analogy to make, since the balls make indentations because of
>gravity, so you are using gravity to explain gravity.
I'm sorry. Whoever it was Ken flamed (way to drop those credit-bars,
cap'n!) has oversimplified and blown away the truly misunderstanding...
The rubber-sheet thing is one of Einstein's, and if you have any
imagination you see that it's a degenerate explanation used to let
non-physicists understand a situation whose accurate description
requires an understanding of potential wells and such other things as
Lagrangians and Lyapunov and...
Anyway, the easiest way to get rid of Ken's gravity-problem is to do
this nicht-viel-gedankenexperiment in free space, and sandwich the
balls between two rubber sheets. The bigger the ball, the bigger
the bi-labial bulge. The existence of another ball in a neighborhood
of the first will produce a direction in which there is less force
pulling the rubber sheets together, because the sheets are being pulled
apart by the other ball. Like aerodynamic lift, the balls
get sucked into the region with the lower pressure. Et voila!
Gravity!
Next step: make it three-dimensionally isotropic. No more rubber
sheets. Generalize to objects of varying, or even nonhomogeneous,
densities. (Hint: Think fundamental units of matter.) Now throw it
all out because energy is matter and matter is energy, and how are you
to explain _that_ to your girlfriend the Economics major?
Now you know why it's best to have mathematics on your side.
--Blair
"Rose (holding up flash-card):
Okay, Pasquale, this is an 'A'.
Pasquale: Why?"
car...@s.cs.uiuc.edu
RE: Gravitional Singularities -
One thing that physicists seem to gloss over is that (to _my understanding_)
there aren't any black holes, since they can't have collapsed yet. Because
of gravitic time dilation, it takes (to an observer in "flat" space) an
infinite amount of time to fall to the event horizon. So the mass that's
collapsing to form the black hole takes this amount of time also, so if the
universe if finitely old, how can we have _actual_ black holes? I suppose
that the mass that starts out inside the event horizon could be thought of
that way. Any astro-physicist want to enlighten an ignorant seeker-of-truth?
Alan M. Carroll "And there you are
car...@s.cs.uiuc.edu Saying 'We have the Moon, so now the Stars...'"
CS Grad / U of Ill @ Urbana ...{ucbvax,pur-ee,convex}!s.cs.uiuc.edu!carroll
Jerry Robertson
>Do gravity waves have mass? If not, how can they have energy?
Think about light. It doesn't have mass, but it certainly has energy.
Jerry Robertson
robe...@umn-cs.cs.umn.edu
The Kookman
Bill Wyatt
> In article <12...@umn-cs.CS.UMN.EDU> robe...@umn-cs.cs.umn.edu (Jerry
> Robertson) writes:
>>Think about light. It doesn't have mass, but it certainly has energy.
>
> calculate the mass knowing the energy:
>
> e = mc^2
No, light (photons, really) have MOMENTUM, not mass. It's not the same
thing for a zero-rest-mass particle.
Bill Wyatt, Smithsonian Astrophysical Observatory
UUCP : {husc6,cmcl2,mit-eddie}!harvard!cfa!wyatt
ARPA: wy...@cfa.harvard.edu
SPAN: cfa::wyatt BITNET: wyatt@cfa
Hal Lillywhite
>question: Do gravity waves have mass? If not, how can they have energy?
>If so, how can they escape from a black hole?
1. Do gravity waves have mass? I'm no general relativity expert
but I believe in that theory mass is nothing more than the curvature
of space-time. Gravity waves would be waves in space-time and
therefore have mass. You are right that they cannot carry energy
unless they have mass.
2. How can they escape from a black hole? Probably they don't but
are generated near the black hole as it moves. eg. consider 2 black
holes orbiting each other. This will cause all sorts of
oscillations in space-time in the vicinity which according to theory
will radiate away as gravity waves. Any waves which escape will
probably come from outside the horizon of the black hole.
If a black hole dissappeared we'd know it (assuming we knew the hole
was there to start with) by the change in gravity outside where the
black hole was.
Craig Eisler
>
>Light has both mass and energy. You can use Einstein's equation to
>calculate the mass knowing the energy:
>
> e = mc^2
You are very wrong. Photons (light) have no mass. The "m" in
e = m c^2 is the relativistic mass, given by
m = m0/(sqrt( 1 - (u/c)^2 ) ),
where u is the velocity of the moving particle and m0 is the rest mass.
If photons had a rest mass, their relativistic mass would always be
infinite, since in the case of a photon, u=c.
However, photons have momentum. A useful form of e=mc^2 is
e^2 = (pc)^2 + (m0 c^2)^2
so for a photon, e = pc.
Sorry, I'm rambling...
To sum up, photons have NO MASS. If they did, they could not move
at the speed of light.
craig
Guangliang He
=From article <72...@bsu-cs.bsu.edu>, by dh...@bsu-cs.bsu.edu (Rahul Dhesi):
=> In article <12...@umn-cs.CS.UMN.EDU> robe...@umn-cs.cs.umn.edu (Jerry
=> Robertson) writes:
=>>Think about light. It doesn't have mass, but it certainly has energy.
=>
=> Light has both mass and energy. You can use Einstein's equation to
=> calculate the mass knowing the energy:
=>
=> e = mc^2
=No, light (photons, really) have MOMENTUM, not mass. It's not the same
=thing for a zero-rest-mass particle.
=
Well, talking about rest-mass of photon does not make much sense at all
since one can never find a reference system in which the photon is rest.
-----------------------------------------------------------------------
|
USMAIL: Guangliang He | INTERNET: g...@PHYSICS.ORST.EDU
Department of Physics | g...@jacobs.CS.ORST.EDU
Oregon State University | BITNET: he...@orstvm.bitnet
Corvallis, OR 97331 | PHONE: (503) 754-4631
|
-----------------------------------------------------------------------
Ralph L. Place
Photons have no rest mass but they do have mass that exhibits both
dynamics (F=dp/dt) via momentum considerations and gravitiational
effects as verified by star light being deflected by large masses such
as the Sun.
From E = mc^2 = hf we obtain the relativistic mass, m = h/c*wavelength.
This can also be written as mc = p = h/wavelength, which, of course is
the de Broglie formula.
R. Place
Mark Sobolewski
>> In article <12...@umn-cs.CS.UMN.EDU> robe...@umn-cs.cs.umn.edu (Jerry
>> Robertson) writes:
>>>Think about light. It doesn't have mass, but it certainly has energy.
>> Light has both mass and energy. You can use Einstein's equation to
>> calculate the mass knowing the energy:
>> e = mc^2
>No, light (photons, really) have MOMENTUM, not mass. It's not the same
>thing for a zero-rest-mass particle.
The formula goes: E^2 = (m * c^2)^2 + (p * c)^2
(rest mass) (momentum(p))
As can be seen, as rest mass (m) approaches zero, momentum (p)= E/c.
This has been verified experimentally. (Actually, this was what Einstein
got the Nobel Prize for).
Experimentally, a lower limit m for a photon was calculated
back in my Nuclear physics class as something like 10^(-51)kg.
Something to do with the voyager probe. Does anyone remember how
this was done?
>Bill Wyatt, Smithsonian Astrophysical Observatory
------------------------------------------------------------------------------
Mark Sobolewski "If you're going to lose anyway, you might as well win."
sobl...@gondor.cs.psu.edu #include<std.disclaimer>
Maarten Litmaath
\... As can be seen, as rest mass (m) approaches zero, momentum (p)= E/c.
\This has been verified experimentally. (Actually, this was what Einstein
\got the Nobel Prize for).
Wrong. He got it for his theory explaining the photo-electric effect.
--
"`Goto considered harmful' considered |Maarten Litmaath @ VU Amsterdam:
harmful" considered harmful! |ma...@cs.vu.nl, mcvax!botter!maart
Hal Lillywhite
>>Light has both mass and energy. You can use Einstein's equation to
>>calculate the mass knowing the energy:
>>
>> e = mc^2
>
>You are very wrong. Photons (light) have no mass. The "m" in
>e = m c^2 is the relativistic mass, given by
>
> m = m0/(sqrt( 1 - (u/c)^2 ) ),
>
>where u is the velocity of the moving particle and m0 is the rest mass.
>If photons had a rest mass, their relativistic mass would always be
>infinite, since in the case of a photon, u=c.
Can we try, once and for all, to bring this "photons have/don't have
mass" thing to an end? Let's stick with generally accepted physics,
document our assertions either with references or with derivations
from known physics, etc.
First, let's agree (I hope) that rest mass is different from moving
mass, and in particular different from mass of an object moving with
the speed of light. Notice that the denominator in the Lorentz
equation, the infamous
m = m0/(sqrt( 1 - (u/c)^2 ) )
goes to zero as u approaches c. Therefore zero rest mass (which in
fact the photon has) gives a relativistic mass of zero divided by
zero! We have to use L'Hospital's rule to calculate relativistic
mass at u=c. However we have no equation for how m0 behaves at
other that light speed so I don't know how to do this. (If you are
not familiar with this procedure check any basic calculus text. It
really doesn't matter since we can't use it anyway.) Suffice it to
say that the Lorentz equation allows an object with zero rest mass
to have non-zero mass if it moves at the speed of light.
If we agree to that we can agree that a photon can possibly have
mass. In fact relativity demands that it have mass
e = mc^2 ==> m = e/c^2
if it has energy. I think you will find that any accepted book on
modern physics or optics, if it deals with photon mass at all, gives
this relationship or some version thereof. For example my old
optics text INTRODUCTION TO MODERN OPTICS by Grant R. Fowles, 1968
(am I dating myself?) Holt, Rinehart, & Winston discusses this on
page 216. The energy of the photon is determined from its frequency
e = hf =mc^2 ==> m=hf/c^2
There have been a few tables published which carelessly
list the mass of the photon as zero without specifying that this is
rest mass but this is a lack of information, not positive
information that a moving photon has no mass.
So, unless you can cite a reference or otherwise back up your
assertion, please, no more massless photons.
Michael Jennings
>
> You are very wrong. Photons (light) have no mass. The "m" in
> e = m c^2 is the relativistic mass, given by
> m = m0/(sqrt( 1 - (u/c)^2 ) ),
>
> where u is the velocity of the moving particle and m0 is the rest mass.
> If photons had a rest mass, their relativistic mass would always be
> infinite, since in the case of a photon, u=c.
NO!!!!!! A photon has zero rest mass. m0 is zero and is on the numerator of the equation. The denominator is also zero . Therefore, we get m = 0/0. This is undefined. All it means is that this equation is not relevant here. What is relevant is the standard E = mc^2. m is the total relativistic mass ( which for any particle consists of the mass equivalents of the rest energy (here zero) and the kinetic energy (here not zero). The relativistic expression for momentum is E = mv, where v is velocity and m is r
elativistic mass. A photon's Energy and velocity are both finite and non zero. Therefore the mass must be finite and non zero.
> To sum up, photons have NO MASS. If they did, they could not move
> at the speed of light.
They have no rest mass. If they did, then they could not move at the speed of light. This is not the same thing as saying that they have no mass.
Michael.
--
'Strangely, the only thing that went through the mind of the bowl of Petunias
was "Oh No, Not again".' - Douglas Adams.
Michael Jennings
University of Wollongong, Australia
Daniel S. Riley
>In article <15...@cfa.cfa.harvard.EDU> wy...@cfatst.HARVARD.EDU (Bill Wyatt) writes:
>=From article <72...@bsu-cs.bsu.edu>, by dh...@bsu-cs.bsu.edu (Rahul Dhesi):
>=> calculate the mass knowing the energy:
This isn't really a valid application of this formula. If m means rest
mass, then this equation is only valid in the rest frame of the particle,
which doesn't exist for a photon. If m means \gamma times the rest mass,
then the result isn't well-defined for a photon, which has 0 rest mass
and \gamma of 1/0.
>=No, light (photons, really) have MOMENTUM, not mass. It's not the same
>=thing for a zero-rest-mass particle.
This is why high energy physicists, at least when we try to be precise,
never talk about mass. The rest mass and the momentum of a particle are
well defined, but mass can mean different things in different contexts...
rest mass or \gamma m. Usually, mass is taken to mean rest mass, but it
can be ambiguous. Regardless, photons have no mass.
>Well, talking about rest-mass of photon does not make much sense at all
>since one can never find a reference system in which the photon is rest.
It's true that there is no rest frame for a photon, but a rest mass can
still be formally calculated from the momentum four-vector,
m^2 = E^2 - p^2 (and some factors of c^2 that I've dropped).
For a photon, this gives a rest-mass of zero. Hence, the photon is
a massless particle.
-Dan Riley (ri...@tcgould.tn.cornell.edu, cornell!batcomputer!riley)
-Wilson Lab, Cornell U.
Blair P. Houghton
> The formula goes: E^2 = (m * c^2)^2 + (p * c)^2
> (rest mass) (momentum(p))
>
> As can be seen, as rest mass (m) approaches zero, momentum (p)= E/c.
>This has been verified experimentally. (Actually, this was what Einstein
>got the Nobel Prize for).
Wrongo. He got the Nobel for the Photoelectric Effect, never for any
work on relativity.
--Blair
"..or was it the Tobel Bombsite?"
wyant
>> The formula goes: E^2 = (m * c^2)^2 + (p * c)^2
>> (rest mass) (momentum(p))
>>
>>This has been verified experimentally. (Actually, this was what Einstein
>>got the Nobel Prize for).
>
> Wrongo. He got the Nobel for the Photoelectric Effect, never for any
> work on relativity.
The story which was passed on to me was that the Nobel committee wanted
to give Albert a prize for special relativity, but SR was still too
controversial at the time. So, they gave it to him for a safer paper.
Note that the criterion for receiving a Nobel has changed, both from what
Nobel intended and what was in vogue at the time that Einstein received one.
Instead of a significant discovery to aid mankind (personkind?), Nobels are
now also awarded to individuals who have made a significant contribution to
their field of science.
Patrick Wyant
Engineering Physics Lab
E.I. du Pont de Nemours & Co.
Wilmington, DE 19880-0357
*!uunet!eplrx7!wyant
rj...@vax5.cit.cornell.edu
>>Do gravity waves have mass? If not, how can they have energy?
>
>Think about light. It doesn't have mass
Yes, it does. What it DOESN'T have is REST MASS. ALL energy has mass.
The RepoMan (Ron Reposh)
rj...@vax5.cit.cornell.edu
rre...@ionvax.tn.cornell.edu
Donald Benson
from the orbit of a planet. Imagine your billiard ball orbiting a bowling
ball on a rubber sheet. The interactions are elastic, and at steady state, the
energy out equals energy in. For instance, I don't think the moon would loose
orbital energy circling an atmosphere-less solid earth. However, because it
produces tides which give up energy in friction, energy is extracted from the
moon's motion. (Tidal-power advocates note: the disaster you would bring upon
us, the accelerated crashing of luna upon terra, is worse than any Chernobyl!)
Now, until steady state is reached (and return reflections of gravitational
waves reaching the edges of the universe <or wrapping around?> have come back),
kinetic energy might be reduced.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
My radical ideas would never be accepted by a staid company like this.
DonB
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Jeff Hunter
>
> RE: Gravitional Singularities -
> One thing that physicists seem to gloss over is that (to _my understanding_)
> there aren't any black holes, since they can't have collapsed yet. Because
> of gravitic time dilation, it takes (to an observer in "flat" space) an
> infinite amount of time to fall to the event horizon. So the mass that's
1) this is relativity son. "ain't collapsed yet" is thinkin like a classical,
ah say, classical Newtonian. We don't give a fig for absolute
space and time no more (That's a funny. "don't give a fig". get it?
All right, don't smart-mouth me son.) If Ah pushed you into onea
them thar holes you'd hit a singularity before you could yell
"that's all folks". (Ok, ok, maybe longer if it were a big one.
Sheesh, kids these days.)
2) Anyrate you said "black hole". Iffen its a hole, and light caint get
out then it's black. Who cares, ah say, who cares about what it
looks like inside? (All right, Hawking says it's not quite black
but he's also the guy who proved there's a singularity inside.)
--
___ __ __ {utzoo,lsuc}!censor!jeff (416-595-2705)
/ / /) / ) -- my opinions --
-/ _ -/- /- No one born with a mouth and a need is innocent.
(__/ (/_/ _/_ Greg Bear
Rahul Dhesi
(Daniel S. Riley) writes:
>...mass can mean different things in different contexts...
When I say e = mc^2, I mean exactly that. Find the speed of light,
find the energy of the photon, and you can find its mass.
On off days, however, I claim that "mass" means "mass when at rest",
and "velocity" means "velocity when at rest", and "momentum" means
"momentum when at rest". I like these definitions because they are so
nice and constant, and independent of context, and so easy to use in
equations!
--
Rahul Dhesi <dh...@bsu-cs.bsu.edu>
UUCP: ...!{iuvax,pur-ee}!bsu-cs!dhesi
Dave Hiebeler
> do photons
> have gravity? I.e., if you had an extremely intense laser burst shot out
> into space, would it exert a gravitational pull on masses that it passed by?
I'd guess that it would... if something like a star exerts
a gravitational force on the beam of light, I'd think the
beam of light would extert a force on the star. You should
be able to calculate the center of mass between the two, and
all that...
And I have no doubt that if I'm mistaken, I'll be severely
corrected. :-)
--
Dave Hiebeler Internet: hieb...@cs.rpi.edu (preferred address)
Center for Nonlinear Studies hieb...@cardinal.lanl.gov
MS B258
Los Alamos National Laboratory / Los Alamos, NM 87545
Joe Voros
[relativistic Energy/mass/momentum formula deleted]
> As can be seen, as rest mass (m) approaches zero, momentum (p)= E/c.
> This has been verified experimentally. (Actually, this was what Einstein
> got the Nobel Prize for).
*Sigh*
There is bound to be a "yes it is","no it isn't" argument about
what he won the Nobel prize for. Thus, to curtail this, let's see
what the _Swedish Academy_ thinks they gave it to him for ;-)
On November 10, 1922, Professor Christopher Aurivillius, the
secretary of the Swedish Academy of Sciences, wrote to Einstein:
"As I have already informed you by telegram, in its meeting held
yesterday the Royal Academy of Sciences decided to award you last
year's [1921] Nobel prize for physics, in consideration of your
work on theoretical physics and in particular for your discovery
of the law of the photoelectric effect, BUT WITHOUT TAKING INTO
ACCOUNT THE VALUE WHICH WILL BE ACCORDED YOUR RELATIVITY AND
GRAVITATION THEORIES after these are confirmed in the future."
[my capitals, for emphasis]
Personally, I'll stick by what the people who actually *gave* the
prize think. Will you? ;-)
joe
+--------------------------------------------+------------------------+
| "In theoretical physics we use very simple | Joe Voros |
| tools: pencil and paper, eraser, chair and | Physics Dept, Monash |
| table. More important than any of these is | University, Australia. |
| the wastebasket." Murray Gell-Mann | jo...@monash.edu.au |
+--------------------------------------------+------------------------+
Gary Murphy
>2) Anyrate you said "black hole". Iffen its a hole, and light caint get
> out then it's black. Who cares, ah say, who cares about what it
> looks like inside? (All right, Hawking says it's not quite black
> but he's also the guy who proved there's a singularity inside.)
>
If I understand Hawking's latest 'imaginary time' proposition (Brief History of Time), Hawking is ALSO the guy who proposes that singularities are not neccessarily singular (due to quantum effects). As for what's inside, to the (astro)Physicist, the question is more "what it is _doing_ inside" - the conditions should support Grand Unified Field interactions.
Although 'black' in the sense that the interior is hidden, what happens when a virtual pair is created at the threshold, and one side leans closer to the 'hole' while the other side escapes? Hawking is also the guy who proposed black-hole evaporation.
--
Gary Murphy | 3755 Riverside Dr. | decvax!utzoo!dciem!
Cognos Incorporated | P.O. Box 9707 | nrcaer!cognos!garym
(613) 738-1338 x5537 | Ottawa, Ontario |
| CANADA K1G 3N3 | THINK GOOD!
Bill Wyatt
> [...]
> Can we try, once and for all, to bring this "photons have/don't have
> mass" thing to an end? Let's stick with generally accepted physics,
> document our assertions either with references or with derivations
> from known physics, etc.
>
> First, let's agree (I hope) that rest mass is different from moving
> mass, and in particular different from mass of an object moving with
> the speed of light. Notice that the denominator in the Lorentz
> equation, the infamous
>
> m = m0/(sqrt( 1 - (u/c)^2 ) )
>
> goes to zero as u approaches c. Therefore zero rest mass (which in
> fact the photon has) gives a relativistic mass of zero divided by
> zero!
> mass. In fact relativity demands that it have mass
>
> e = mc^2 ==> m = e/c^2
>
> if it has energy.
WRONG (see below).
> I think you will find that any accepted book on
> modern physics or optics, if it deals with photon mass at all, gives
> this relationship or some version thereof.
Presumeably it gives the CORRECT relationship, which is:
E**2 = (m c)**4 + (p c)**2
where m is the REST MASS (should be written `m sub zero'),
and p is the MOMENTUM.
For a PHOTON, E = pc , *NOT* mc**2 !!!
For ordinary matter, p = mv, (not rest mass, this time), so AT REST,
the equation is E = mc**2.
(A little knowledge is a dangerous thing...)
Bill Wyatt, Smithsonian Astrophysical Observatory
UUCP : {husc6,cmcl2,mit-eddie}!harvard!cfa!wyatt
ARPA: wy...@cfa.harvard.edu
SPAN: cfa::wyatt BITNET: wyatt@cfa
Bill Wyatt, Smithsonian Astrophysical Observatory
Geoff HUNTER
photo-electric effect - in terms of quanta of light, which G.N.Lewis
named "photons" much later - in 1926. So while our Albert is best known
for "relativity" his Nobel prize was for something else.
Geoffrey Hunter, Chemistry Department, York University, Toronto.
Paul Rodman
>whether it has mass. How about this unambiguous question instead: do photons
>have gravity? I.e., if you had an extremely intense laser burst shot out
>into space, would it exert a gravitational pull on masses that it passed by?
>object with (relativistic) mass e/c^2 ?
> Hopefully someone can clarify this situation.
Well, photons certainly FEEL the effects of gravity, their path being
bent by stars, etc. So doesn't that imply the photon also pulling the other way
as well? (Or do we have forces that act only on on guy?!)
Paul K. Rodman
rod...@mfci.uucp
__... ...__ _.. . _._ ._ .____ __.. ._
Hal Lillywhite
>Okay, so the photon has momentum, but it's an ambiguous question to ask
>whether it has mass. How about this unambiguous question instead: do photons
>have gravity? I.e., if you had an extremely intense laser burst shot out
>into space, would it exert a gravitational pull on masses that it passed by?
>Wouldn't a beam with energy e have the same gravitational pull as an
>object with (relativistic) mass e/c^2 ?
First let's consider the inverse question, do the masses light
passes by exert gravitational pull on the light? I believe this was
answered in an experiment to test relativity, looking at light from
a star behind the sun during an eclipse (I think in about 1926,
anyone know an exact reference?). The star was visible indicating
that the light from it was bent by the sun's gravitation. Thus
gravity acts on light to "pull" it toward a massive object. From
Newton's law if the sun exerts a force on the light the light must
exert an equal and opposite force on the sun. The only way out of
this would be to claim that it takes no force to move light (bend it
out of its path). This would require that light be massless.
As an added example, let's do the following thought experiment to
actually "weigh" light. Consider a box with perfectly reflecting
inside walls (we can get NICE equipment for thought experiments).
We evacuate the box and place it in a gravitational field on a set
of very accurate scales so we can weigh it. Record the empty weight
and introduce light (photons) into the box. This being a thought
experiment we introduce the light so it is moving exactly
vertically and bounces between the top and bottom of the box.
General relativity indicates that there will be a gravitational red
shift between the top and bottom of the box, that is the frequency
of the light will be slightly lower at the top than at the bottom.
(Anybody out there know enough general relativity to calculate
this?) Thus the photon momentum p=hf/c will be less at the top than
at the bottom.
Each time a photon hits the top or bottom in gives an impulse
2p=2hf/c to that surface and hence to the box. Of course it pushes
the bottom downward and the top upward. Since each photon will
bounce off the top exactly as often as off the bottom the number of
such impulses is the same for top and bottom. However the magnitude
is different. Because the momentum is greater at the bottom than
at the top the impulse will be greater there for each reflection.
This results in a net downward force which we can measure with our
scales. Conclusion: light has weight!
If the light did not enter the box vertically, we will still see a
similar effect because the gravitational bending of its trajectory
would cause it to follow a path somewhat like a bouncing ball,
bouncing off the floor at a different angle than off the ceiling
(and of course there would still be a red shift). This would also
introduce the possibility that there could be more reflections from
the floor than from the ceiling. Again we could, in theory, measure
the weight of light.
Again, if the light shows up as an increase in weight of the box
(force from gravity on the box) Newton's law says there is an equal
and opppsite force from the light acting on the gravity (or its
source). Thus light exerts a gravitational attraction on other
objects!
BTW, we now have light which has energy, is acted upon by
gravitation and acts on massive objects through gravitation. If it
doesn't have mass, how do you define mass?
Michael Frank
>still be formally calculated from the momentum four-vector,
>m^2 = E^2 - p^2 (and some factors of c^2 that I've dropped).
>For a photon, this gives a rest-mass of zero. Hence, the photon is
>a massless particle.
Okay, so the photon has momentum, but it's an ambiguous question to ask
whether it has mass. How about this unambiguous question instead: do photons
have gravity? I.e., if you had an extremely intense laser burst shot out
into space, would it exert a gravitational pull on masses that it passed by?
Wouldn't a beam with energy e have the same gravitational pull as an
object with (relativistic) mass e/c^2 ?
Hopefully someone can clarify this situation.
--
Michael Frank "The Ear-God" Undergrad, Symbolic Systems. I speak for no one.
bug...@portia.stanford.edu Attempting AI, fencing, juggling. Enjoy swimming,
Box 6536, Stanford CA 94309 Pink Floyd, Nanotech, Star Trek, some philosophy.
(415) EAR-0-GOD "I listen." Foozball wizard. Own Amiga. Would like a Sun-4.
Bob Myers
>and "velocity" means "velocity when at rest", and "momentum" means
>"momentum when at rest". I like these definitions because they are so
>nice and constant, and independent of context, and so easy to use in
>equations!
?????????????????????????????????????????????????????????????
Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not
Ft. Collins, Colorado | those of my employer or any other
{the known universe}!hplabs!hpfcla!myers | sentient life-form on this planet.
Daniel S. Riley
>>It's true that there is no rest frame for a photon, but a rest mass can
>>still be formally calculated from the momentum four-vector,
>>m^2 = E^2 - p^2 (and some factors of c^2 that I've dropped).
>>For a photon, this gives a rest-mass of zero. Hence, the photon is
>>a massless particle.
>
>Okay, so the photon has momentum, but it's an ambiguous question to ask
>whether it has mass. How about this unambiguous question instead: do photons
>have gravity?
Fine unambiguous question. Hopefully unambiguous answer: Yes.
An unambiguous statement is that, in general relativity, all energy
gravitates. Any energy has gravity. This is one of things that makes
general relativity complicated, and quantum gravity doubly complicated:
since any energy gravitates (i.e., has gravity), gravity itself has
gravity--the field couples to itself. If you imagine the quantum case,
the little gravitons that get exchanged to transmit gravity have their
own little gravitational fields, with the associated gravitons. It's a
non-linear theory, sort of like what electromagnetism would be like if
the photon had an electric charge, and thus could emit more photons.
Very awkward.
>Wouldn't a beam with energy e have the same gravitational pull as an
>object with (relativistic) mass e/c^2 ?
Yes, it would. What we're really arguing here is semantics. My main
point before was that there are two things you could call the mass--there's
the energy, and there's the rest mass (which is well-defined in *any*
reference frame--a particle doesn't have a rest mass only in it's rest
frame, it has a rest mass in *any* frame, which you can trivially
calculate from the momentum four-vector). In my experience (as a
physicist who has done some GR and high energy physics), "mass" is more
commonly used to refer to the rest mass, but it is an ambiguous usage.
Anyway, it's just a usage question, not a fundamental law.
> Hopefully someone can clarify this situation.
I tried.
-Dan Riley (ri...@tcgould.tn.cornell.edu, cornell!batcomputer!riley)
-Wilson Lab, Cornell U.
>(415) EAR-0-GOD "I listen." Foozball wizard. Own Amiga. Would like a Sun-4.
^^^ ^^^^^
I've got an Amiga and a DECstation 3100. I get more work done on the DEC,
but the Amiga is more fun.
Felix Lee
Excuse my ignorance, but isn't what Mark describes the photoelectric
effect? The experimental verification of photons' momentum?
--
Felix Lee fl...@shire.cs.psu.edu *!psuvax1!shire!flee
Steven Den Beste
What I find somewhat interesting about all of this is that people are making
comments of the form "Photons have no rest mass - see, this formula
here says so."
I just wanted to point out that that the formula is only a model. You cannot
reference the formula as absolute proof of the characteristics of reality. What
you should be saying is more like "Our current model implies that photons have
no rest mass."
An even bigger caveat about the question "Are gravitons affected by gravity?"
Someone made a big extrapolation that ran something like this:
1. Gravitons carry energy
2. Therefore gravitons must manifest as having mass in some cases
3. Therefore gravitons are themselves affected by gravity.
I caution you to remember the following: So far as I know, all attempts to try
to unify gravity with the other forces have failed. The "unified field theory"
is not an accomplished goal, it is the Holy Grail - lots of physics paladins
are out there seeking it, but none have found it.
We don't even know that gravity has a mediating particle, as the other forces
have. And if it does, we don't know that it follows the same rules as the other
particles. Thus the "proof" above is premature at best.
Let's just keep in mind that our theories are not necessarily complete nor
perfect, eh?
Steven C. Den Beste, BBN Communications Corp., Cambridge MA
denb...@bbn.com(ARPA/CSNET/UUCP) harvard!bbn.com!denbeste(UUCP)
Daniel S. Riley
>What I find somewhat interesting about all of this is that people are making
>comments of the form "Photons have no rest mass - see, this formula
>here says so."
You've got to have some definition of rest mass. Then you go out and
measure it. If you want to read about some upper limits on the photon
rest mass, read the introduction to the latest edition of Jackson (the
standard graduate E&M text). It's *small*. But you do need a formula
in there somewhere, or you don't know what you're talking about. And
the one I cited is fundamental to relativity (which is fundamental to
most modern formulations of E&M), so it seemed like a good choice.
I agree that you've got to make some connection to reality somewhere
along the way.
>An even bigger caveat about the question "Are gravitons affected by gravity?"
>Someone made a big extrapolation that ran something like this:
> 1. Gravitons carry energy
> 2. Therefore gravitons must manifest as having mass in some cases
Step 2 here is uneccesary--if it's got energy, then it has gravity, according
to GR. If you want a reference, go see Misner, Thorne and Wheeler,
chapter 35, "Propagation of Gravitational Waves", which discusses (briefly)
the spacetime curvature due to the energy of gravitational waves, or chapter
17, "How Mass-Energy Generates Curvature". (This is a less than ideal
reference, since MTW is pretty impenetrable for the uninitiated. However,
it's what I have handy, and it's the "Bible" of GR.)
> 3. Therefore gravitons are themselves affected by gravity.
>
>I caution you to remember the following: So far as I know, all attempts to try
>to unify gravity with the other forces have failed. The "unified field theory"
>is not an accomplished goal, it is the Holy Grail - lots of physics paladins
>are out there seeking it, but none have found it.
>
>We don't even know that gravity has a mediating particle, as the other forces
>have. And if it does, we don't know that it follows the same rules as the other
>particles. Thus the "proof" above is premature at best.
Several comments:
1) GR is a non-linear field theory--gravity does couple to itself in
classical general relativity. If there is a quantum field theory
which corresponds to GR, then it will have to be non-linear. At least
in the article I posted on the subject, the main point was that,
within the framework of GR, gravity has gravity. I invoked the
quantum case just on the off chance that it might help people to have
a particle around, since the idea that a field can have gravity can
take some getting used to.
2) A quantum theory of gravity isn't the same thing as a unified field
theory. It's generally assumed that a true unified field theory will
include quantum gravity, but you can have a quantum field theory of
gravity without reference to any of the other fundamental forces.
In fact, lots of people have worked on quantum gravity, and it's a hairy
mess, because consistency with GR requires that the graviton be spin-2
(at least) and have gravity.
3) Which brings to me point 3: the discussion was within the framework
of general relativity. If we assume that any theory of quantum gravity
has to be consistent with GR, we can make some definite statements about
the structure of any quantum gravity theory just from the mathematical
structure of GR, without needing to actually have a working quantum
gravity in front of us.
4) A lot of physicists will be very surprised if there isn't a quantum
theory of gravity--it sure looks like a field theory, so it's natural
to expect that there exists a quantum field theory which correspons to
the classical field theory. Of course, surprises do happen, so your
point that there isn't a working theory is a good one.
Matt Crawford
[ A photon of frequency f is bouncing vertically inside a reflecting box
in a gravitational field ... ]
) General relativity indicates that there will be a gravitational red
) shift between the top and bottom of the box, that is the frequency
) of the light will be slightly lower at the top than at the bottom.
) (Anybody out there know enough general relativity to calculate
) this?) Thus the photon momentum p=hf/c will be less at the top than
) at the bottom.
delta f / f = GM/c^2 ( 1/r1 - 1/r2 ), where M = mass of attracting body.
Net downward momentum given to box by one pair of bounces is
delta p = (2h/c) delta f.
Rate of pairs of bounces per second is 1/t = c/2(r2-r1).
Momentum transfer per second = net force is
F = (delta p)/t = (hGMf/c^2)/(r1*r2).
If you make the box small, so r1 = r2 approximately, then this
is the same as the gravitational force on a particle of mass hf/c^2.
(Recall that the energy of the photon is E=hf.)
________________________________________________________
Matt Crawford ma...@oddjob.uchicago.edu
Bob Knighten
gravitational. In the years preceeding his 1916 paper on General Relativity
Einstein addressed the question of both the inertial and gravitational mass of
energy in two lovely papers:
Ist die Traegheit eines Koerpers von seinem Energieghalt abhaengig?
Annalen der Physik, 17, 1905
[Does the inertia of a body depend upon its energy content?]
Ueber den Einfluss de Schwerkraft aut die Ausbreitung des Lichtes
Annalen der Physik, 35, 1911
[On the influence of gravitation on the propagation of light]
English translations of both of these papers are reprinted in the book
The Principle of Relativity (Methuen and Company, 1923) which is readily
available in a Dover reprint. These two papers are easier reading than the
treatment of the same topic that you will find in most physics text books.
News Manager
observer(system of reference?) of course C is constant as reference at
origin and everywhere else for no accelaration and no gravition. But
when there is gravity or equivalent accelation then gradient of C creates
gravitation and various gravitational potential correspond to various C.
If so, Then masses or equivalent package of energies are atracted by gravity
because they lose energy as they move in lower C (E=(mc)*c). For objects
the enegy is converted to velocity and for photon converted to shorter
wavelength not higher frequency. Light like any other object having energy
creates a gravity field by forcing the C lower where it is located.
The lower C the lower is energy E=p*c. But purturbing C in space requires
energy so an equilibrium is achieved by lowering C proportional to m*C.
From: fto...@charming.nrtc.northrop.com (Farhad Towfiq)
Path: charming!ftowfiq
The Gravity Waves according to this theory are then vibrations in the
speed of light C, and energies of any kind, either light or electrostatic fiels
or anything else are treated equal.
Any idea?
Paul Rodman
>
>Okay, so the photon has momentum, but it's an ambiguous question to ask
>whether it has mass. How about this unambiguous question instead: do photons
>into space, would it exert a gravitational pull on masses that it passed by?
>object with (relativistic) mass e/c^2 ?
david.appell
In article <36...@vice.ICO.TEK.COM> ha...@vice.ICO.TEK.COM (Hal
Lillywhite) writes:
# In article <23...@Portia.Stanford.EDU> bug...@Portia.Stanford.EDU (Michael Frank) writes:
# >Okay, so the photon has momentum, but it's an ambiguous question to ask
# >whether it has mass. How about this unambiguous question instead: do photons
# >have gravity? I.e., if you had an extremely intense laser burst shot out
# >into space, would it exert a gravitational pull on masses that it passed by?
# >Wouldn't a beam with energy e have the same gravitational pull as an
# >object with (relativistic) mass e/c^2 ?
# [Thought experiment deleted, which claims to show that light has
# "weight"]
#
# BTW, we now have light which has energy, is acted upon by
# gravitation and acts on massive objects through gravitation. If it
# doesn't have mass, how do you define mass?
Let's get relativistic, like we should. For any particle
p^2 = m^2
where p^2 is the square of that particle's 4-momentum. m is a constant,
*the* mass. For a photon, m=0. All this talk about "rest mass" and
"relativistic mass" is at bottom an attempt to make the kinetic
equations of motion appear as classical as possible -- but they're
inherently not here. In other words, a particle's mass does *not*
increase as it's speed increases -- we use those words so that the
equations of motion look "the same" to us. What changes are the actual
equations, from the classical ones at v=0, to relativistic ones as v
nears c.
As for the gravitational attraction of a photon: Einstein's equation
of gravitation is (*quite* loosely)
geometric distribution of spacetime = energy distribution of spacetime
The right-hand side of this equation is the stress-energy tensor, which
is nonzero even for a massless (m=0) particle, because such a particle
has a nonzero energy. That's enough for a gravitational effect.
I'm not sure how you would even think of the flat space
(nonrelativistic) limit of this for a photon, since there is no such
thing for a photon....
-- David Appell
...att!hou2d!appell
Radford Neal
> ... Photons (light) have no mass. The "m" in
> e = m c^2 is the relativistic mass, given by
>
> m = m0/(sqrt( 1 - (u/c)^2 ) ),
>
> where u is the velocity of the moving particle and m0 is the rest mass.
> If photons had a rest mass, their relativistic mass would always be
> infinite, since in the case of a photon, u=c.
The posters on this topic are clearly missing each other's points. You
can't answer this question without a definition of "mass".
If two photons are sent on parallel courses, starting out a metre apart,
will they gradually approach each other due to mutual gravitational
attraction?
If the answer is "yes", then photons have at least "gravitational" mass.
Note that this is a somewhat different question from asking whether light
gets bent by the sun.
I expect that the answer _is_ yes, and that photons _do_ have gravitational
mass. Otherwise, when an electron and a positron annihalated (producing
only photons), there would be a sudden, discontinuous, drop in gravitational
field. This doesn't seem right, somehow...
Radford Neal
Paul Rodman
>energy out equals energy in. For instance, I don't think the moon would loose
>orbital energy circling an atmosphere-less solid earth. However, because it
>produces tides which give up energy in friction, energy is extracted from the
>moon's motion.
Isn't the energy extracted from the earth's rotation? I belive the
Moon is moving AWAY from the earth , not toward it, at a small rate
(<1Meter/century??) due to tidal interations. If energy was
extracted from the Moon's motion the moon would spiral in, and I have
this piece of information stuck in my brain (possible in error) that
the Moon is actually spiraling out... anyone know?
pkr
Dave Raggett
> Moon is moving AWAY from the earth , not toward it, at a small rate
> (<1Meter/century??) due to tidal interations. If energy was
> extracted from the Moon's motion the moon would spiral in, and I have
> this piece of information stuck in my brain (possible in error) that
> the Moon is actually spiraling out... anyone know?
Think conservation laws - conservation of angular momentum means
that as tides slow the Earth's rotation, the angular momentum is
converted into the relative orbital motion of the Earth-Moon system.
A dimensional analysis implies that orbital angular momentum increases
as the square root of the separation. Therefore tidal forces
will cause the Earth-Moon distance to increase!
-----------------------------------------------------------------------------
Dave Raggett @ HP Labs | Phone: +44 272 799910 x 24046
Information Systems Centre | d...@hpl.hp.co.uk
Bristol | dsr%hplb...@ukc.ac.uk
U.K. | d...@hplb.hpl.hp.com
Hans Aberg
>secretary of the Swedish Academy of Sciences, wrote to Einstein:
>"As I have already informed you by telegram, in its meeting held
>yesterday the Royal Academy of Sciences decided to award you last
>year's [1921] Nobel prize for physics, in consideration of your
>work on theoretical physics and in particular for your discovery
>of the law of the photoelectric effect, BUT WITHOUT TAKING INTO
>ACCOUNT THE VALUE WHICH WILL BE ACCORDED YOUR RELATIVITY AND
>GRAVITATION THEORIES after these are confirmed in the future."
As far as I remember this is indeed correct. Einstein officially
got the Nobel prize for his works on the photo-electric effect, and
also the Brownian motion, his relativity and gravitation works
deliberately being excluded.
I think this was a very smart move by the Nobel committee, because at
that time, relativity was still a very controversial topic. By doing
what they actually did, they could without arguments hand him the prize.
Hans Aberg, Mathematics
ab...@math.rutgers.edu
David Bofinger, Theoretical Physics, RSPhysS ANU
writes:
> Presumeably it gives the CORRECT relationship, which is:
>
> E**2 = (m c)**4 + (p c)**2
Presumably it does indeed give the correct relationship:
2 2 4 2 2
E = m c + p c
2
which reduces to E=mc for p=0
> where m is the REST MASS (should be written `m sub zero'),
Any practicing physicist will be perfectly happy with m. This distinction
between mass and rest mass really should be carried on in
aus.science.linguistics, its only a semantic difference. Basically the concept
of relativistic mass is pretty useless for anything important (though it
may be a handy pedagogical analogy); contrary to popular belief, the equation
F=ma doesn't work at relativistic speeds, even if the m is relativistic
mass. So physicists only use rest mass, and then drop the word `rest' (after
all, if its the only sort of mass there is then the qualifier's redundant,
no ? :-) ). Some sciences confuse laypeople by using long complicated words.
Physics uses everyday words like colour and flavour, and totally changes
their meanings. This is even more effective. ;-)
> and p is the MOMENTUM.
>
> For a PHOTON, E = pc , *NOT* mc**2 !!!
>
> For ordinary matter, p = mv, (not rest mass, this time), so AT REST,
What ?? `Ordinary Matter' ?? Is there some other kind ?
p=mv is an accurate _approximation_ for v<<c, it has nothing to do with what
sort of particle we are discussing.
> the equation is E = mc**2.
> (A little knowledge is a dangerous thing...)
But it sure beats ignorance :-)
______________________________________________________________________________
David Bofinger ACSNet: dxb...@phys0.anu.oz
Australia Post: Department of Theoretical Physics
Research School of Physical Sciences
Australian National University
"Don't walk. Don't walk. Don't walk. Walk Now. Walk Now. Walk Now. Walk Now."
-Blade Runner
Paul Rodman
>
>I dug out my ol' astronomy book, remembering that there was something about
>this in there. Here is what the book had to say :
>"The gravatational pull of this bulge produces a small but constant force
> tugging the Moon ahead of its orbit , speeding up its motion very slightly.
> As the Moon's speed of revolution around the Earth increases, it moves
> gradually into more and more distant orbits around the Earth. In other words,
>the Moon is very slowly, but relentlessly spiraling away from the Earth at
>a rate of about 4 cm per year. "
>
Thanks for looking this up. I didn't think the deflection of the tidal
bulge accounted for that, I thought it caused the precession of the
equinoxes, I guess it causes both.
Of course the effect is VERY small as the bulge on the other side of the
earth, which is pulling the moon from behind, acts against the other in the
usual tidal-force way, making the effect a 1/r**3 thing, so its very small.
Its also interesting that not only is it required that the earth
rotate faster than the lunar month, but the bulge *must* be of tidal
origin, i.e. a rotating earth with a dipole moment (i.e. "barbell"
shape) would not have the same effect, as the bulges would change
their axis with respect to the earth-moon axis constantly, pulsing the
moon both forward and back, but with the net force being zero (I
think)....Fun stuff.
Jason Gabler
> Moon is moving AWAY from the earth , not toward it, at a small rate
> (<1Meter/century??) due to tidal interations. If energy was
> extracted from the Moon's motion the moon would spiral in, and I have
> this piece of information stuck in my brain (possible in error) that
> the Moon is actually spiraling out... anyone know?
I dug out my ol' astronomy book, remembering that there was something about
this in there. Here is what the book had to say :
"The Earth rotates about its axis faster than the Moon revolves around the
Earth. Therefore as the Moon produces a tidal bulge in the oceans, the
Earth's rotation carries that bulge around the planet -- that is, the high tidemoves on ahead of the moon Eather than staying just below it.
-----<-------<-----
\/_ MM \
MM
oooooooo Please excuse the crudeness...
ooEEEEo The 'E's make up the Earth.
oEEEEEEo The 'o's are the exaggeratted
oEEEEoo tidal bulges. The 'M's make
ooooooo up the moon. And the 'arrow'
shows the Moon's path of revolution and the Earth's rotation. Notice how
the tidal bulge is 'ahead' of the moon.
"The gravatational pull of this bulge produces a small but constant force
tugging the Moon ahead of its orbit , speeding up its motion very slightly.
As the Moon's speed of revolution around the Earth increases, it moves
gradually into more and more distant orbits around the Earth. In other words,
the Moon is very slowly, but relentlessly spiraling away from the Earth at
a rate of about 4 cm per year. "
Text and "drawing" adapted from:
"Universe" by William J. Kaufmann, III
c. 1985
Hope that answers your question!
p.s. If you'd like to flame the "drawing", please do so via e-mail. ;)
Jason Gabler jygabler@ucdavis
Paul Rodman
>> Isn't the energy extracted from the earth's rotation? I belive the
>> Moon is moving AWAY from the earth , not toward it, at a small rate
>
>that as tides slow the Earth's rotation, the angular momentum is
>converted into the relative orbital motion of the Earth-Moon system.
>
Yeah, thats the easy part, now can somebody remember HOW the earth's rotation
pumps the moon to a higher orbit? I recall that the axis of the earth's
deformation is not quite aligned with the earth-moon axis....due to
friction it's slightly canted in the direction of the earths rotation, but
I think this only creates the precession of the equinoxes, not the moons
outward spiral.....anyone know?
William H. Jefferys
#Yeah, thats the easy part, now can somebody remember HOW the earth's rotation
#pumps the moon to a higher orbit? I recall that the axis of the earth's
#deformation is not quite aligned with the earth-moon axis....due to
#friction it's slightly canted in the direction of the earths rotation, but
#I think this only creates the precession of the equinoxes, not the moons
#outward spiral.....anyone know?
The rotation of the Earth drags the tidal bulges a little ahead of the
Moon's position. This provides a net torque on the Moon, causing
it to gain energy and move outwards. The energy for this comes from
the Earth's rotation.
Many elementary astronomy texts have a diagram showing this.
Bill Jefferys
--
Glend. I can call spirits from the vasty deep.
Hot. Why, so can I, or so can any man; But will they come when you
do call for them? -- Henry IV Pt. I, III, i, 53
Jeff Hunter
> I don't thing the radiation of "gravitational waves" in itself saps energy
> from the orbit of a planet. Imagine your billiard ball orbiting a bowling
> ball on a rubber sheet. The interactions are elastic, and at steady state, the
> energy out equals energy in. For instance, I don't think the moon would loose
Only if you use the Newtonian assumption that gravity moves at
lightspeed. If you include GR then there is a loss. I think I recall that
for the Sun-Jupiter system the radiation is about 5 watts, so it is
obviously swamped by tidal effects, etc...
I thought that there had been experimental measurement of grav. rad.,
but a quick check tells me that the "most promising candidate" requires
several more years of observation to get an accuratte measure.
--
___ __ __ {utzoo,lsuc}!censor!jeff (416-595-2705)
/ / /) / ) -- my opinions --
-/ _ -/- /- No one born with a mouth and a need is innocent.
(__/ (/_/ _/_ Greg Bear
Jeff Hunter
In article <61...@ulysses.UUCP>, ga...@ulysses.UUCP (Gary Murphy) writes:
> If I understand Hawking's latest 'imaginary time' proposition (Brief History
> of Time), Hawking is ALSO the guy who proposes that singularities are not
> neccessarily singular (due to quantum effects).
Sort-of. He found a sensible alternate choice of parameters in which
the singularity is no longer pathological. (space-i-c-time rather than
space-time.) However try saying that while pretending to be a giant chicken :-)
> Although 'black' in the sense that the interior is hidden, what happens
> when a virtual pair is created at the threshold, and one side leans closer
> to the 'hole' while the other side escapes? Hawking is also the guy who
> proposed black-hole evaporation.
>
Well I *did* say "All right, Hawking says it's not quite black", but
since then I ran into an interesting stat. A solar mass black hole would
have a Hawking temperature of 10e-7 K, which is much smaller than the
blackness of space (blackbody at 3K), or the blackness of a cave (~300K).
So yup, a big enough black hole can be spectacularly black. (Wanna build a
really cold refridgerator? :)
David Bofinger, Theoretical Physics, RSPhysS ANU
> In article <525...@hpcuhb.HP.COM> do...@hpcuhb.HP.COM (Donald Benson) writes:
>
>>energy out equals energy in. For instance, I don't think the moon would loose
>>orbital energy circling an atmosphere-less solid earth. However, because it
>>moon's motion.
>
> Isn't the energy extracted from the earth's rotation? I belive the
> Moon is moving AWAY from the earth , not toward it, at a small rate
> (<1Meter/century??) due to tidal interations. If energy was
> extracted from the Moon's motion the moon would spiral in, and I have
> this piece of information stuck in my brain (possible in error) that
> the Moon is actually spiraling out... anyone know?
The point is that we need to conserve angular momentum, as well as energy.
Originally, both the Earth and Moon rotated relative to their revolution.
Tidal forces tend to `phase-lock' their rotation so that the Earth-Moon
system rotated as a rigid body. Once this has happened, there will be no
travelling tidal bulge on either world, and hence no friction to induce energy
losses. At the moment, the Moon has been phase-locked (we see only one face),
but the Earth still rotates. Friction due to the tides around places like
the Bering sea will gradually make the Earth a single-face world (with respect
to the Moon) too.
Given that, you can see that a lot of angular momentum will no longer be
locked up in rotation. The only place this can be absorbed is the orbital
angular momentum of the combined system, and the way that works is by
increasing the moment of inertia of the system, hence the increase in
separation.
The rigid body will then interact with the rest of the solar system much
the way that the Earth does with the Moon today. This _will_ lead (eventually,
I don't know how long it takes) to the loss of angular momentum from the
system, and the Moon will begin to approach the Earth again (and break up
when it reaches the Roche(?) limit, I suppose - so we lose our planet's most
famous tourist attraction :-) )
John E Van Deusen III
>
> precession of the equinoxes ...
The precession of the equinoxes is caused by the attraction of the sun
and moon on the *equatorial* bulge, (J2). There is a torque in the same
sense at both solstices and no torque at the equinoxes.
I am curious whether the J3 spherical harmonic that causes the earth to
be pear shaped, and thus causes the tidal bulge in the southern
hemisphere to be about 20 meters closer to the moon, would be sufficient
to induce a precession of 2 to 5 degrees with respect to the earth's
axis and a period of 40 to 100 thousand years.
--
John E Van Deusen III, PO Box 9283, Boise, ID 83707, (208) 343-1865
uunet!visdc!jiii
Craig Eisler
>mass, and in particular different from mass of an object moving with
>the speed of light. Notice that the denominator in the Lorentz
>equation, the infamous
>
> m = m0/(sqrt( 1 - (u/c)^2 ) )
>
>fact the photon has) gives a relativistic mass of zero divided by
>zero! We have to use L'Hospital's rule to calculate relativistic
>mass at u=c. However we have no equation for how m0 behaves at
>other that light speed so I don't know how to do this. (If you are
>not familiar with this procedure check any basic calculus text. It
>really doesn't matter since we can't use it anyway.) Suffice it to
>say that the Lorentz equation allows an object with zero rest mass
>to have non-zero mass if it moves at the speed of light.
You us to stick to "generally accepted physics.". Why don't you stick
to generally accepted mathematics?????
Perhaps you should go back to this basic calculus text and
review "L'Hospital's" [sic] rule. m0 is a constant. Not a function
of the velocity u. You cannot apply that rule here.
Since you are talking about limits (l'hopital's rule is used to compute
a limit), let's examine the limit of m as u -> c. The limit
is the value to which the function tends, but not necessarily the
function value at the point. For all values of u<c, m is 0.
Letting u get closer and closer to c, m is still 0. In fact, the limiting
value is 0.
So your argument is worthless.
On another note:
I went digging through the citation index looking for articles published
on photon mass. I discovered that current research is pointing towards
photons in fact having mass. Apparently there is recent experimental
evidence to suggest that photons have mass, and people are publishing
papers like mad trying to justify this using quantum electrodynamics.
Now, I only have passing knowledge of QED. Anyone out there know
more about this?
craig
--
Craig Eisler, Applied Math, University of Waterloo
"Either way, I'm afraid to try it" - Calvin.
Kurt Sonnenmoser
> > As can be seen, as rest mass (m) approaches zero, momentum (p)= E/c.
> >This has been verified experimentally. (Actually, this was what Einstein
> >got the Nobel Prize for).
> Wrongo. He got the Nobel for the Photoelectric Effect, never for any
> work on relativity.
Not quite. Let me quote (without permission) one of my teachers in
theoretical physics (N. Straumann) commenting on this in his lecture
on Quantum Mechanics (my translation from German):
"The work Einstein got the Nobel Prize for was:
`Ueber einen die Erzeugung und Verwandlung des Lichtes betreffenden
heuristischen Gesichtspunkt', Ann. Phys. 17, 132 (1905)
(`On a heuristic aspect concerning creation and transformation of
light')
It is impossible to eradicate the opinion of many of the present
generation of physicists that this paper of Einstein's was mainly
concerned with an explanation of the Photoelectric Effect.
As a matter of fact, the measurements of this effect were too
inexact at that time to base the Photon-Hypothesis on them."
To quote Einstein from that paper (after analyzing Planck's radiation law):
"Monochromatic radiation (frequency f) of low density (...) from the
point of view of Statistical Mechanics looks as if it were composed
of independent energy quanta of magnitude hf." (h = Planck's constant)
That was the revolutionary point: THE PARTICLE ASPECT OF LIGHT.
The paper is further concerned with the application of this idea
to different phaenomena: the Photoelectric Effect, among others.
From article <3...@eplrx7.UUCP>, by wy...@eplrx7.UUCP (wyant):
> The story which was passed on to me was that the Nobel committee wanted
> to give Albert a prize for special relativity, but SR was still too
> controversial at the time. So, they gave it to him for a safer paper.
Nevertheless, the success of Einstein's Photon-Hypothesis led De Broglie
to the postulate of matter waves for which Schroedinger found his famous
equation some years later, a discovery that marks the birth of modern
Quantum Mechanics.
It's hard for us today to appreciate the boldness of Einstein's postulate.
Remember the truly overwhelming success of Maxwell's wave theory of light.
And nobody in the world was talking about wave-particle-duality at 1905!
Actually, the Photon-Hypothesis was only GENERALLY accepted after the
discovery of the Compton Effect in 1923 (two years after he got the Nobel
Prize). Even Planck himself first refused to believe in the reality of
light quanta.
So even if the story of the intended Prize for Relativity was true, the
above paper represents an essential break-through in the discovery of
Quantum Mechanics, a theory that in my opinion is even more counter-
intuitive than Relativity.
--
---------------------------------------- UUCP:...mcvax!cernvax!forty2!ks
Kurt Sonnenmoser |
Institut fuer Theoretische Physik |
Universitaet Zuerich, Switzerland | This is NOT a disclaimer
Rahul Dhesi
Neal) writes:
>The posters on this topic are clearly missing each other's points. You
>can't answer this question without a definition of "mass".
Being a very simple human being, I consider "mass" to be a way of
saying "weight" without limiting myself to a specific planet. For
centuries people have used this definition; let's not start changing
it now.
The question is simple: We ask how much it weighs, and if the answer
is greater than zero, then it has mass.
--
Rahul Dhesi <dh...@bsu-cs.bsu.edu>
UUCP: ...!{iuvax,pur-ee}!bsu-cs!dhesi
Career change search is on -- ask me for my resume
John Sparks
<23...@Portia.Stanford.EDU> <8...@m3.mfci.UUCP>
Sender:
Reply-To: spa...@corpane.UUCP (John Sparks)
Followup-To:
Distribution:
Organization: Corpane Industries, Inc.
Keywords:
In article <8...@m3.mfci.UUCP> rod...@mfci.UUCP (Paul Rodman) writes:
>
>Well, photons certainly FEEL the effects of gravity, their path being
>bent by stars, etc. So doesn't that imply the photon also pulling the other
way
>as well? (Or do we have forces that act only on one guy?
[I readily admit not being a physicist or anything close, so if this sounds
completley ignorant, please correct me, I am just trying to learn something
here :) ]
1> Gravity is expressed as acceleration in meters per second right?
2> Time dilates at the speed of light to almost a standstill, right?
So at the speed of light (at least from the reference of the light beam)
There can't be any gravity because there isn't any time for anything to
be accellerated into the lights gravity well.
Does this make any sense? Or am I completly out in left field? should
us electronics folx stick with electronics?
Just trying to learn sumthin'
--
John Sparks | {rutgers|uunet}!ukma!corpane!sparks | D.I.S.K. 24hrs 1200bps
[not for RHF] | spa...@corpane.UUCP | 502/968-5401 thru -5406
The next sentence is true. The previous sentence is false.
Michael Beaton
What is the gravitational effect of something small, say a baseball, moving
past you at a speed such that its relativistic mass is equal to, for instance,
the rest mass of the Earth!?
I await appropriate <FLAMES> for triviality!!!
------------------------------------------------------------------------------
m...@uk.ac.ed.aipna
------------------------------------------------------------------------------
Otto J. Makela
answers to it from different physicists):
Is the mass increase at relativistic speeds in special relativity
visible to outside observers, or is it simply a model artifact ?
(Did I phrase this correctly ?)
(visions of whole planetary systems being disturbed by passing photon-drive
spaceship accelerated to 99.9% c...)
Otto J. Makela (with poetic license to kill), University of Jyvaskyla
InterNet: mak...@tukki.jyu.fi, BitNet: MAKELA...@FINJYU.BITNET
BBS: +358 41 211 562 (V.22bis/V.22/V.21, 24h/d), Phone: +358 41 613 847
Mail: Kauppakatu 1 B 18, SF-40100 Jyvaskyla, Finland, EUROPE
Jim Meritt
}About this mass of photons, etc. thing...
}What is the gravitational effect of something small, say a baseball, moving
}past you at a speed such that its relativistic mass is equal to, for instance,
}the rest mass of the Earth!?
}I await appropriate <FLAMES> for triviality!!!
Fub that. I want a few opinions on what happens when the relativistic
mass of the object gets so high that its relativistically contracted
dimension is less than its gravitational radius i.e. a black hole
from the outside but only in the direction of travel? NNNaahhhhhhhhh.
What WOULD happen?
Use Einstein's elevator for the discussion...
........................................................................
The above was test data, and not the responsibility of any organization.
j...@aplvax.jhuapl.edu - or - j...@aplvax.uucp - or - meritt%aplvm.BITNET