🚀 go-pugleaf

In article <230@sri-arpa.ARPA> Joe writes:
>
>   Maybe someone out there can answer this one quickly. I hope its
>not too fundamental or repetitous.
>
>   Actually, now that I think about it, its really two questions:
>
>   o  Is there a difference between inertial mass and gravitational
>      mass. (I believe that is the same as asking,"is there a 
>      difference between the effects of gravity and the effects
>      of being in an accellerating frame of reference?" But
>      Im sure Im making a drastic generalization.)

	The basic question is, is the ratio between inertial mass
and gravitational mass the same for everything?  If not, you would be
able to detect this by observing different rates of acceleration of
objects in a gravitational field.  This would be very problematic
because states of free fall would be impossible to define; your lab
might be free falling at a different acceleration than your lab
equipment, so you would be able to measure forces in your free falling
lab that operated on different materials to different degrees.

	If so, however, you can equate the two for convenience and
just talk about "mass".  This is in fact the case.  From the point of
view of general relativity, objects simply follow geodesics; no
matter what the composition of the object (no matter what its mass)
it will follow the same geodesics through space-time.  (This is not
perfectly true for macroscopic objects, since they have extent they
are acted upon by tidal forces which may change their trajectory
somewhat.  However, in the limit as the objects become small
in extent the assertion is true, within the framework of general
relativity.)  This is assuming free fall in vacuum, of course.

	Recently some physicists were going over the data of
old experiments trying to establish the equivalence of inertial
and gravitational mass and claimed to have found slight
differences between atoms of different atomic number.  They attributed
this to an effect related to a quantity they called "hypercharge."
Apparently their result was flawed by a sign error, however,
so it is very dubious at the moment.

>   o  Given the old "accellerating-elevator-in-space" experiment,
>      how can one within the elevator tell the whether he is
>      accellerating or under the influence of gravity?

	You could determine this by trying to measure tidal
forces (i.e., the gravitational field is not uniform, but varies
slightly through the elevator; this causes tidal forces).  Again,
however, in the limit as the elevator becomes small the two systems
are identical from the point of view of the people inside.

				-Mitsu

In article <230@sri-arpa.ARPA> JDM%SMVL%rca.com@CSNET-RELAY.ARPA writes:
    o  Is there a difference between inertial mass and gravitational
       mass. (I believe that is the same as asking,"is there a 
       difference between the effects of gravity and the effects
       of being in an accellerating frame of reference?" But
       Im sure Im making a drastic generalization.)

No difference.

    o  Given the old "accellerating-elevator-in-space" experiment,
       how can one within the elevator tell the whether he is
       accellerating or under the influence of gravity?

You can't tell the difference.  This is the heart of general
relativity.


-- 
- Joe Buck 	{hplabs,ihnp4}!oliveb!epimass!jbuck
  Entropic Processing, Inc., Cupertino, California

In article <230@sri-arpa.ARPA> JDM%SMVL%rca.com@CSNET-RELAY.ARPA writes:
>   o  Given the old "accellerating-elevator-in-space" experiment,
>      how can one within the elevator tell the whether he is
>      accellerating or under the influence of gravity?

Well, in the elevator, the force would be exactly perpendicular to the floor
of the elevator, whereas in an elevator on earth would always have the force
aimed slightly towards the center of the elevator due to the shape of the
gravitational field.  But I think that this question is exactly what general
relativity is all about: I think one of the central thrusts was that you
couldn't tell inertia from gravity.  You could determine the shape of the
field but not any other difference (which brings up the question of whether
or not you could construct a gravity field where the force was perpendicular
to the floor - I would think so).
-- 
James R. Van Artsdalen    ...!ut-ngp!utastro!osi3b2!james    "Live Free or Die"

> In article <230@sri-arpa.ARPA> JDM%SMVL%rca.com@CSNET-RELAY.ARPA writes:
 ... 
>     o  Given the old "accellerating-elevator-in-space" experiment,
>        how can one within the elevator tell the whether he is
>        accellerating or under the influence of gravity?
> 
> You can't tell the difference.  This is the heart of general
> relativity.
> 
> 
> -- 
> - Joe Buck 	{hplabs,ihnp4}!oliveb!epimass!jbuck
>   Entropic Processing, Inc., Cupertino, California

I've wondered about one "difference" which I once read about.

Hold a ball in each hand, say 1 meter apart, and level with the floor.
Then let go of them, and they will "drop" to the floor - mark where
they hit.  In the accelerating-at-1g-elevator-in-space they will hit 
exactly the same distance apart.  On the 1-g-surface-of-earth
they will be slightly closer together because they fall along
radii converging at the center of mass of the earth.

What is the loophole here?  Can gravity produce parallel paths
of the falling balls, or does the elevator test only allow one ball,
or ... ?  (Please help - I'd hate to discard general relativity
because of this. :-)

--henry schaffer  n c state univ  (not physics dept)

> >     o  Given the old "accellerating-elevator-in-space" experiment,
> >        how can one within the elevator tell the whether he is
> >        accellerating or under the influence of gravity?
> > 
> > You can't tell the difference.  This is the heart of general
> > relativity.
> 
> I've wondered about one "difference" which I once read about.
> 
> Hold a ball in each hand, say 1 meter apart, and level with the floor.
> Then let go of them, and they will "drop" to the floor - mark where
> they hit.  In the accelerating-at-1g-elevator-in-space they will hit 
> exactly the same distance apart.  On the 1-g-surface-of-earth
> they will be slightly closer together because they fall along
> radii converging at the center of mass of the earth.
> 
> What is the loophole here?  Can gravity produce parallel paths
> of the falling balls, or does the elevator test only allow one ball,
> or ... ?  (Please help - I'd hate to discard general relativity
> because of this. :-)


Two other ways:

2) weigh the ball (or time its acceleration) near the ceiling and
   near the floor.  If there's a gravity gradient, you're on a planet, if
   not, you're in an acceleration.

3) Tell the elevator operator "I'll give you this nice ball if you tell
   me if this elevator is accelerating or not"  :-)


-- 
		-ed falk, sun microsystems
			falk@sun.com
			sun!falk

Joe Buck sez:
>In article <230@sri-arpa.ARPA> JDM%SMVL%rca.com@CSNET-RELAY.ARPA writes:
>>    o  Given the old "accellerating-elevator-in-space" experiment,
>>       how can one within the elevator tell the whether he is
>>       accellerating or under the influence of gravity?
>
>You can't tell the difference.  This is the heart of general
>relativity.

Actually you can tell the difference, as I am sure Joe knows.
Sufficiently sensitive instruments will detect the slight
variations from place to place in a gravitational field.  If the
accuracy of the instrument is known in advance, the elevator can
be made small enough that the variation will be undetectable.
Example: a device which can detect variations of one part in a
million, used inside an "elevator" 3 meters across, will not be
able to distinguish between a 1 g acceleration and the earth the
gravitational field at the earth's surface.
_____________________________________________________
Matt		University	crawford@anl-mcs.arpa
Crawford	of Chicago	ihnp4!oddjob!matt

	Physics are good for you!

If the elevator is accelerating, then after a long, long time its speed will
reach the speed of light then pass it.  I know that the problem here
has to do with the fact that F=mA is false when the speed approches 'c'.
In fact F approches infinity near 'c'.  So with a constant (or even increasing
but finite) force, wouldn't the acceleration necessarly slow down as the rider
approched 'c' and hence notice that he was in an elevator? 

In other words, with a constant acceleration eventually you would be going
at a speed of 'c' with no more acceleration therefore losing the 'gravity'
feeling.  Then is it impossible to simulate gravity via acceleration for
a 'long' period of time?

< If the elevator is accelerating, then after a long, long time its speed will
< reach the speed of light then pass it.  I know that the problem here
< has to do with the fact that F=mA is false when the speed approches 'c'.
< In fact F approches infinity near 'c'.  So with a constant (or even increasing
< but finite) force, wouldn't the acceleration necessarly slow down as the rider
< approched 'c' and hence notice that he was in an elevator? 
< In other words, with a constant acceleration eventually you would be going
< at a speed of 'c' with no more acceleration therefore losing the 'gravity'
< feeling.  Then is it impossible to simulate gravity via acceleration for
< a 'long' period of time?
-----------
No.  Constant accelleration means constant in the instantaneous frame of
reference of the elevator passenger.  Such acceleration can be maintained
indefinitely.  To a non-acclelerating observer, the acceleration appears
to decrease in magnitude as the elevator velocity asymptotically approaches
'c'.  This phenomenon is a simple consequence of Lorentz contraction
and time dilation.  The elevator passenger feels constant acceleration,
but never reaches a velocity 'c' with respect to an observer in any
inertial frame.
-- 
Bill Tanenbaum - AT&T Bell Labs - Naperville IL  ihnp4!ihlpg!tan

In article <1388@trwrb.UUCP> galins@trwrb.UUCP (Joseph E. Galins) writes:
>In other words, with a constant acceleration eventually you would be going
>at a speed of 'c' with no more acceleration therefore losing the 'gravity'
>feeling.

No, acceleration at a constant rate "forever" will never exceed the
speed of light (measured with respect to the "stationary" reference frame
from which you started).  (For simplicity let's keep the discussion at
the special-relativistic level and leave cosmology out of it.)

Using primes (') for the accelerating object and no-primes-attached for
the stationary reference frame:

	dv' = a' * dt'		for acceleration a' felt by object

	Beta(t'+dt') = Beta(t') VADD dv'/c

where Beta is the velocity of the traveler measured in the stationary
frame divided by c, and VADD is relativistic velocity-combination

	Beta[A wrt C] = Beta[A wrt B] VADD Beta[B wrt C]
		      =	(Beta[A wrt B] + Beta[B wrt C])
			/ (1 + Beta[A wrt B] * Beta[B wrt C])

One also needs to take into account the relationship between time
measures between the two frames ("time dilation")

	dt' = dt / Gamma

where

	Gamma = sqrt( 1 - Beta^2 )

Combining all this, one gets

	Beta(t'+dt') - Beta(t') = a' * dt'
				  * (1 - Beta(t'+dt') * Beta(t')) / c

In the limit dt' -> 0 this becomes

	dBeta(t') = a' * (1 - Beta(t')^2) * dt'

which (using the initial condition Beta(0)==0) integrates to


	Beta(t') = tanh( a' * t' / c )

[This asymptotically approaches 1, i.e. speed of light, thereby proving
the claim made in my first sentence.]

To finish the analysis:

	dx(t') = v(t') * dt

	       = sinh( a' * t' / c ) * c * dt'

To compute the distance D traveled in ship-board time T', when half
the time is spent decelerating at -a', integrate dx from 0 to D/2
(corresponding to t' from 0 to T'/2; Symmetry! Is The Way Things Have
To Be -- Jane Siberry) to obtain

	D(a',T') = 2 * c^2 * (cosh( a' * T' / (2 * c) ) - 1) / a'

Using light-years, gees, and years for units,

	D(a',T') = 1.938 * (cosh( 0.516 * a' * T' ) - 1) / a'

This is the formula I used to calculate travel distance in my other
posting.

In article <1388@trwrb.UUCP> galins@trwrb.UUCP (Joseph E. Galins) writes:
>If the elevator is accelerating, then after a long, long time its speed will
>reach the speed of light then pass it.

No, the speed will approach c but not exceed.

>					...I know that the problem here
>has to do with the fact that F=mA is false when the speed approches 'c'.

That's correct.

>In fact F approches infinity near 'c'.

No, don't forget that F drives the elevator to simulate the gravitation.
It has to be constant and equal to Fg = m0 g (m0 = rest mass).

>				...So with a constant (or even increasing
>but finite) force, wouldn't the acceleration necessarly slow down as the rider
>approched 'c' and hence notice that he was in an elevator? 

There won't be an acceleration, but you can't detect that. Hence you can't
say you are in an elevator. I'll explain that in a minute.

>In other words, with a constant acceleration eventually you would be going
>at a speed of 'c' with no more acceleration therefore losing the 'gravity'
>feeling...

You probably mean 'constant force' not 'constant acceleration' but anyway,
didn't you say F=ma is false for high speeds? Acceleration causes 'gravity'
feeling (F) only at low speeds.

You have to be careful not to mix Newtonian mechanics and relativistic
mechanics.  For velocities comparable to c you have to forget everything
intuitive, all your experiences. Any time you make an inference you better
ask yourself: "Did I derive this from the relativistic laws or do I rely on
ANYTHING else"? It is difficult task and many people won't accept it.

According to Newtonian mechanics, F = ma, therefore if a constant force
acts on a body for a long time, the velocity will increase infinitely.
We know, that it is not true. Relativistic physics says F = dp/dt
where p = mv = GAMMA m0 v, GAMMA = (1-(v/c)^2)^-1/2, m0 = rest mass.
The formula says the following:
	1. The force depends on the change of momentum (p)
	   i.e. BOTH mass and velocity.
	2. If you apply a constant force, the momentum will
	   increase infinitely.

You can see that for low velocities (v<<c) mass is approximately constant
and the force (gravity feeling) depends only on acceleration. For very
high velocities (v -> c) there is almost no acceleration but the momentum
keeps increasing because the mass keeps increasing. Since the momentum or
more precisely its change is what determines the force, you won't feel any
difference. So you are right when you say there won't be any acceleration,
but that has nothing to do with a gravity feeling.

zdenek

P.S.
But I think you'd find out anyway. The human body starts to act really
funny when it moves at the speed of light. The strongest effects are on
the digestive tract.

 -------------------------------------------------------------------------
 Men are four:
 He who knows and knows that he knows, he is wise - follow him;
 He who knows and knows not that he knows, he is asleep - wake him;
 He who knows not and knows that he knows not, he is simple - teach him;
 He who knows not and knows not that he knows not, he is a fool - shun him!

 zdenek@CS.COLUMBIA.EDU  or 	...!seismo!columbia!cs!zdenek
 Zdenek Radouch, 457 Computer Science, Columbia University,
 500 West 120th St., New York, NY 10027

>Joe Buck sez:
>>In article <230@sri-arpa.ARPA> JDM%SMVL%rca.com@CSNET-RELAY.ARPA writes:
>>>    o  Given the old "accellerating-elevator-in-space" experiment,
>>>       how can one within the elevator tell the whether he is
>>>       accellerating or under the influence of gravity?
>>
>>You can't tell the difference.  This is the heart of general
>>relativity.
>
In article <1529@oddjob.UUCP> matt@oddjob.UUCP (Matt Crawford) writes:
>Actually you can tell the difference, as I am sure Joe knows.
>Sufficiently sensitive instruments will detect the slight
>variations from place to place in a gravitational field.  If the
>accuracy of the instrument is known in advance, the elevator can
>be made small enough that the variation will be undetectable.
>Example: a device which can detect variations of one part in a
>million, used inside an "elevator" 3 meters across, will not be
>able to distinguish between a 1 g acceleration and the earth the
>gravitational field at the earth's surface.

I didn't know the General R. had a heart, although perhaps Dr. Who 
makes up for it!

I think in one case as time went on time dilation would begin
setting in as well as increasing contraction (spatial) along the
line of acceleration. Also in the "static" case the vertical height
of the elevator is important since the "Mossbauer effect" could 
reveal a "time rate (field) gradient", if it were present.  By
checking the "frequency" spectroscopy of sharp line emmissions 
in the "horizontal" plane of the elevator using a grating spectro-
meter and comparing it with the standard (reference on a known 
platform), perhaps the time dilation due to increasing velocity 
would become apparent although the spacing of the grating rulings 
should not be effected, unless rotated.

Then again may be we could open the door and look out!

        For those really long term acceleration experiments,
          Use PLASMAK(TM) fusion astro propulsion drive. 
+---------------------------------------------------------+--------+
| Paul M. Koloc, President: (301) 445-1075                | FUSION |
| Prometheus II, Ltd.; College Park, MD 20740-0222        |  this  |
| {umcp-cs | seismo}!prometheus!pmk; pmk@prometheus.UUCP  | decade |
+---------------------------------------------------------+--------+

In article <230@sri-arpa.ARPA> JDM%SMVL%rca.com@CSNET-RELAY.ARPA writes:
>
>   Maybe someone out there can answer this one quickly. I hope its
>not too fundamental or repetitous.
>
>   Actually, now that I think about it, its really two questions:
>
>   o  Is there a difference between inertial mass and gravitational
>      mass. (I believe that is the same as asking,"is there a 
>      difference between the effects of gravity and the effects
>      of being in an accellerating frame of reference?" But
>      Im sure Im making a drastic generalization.)
There is none.
>   o  Given the old "accellerating-elevator-in-space" experiment,
>      how can one within the elevator tell the whether he is
>      accellerating or under the influence of gravity?
>
Locally, one cannot.  Key word there is locally.  If one can make 
observations over a finite area, then one can tell, if the gravitational
field is varies with position.  That is, if one measures tidal forces,
it is a gravitational field, not acceleration.   This in not a contradiction
of my first answer, or of the equivalence principle -- these state that 
the effects on a point object of a gravitational field and an acceleration
are identical;  however tidal forces can not be measured at a single point.
A formal statement contains incantations about "motions on a geodesic"
and "ability to locally transform the metric into Minkowski form", but
I don't think that such a statement would be particularly enlightening.
>
>    Any correct answer to these questions will end a heated argument
>that is rapidly degenerating into World War III. Can someone save
>the world?
>
>                            Thanks in Advance
>                                Joe


-- 
--------------------------------------------------------------------------------
Courtenay Footman		ARPA:	cpf@lnssun1.tn.cornell.edu
Lab. of Nuclear Studies		Usenet:	cornell!lnssun1!cpf
Cornell University		Bitnet: cpf%lnssun1.tn.cornell.edu@WISCVM.BITNET

In <1388@trwrb.UUCP> galins@trwrb.UUCP (Joseph E. Galins) writes:
>If the elevator is accelerating, then after a long, long time its speed will
>reach the speed of light then pass it.  I know that the problem here
>has to do with the fact that F=mA is false when the speed approches 'c'.
>In fact F approches infinity near 'c'.  So with a constant (or even increasing
>but finite) force, wouldn't the acceleration necessarly slow down as the rider
>approched 'c' and hence notice that he was in an elevator? 
>
>In other words, with a constant acceleration eventually you would be going
>at a speed of 'c' with no more acceleration therefore losing the 'gravity'
>feeling.  Then is it impossible to simulate gravity via acceleration for
>a 'long' period of time?

	No.  The point is, the acceleration is always "felt" in the
frame of reference of the person in the elevator.  However, this frame
of reference is continually accelerating (it is perhaps pedantically
more desirable to say the person in the elevator is instantaneously
at rest in frames of reference which are moving at continally greater
velocities relative to some fixed reference 'lab' frame.)

	To put it more simply, the acceleration as observed in a fixed
lab frame is continually decreasing; it decreases until it asymptotically
approaches zero.  However, the principle of special relativity states
that all reference frames are equivalent (where by "reference frame" I
mean non-accelerating coordinate frames).  That is, in a frame which
is moving relative to the lab frame "in the same direction" as the
elevator would register a "faster" acceleration.  So the observed accelration
depends on the reference frame in which you are measuring the
velocities and times.

	However, from the point of view of someone in the elevator,
the only accelration which is relevant is the one measure in the
frame in which s/he is instantaneously at rest.  That is, in the frame
which is "moving along" with the elevator so that at the instant of
measurement the elevator seems to be at rest.  Defined in this way
(this would be the acceleration "felt" by the observer in the elevator)
the acceleration could certainly remain constant for indefinite
periods of time (of course limitations of hydrogen gas drag and micro-
meteorite penetration would cause you to think twice and then again
before trying such a stunt.)  A quick calculation shows that,
for example, after accelerating for 1 year at 1 gravity you can approach
the speed of light to well within 1% (as measured from Earth.)

				-Mitsu

In article <590@husc6.HARVARD.EDU> hadeishi@husc4 (mitsuharu hadeishi) writes:

>A quick calculation shows that,
>for example, after accelerating for 1 year at 1 gravity you can approach
>the speed of light to well within 1% (as measured from Earth.)

  According to my quick calculation, g = 1.0325 ly/year^2, and so after
one year at one g, one is going tanh(1.0325) = .775c.

ucbvax!brahms!gsmith    Gene Ward Smith/UCB Math Dept/Berkeley CA 94720
 This posting was made possible by a grant from the Mobil Corporation

In article <187@cartan.Berkeley.EDU> gsmith@brahms (Gene Ward Smith) writes:
>In article <590@husc6.HARVARD.EDU> hadeishi@husc4 (mitsuharu hadeishi) writes:
>
>>for example, after accelerating for 1 year at 1 gravity you can approach
>>the speed of light to well within 1% (as measured from Earth.)
>
>  According to my quick calculation, g = 1.0325 ly/year^2, and so after
>one year at one g, one is going tanh(1.0325) = .775c.

When one hits a velocity of sqrt(.5) =~ .7071c then the Lorentz equation
gives a dilation of sqrt(2)=1.414, so the "ground made good" as seen by
someone in the spaceship is exactly 1.00c.

Now .7071 isn't .775 to within a percent, but perhaps this is what was meant.

    /Alan Paeth

In article <523@husc6.HARVARD.EDU> hadeishi@husc4.UUCP (mitsuharu hadeishi) writes:
>	Recently some physicists were going over the data of
>old experiments trying to establish the equivalence of inertial
>and gravitational mass and claimed to have found slight
>differences between atoms of different atomic number.  They attributed
>this to an effect related to a quantity they called "hypercharge."
>Apparently their result was flawed by a sign error, however,
>so it is very dubious at the moment.


Actually the "sign error" in the original analysis as now been resolved.
There was a sign *discrepancy* between the result implied by the Eotvos
experiment (the 1908 experiment you refer to), when one analyzes the
experimental result in terms a spherical rotating Earth, and the result implied
by the geophysical result of Frank Stacey.  For purposes of analyzing such
experiments as the Eotvos experiment, this simple model of the Earth appears
inadequate (at least if one is discussing the possibility of an intermediate
range force [one whose range is on the order of a few hundred meters]).
When a more realistic model is employed (one which takes into account the
local matter distribution), there no longer appears to be any real
problems with the analysis.

Of course the real question at this time is whether such an effect actually
exists in nature, but this can only be resolved by ongoing experiments
(such as three which are underway here at UW).  An interesing article
on the current status of the "fifth force" may be found in the latest
Physics Today.


Carrick Talmadge

In article <1388@trwrb.UUCP>, galins@trwrb.UUCP (Joseph E. Galins) writes:
> If the elevator is accelerating, then after a long, long time its
> speed will reach the speed of light then pass it.

Not if current theories are correct.

> I know that the problem here has to do with the fact that F=mA is
> false when the speed approches 'c'.

Well, this depends on how you interpret the symbols.  As I recall, if
you interpret m as relativistic mass then it is true.

> In fact F approches infinity near 'c'.

Really?  Where does the energy to exert such a force come from?

> So with a constant (or even increasing but finite) force, wouldn't
> the acceleration necessarly slow down as the rider approched 'c' and
> hence notice that he was in an elevator?

Yes, EXCEPT that as the rider views it, the acceleration is constant.
Time dialation, remember.  It's the outside observer that sees
acceleration as slowing down.

> In other words, with a constant acceleration eventually you would be
> going at a speed of 'c' with no more acceleration therefore losing
> the 'gravity' feeling.

(1) You can't reach c.
(2) The rider still feels the acceleration as proportional to the
force; it's just the observer with respect to whom the elevator is
moving near c (the "stationary" observer) that sees the acceleration
slack off.

> Then is it impossible to simulate gravity via acceleration for a
> 'long' period of time?

Not so.  The only difference between "gravity as created by a mass" and
"gravity as created by acceleration" (I use the quotes because gravity
*is* acceleration, but I don't see any simple, brief, and more accurate
way of putting it) is the structure of the field (ie, parallel versus
converging lines of force....).

Note another difference.  If the elevator rider measures his velocity
with respect to "the universe", whatever that means, he will find it to
be steadily increasing.  The guy on the planet will not.

For "the universe", picky theorists may substitute, say, the background
black-body radiation (a suspiciously absolute frame of reference, if
you ask me!).

					der Mouse

USA: {ihnp4,decvax,akgua,utzoo,etc}!utcsri!mcgill-vision!mouse
     think!mosart!mcgill-vision!mouse
Europe: mcvax!decvax!utcsri!mcgill-vision!mouse
ARPAnet: think!mosart!mcgill-vision!mouse@harvard.harvard.edu

	Aren't you glad you don't shave with Occam's Razor?

So, from inside a small room can one tell the difference between the
room being in a gravity field and being linearly accelerated, or what?

No.  And the "look for tides" or "look for mossbauer effect" or whatever
won't help you, since you don't know the mass distribution around your
room in either scenario.  With a suitable surrounding mass distribution,
tides (and other spherical gravity field effects) can be reduced to
arbitrarily small effects in the gravity case.  And even then, "finding
a tidal effect proves you aren't accelerating" is incorrect, since the
"tides" might be provided by cleverly placed auxiliary masses for the
acceleration case.

Further, you can't wait and try to observe special relativistic effects
either as suggested here:

> zdenek@heathcliff.columbia.edu (Zdenek Radouch)
> But I think you'd find out anyway. The human body starts to act really
> funny when it moves at the speed of light. The strongest effects are on
> the digestive tract.

because there are no such "funny effects" on the body.  Physics (and
chemistry, and digestion) are supposedly invariant across reference
frames.  (Zdenek's suggestion is incongruous, especially the reference
to the digestive tract, and may therefore be a joke.  If so, obviously I
didn't get it.)

And, for the same reason, these suggestions won't work either:

> pmk@prometheus.UUCP (Paul M Koloc)
> I think in one case as time went on time dilation would begin
> setting in as well as increasing contraction (spatial) along the
> line of acceleration.

Perhaps so.  But how would one measure it from inside the room?
(Hint: one could not.)

Certainly this attempt fails:

> By
> checking the "frequency" spectroscopy of sharp line emmissions 
> in the "horizontal" plane of the elevator using a grating spectro-
> meter and comparing it with the standard (reference on a known 
> platform), perhaps the time dilation due to increasing velocity 
> would become apparent although the spacing of the grating rulings 
> should not be effected, unless rotated.

because all measuring instruments will be similarly time-and-space
"distorted", so that they will insist that all is as it was at rest.
Even the slight cheat of "looking outside" (by refering to measurements
taken on a "known platform") doesn't help, since (from inside the room)
the room should have all the characteristics of the "known platform" (if
it too was at rest relative to the measuring devices).

The only (ah say ONLY!) way to tell is:

> [...] we could open the door and look out!

--
"Pwease, mistew ewevatow opewatow, I've twied EVEWYTHING...
 I'll give you aww this vawuabwe scientific equipment if you wouwd just
 teww me what's WEAWWY going on!"

"Why SOIT'NY m'boy!  It's REALLY another puzzle in sci.physics!"

"AAAAAAAAUUUUUUUUGGGGGGGGHHHHHHHHH!!!!!!!!"
-- 
Wayne Throop      <the-known-world>!mcnc!rti-sel!dg_rtp!throopw

IN article <546@mcgill-vision.UUCP>, mouse@mcgill-vision.UUCP (der Mouse) wrote:
> In article <1388@trwrb.UUCP>, galins@trwrb.UUCP (Joseph E. Galins) writes:
> > So with a constant (or even increasing but finite) force, wouldn't
> > the acceleration necessarly slow down as the rider approched 'c' and
> > hence notice that he was in an elevator?
> 
> Yes, EXCEPT that as the rider views it, the acceleration is constant.
> Time dialation, remember.  It's the outside observer that sees
> acceleration as slowing down.

Hmmm...  I take it you mean that if the rider measures acceleration in terms
of the gravity the rider feels, then it is constant.  But if the rider
measures acceleration in terms of the observer that was left behind?  I
guess it all depends on how you want to define acceleration maybe?  After all,
since the rider can't see the observer recede faster than the speed of light,
eventually the change in velocity between the rider and observer would have
to fall off.

> Note another difference.  If the elevator rider measures his velocity
> with respect to "the universe", whatever that means, he will find it to
> be steadily increasing.  The guy on the planet will not.

Huh?  Are you suggesting that the observer could view the rider as having no
acceleration even while the rider viewed acceleration?  I would think the guy
on the planet would still measure the _velocity_ as increasing, but the
acceleration as decreasing...

> For "the universe", picky theorists may substitute, say, the background
> black-body radiation (a suspiciously absolute frame of reference, if
> you ask me!).

Ah, but absolute relative to what?  :-)

One other thing that confuses me about all of this.  The rider takes off from
the planet at relativistic speeds, and the observer on the planet sees the
rider "suffer" time dilation: the rider appears to move more slowly through
time.  I think I see where this comes from.  But what does the rider see of
the planet?  Since the planet is also receding relativisticly from the rider,
the rider observes the planet suffering the same fate: time dilation by the
same factor as the planetary observer measured.

Now, when the rider arrives at the destination, the rider must conclude that
the rider's clocks are *ahead* of the planet's, since the rider clearly observed
the planet's clocks "slow down".  Similarly the planetary observer must also
conclude that the rider's clocks are *behind* the planetary clocks.  Clearly
I'm missing something rather fundemental, but it seems to me that I have to be
able to view the situation from both the rider's standpoint and the planet's
and get the same answers (that's what relativity is about in this example?).
When the rider returns to the planet, every cutsey story says the rider is
much "younger" than the contemporaries left behind because of time dilation.
Yet somehow I much account for the fact that the rider observed the rider's
contemporaries age slower than the rider as the rider left the planet...

Not very well put, but I think the dilema is stated.  My high school physics
isn't letting me in on the secret...
-- 
James R. Van Artsdalen    ...!ut-ngp!utastro!osi3b2!james    "Live Free or Die"

> james@reality1.UUCP (james)
>> mouse@mcgill-vision.UUCP (der Mouse)
>>> galins@trwrb.UUCP (Joseph E. Galins)

>>> So with a constant (or even increasing but finite) force, wouldn't
>>> the acceleration necessarly slow down as the rider approched 'c' and
>>> hence notice that he was in an elevator?
>> Yes, EXCEPT that as the rider views it, the acceleration is constant.
> But if the rider
> measures acceleration in terms of the observer that was left behind?

How would you suggest the rider do this without looking outside?

>> Note another difference.  If the elevator rider measures his velocity
>> with respect to "the universe", whatever that means, he will find it to
>> be steadily increasing.  The guy on the planet will not.

Well, one thing it means is looking outside the elevator.

> One other thing that confuses me about all of this.
> [description of classic twin "paradox", omitted]
> My high school physics isn't letting me in on the secret...

My college physics DOES let me in on the secret... but just how many
times should I post the same thing over and over?  Well, maybe just once
more, since this aspect of it hasn't come up in this particular
newsgroup for a while (though you'd already know the answer if you were
paying close attention to the "instant message channel" exchange).  This
is a repeat of a talk.philosophy.misc fragment, with a small correction
pointed out to me by ucbvax!brahms!weemba.

We have a stay-at-home A, and a passing traveler, B.  A says something
like "First, take the event on B's timeline that I consider simultaneous
with 'now' on my timeline.  Second, take another event on B's timeline
that I consider simultaneous with a point N seconds ago on my timeline.
B will think that M seconds have passed between these two events, and M
will be less than N." Fair enough.  But change the B's to A's in the
above statement, and it must be *just* *as* *true* when B says it.  How
can this be?  Well, note that the point on A's timeline M seconds ago
(according to A) doesn't *HAVE* to be simultaneous with the point on B's
timeline of M seconds ago (according to B).  How does it all work? See
the diagram below.  Events are 1, 2, 3, 4 and 5.  Worldlines are A and
B.  A and B cross at 3, so event 3 is on both A's and B's timeline.

        |                .     .          .   .
        |                .    .           .  .        *
        |                .   .            . .     *
   ^    |                .  .             ..  *
   |    |       A ****** 1 2 ************ 3 ************
 space  |                ..           *  ..
        |                .        *     . .
        |               ..    *        .  .
        |              . 4*           .   .
        |             5  .           .    .
        |       B *  .   .          .     .
        +-------------------------------------------------
                           time -->

Note that in this diagram, events located along the vertical dotted
lines are considered to be all simultaneous by A, and events located
along the slanted dotted lines are considered to be simultaneous by B.
This is a space-time geometric cosequence of the
generally-given-as-algebraic "laws" of special relativity.

So what have we got?  Well, the time interval 1-3 is what A considers N
seconds.  A considers 1 and 4 to be simultaneous.  The interval 5-3 is
what B considers N seconds.  B considers 5 and 2 simultaneous.  We can
see that A considers 2-3 to be less than N seconds long, B considers 4-3
to be less than N seconds long.  So, lo and behold, A thinks that B is
slower, and B thinks that A is slower.  And all because space and time
are (to a limited extent) interchangeable, just as various spatial
dimensions are.  Fancy that!

Note that all these relationships stay the same from any point of view,
whether A's, B's, or somebody else going a different constant speed.  Of
course, none of these others will agree that 1-3 or 5-3 is N seconds
long, but so what?  Even A doesn't think that 5-3 is N seconds (A thinks
it is longer!), nor does B think that 1-3 is N seconds (again, B thinks
it is longer!).

Ok.  Fine.  But how does all this hoo-hah relate to the twin "paradox"?
After all, either A or B has to be "really" slower, right?  Wrong.  As B
travels away from A, B considers events on slanted lines to be
"simultaneous".  When B accelerates to stop at some distance from A, B
has changed reference frames, and now considers events on vertical lines
to be "simultaneous".  Thus, for any stretch of the trip with no
acceleration (and with B and A moving relative to each other), both B
and A think that the other has slowed down.  But during B's
acceleration, B skips large parts of A's history, because B's line
through space-time indicating what is the "present moment" rotates, and
sweeps along a large part of A's history.  The same occurs when B
accelerates to return home.

The difference in the twin's timelines is that A didn't change inertial
frames, and B did.  Simple as that.

--
Our tool is diagram and explanation... uh, two, our TWO tools are
diagrams and explanataions, and appeal to authority... uh THREE our
THREE tools are diagrams, explanations, appeal to authority, and...
AMONG our tools ARE: diagrams, explanations, appeal to authority, and a
fanatical devotion to Physics!
                --- (NOBODY expects the RELATIVISTIC INQUISITION!)
-- 
Wayne Throop      <the-known-world>!mcnc!rti-sel!dg_rtp!throopw

In article <706@dg_rtp.UUCP> throopw@dg_rtp.UUCP (Wayne Throop) writes:
>> [somebody about twin paradox]
>> My high school physics isn't letting me in on the secret...
>
>My college physics DOES let me in on the secret... but just how many
>times should I post the same thing over and over?  Well, maybe just once
>more...

I don't think the problem with understanding relativistic physics is
as simple as you put it. To be more precise I don't think there are too
many people around who DO KNOW how to use space-time diagrams and at the
same time CAN'T EXPLAIN the twin paradox.
My guess is that there are two big groups with the following comments:

1. Why the hell does he repeat this over ond over?
2. Why the hell does he repeat this over ond over?

The first group is annoyed because they don't learn anything new, whereas
the second group is annoyed because they don't learn anything new, too.

Oh, I almost forgot to say that the first group understands the problem
and the second doesn't. So I am not sure about your "one more explanation".

zdenek

P.S.
Could anybody who didn't understand the twin paradox and does understand
it now as a result of Wayne's article please raise hand and say I'm wrong?

 zdenek@cs.columbia.edu  or 	...!seismo!columbia!cs!zdenek