JDM%SMVL%rca.com@CSNET-RELAY.ARPA (10/21/86)
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.)
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?
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
JoeKATZ@venera.isi.edu (10/23/86)
From: Alan R. Katz <KATZ@venera.isi.edu> Forgive the brevity but, The answer to the first question is NO and the answer to the second is HE CAN'T. (Actually he can, but for a "perfect" pointlike elevator, he can't). Alan (Katz@ISI.EDU) -------
hadeishi@husc4.harvard.edu (mitsuharu hadeishi) (10/24/86)
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
jbuck@epimass.UUCP (Joe Buck) (10/25/86)
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, Californiajames@reality1.UUCP (james) (10/26/86)
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"
hes@ecsvax.UUCP (Henry Schaffer) (10/27/86)
> 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)
falk@sun.UUCP (10/28/86)
> > 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
matt@oddjob.UUCP (Matt Crawford) (10/28/86)
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!
galins@trwrb.UUCP (Joseph E. Galins) (10/29/86)
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?
tan@ihlpg.UUCP (Bill Tanenbaum) (10/29/86)
< 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
gwyn@brl-smoke.ARPA (Doug Gwyn ) (10/30/86)
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.
zdenek@heathcliff.columbia.edu (Zdenek Radouch) (10/30/86)
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
pmk@prometheus.UUCP (Paul M Koloc) (10/31/86)
>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 | +---------------------------------------------------------+--------+
cpf@batcomputer.TN.CORNELL.EDU (Courtenay Footman) (11/01/86)
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
hadeishi@husc4.harvard.edu (mitsuharu hadeishi) (11/05/86)
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
gsmith@brahms (Gene Ward Smith) (11/06/86)
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
awpaeth@watcgl.UUCP (Alan W. Paeth) (11/07/86)
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
ephraim@uw-larry (Carrick Talmadge) (11/08/86)
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